Beginning Algebra (Available 2010 Titles Enhanced Web Assign)

Rectangle Triangle w a c h l A = lw P = 2l + 2w A P l w area perimeter length width Parallelogram b A = 1 bh...

Author: Jerome E. Kaufmann | Karen L. Schwitters

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Rectangle

Triangle

w

a

c

h

l A = lw

P = 2l + 2w A P l w

area perimeter length width

Parallelogram

b

A = 1 bh 2 A area P perimeter b base h altitude a, c sides

P=a+b+c

Right Triangle

h

c

b A = bh A area b base h altitude (height)

Trapezoid

b

a Pythagorean theorem: a2 + b2 = c2 a, b legs c hypotenuse

Circle

b1 h r b2 1 A = h(b1 + b2) 2 A area b1, b2 bases h altitude

A = πr2

C = 2πr

A area C circumference r radius

Cube

Right Circular Cylinder

Rectangular Prism b

r

s s

s V = s3

S = 6s2

V S s

h

c

volume total surface area length of a side

a V = abc V S a b c

Sphere

S = 2ac + 2 ab + 2bc volume total surface area width length height

Pyramid

V = πr2h

S = 2πr2 + 2πrh

V S r h

volume total surface area radius altitude (height)

Right Circular Cone

h

r

base V = 4 πr3 3 V S r

S = 4πr2 volume total surface area radius

V B h

V = 1 Bh 3 volume area of base altitude (height)

s

h

r

V = 1 πr2h 3 V S r h s

S = πr2 + πrs

volume total surface area radius altitude (height) slant height

BEGINNING ALGEBRA Jerome E. Kaufmann Karen L. Schwitters SEMINOLE COMMUNITY COLLEGE

Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States

Beginning Algebra Jerome E. Kaufmann Karen L. Schwitters

© 2010 Brooks/Cole, Cengage Learning ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher.

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Printed in the United States of America 1 2 3 4 5 6 7 12 11 10 09 08

Contents 1

Basic Concepts of Arithmetic and Algebra 1.1 Numerical and Algebraic Expressions

2

1.2 Addition, Subtraction, Multiplication, and Division of Integers 1.3 Use of Properties

11

24

1.4 Prime and Composite Numbers 1.5 Rational Numbers

1

33

41

1.6 Decimals and Real Numbers

55

Chapter 1 Summary 67 Chapter 1 Review Problem Set 71 Chapter 1 Test 73

2

Equations, Inequalities, and Problem Solving 2.1 Solving First-Degree Equations

76

2.2 Translating from English to Algebra 2.3 Equations and Problem Solving

82

89

2.4 More on Solving Equations and Problem Solving

96

2.5 Equations Involving Parentheses and Fractional Forms 2.6 Inequalities

75

103

113

2.7 Inequalities, Compound Inequalities, and Problem Solving

121

Chapter 2 Summary 130 Chapter 2 Review Problem Set 133 Chapter 2 Test 136 Chapters 1–2 Cumulative Practice Test 138

3

Ratio, Percent, Formulas, and Problem Solving 3.1 Ratio, Proportion, and Percent

140

3.2 More on Percents and Problem Solving 3.3 Formulas: Geometric and Others 3.4 Problem Solving

139

149

156

166

3.5 More about Problem Solving

174

Chapter 3 Summary 181 Chapter 3 Review Problem Set 184 Chapter 3 Test 187 Chapters 1–3 Cumulative Review Problem Set 189

v

vi

Contents

4

Exponents and Polynomials 4.1 Addition and Subtraction of Polynomials 4.2 Multiplying Monomials

192

199

4.3 Multiplying Polynomials

206

4.4 Dividing by Monomials

215

4.5 Dividing by Binomials

191

219

4.6 Zero and Negative Integers as Exponents

224

Chapter 4 Summary 234 Chapter 4 Review Problem Set 237 Chapter 4 Test 239 Chapters 1– 4 Cumulative Review Problem Set 240

5

Factoring, Solving Equations, and Problem Solving 241 5.1 Factoring by Using the Distributive Property 5.2 Factoring the Difference of Two Squares

242

250

5.3 Factoring Trinomials of the Form x bx c 256 2

5.4 Factoring Trinomials of the Form ax2 bx c

265

5.5 Factoring, Solving Equations, and Problem Solving

270


6

Rational Expressions and Rational Equations 6.1 Simplifying Rational Expressions 286 6.2 Multiplying and Dividing Rational Expressions 6.3 Adding and Subtracting Rational Expressions

291 296

6.4 More on Addition and Subtraction of Rational Expressions 6.5 Rational Equations and Problem Solving

313

6.6 More Rational Equations and Problem Solving

320

Chapter 6 Summary 330 Chapter 6 Review Problem Set 333 Chapter 6 Test 335 Chapters 1– 6 Cumulative Review Problem Set 337

7

Coordinate Geometry 7.1 Cartesian Coordinate System

340

7.2 Graphing Linear Equations and Applications 7.3 Graphing Linear Inequalities

339

360

350

303

285

Contents 7.4 Slope of a Line

365

7.5 Writing Equations of Lines

374


8

Systems of Equations 8.1 Solving Linear Systems by Graphing 8.2 Elimination-by-Addition Method 8.3 Substitution Method

391

392

398

406

8.4 3 3 Systems of Equations

417


9

Roots and Radicals 9.1 Roots and Radicals

435

436

9.2 Simplifying Radicals

444

9.3 Simplifying Radicals of Quotients

451

9.4 Products and Quotients Involving Radicals 9.5 Solving Radical Equations

458

464

Chapter 9 Summary 472 Chapter 9 Review Problem Set 476 Chapter 9 Test 478 Chapters 1–9 Cumulative Review Problem Set 480

10

Quadratic Equations 10.1 Quadratic Equations

484

10.2 Completing the Square 10.3 Quadratic Formula

483

493

499

10.4 Solving Quadratic Equations—Which Method? 10.5 Solving Problems Using Quadratic Equations

504

509


Practice Your Skills Solutions 521 Answers to Odd-Numbered Problems and All Chapter Review, Chapter Test, and Cumulative Review Problems 575

Index

I-1

vii

viii

Chapter 1

Basic Concepts of Arithmetic and Algebra

Preface

When preparing Beginning Algebra, First Edition, we attempted to preserve the features that made the previous editions of our hardcover series successful. At the same time, we made a special effort to incorporate many changes and improvements, suggested by reviewers, to create a book that would serve the needs of students and instructors who prefer a paperback text. In our experience, instructors who prefer a paperback are more interested in reinforcing skills through practice. Hence, the text has a structured pedagogy that includes in-text practice exercises, detailed examples, learning objectives, an extensive selection of problem-set exercises, and wellorganized end-of-chapter reviews and assessments. This text was written for students who have never had a beginning algebra course and for students who need a refresher before taking additional mathematics courses. The basic concepts of beginning algebra are presented in a simple, straightforward manner. The structure for explaining mathematical techniques and concepts has proven successful. Concepts are developed through examples, continuously reinforced through additional examples, and then applied in problem-solving situations. A common thread runs throughout this book: learn a skill, use the skill to help solve equations, and then use the equations to solve the application problems. The examples are the “learning” portion, the Practice Your Skill in-text problems are the “using” portion, and the Apply Your Skill examples are the “apply” portion. This thread influences many of the decisions we made in preparing this text. Early chapters are organized to start the book at the appropriate mathematical level with the right amount of review. We have added Learning Objectives to the section openers and repeat those learning objectives within the sections and exercise sets. In every example, we added a practice problem, Practice Your Skill, to provide an exercise that will immediately reinforce the skill presented in the example. We have added Concept Quizzes before each problem set to assess student’s mastery of the mathematical ideas and vocabulary presented in the section. By broadening the topics in the problem sets, we show students that mathematics is part of everyday life. Problems and examples include references to career areas such as the electronics, mechanics, and healthcare fields. By strengthening the examples, we give students more support with problem solving in the main text. Further, we have designed the structure of the problem sets to stress learning outcomes and easy student access to the objectives. The exercises in problem sets are grouped by learning objectives. To recap the chapter and its learning outcomes, the examples in the chapter summary are grouped by learning objectives in a grid format. Using the learning objectives to organize the problem sets and the end-of-chapter summary grid gives students a strong sense of the objectives for the topics.

Key Features to the Series

viii

•

The table of contents is organized to present the standard basic algebra topics and provide review at the beginning of the book. Chapter 1 takes a fresh approach to review topics, such as integers, rational numbers, and the transition into algebra. Chapter 2 progresses on to equation solving and problem solving.

•

This book takes a practical approach to problem solving. Instructors will notice how much we stress a practical way to learn to solve problems in Chapters 2 and 3. We bring in problem solving early, and we stress problem solving often. (After work with equations and inequalities in Chapter 2, we offer Chapter 3 and its many applications as a way of showing the basics of algebra and problem solving in action.)

Preface

ix

•

We think that students will find the powerful ideas of equations of lines and systems of equations more understandable if we separate the many ideas and techniques into two chapters. Chapter 7, therefore, centers on linear equations, whereas Chapter 8 follows up by expanding these ideas into systems of equations. For courses that do not include systems of equations, separating the topics allows graphing to be presented in an entire complete chapter.

•

The structure of the main text and exercise sets centers on learning objectives and learning outcomes. A list of learning objectives opens each section. The objectives are repeated as subheads within the section to organize the material, and the exercises are grouped by objective so that a instructor can easily see which concepts a student has mastered. At the end of the sections, a concept quiz reviews the “big ideas” of the section. •

Expressly for this series, Practice Your Skill problems have been added to the worked examples as a way of enhancing the material. On-thespot problems help students to master the content of the examples more readily. Worked-out solutions to the practice problems are located at the back of the text.

EXAMPLE 2

The difference of two numbers is 14. The smaller number is n. What is the larger number?

Solution Because the smaller number plus the difference must equal the larger number, we can represent the larger number by n 14.

PRACTICE YOUR SKILL The difference of two numbers is 20. The smaller number is x. Write an expression for the larger number.

•

Concept Quizzes are included, immediately preceding each section problem set. These problems predominantly rely on the true/false format that allows students to check their understanding of the mathematical concepts introduced in the section. Users have reacted very favorably to concept quizzes, and they indicated that they used the problems for many different purposes.

CONCEPT QUIZ

For Problems 1– 8, answer true or false. 1. A binomial that has two perfect square terms that are subtracted is called the difference of two squares. 2. The sum of two squares is factorable using integers. 3. When factoring it is usually best to look for a common factor first. 4. The polynomial 4x2 y2 factors into (2x y)(2x y). 5. The completely factored form of y4 81 is (y2 9)(y2 9). 6. The solution set for x2 16 is { 4}. 7. The solution set for 5x3 5x 0 is { 1, 0, 1}. 4 8. The solution set for x 9x2 0 is { 3, 0, 3}.

•

We wanted an easy-to-use—and information-rich— chapter summary in grid format. This highly structured recap organizes the chapter content for easy accessibility. The summary grid includes Objective in the leftmost column as the organizing information. To the right of the objective, a Summary column recaps the mathematical technique or concept in simple language. To reinforce the objective, we offer a new example in the Example column, and to reinforce the need to practice, we list the appropriate Chapter Review Problems for the objective in the rightmost column. We think that this new way of organizing the information will attract student interest and offer a valuable feature to instructors.

x

Preface

Chapter 3 Summary OBJECTIVE

SUMMARY

Solve proportions.

A statement of equality between two ratios is a proportion. In a proportion, the cross products are equal; that is to say: c a If , then ad bc, b d when b 0 and d 0.

Solve word problems using proportions.

CHAPTER REVIEW PROBLEMS

EXAMPLE

A variety of word problems can be set up and solved using proportions.

Solve

1

x 3

x . 2

Problems 1–10

Solution

1 x , then 3 2 2 1x 12 3x 2 x 2 3x 2 x The solution set is { 2}. If

x

Problems 11–18

The scale on a blueprint shows that 1 inch represents 8 feet. If the length of the house on the blueprint is 4.5 inches, what is the length of the actual house? Solution

Set up the proportion 4.5 1 and solve for x. x 8 x 4.5(8) 36 The house is 36 feet long.

•

Answer boxes for the Practice Your Skill in-text problems and Concept Quiz questions are conveniently placed at the end of the section problem sets. By doing so, we encourage students to study a worked example, practice an on-the-spot problem, and find the answer without having to search the appendices at the back of the book.

Answers to the Example Practice Skills 1. 7a2

2. 2m2n3

3. 3a2(5

7. (a) (a 3)(b 5) (b) (x 12. The length is 12.

7a4) 4. 2m3n(4n 4)(y

m3) 5. 8y4(6y4

2) 8. {0, 7}

9. {0, 15}

2y2 3) 6. 8b2(b 1) 2 , 0 f 11. { 5, 2} 5

10. e

Answers to the Concept Quiz 1. False

2. True

3. False 4. True

5. False

6. False

7. False

8. True 9. True

10. False

Problem-Solving Approach As mentioned, a common thread you will see throughout this text—and in all the texts we currently publish—is our well-known problem-solving approach. We keep students focused on problem solving by understanding and applying three easy steps: learn a skill, use the skill to solve equations and inequalities, and then use equations and inequalities as problem-solving tools. This straightforward approach has been the inspiration for many of the features in this text. Learn a Skill. Algebraic skills are demonstrated in the many workedout examples. Learning the skill is immediately reinforced with the Practice Your Skill problem within the example. Use a Skill. Newly acquired skills are used as soon as possible to solve equations and inequalities. Therefore, equations and inequalities are introduced early in the text and then used throughout in a large variety of problem-solving situations. Use equations and inequalities as problem-solving tools. Many word problems are scattered throughout the text. These problems deal with a large variety of applications and constantly show the connections between mathematics and the world around us. Problem-solving suggestions are offered throughout, with special discussion in several sections.

Preface

EXAMPLE 4

xi

Apply Your Skill An investor invests a certain amount of money at 3%. Then he finds a better deal and invests $5000 more than that amount at 5%. His yearly income from the two investments is $650. How much did he invest at each rate?

Solution Let x represent the amount invested at 3%. Then x invested at 5%. 13% 21x 2 0.03x

15% 21x

50002

0.051x

50002

650

650

3x

51x

50002

65,000

3x

5x

25,000

65,000

8x

25,000

65,000

8x

40,000

x Therefore, x

5000

5000 represents the amount

5000

10,000.

He invested $5000 at 3% and $10,000 at 5%.

PRACTICE YOUR SKILL Mrs. Stewart invested a certain amount of money at 6%. Then she found a second investment that paid 9%. Because that second investment carried more risk, she invested $2000 less than the amount she invested at 6%. Her yearly income from the two investments is $870. Find the amount invested in each investment.

Other Special Features •

Many examples contain helpful explanations in the form of line-by-line annotations indicated in blue and placed alongside the solution steps. Also, some examples contain remarks with added information below the example solution. Many examples contain check steps, which verify that an answer is correct and encourage students to check their work, a valuable habit. These checks are accompanied by a checkmark icon and are located at the end of an example.

•

As recommended by the American Mathematical Association of Two-Year Colleges, many basic geometric concepts are integrated in problem-solving settings. The following geometric concepts are presented in problem-solving situations: complementary and supplementary angles, the sum of the measures of the angles of a triangle equals 180°, area and volume formulas, perimeter and circumference formulas, ratio, proportion, Pythagorean theorem, isosceles right triangle, and 30°– 60° right triangle relationships.

•

Every chapter opener is followed by an Internet Project. These projects ask the student to conduct an Internet search on a topic that is relevant to the mathematics presented in that chapter. Many of the projects cover topics concerning the history of math or famous mathematicians.

•

Problems called Thoughts into Words are included in every problem set except the review exercises. These problems are designed to encourage students to express in written form their thoughts about various mathematical ideas.

•

Problems called Further Investigations appear in many of the problem sets. These are designed to lead into upcoming topics or to be slightly more complex. These problems encompass a variety of ideas: some exhibit different approaches to topics covered in the text, some are proofs, some bring in supplementary topics and relationships, and some are more challenging problems. These problems add variety and flexibility to the problem sets and to the classroom experience, but

xii

Preface

they can be omitted entirely without disrupting the continuity pattern of the text. •

Every chapter includes a Chapter Summary, Chapter Review Problem Set, and Chapter Test. In addition, Cumulative Review Problem Sets are placed after Chapters 3, 4, 6, and 9. The cumulative reviews help students retain essential skills.

•

All the answers for the Chapter Review Problem Sets, Chapter Tests, and Cumulative Review Problem Sets appear in the back of the text, along with answers to the odd-numbered problems.

•

We think this text has exceptionally pleasing design features, including the functional use of color. The open format makes the flow of reading continuous and easy. In this design, we hope to capture the spirit of the way we present information: open, clean, friendly, and accessible.

Ancillaries For the Instructor Annotated Instructor’s Edition. (0-495-38829-7) In the AIE, answers are printed next to all respective exercises. Graphs, tables, and other answers appear in an answer section at the back of the text. Problems that are available in electronic form in Enhanced WebAssign are identified by a bulleted problem number. To create an assessment, whether a quiz, homework assignment, or test, the instructor can select the problems by problem number from the identified problems in the text. Complete Solutions Manual. (0-495-38824-6) Laurel Fischer The Complete Solutions Manual provides worked-out solutions to all of the problems in the text. Test Bank. (0-495-38825-4) John Garlow, Tarant County College The Test Bank includes eight tests per chapter as well as three final exams. The tests are made up of a combination of multiple-choice, free-response, true/false, and fill-inthe-blank questions. The Test Bank has added focus on learning objectives and outcomes. Power Lecture CD-ROM with ExamView and JoinIn™. New! (0-495-38822-X) This CD-ROM provides the instructor with dynamic media tools for teaching. Create, deliver, and customize tests (both print and online) in minutes with ExamView® Computerized Testing Featuring Algorithmic Equations. JoinIn™ Student Response System allows you to pose book specific questions and display students’ answers seamlessly within the Microsoft® PowerPoint® slides of your own lecture, in conjunction with the “clicker” hardware of your choice. Easily build solution sets for homework or exams using Solution Builder’s online solutions manual. Microsoft® PowerPoint® lecture slides, figures from the book, and Test Bank, in electronic format, are also included on this CD-ROM. Enhanced WebAssign. (0-495-38827-0) WebAssign, the most widely used homework system in higher education, allows you to assign, collect, grade, and record homework assignments via the Web. Through a partnership between WebAssign and Cengage Learning Brooks/Cole, this proven homework system has been enhanced to include links to textbook sections, video examples, and problem-specific tutorials.

Preface

xiii

Text-Specific Videos. New! (0-495-38828-9) Rena Petrello, Moorpark College These highly praised videos feature valuable 10- to 20-minute demonstrations of nearly every learning objective lesson covered in the text. They may be used as a supplement to classroom learning or as the primary content for an online student. Videos will be available by DVD and online download.

For the Student Student Solutions Manual. (0-495-38823-8) Laurel Fischer The Student Solutions Manual provides worked-out solutions to the odd-numbers problems in the text and worked-out solutions for the Chapter Review Problem Sets, Chapter Tests, and Cumulative Review Problem Sets. Text-Specific Videos. New! (0-495-38828-9) Rena Petrello, Moorpark College These highly praised videos feature valuable 10- to 20-minute demonstrations of nearly every learning objective lesson covered in the text. They may be used as a supplement to classroom learning or as the primary content for an online student. Videos will be available by DVD and online download.

Acknowledgments We would like to take this opportunity to thank the following people who served as reviewers for the first edition of this project: Kochi Angar Nash Community College

Margaret Hathaway Kansas City Kansas Community College

Amir Fazi Arabi Central Virginia Community College

Louis C. Henderson, Jr. Coppin State University

Sarah E. Baxter Gloucester County College

Patricia Horacek Pensacola Junior College

Annette Benbow Tarrant County College

Kelly Jackson Camden County College

A. Elena Bogardus Camden County College

Tom Johnson University of Akron

Dorothy Brown Camden County College

Elias M. Jureidini Lamar State College

Terry F. Clark Keiser University

Carolyn Krause Delaware Tech and Community College

Linda P. Davis Keiser University

Patricia Labonne Cumberland County College

Archie Earl Norfolk State University

Dottie Lapre Nash Community College

Arlene Eliason Minnesota School of Business

Bruce H. Laster Keiser University

Deborah D. Fries Wor-Wic Community College

Maria Luisa Mendez Laredo Community College

Nathaniel Gay Keiser University

Sunny Norfleet St. Petersburg College

xiv

Preface

Ann Ostberg Grace University

Daryl Schrader St. Petersburg College

Armando I. Perez Laredo Community College

Lee Ann Spahr Durham Technical College

Peter Peterson John Tyler Community College

Janet E. Thompson University of Akron

Vien Pham Keiser University

Mary Lou Townsend Wor-Wic Community College

Maria Pickle St. Petersburg College

Bonnie Filer-Tubaugh University of Akron

Sita Ramamurti Trinity Washington University

Susan Twigg Wor-Wic Community College

Denver Riffe National College-Bluefield We are very grateful to the staff of Brooks/Cole, especially Gary Whalen, Kristin Marrs, Laura Localio, Greta Kleinert, and Lynh Pham, for their continuous cooperation and assistance throughout this project. We would also like to express our sincere gratitude to Fran Andersen and to Hal Humphrey. They continue to make life as an author so much easier by carrying out the details of production in a dedicated and caring way. Additional thanks are due to Arlene Kaufmann who spends numerous hours reading page proofs. Jerome E. Kaufmann Karen L. Schwitters


1 1.1 Numerical and Algebraic Expressions 1.2 Addition, Subtraction, Multiplication, and Division of Integers 1.3 Use of Properties 1.4 Prime and Composite Numbers 1.5 Rational Numbers

Corbis RF/Alamy

1.6 Decimals and Real Numbers

■ Positive and negative integers are used to represent golf scores. A positive integer means over par and a negative integer means under par.

K

arla started 2008 with $2500 in her savings account, and she planned to save an additional $150 every month of 2008. If we disregard any accumulated interest, the numerical expression 2500 12(150) represents the amount in her savings account at the end of 2008. The numbers 2, 1, 3, 1, and 4 are Woody’s scores relative to par for five rounds of golf. The numerical expression 2 (1) (3) 1 (4) can be used to determine where Woody stands relative to par at the end of the five rounds. The temperature at 4 A.M. was 14°F. By noon the temperature had increased by 23°F. The numerical expression 14 23 gives the temperature at noon. In the first two chapters of this text, the concept of a numerical expression is used as a basis for reviewing addition, subtraction, multiplication, and division of various kinds of numbers. Then the concept of a variable enables us to move from numerical expressions to algebraic expressions—that is, to start the transition from arithmetic to algebra. Keep in mind that algebra is simply a generalized approach to arithmetic. Many algebraic concepts are extensions of arithmetic ideas. Therefore, we will build on your knowledge of arithmetic to help you with the study of algebra.

Video tutorials for all section learning objectives are available in a variety of delivery modes.

1

I N T E R N E T

P R O J E C T

Do an Internet search on fractions in the stock market. When the New York Stock Exchange began in 1792, the Buttonwood Agreement determined the basis for the stock values. What fraction was used for the basis or spread for stock prices when the NYSE began? Is this currently the basis for stock prices? How did the Common Cents Stock Pricing Act of 1997 change the pricing of stocks? Why do you think the basis for stock prices was changed?

1.1

Numerical and Algebraic Expressions OBJECTIVES 1

Learn the Basic Vocabulary and Symbols for Sets

2

Simplify Numerical Expressions According to the Order of Operations

3

Evaluate Algebraic Expressions Algebraic Expressions In arithmetic, we use symbols such as 4, 8, 17, and p to represent numbers. We indicate the basic operations of addition, subtraction, multiplication, and division by the symbols , , , and , respectively. For example, we can write the indicated sum of eight and four as 8 4; 8 4 is a numerical expression. In algebra, the concept of a “variable” provides the basis for generalizing. By using x and y to represent any number, we can use the expression x y to represent the indicated sum of any two numbers. The x and y in this expression are called variables, and the phrase x y is called an algebraic expression. We commonly use letters of the alphabet such as x, y, z, and w as variables. The key idea is that they represent numbers; therefore, as we review various operations and properties pertaining to numbers, we are building the foundation for our study of algebra. Many of the notational agreements made in arithmetic are extended to algebra with a few slight modifications. The following chart summarizes the notational agreements pertaining to the four basic operations. Note the variety of ways to write a product by using parentheses to indicate multiplication. Actually, the ab form is the simplest and the form probably used most often; expressions such as abc, 6x, and 7xyz all indicate multiplication. Also note the various forms for indicating division; the c fractional form, , is generally used in algebra, although the other forms do serve a d purpose at times. Operation

Arithmetic

Algebra

Addition Subtraction

46 72

xy wz

Multiplication

98

Division

8 8 2, , 2冄8 2

a b, a(b), (a)b, (a)(b), or ab c c d, , or d冄 c d

Vocabulary The sum of x and y The difference of w and z The product of a and b The quotient of c and d

1 Learn the Basic Vocabulary and Symbols for Sets As we review arithmetic ideas and introduce algebraic concepts, it is convenient to use some of the basic vocabulary and symbols associated with sets. A set is 2

1.1

Numerical and Algebraic Expressions

3

a collection of objects, and the objects are called elements or members of the set. In arithmetic and algebra, the elements of a set are often numbers. To communicate about sets, we use set braces, { }, to enclose the elements (or a description of the elements), and we use capital letters to name sets. For example, we can represent a set A, which consists of the vowels of the English alphabet, in these ways: A {vowels of the English alphabet}

Word description

A {a, e, i, o, u}

List or roster description

We can modify the listing approach if the number of elements is quite large. For example, all the letters of the alphabet can be listed as {a, b, c, . . . , z} We simply begin by writing enough elements to establish a pattern, and then the three dots indicate that the set continues in that pattern. The final entry indicates the last element of the pattern. If we write {1, 2, 3, . . .} the set begins with the counting numbers, 1, 2, and 3. The three dots indicate that it continues in a like manner forever; there is no last element. A set that consists of no elements is called the null set (written ). Two sets are said to be equal if they contain exactly the same elements. For example, {1, 2, 3} {2, 1, 3} because both sets contain the same elements; the order in which the elements are written doesn’t matter. The slash mark through the equality symbol denotes not equal to. Thus if A {1, 2, 3} and B {1, 2, 3, 4}, we can write A B, which we read as “set A is not equal to set B.”

2 Simplify Numerical Expressions According to the Order of Operations Now let’s simplify some numerical expressions that involve the set of whole numbers — that is, the set {0, 1, 2, 3, . . .}.

EXAMPLE 1

Simplify 8 7 4 12 7 14.

Solution The additions and subtractions should be performed from left to right in the order in which they appear. 8 7 4 12 7 14 15 4 12 7 14 11 12 7 14 23 7 14 16 14 30 Thus 8 7 4 12 7 14 simplifies to 30.

▼ PRACTICE YOUR SKILL Simplify 5 3 12 2 1.

■

4

Chapter 1


EXAMPLE 2

Simplify 7(9 5).

Solution The parentheses indicate the product of 7 and the quantity 9 5. Perform the addition inside the parentheses first, and then multiply by 7. 719 52 71142 98 Thus 7(9 5) simplifies to 98.

▼ PRACTICE YOUR SKILL Simplify 12(6 4).

EXAMPLE 3

■

Simplify (7 8) (4 1).

Solution First, perform the operations inside the parentheses. Then divide the two quantities. 17 82 14 12 15 3 5 Thus (7 8) (4 1) simplifies to 5.

▼ PRACTICE YOUR SKILL Simplify (9 1) (9 5).

■

78 . We don’t 41 need parentheses in this case, because the fraction bar indicates that the sum of 7 and 8 is to be divided by the difference 4 1. A problem may, however, contain both parentheses and fraction bars, as the next example illustrates. We frequently express a problem like Example 3 in the form

EXAMPLE 4

Simplify

(4 2)(7 1) 7 . 9 10 3

Solution First simplify above and below the fraction bars, and then proceed to evaluate as follows: (4 2)(7 1) 9

(6)(6) 7 7 10 3 9 7

36 1415 9

▼ PRACTICE YOUR SKILL Simplify

18 52 11 32 6

12 . 11 7

■

1.1

EXAMPLE 5


5

Simplify 7 9 5.

Solution If there are no parentheses to indicate otherwise, multiplication takes precedence over addition. First perform the multiplication, and then do the addition. 7 # 9 5 63 5 68 Thus 7 # 9 5 simplifies to 68.

▼ PRACTICE YOUR SKILL Simplify 16 # 2 6 .

■

(Compare Example 2 and Example 5, and note the difference in meaning.)

EXAMPLE 6

Simplify 8 4 3 14 2.

Solution The multiplication and division should be done first in the order in which they appear, from left to right. Then perform the addition and subtraction in the order in which they appear. 8 4 # 3 14 2 8 12 7 20 7 13 Therefore, 8 4 # 3 14 2 simplifies to 13.

▼ PRACTICE YOUR SKILL Simplify 3 5 # 2 10 2 . EXAMPLE 7

■

Simplify 18 3 2 8 10 2.

Solution If we perform the multiplications and divisions first, in the order in which they appear, and then do the additions and subtractions, our work takes on the following format: 18 3 2 8 10 2 6 2 80 2 12 40 52 Thus 18 3 # 2 8 # 10 2 simplifies to 52.

▼ PRACTICE YOUR SKILL Simplify 24 6 # 2 4 # 8 2 . EXAMPLE 8

■

Simplify 5 6[2(3 9)].

Solution We use brackets for the same purpose as parentheses. In such a problem, we need to simplify from the inside out, performing the operations in the innermost parentheses first.

6

Chapter 1


5 63213 92 4 5 63 21122 4 5 63 244 5 144 149

▼ PRACTICE YOUR SKILL Simplify 12 4冤512 82冥 .

■

Exponents are used in both arithmetic and algebra to indicate repeated multiplications. For example, we can write 5 # 5 # 5 as 53, and the 3 indicates that 5 is to be used as a factor three times. For the expression 53, the 5 is referred to as the base and the 3 is called the exponent. The exponent is also sometimes referred to as a power. The expression 53 can be read “5 raised to the 3 power.” Exponents are also used in algebraic expressions when the base is a variable. For example, x4 means that x will be used as a factor four times, so x4 x # x # x # x. An exponent of 1 is usually not written. For example, y1 would just be written as y. When simplifying numerical expressions, exponential expressions should be evaluated before performing the operations of multiplication, division, addition, or subtraction. The following example presents an expression with exponents that needs simplifying.

EXAMPLE 9

Simplify 62 # 5 24.

Solution First simplify the expressions with exponents by performing their repeated multiplications. Then do the indicated multiplication followed by the addition. 62 # 5 24 16 # 62 # 5 12 # 2 # 2 # 22 36 # 5 16 180 16 196

▼ PRACTICE YOUR SKILL Simplify 3 # 52 7 # 23 .

■

Let us now summarize the ideas presented in the preceding examples on simplifying numerical expressions. When we simplify a numerical expression, the operations should be performed in the order listed below:

Order of Operations 1. Perform the operations inside the symbols of inclusion (parentheses and brackets) and above and below each fraction bar. Start with the innermost inclusion symbol. 2. Evaluate all exponential expressions. 3. Perform all multiplications and divisions in the order in which they appear from left to right. 4. Perform all additions and subtractions in the order in which they appear from left to right.

1.1


7

3 Evaluate Algebraic Expressions We can use the concept of a variable to generalize from numerical expressions to algebraic expressions. Each of the following is an example of an algebraic expression: 3x 2y

5a 2b c

7(w z)

a2 b2

5d 3e 2c d

2xy 5yz

(x y)(x y)

9x2 y2

An algebraic expression takes on a numerical value whenever each variable in the expression is replaced by a specific number. For example, if x is replaced by 9 and z by 4, the algebraic expression x z becomes the numerical expression 9 4, which simplifies to 5. We say that x z has a value of 5 when x equals 9 and z equals 4. The value of x z, when x equals 25 and z equals 12, is 13. The general algebraic expression x z has a specific value each time x and z are replaced by numbers. Consider the next examples, which illustrate the process of finding the value of an algebraic expression. This process is often referred to as evaluating algebraic expressions.

EXAMPLE 10

Find the value of 3x 2y when x is replaced by 5 and y by 17.

Solution The following format is convenient for such problems: 3x 2y 3152 21172

when x 5 and y 17

15 34 49

▼ PRACTICE YOUR SKILL Evaluate 4x 3y when x 5 and y 2.

■

In Example 10, for the algebraic expression, 3x 2y, note that the multiplications “3 times x” and “2 times y” are implied without the use of parentheses. The algebraic expression switches to a numerical expression when numbers are substituted for variables; in that case, parentheses are used to indicate the multiplication. We could also use the raised dot to indicate multiplication; that is, 3(5) 2(17) could be written as 3 5 2 17. Furthermore, note that once we have substituted numbers for variables, we can start the process of simplifying numerical expressions.

EXAMPLE 11

Find the value of 12a 3b when a 5 and b 9.

Solution 12a 3b 12152 3192

when a 5 and b 9

60 27 33

▼ PRACTICE YOUR SKILL Evaluate 10a 5b when a 8 and b 3.

■

8

Chapter 1


EXAMPLE 12

Evaluate 4xy 2xz 3yz when x 8, y 6, and z 2.

Solution 4xy 2xz 3yz 4(8)(6) 2(8)(2) 3(6)(2) when x 8, y 6, and z2 192 32 36 188

▼ PRACTICE YOUR SKILL Evaluate 3xy 2x 4xyz when x 6, y 2, and z 4.

EXAMPLE 13

Evaluate

■

5c d for c 12 and d 4. 3c d

Solution 5(12) 4 5c d 3c d 3(12) 4

60 4 36 4

64 32

for c 12 and d 4

2

▼ PRACTICE YOUR SKILL Evaluate

EXAMPLE 14

7c d when c 4 and d 2. 2c d

■

Evaluate (2x 5y)(3x 2y) when x 6 and y 3.

Solution (2x 5y)(3x 2y) (2 6 5 3)(3 6 2 3) when x 6 and y3 (12 15)(18 6) (27)(12) 324

▼ PRACTICE YOUR SKILL Evaluate (x 3y)(4x 2y) when x 8 and y 3.

EXAMPLE 15

■

Evaluate 5x2 3x 7 when x 4.

Solution It is always a good practice to use parentheses when substituting a value for a variable. This is especially true when the variable is raised to an exponent.

1.1


9

5x2 3x 7 5142 2 3142 7 when x 4 51162 3142 7 80 12 7 68 7 75

▼ PRACTICE YOUR SKILL Evaluate a2 2ab b2 for a 7 and b 4.

EXAMPLE 16

■

bh , where b is the length of the base of 2 the triangle, and h represents the length of the height of the triangle. Use the formula bh to determine the value of A when b 14 and h 5. A 2

The formula for the area of a triangle is A

Solution A

14 # 5 bh 2 2 35

when b 14 and h 5

▼ PRACTICE YOUR SKILL Use the formula A

CONCEPT QUIZ

bh to determine A when b 15 and h 4. 2

■

For Problems 1–10, answer true or false. 1. The expression ab indicates the sum of a and b. 2. Any of the following notations, (a)b, a b, a(b), can be used to indicate the product of a and b. 3. The phrase “2x y 4z” is called an algebraic expression. 4. A set is a collection of objects, and the objects are called “terms.” 5. The sets {2, 4, 6, 8} and {6, 4, 8, 2} are equal. 6. The set {1, 3, 5, 7, . . .} has a last element of 99. 7. The null set has one element. 8. To evaluate 24 6 2, the first operation that should be performed is to multiply 6 times 2. 9. To evaluate 6 8 3, the first operation that should be performed is to multiply 8 times 3. 10. The algebraic expression 2(x y) simplifies to 24, if x is replaced by 10 and y is replaced by 0.

Problem Set 1.1 1 Learn the Basic Vocabulary and Symbols for Sets For Problems 1– 4, given the word description of the set, write its list description as a set. 1. A {even whole numbers less than 5} 2. B {even whole numbers} 3. C {odd whole numbers}

4.

D {states in the United States that begin with the letter T}

For Problems 5 – 8, insert the symbol for equal or not equal to make each of the following a true statement. 5. {1, 2, 3, 4, 5}

{1, 2, 3, 4, 5, . . .}

6. {2, 4, 6, . . .}

{6, 4, 2}

7. {7, 5, 3, 1}

{1, 3, 5, 7}

8. {a, b, c, }

{a, b, c, . . .}

10

Chapter 1


2 Simplify Numerical Expressions According to the Order of Operations For Problems 9 – 48, simplify each of the numerical expressions. 9. 9 14 7

10. 32 14 6

11. 7(14 9)

12. 8(6 12)

13. 16 5 7

14. 18 3(5)

15.

3 # 42

16. 2 # 5

51. 16a 9b

for a 3 and b 4

52. 14a 5b

for a 7 and b 9

53. 4x 7y 3xy

for x 4 and y 9

54. x 8y 5xy

for x 12 and y 3

55. x2 2x 6

for x 5

56. x2 3x 1

for x 8

3

57. 14xz 6xy 4yz

17. 4(12 9) 3(8 4)

18. 7(13 4) 2(19 11)

19. 4(7) 6(9)

20. 8(7) 4(8)

21. 6 42

22. 20 42

59. y2 2xy x2

23. 6 7 5 8 3 9

24. 8(13) 4(9) 2(7)

60. x2 y2

25. (6 9)(8 4)

26. (15 6)(13 4)

27. (7 4)

28. (3 4)

2

58. 9xy 4xz 3yz

29. 6 4[3(9 4)]

30. 92 3[2(5 2)]

31. 16 8 4 36 4 2

32. 7 8 4 72 12

33.

8 12 9 15 4 8

34.

19 7 38 14 6 3

35. 56 [3(9 6)]

36. 17 2[3(4 2)]

37. 7 # 4 # 2 8 14

38. 14 7 # 8 35 7 # 2

39. 32 8 2 24 6 1

3

45. 83

47.

12(7 4) 9

4(12 7) 5

54 n n 3

62.

n 60 n 4 n 6

63.

y 16 50 y 6 3

64.

90 w w 57 9 7

for n 9 for n 12

65. (x y)(x y)

for y 8 for w 6

for x 8 and y 3 for x 7 and y 4

68. (3a b)(7a 2b)

42. 5 8 4 8 4 3 6

for x 10 and y 2

67. (5x 2y)(3x 4y)

41. 4 9 12 18 2 3

6(8 3)

for x 3 and y 4

66. (x 2y)(2x y)

40. 48 12 7 2 2 1

43.

for x 7, y 3, and z 2

61.

2

for x 8, y 5, and z 7

44.

3(17 9) 4

46. 78

9(16 7) 3

6(21 9) 4

4·65·3 7·96·5 72·3 3·58·2

9·64 7·84 48. 5 · 8 10 6 · 5 20

3 Evaluate Algebraic Expressions For Problems 49 –72, evaluate each algebraic expression for the given values of the variables. 49. 7x 4y

for x 6 and y 8

50. 8x 6y

for x 9 and y 5

for x 3 and y 6 for a 5 and b 7

69. 6 3[2(x 4)]

for x 7

70. 9 4[3(x 3)]

for x 6

71. 81 2[5(n 4)]

for n 3

72. 78 3[4(n 2)]

for n 4

bh For Problems 73 –78, find the value of for each set of 2 values for the variables b and h. 73. b 8 and h 12

74. b 6 and h 14

75. b 7 and h 6

76. b 9 and h 4

77. b 16 and h 5

78. b 18 and h 13

79. You should be able to do calculations like those in Problems 9 – 48 both with and without a calculator. Do Problems 9 – 48 with your calculator. Be sure to use the parentheses and exponent keys when appropriate.

1.2

Addition, Subtraction, Multiplication, and Division of Integers

11

THOUGHTS INTO WORDS 80. Explain the difference between a numerical expression and an algebraic expression.

81. Your friend keeps getting an answer of 45 when simplifying 3 2(9). What mistake is he making, and how would you help him?

FURTHER INVESTIGATIONS 84. Insert parentheses so that 36 12 3 3 6 2 is equal to 38.

Grouping symbols can affect the order in which the arithmetic operations are performed. For the following problems, insert parentheses so that the expression is equal to the given value.

85. Insert parentheses so that 36 12 3 3 6 2 is equal to 55.

82. Insert parentheses so that 36 12 3 3 6 2 is equal to 20. 83. Insert parentheses so that 36 12 3 3 6 2 is equal to 50.

Answers to the Example Practice Skills 1. 11 2. 24 3. 2 15. 121 16. 30

4. 5

5. 38

6. 8

7. 24

8. 212

9. 19

10. 26

11. 65

12. 216

13. 5

14. 442

7. False 8. False

9. True

10. False


1.2

2. True 3. True

4. False 5. True

6. False

Addition, Subtraction, Multiplication, and Division of Integers OBJECTIVES 1

Know the Names for Certain Sets of Integers

2

Add Integers

3

Subtract Integers

4

Multiply Integers

5

Divide Integers

6

Simplify Numerical Expressions Involving Integers

7

Evaluate Algebraic Expressions for Integer Values

8

Use Integers to Solve Application Problems

1 Know the Names for Certain Sets of Integers “A record temperature of 35° below zero was recorded on this date in 1904.” “The IMDigital stock closed down 3 points yesterday.” “On a first-down sweep around left end, Faulk lost 7 yards.” “The West Coast Manufacturing Company reported assets of 50 million dollars and liabilities of 53 million dollars for 2007.” These statements illustrate our need for negative numbers.

12

Chapter 1


The number line is a helpful visual device for our work at this time. We can associate the set of whole numbers with evenly spaced points on a line as indicated in Figure 1.1. For each nonzero whole number, we can associate its negative to the left

0

1

2

3

4

5

Figure 1.1

of zero; with 1 we associate 1, with 2 we associate 2, and so on, as indicated in Figure 1.2. The set of whole numbers, along with 1, 2, 3, and so on, is called the set of integers.

−5

−4

−3

−2

−1

0

1

2

3

4

5

Figure 1.2

The following terminology is used with reference to the integers: {. . . , 3, 2, 1, 0, 1, 2, 3, . . .} {1, 2, 3, 4, . . .} {0, 1, 2, 3, 4, . . .}

Integers Positive integers Nonnegative integers

{. . . , 3, 2, 1}

Negative integers

{. . . , 3, 2, 1, 0}

Nonpositive integers

The symbol 1 can be read as “negative one,” “opposite of one,” or “additive inverse of one.” The opposite-of and additive-inverse-of terminology is very helpful when working with variables. For example, reading the symbol x as “opposite of x” or “additive inverse of x” emphasizes an important issue. Because x can be any integer, x (the opposite of x) can be zero, positive, or negative. If x is a positive integer, then x is negative. If x is a negative integer, then x is positive. If x is zero, then x is zero. These statements can be written and illustrated on the number lines as in Figure 1.3. If x 3, then x (3) 3. If x 3, then x (3) 3. If x 0, then x (0) 0.

x −4 −3 −2 −1

0

1

2

3

4

0

1

2

3

4

1

2

3

4

x −4 −3 −2 −1

x −4 −3 −2 −1

0

Figure 1.3

From this discussion we can recognize the following general property:

Property 1.1 If a is any integer, then (a) a (The opposite of the opposite of any integer is the integer itself.)

1.2


13

2 Add Integers The number line is also a convenient visual aid for interpreting the addition of integers. Consider the following examples and their number-line interpretations as shown in Figure 1.4.

Problem

Number line interpretation 3

32

2

−5 −4 −3 −2 −1 0 1 2 3 4 5 3

3 (2)

3 2

3 (2)

3 (2) 1

−3

−5 −4 −3 −2 −1 0 1 2 3 4 5 −2

325

−2

−5 −4 −3 −2 −1 0 1 2 3 4 5 2

Sum

3 2 1

−3

−5 −4 −3 −2 −1 0 1 2 3 4 5

3 (2) 5

Figure 1.4

Once you get a feel for movement on the number line, simply forming a mental image of this movement is sufficient. Consider the next addition problems, and mentally picture the number-line interpretation. Be sure that you agree with all of our answers. 5 (2) 3

6 4 2

8 11 3

7 (4) 11

5 9 4

9 (2) 7

14 (17) 3

0 (4) 4

6 (6) 0

The last example illustrates a general property that you should note: Any integer plus its opposite equals zero.

Remark: Profits and losses pertaining to investments also provide a good physical model for interpreting the addition of integers. A loss of $25 on one investment, along with a profit of $60 on a second investment, produces an overall profit of $35. We can express this as 25 60 35. You may want to check the preceding examples using a profit and loss interpretation. Even though all problems that involve the addition of integers could be done by using the number-line interpretation, it is sometimes convenient to give a more precise description of the addition process. For this purpose, we need to consider briefly the concept of absolute value. The absolute value of a number is the distance between the number and zero on the number line. For example, the absolute value of 6 is 6. The absolute value of 6 is also 6. The absolute value of 0 is 0. Vertical bars on either side of a number denote absolute value. Thus we write 060 6

0 6 0 6

000 0

Note that the absolute value of a positive number is the number itself, but the absolute value of a negative number is its opposite. Thus the absolute value of any number except 0 is positive.

14

Chapter 1


We can describe the process of adding integers by using the concept of absolute value.

Addition of Integers: Two Positive Integers The sum of two positive integers is the sum of their absolute values. (The sum of two positive integers is a positive integer.) 43 54 0 43 0 0 54 0 43 54 97

Addition of Integers: Two Negative Integers The sum of two negative integers is the opposite of the sum of their absolute values. (The sum of two negative integers is a negative integer.) (67) (93) (0 67 0 0 93 0 ) (67 93) 160

Addition of Integers: One Positive and One Negative Integer The sum of a positive integer and a negative integer can be found by subtracting the smaller absolute value from the larger absolute value and giving the result the sign of the original number that has the larger absolute value. If the integers have the same absolute value, then their sum is zero. 82 (40) 0 82 0 0 40 0 82 40 42 74 (90) (0 90 0 0 74 0) (90 74) 16 (17) 17 0 17 0 0 17 0 17 17 0

Addition of Integers: Zero and Another Integer The sum of zero and any integer is the integer itself.

0 (46) 46 72 0 72

1.2

EXAMPLE 1


15

Add the integers to find their sum. (a) 18 (56) (e) 114 48

(b) 71 (32) (f) 45 (73)

(c) 64 (49) (g) 46 (46)

(d) 56 93 (h) 0 (81)

Solution (a) 18 1562 1 0 18 0 056 0 2 118 562 74

(b) 71 1322 1 0 71 0 032 0 2 171 322 103 (c) 64 1492 064 0 0 49 0 64 49 15 (d) 56 93 0 93 0 056 0 93 56 37

(e) 114 48 1 0 114 0 048 0 2 1114 482 66 (f) 45 1732 1 0 73 0 0 45 0 2 173 452 28 (g) 46 1462 0

(h) 0 1812 81

▼ PRACTICE YOUR SKILL Find the sum. (a) 24 (32)

(b) 40 75

(c) 86 (31)

(d) 15 (15)

■

3 Subtract Integers The following examples illustrate a relationship between addition and subtraction of whole numbers: 725

because 2 5 7

963

because 6 3 9

514

because 1 4 5

This same relationship between addition and subtraction holds for all integers: 5 6 1

because 6 112 5

4 9 13 because 9 1132 4

3 172 4

8 132 11

because 7 4 3 because 3 11 8

Now consider a further observation: 5 162 1

5 6 1

and

4 9 13

and

4 192 13

and

3 7 4

3 172 4

8 132 11

and

8 3 11

The previous examples help us realize that we can state the subtraction of integers in terms of the addition of integers. More precisely, a general description for the subtraction of integers follows:

Subtraction of Integers If a and b are integers, then a b a (b).

16

Chapter 1


It may be helpful for you to read a b a (b) as “a minus b is equal to a plus the opposite of b.” Every subtraction problem can be changed into an equivalent addition problem, as illustrated by the next example.

EXAMPLE 2

Subtract the integers. (a) 6 13

(b) 9 (12)

(c) 8 13

(d) 7 (8)

Solution (a) 6 13 6 1132 7 (b) 9 1122 9 12 21

(c) 8 13 8 1132 21 (d) 7 182 7 8 1

It should be apparent that addition of integers is a key operation. Being able to add integers effectively is indispensable for further work in algebra.

▼ PRACTICE YOUR SKILL Subtract the integers. (a) 20 (38)

(b) 35 75

(c) 81 (31)

(d) 25 (25)

■

4 Multiply Integers Multiplication of whole numbers may be interpreted as repeated addition. For example, 3 4 means the sum of three 4s; thus 3 4 4 4 4 12. Consider the following examples, which use the idea of repeated addition to find the product of a positive integer and a negative integer. 3(2) 2 (2) (2) 6 2(4) 4 (4) 8 4(1) 1 (1) (1) (1) 4 Note the use of parentheses to indicate multiplication. Sometimes both numbers are enclosed in parentheses; in this case we would have (3)(2). When multiplying whole numbers, we realize that the order in which we multiply two factors does not change the product; in other words, 2(3) 6 and 3(2) 6. Using this idea, we can now handle a negative integer times a positive integer: (2)(3) (3)(2) (2) (2) (2) 6 (3)(2) (2)(3) (3) (3) 6 (4)(3) (3)(4) (4) (4) (4) 12 Finally, let’s consider the product of two negative integers. The following pattern helps us with the reasoning for this situation. 4(3) 12 3(3) 9 2(3) 6 1(3) 3 0(3) 0 (1)(3) ?

The product of zero and any integer is zero

1.2


17

Certainly, to continue this pattern, the product of 1 and 3 has to be 3. This type of reasoning helps us to realize that the product of any two negative integers is a positive integer. Using the concept of absolute value, we can now precisely describe the multiplication of integers:

Multiplying Integers 1. The product of two positive integers or two negative integers is the product of their absolute values. 2. The product of a positive integer and a negative integer (either order) is the opposite of the product of their absolute values. 3. The product of zero and any integer is zero.

The next example illustrates this description of multiplication.

EXAMPLE 3

Find the product of the integers. (a) (5)(2) (e) (0)( 28)

(b) (7)(6)

(c) (8)(9)

(d) (14)(0)

Solution (a) 152 122 05 0 (b) 172162 1 0 7 0

02 0 5 # 2 10

# #

(c) 182192 1 08 0

06 0 2 17 # 62 42

#

09 0 2 18 # 92 72

(d) 1142 102 0 (e) 1021282 0

▼ PRACTICE YOUR SKILL Find the product. (a) (6)(2)

(b) (1)(5)

(c) (0)(8)

(d) (4)(6) ■

This example showed a step-by-step process for multiplying integers. In reality, however, the key issue to remember is whether the product is positive or negative. In other words, we need to remember that the product of two positive integers or two negative integers is a positive integer, and that the product of a positive integer and a negative integer (in either order) is a negative integer. Then we can avoid the step-by-step analysis and simply write the results as follows: 172192 63 182172 56 152162 30 1421122 48

18

Chapter 1


5 Divide Integers By looking back at our knowledge of whole numbers, we can get some guidance for our work with integers. We know, for example, that

8 4, because 2 4 8. In other 2

words, we find the quotient of two whole numbers by looking at a related multiplication problem. In the following examples, we have used this same link between multiplication and division to determine the quotients: 8 4 2

because 122142 8

10 2 5

because 152122 10

12 3 4

because 142132 12

0 0 6

because 162102 0

9 is undefined because no number times 0 produces 9 0 0 is undefined 0

because any number times 0 equals 0. Remember that division by zero is undefined!

Dividing Integers 1. The quotient of two positive integers or two negative integers is the quotient of their absolute values. 2. The quotient of a positive integer and a negative integer (or a negative and a positive) is the opposite of the quotient of their absolute values. 3. The quotient of zero and any nonzero integer (zero divided by any nonzero integer) is zero.

The next example illustrates this description of division.

EXAMPLE 4

Find the quotient of the integers. (a)

8 4

(b)

14 2

(c)

15 3

(d)

0 4

Solution (a) (b)

08 0 8 8 2 4 04 0 4

0 14 0 14 14 a b a b 7 2 02 0 2

(c)

015 0 15 15 a b a b 5 3 03 0 3

(d)

0 0 4

1.2


19

▼ PRACTICE YOUR SKILL Find the quotient. (a)

18 6

(b)

0 8

(c)

50 2

(d)

80 4

■

For practical purposes, it is important to determine whether the quotient is positive or negative. The quotient of two positive integers or two negative integers is positive, and the quotient of a positive integer and a negative integer or of a negative integer and a positive integer is negative. We can then simply write the quotients as follows without showing all of the steps: 18 3 6

24 2 12

36 4 9

Remark: Occasionally people use the phrase “two negatives make a positive.” We hope they realize that the reference is to multiplication and division only; in addition, the sum of two negative integers is still a negative integer. It is probably best to avoid such imprecise statements.

6 Simplify Numerical Expressions Involving Integers Now we can simplify numerical expressions involving any or all of the four basic operations with integers. Keep in mind the agreements on the order of operations we stated in Section 1.1.

EXAMPLE 5

Simplify 4132 7182 3192 .

Solution 4132 7182 3192 12 1562 1272 12 56 1272 41

▼ PRACTICE YOUR SKILL Simplify 5(3) 4(2) 7(1).

EXAMPLE 6

Simplify

■

8 4(5) . 4

Solution 8 4152 4

8 20 4

28 4

7


20 4122 3

.

■

20

Chapter 1


7 Evaluate Algebraic Expressions for Integer Values Evaluating algebraic expressions often involves using two or more operations with integers. The final examples of this section illustrate such situations. Be sure to use parentheses when substituting values for the variables.

EXAMPLE 7

Find the value of 3x 2y when x 5 and y 9.

Solution 3x 2y 3(5) 2(9)

when x 5 and y 9

15 (18) 3

▼ PRACTICE YOUR SKILL Find the value of 4x 5y when x 2 and y 6.

EXAMPLE 8

■

Evaluate 2a 9b for a 4 and b 3.

Solution 2a 9b 2(4) 9(3)

when a 4 and b 3

8 (27) 35

▼ PRACTICE YOUR SKILL Evaluate 3a 4b when a 1 and b 5.

EXAMPLE 9

The formula C

51F 322 9

Celsius. Find the value of

■

is used to convert temperatures from Fahrenheit to

51F 322 9

when F 5.

Solution Evaluate

51F 322 9

51F 322 9

when F 5.

515 322 9 51272 9

when F 5 135 15 9

▼ PRACTICE YOUR SKILL Find the value of

51F 322 9

when F 4.

■

1.2


21

8 Use Integers to Solve Application Problems The addition, subtraction, multiplication, and division of integers can be used to solve problems that involve both positive and negative numbers. In business situations, profits are represented by positive numbers, and losses are represented by negative numbers. In sports such as football, a loss of 5 yards is represented as 5 yards, and in golf being 2 strokes under par would be represented as 2. In the problem set, you will apply your skill in adding, subtracting, multiplying, and dividing integers to solve application problems as shown in the next two examples.

EXAMPLE 10

Apply Your Skill

Creatas Images/Jupiter Images

A nutrition and fitness instructor asked her clients to weigh in and record their weight loss or gain in pounds every Friday over a six-week period. Given the records for one of the clients (Dominic), determine his overall weight loss or gain.

Dominic

Aug. 12

Aug. 19

Aug. 26

Sept. 2

Sept. 9

Sept. 16

loss of 3

gain of 1

loss of 5

loss of 2

gain of 3

loss of 4

Solution Let’s use negative numbers to represent the losses and positive numbers to represent the gains. Then Dominic’s weekly records would be represented by 3, 1, 5, 2, 3, and 4. Adding the numbers will give the overall loss or gain. 3 112 152 122 132 142 10 So over the six weeks Dominic lost 10 pounds.

▼ PRACTICE YOUR SKILL A boat manufacturer reported that the company incurred a loss of $835,000 in 2004, a gain of $320,000 in 2005, a gain of $410,000 in 2006, and a loss of $120,000 in 2007. Use the addition of integers to describe this situation, and determine the boat manufacturer’s total loss or gain for the four-year period. ■

Comstock Images/Jupiter Images

EXAMPLE 11

Apply Your Skill In a local amateur poker tournament, Dwayne lost $25 for three hands, he lost $14 for two hands, and he won $10 for four hands in the first round. Determine the average amount Dwayne won or lost for his hands in the first round of play.

Solution Let’s use negative numbers to represent the losses and positive numbers to represent the wins. Dwayne’s first round would be represented by 3(25) 2(14) 4(10). To find the average we would divide this expression by 9, the total number of hands played. 31252 21142 41102 9

75 28 40 9 63 9 7

So, on average, Dwayne lost $7 per hand in the first round.

22

Chapter 1


▼ PRACTICE YOUR SKILL In a golf tournament to raise funds for scholarships, Kay shot 2 strokes under par on four holes, 1 stroke under par on three holes, and 3 strokes over par on two holes. Use multiplication and addition of integers to describe this situation and to determine Kay’s score relative to par for the tournament. ■

CONCEPT QUIZ

For Problems 1–10, answer true or false. 1. The number zero is considered a positive integer. 2. The absolute value of a number is the distance on the number line between the number and one. 3. The 冟4冟 is 4. 4. The opposite of 5 is 5. 5. a minus b is equivalent to a plus the opposite of b. 6. The product of a positive integer and a negative integer is a positive integer. 7. When multiplying three negative integers, the product is negative. 8. The rules for adding integers and the rules for multiplying integers are the same. 9. The quotient of two negative integers is negative. 10. The product of zero and any nonzero integer is zero.

Problem Set 1.2 1 Know the Names for Certain Sets of Integers For Problems 1– 4, match the description with the set of numbers. 1. {. . . , 3, 2, 1}

A. Positive integers

2. {1, 2, 3, . . .}

B. Negative integers

3. {0, 1, 2, 3, . . .}

C. Nonnegative integers

4. {. . . , 3, 2, 1, 0}

D. Nonpositive integers

2 Add Integers For Problems 5 –24, find each sum.

3 Subtract Integers For Problems 25 – 44, subtract as indicated. 25. 3 8

26. 5 11

27. 4 9

28. 7 8

29. 5 (7)

30. 9 (4)

31. 6 (12)

32. 7 (15)

33. 11 (10)

34. 14 (19)

35. 18 27

36. 16 25

37. 34 63

38. 25 58

39. 45 18

40. 52 38

41. 21 44

42. 26 54

43. 53 (24)

44. 76 (39)

5. 17 (9)

6. 16 (5)

7. 8 (19)

8. 9 (14)

9. 7 (8)

10. 6 (9)

For Problems 45 –58, add or subtract as indicated.

11. 15 8

12. 22 14

45. 6 8 9

46. 5 9 4

13. 13 (18)

14. 15 (19)

47. 4 (6) 5 8

48. 3 8 9 (6)

15. 27 8

16. 29 12

49. 5 7 8 12

50. 7 9 4 12

17. 32 (23)

18. 27 (14)

51. 6 4 (2) (5)

19. 25 (36)

20. 34 (49)

52. 8 11 (6) (4)

21. 54 (72)

22. 48 (76)

53. 6 5 9 8 7

23. 34 (58)

24. 27 (36)

55. 11 (14) (17) 18

54. 7 12 14 15 9

1.2


56. 15 20 14 18 9

95. 6x 7y for x 4 and y 6

57. 16 21 (15) (22)

96. 5x 12y for x 5 and y 7

58. 17 23 14 (18)

97.

5x 3y 6

4 Multiply Integers

98.

7x 4y 8

For Problems 59 –70, find the product.

for x 6 and y 4 for x 8 and y 6

99. 5x 6y xy for x 7 and y 7 100. 3x 7y 2xy for x 6 and y 4

59. 5(6)

60. 7(9)

61. (7)(8)

62. (6)(9)

63. (5)(12)

64. (7)(14)

For Problems 101–104, find the value of

65. 14(9)

66. 17(7)

the given values for F.

67. (11)(14)

68. (13)(17)

69. (15)(15)

70. (18)(18)

101. F 59

102. F 68

103. F 14

104. F 4

5 Divide Integers

73.

42 6

74.

72 8

75.

96 8

76.

91 7

78. 1482 132

81. 1722 18

82. 1702 5

80. 102 172

6 Simplify Numerical Expressions Involving Integers For Problems 83 –92, simplify each numerical expression. 85. 132 182 192 152 87. 89.

13 1252 3 71102 6192 4

84. 6132 5192

86. 172 162 142132 88. 90.

for each of

9C 32 for each of 5

105. C 25

106. C 35

107. C 40

108. C 0

8 Use Integers to Solve Application Problems

77. 1562 142

83. 3142 5172

9

the given values for C.

35 72. 5

79. 1192 102

51F 322

For Problems 105 –108, find the value of

For Problems 71– 82, find the quotient. 27 71. 3

23

15 1362 7 6182 41142 8

91. 2132 3142 4152 6172 92. 2142 4152 7162 3192

7 Evaluate Algebraic Expressions for Integer Values

109. The temperature at 5 A.M. was 17°F. By noon the temperature had increased by 14°F. Use the addition of integers to describe this situation, and determine the temperature at noon. 110. The temperature at 6 P.M. was 6°F, and by 11 P.M. the temperature had dropped another 5°F. Use the subtraction of integers to describe this situation, and determine the temperature at 11 P.M. 111. Megan shot rounds of 3 over par, 2 under par, 3 under par, and 5 under par for a four-day golf tournament. Use the addition of integers to describe this situation, and determine how much over or under par she was for the tournament. 112. The annual report of a company contained these figures: a loss of $615,000 for 2000, a loss of $275,000 for 2001, a loss of $70,000 for 2002, and a profit of $115,000 for 2003. Use the addition of integers to describe this situation, and determine the company’s total loss or profit for the four-year period. 113. Suppose that during a five-day period a share of Dell stock recorded the following gains and losses: Monday Lost $2

Tuesday Gained $1

For Problems 93 –100, evaluate each algebraic expression for the given values of the variable.

Thursday Gained $1

Friday Lost $2

93. 7x 5y for x 5 and y 9

Use the addition of integers to describe this situation and to determine the amount of gain or loss for the fiveday period.

94. 4a 6b for a 6 and b 8

Wednesday Gained $3

24

Chapter 1


114. The Dead Sea is approximately thirteen hundred ten feet below sea level. Suppose that you are standing eight hundred five feet above the Dead Sea. Use the addition of integers to describe this situation and to determine your elevation. 115. Monday morning Thad bought 800 shares of a stock at $19 per share. During that week the stock went up $2 per share on one day and dropped $1 per share on each of the other four days. Use multiplication and addition of integers to describe this situation, and determine the value of the 800 shares at closing time on Friday. 116. In one week a small company showed a profit of $475 for one day and a loss of $65 for each of the other four days. Use multiplication and addition of integers to describe

this situation, and determine the company’s profit or loss for the week. 117. At 6 P.M. the temperature was 5°F. For the next four hours the temperature dropped 3° per hour. Use multiplication and addition of integers to describe this situation and to find the temperature at 10 P.M. 118. For each of the first three days of a golf tournament, Jason shot 2 strokes under par. Then for each of the last two days of the tournament, he shot 4 strokes over par. Use multiplication and addition of integers to describe this situation and to determine Jason’s score relative to par for the five-day tournament. 119. Use your calculator to check your answers for Problems 83 –92.

THOUGHTS INTO WORDS 120. The statement 6 (2) 6 2 4 can be read as “negative six minus negative two equals negative six plus two, which equals negative four.” Express each equation in words. (a) 8 1102 2

(b) 7 4 7 142 11

121. The algebraic expression x y can be read “the opposite of x minus y.” Give each expression in words. (a) x y (b) x y (c) x y z 122. Your friend keeps getting an answer of 7 when simplifying the expression 6 (8) 2. What mistake is she making and how would you help her?

(c) 9 1122 9 12 21 (d) 5 162 11

123. Make up a problem that you can solve using 6(4) 24.

Answers to the Example Practice Skills 1. (a) 56 (b) 35 (c) 55 (d) 0 2. (a) 18 (b) 40 (c) 112 (d) 50 3. (a) 12 (b) 5 (c) 0 (d) 24 4. (a) 3 (b) 0 (c) 25 (d) 20 5. 14 6. 4 7. 22 8. 17 9. 20 10. Loss of $225,000 11. 5 under par


1.3

2. False 3. False

4. True 5. True

6. False 7. True

8. False

9. False 10. True

Use of Properties OBJECTIVES 1

Recognize the Properties of Integers

2

Use the Properties of Integers to Simplify Numerical Expressions

3

Simplify Algebraic Expressions

4

Evaluate Algebraic Expressions

1 Recognize the Properties of Integers We will begin this section by listing and briefly commenting on some of the basic properties of integers. We will then show how these properties facilitate manipulation with integers and also serve as a basis for some algebraic computation.

1.3

Use of Properties

25

Commutative Property of Addition If a and b are integers, then abba

Commutative Property of Multiplication If a and b are integers, then ab ba Addition and multiplication are said to be commutative operations. This means that the order in which you add or multiply two integers does not affect the result. For example, 3 5 5 3 and 7(8) 8(7). It is also important to realize that subtraction and division are not commutative operations; order does make a difference. For example, 8 7 7 8 and 16 4 4 16.

Associative Property of Addition If a, b, and c are integers, then (a b) c a (b c)

Associative Property of Multiplication If a, b, and c are integers, then (ab)c a(bc) Our arithmetic operations are binary operations. We only operate (add, subtract, multiply, or divide) on two numbers at a time. Therefore, when we need to operate on three or more numbers, the numbers must be grouped. The associative properties can be thought of as grouping properties. For example, (8 3) 9 8 (3 9). Changing the grouping of the numbers for addition does not affect the result. This is also true for multiplication, as [(6)(5)](4) (6)[(5)(4)] illustrates. Addition and multiplication are associative operations. Subtraction and division are not associative operations. For example, (8 4) 7 3, whereas 8 (4 7) 11. An example showing that division is not associative is (8 4) 2 1, whereas 8 (4 2) 4.

Identity Property of Addition If a is an integer, then a00aa We refer to zero as the identity element for addition. This simply means that the sum of any integer and zero is exactly the same integer. For example, 197 0 0 11972 197.

Identity Property of Multiplication If a is an integer, then a112 11a2 a

26

Chapter 1


We call one the identity element for multiplication. The product of any integer and one is exactly the same integer. For example, 15732 112 11215732 573.

Additive Inverse Property For every integer a, there exists an integer a such that a (a) (a) a 0 The integer a is called the additive inverse of a or the opposite of a. Thus 6 and 6 are additive inverses, and their sum is 0. The additive inverse of 0 is 0.

Multiplication Property of Zero If a is an integer, then a(0) (0)(a) 0 In other words, the product of 0 and any integer is 0. For example, (873)(0) (0)(873) 0.

Multiplicative Property of Negative One If a is an integer, then (a)(1) (1)(a) a The product of any integer and 1 is the opposite of the integer. For example, (1)(48) (48)(1) 48.

Distributive Property If a, b, and c are integers, then a(b c) ab ac The distributive property involves both addition and multiplication. We say that multiplication distributes over addition. For example, 3(4 7) 3(4) 3(7). Because b c b (c), it follows that multiplication also distributes over subtraction. This could be stated as a(b c) ab ac. For example, 7(8 2) 7(8) 7(2). Let’s now consider some examples that use these properties to help with certain types of manipulations.

2

EXAMPLE 1

Use the Properties of Integers to Simplify Numerical Expressions

Find the sum [43 (24)] 24.

Solution In such a problem, it is much more advantageous to group 24 and 24. Thus [43 (24)] 24 43 [(24) 24] 43 0 43

Associative property for addition

1.3

Use of Properties

27

▼ PRACTICE YOUR SKILL Find the sum 37 [37 (8)].

EXAMPLE 2

■

Find the product [(17)(25)](4).

Solution In this problem, it is easier to group 25 and 4. Thus [(17)(25)](4) (17)[(25)(4)]

Associative property for multiplication

(17)(100) 1700

▼ PRACTICE YOUR SKILL Find the product 50[(2)(37)].

EXAMPLE 3

■

Find the sum 17 (24) (31) 19 (14) 29 43.

Solution Certainly we could add in the order in which the numbers appear. However, because addition is commutative and associative, we can change the order, and group in any convenient way. For example, we can add all the positive integers, add all of the negative integers, and then add these two results. It might be convenient to use the vertical format: 17 19 29 43 108

24 31 14 69

108 69 39

▼ PRACTICE YOUR SKILL Find the sum 18 24 (8) (12) 32.

■

For a problem such as Example 3, it might be advisable first to work out the problem by adding in the order in which the numbers appear and then to use the rearranging and regrouping idea as a check. Don’t forget the link between addition and subtraction: A problem such as 18 43 52 17 23 can be changed to 18 (43) 52 (17) (23).

EXAMPLE 4

Simplify (75)(4 100).

Solution For such a problem, it might be convenient to apply the distributive property and then to simplify. Thus (75)(4 100) (75)(4) (75)(100) 300 (7500) 7200

▼ PRACTICE YOUR SKILL Simplify (15)(2 10).

■

28

Chapter 1


EXAMPLE 5

Simplify 19(26 25).

Solution For this problem, we are better off not applying the distributive property but simply adding the numbers inside the parentheses and then finding the indicated product. Thus 19(26 25) 19(1) 19

▼ PRACTICE YOUR SKILL Simplify 19(34 34).

EXAMPLE 6

■

Simplify 27(104) 27(4).

Solution Keep in mind that the distributive property enables us to change from the form a(b c) to ab ac, or from ab ac to a(b c). In this problem we want to use the latter change: 27(104) 27(4) 27[104 (4)] 27(100) 2700

▼ PRACTICE YOUR SKILL Simplify 27(12) 27(2).

■

Examples 4, 5, and 6 demonstrate an important issue. Sometimes the form a(b c) is the most convenient, but at other times the form ab ac is better. A suggestion regarding this issue—a suggestion that also applies to the use of the other properties—is to think first and then decide whether or not you can use the properties to make the manipulations easier.

3

Simplify Algebraic Expressions

Algebraic expressions such as these: 3x

5y

4abc

7xy

z

are called “terms.” A term is an indicated product that may have any number of factors. We call the variables in a term literal factors, and we call the numerical factor the numerical coefficient. Thus in the term 7xy, the x and y are literal factors, and 7 is the numerical coefficient. The numerical coefficient of the term 4abc is 4. Because z 1(z), the numerical coefficient of the term z is 1. Terms that have the same literal factors are called like terms or similar terms. Some examples of similar terms are 3x and 9x 7xy and 15xy

14abc and 29abc 4z , 9z, and 14z

We can simplify algebraic expressions that contain similar terms by using a form of the distributive property. Consider these examples: 3x 5x (3 5)x 8x 9xy 7xy (9 7)xy 2xy

1.3

Use of Properties

29

18abc 27abc (18 27)abc [18 (27)]abc 9abc 4x x (4 1)x

Don’t forget that x 1(x)

5x More complicated expressions might first require that we rearrange terms by using the commutative property.

EXAMPLE 7

Simplify these algebraic expressions: (a) 7x 3y 9x 5y

(b) 9a 4 13a 6

Solution (a) 7x 3y 9x 5y 7x 9x 3y 5y (7 9)x (3 5)y

Commutative property for addition Distributive property

16x 8y (b) 9a 4 13a 6 9a (4) (13a) 6 9a (13a) (4) 6 [9 (13)]a 2

Commutative property for addition Distributive property

4a 2

▼ PRACTICE YOUR SKILL Simplify these algebraic expressions: (a) 6a 3a 7 5a 1

(b) 2y 3x 5x 6y

■

As you become more adept at handling the various simplifying steps, you may want to do the steps mentally and thereby go directly from the given expression to the simplified form. 19x 14y 12x 16y 31x 2y 17ab 13c 19ab 30c 2ab 17c 9x 5 11x 4 x 6 x 3 Simplifying some algebraic expressions requires repeated applications of the distributive property, as the next example demonstrates.

EXAMPLE 8

Simplify these algebraic expressions: (a) 5(x 2) 3(x 4)

(b) 7(y 1) 4(y 3)

(c) 5(x 2) (x 3)

Solution (a) 5(x 2) 3(x 4) 5(x) 5(2) 3(x) 3(4)

Distributive property

5x 10 3x 12 5x 3x 10 12 8x 2

Commutative property

30

Chapter 1


(b) 7(y 1) 4( y 3) 7(y) 7(1) 4(y) 4(3) 7y 7 4y 12

Be careful with this sign.

7y 4y 7 12 11y 5 (c) 5(x 2) (x 3) 5(x 2) 1(x 3)

Remember that a 1a.

5(x) 5(2) 1(x) 1(3) 5x 10 x 3 5x x 10 3 4x 7

▼ PRACTICE YOUR SKILL Simplify these algebraic expressions: (a) 2(a 2) 5(a 6)

(b) 7(4 2b) 4(5 b)

■

After you are sure of each step, you can use a more simplified format. 5(a 4) 7(a 2) 5a 20 7a 14 2a 34 9(z 7) 11(z 6) 9z 63 11z 66 20z 3 (x 2) (x 6) x 2 x 6 8

4 Evaluate Algebraic Expressions Simplifying by combining similar terms aids in the process of evaluating some algebraic expressions. The next examples of this section illustrate this idea.

EXAMPLE 9

Evaluate 8x 2y 3x 5y for x 3 and y 4.

Solution Let’s first simplify the given expression. 8x 2y 3x 5y 11x 3y Now we can evaluate for x 3 and y 4. 11x 3y 11(3) 3(4) 33 (12) 21

▼ PRACTICE YOUR SKILL Evaluate 5a 3b 2a 7b for a 6 and b 5.

■

1.3

EXAMPLE 10

Use of Properties

31

Evaluate 2ab 5c 6ab 12c for a 2, b 3, and c 7.

Solution 2ab 5c 6ab 12c 4ab 17c

When a 2, b 3, and c7

4(2)(3) 17(7) 24 119 143

▼ PRACTICE YOUR SKILL Evaluate 2xy 3x 2y 4xy for x 3 and y 1.

EXAMPLE 11

■

Evaluate 8(x 4) 7(x 3) for x 6.

Solution 8(x 4) 7(x 3) 8x 32 7x 21


15x 11 15(6) 11

When x 6

79

▼ PRACTICE YOUR SKILL Evaluate 6(a 3) 2(a 4) for a 7.

CONCEPT QUIZ

For Problems 1–10, answer true or false. 1. Addition is a commutative operation. 2. Subtraction is a commutative operation. 3. [(2)(3)](7) (2)[(3)(7)] is an example of the associative property for multiplication. 4. [(8)(5)](2) (2)[(8)(5)] is an example of the associative property for multiplication. 5. Zero is the identity element for addition. 6. The integer a is the additive inverse of a. 7. The additive inverse of 0 is 0. 8. The numerical coefficient of the term 8xy is 8. 9. The numerical coefficient of the term ab is 1. 10. 6xy and 2xyz are similar terms.

Problem Set 1.3 1 Recognize the Properties of Integers For Problems 1–12, state the property that justifies each statement. For example, 3 (4) (4) 3 because of the commutative property of addition. 1. 3(7 8) 3(7) 3(8) 2. (9)(17) 17(9)

5. 143(7) 7(143) 6. 5[9 (4)] 5(9) 5(4) 7. 119 119 0 8. 4 (6 9) (4 6) 9 9. 56 0 56 10. 5 (12) 12 5

3. 2 (5 7) (2 5) 7

11. [5(8)]4 5[8(4)]

4. 19 0 19

12. [6(4)]8 6[4(8)]

■

32

Chapter 1


2 Use the Properties of Integers to Simplify Numerical Expressions

41. 6xy x 13xy 4x 42. 7xy 2x xy x

For Problems 13 –30, simplify each numerical expression. Don’t forget to take advantage of the properties if they can be used to simplify the computation.

43. 5x 4 7x 2x 9

13. (18 56) 18

45. 2xy 12 8xy 16

14. 72 [72 (14)]

46. 14xy 7 19xy 6

15. 36 48 22 41 16. 24 18 19 30 17. (25)(18)(4) 18. (2)(71)(50) 19. (4)(16)(9)(25) 20. (2)(18)(12)(5) 21. 37(42 58) 22. 46(73 27) 23. 59(36) 59(64) 24. 49(72) 49(28) 25. 15(14) 16(8) 26. 9(14) 7(16) 27. 17 (18) 19 14 13 17

44. 8x 9 14x 3x 14

47. 2a 3b 7b b 5a 9a 48. 9a a 6b 3a 4b b a 49. 13ab 2a 7a 9ab ab 6a 50. ab a 4ab 7ab 3a 11ab 51. 3(x 2) 5(x 6) 52. 7(x 8) 9(x 1) 53. 5(x 4) 6(x 8) 54. 3(x 2) 4(x 10) 55. 9(x 4) (x 8) 56. (x 6) 5(x 9) 57. 3(a 1) 2(a 6) 4(a 5) 58. 4(a 2) 6(a 8) 3(a 6)

28. 16 14 18 21 14 17

59. 2(m 3) 3(m 1) 8(m 4)

29. 21 22 23 27 21 19

60. 5(m 10) 6(m 11) 9(m 12)

30. 24 26 29 26 18 29 17 10

61. ( y 3) (y 2) (y 6) 7( y 1) 62. ( y 2) (y 4) ( y 7) 2(y 3)

3 Simplify Algebraic Expressions For Problems 31– 62, simplify each algebraic expression by combining similar terms. 31. 9x 14x 32. 12x 14x x 33. 4m m 8m 34. 6m m 17m

4 Evaluate Algebraic Expressions For Problems 63 – 80, simplify each algebraic expression and then evaluate the resulting expression for the given values of the variables. 63. 3x 5y 4x 2y

for x 2 and y 3

64. 5x 7y 9x 3y

for x 1 and y 4

65. 5(x 2) 8(x 6)

for x 6

36. 14y 17y 19y

66. 4(x 6) 9(x 2)

for x 7

37. 4x 3y 7x y

67. 8(x 4) 10(x 3)

for x 5

38. 9x 5y 4x 8y

68. (n 2) 3(n 6)

for n 10

39. 7a 7b 9a 3b

69. (x 6) (x 12)

40. 12a 14b 3a 9b

70. (x 12) (x 14)

35. 9y 5y 7y

for x 3 for x 11

1.4

Prime and Composite Numbers

33

71. 2(x y) 3(x y)

for x 2 and y 7

80. 5x 6x x 7x x 2x

72. 5(x y) 9(x y)

for x 4 and y 4

81. Use a calculator to check your answers for Problems 13 –30.

73. 2xy 6 7xy 8

for x 2 and y 4

74. 4xy 5 8xy 9

for x 3 and y 3

For Problems 82 – 87, state whether the expressions in each problem are equivalent and explain why or why not.

75. 5x 9xy 3x 2xy

for x 12 and y 1

76. 9x xy 4xy x

for x 10 and y 11

77. (a b) (a b)

for a 19 and b 17

78. (a b) (a b)

for a 16 and b 14

79. 3x 7x 4x 2x x

for x 15

82. 15a(x y)

83. (6a 7b) 11c

and

84. 2x 3y 4z

and

2x 4z 3y

85. a 5(x y)

and

(a 5)x y

86. 7x 6( y 2z)

for x 13

5a(3x 3y)

and

87. 9m 8(3p q)

and and

7b (11c 6a)

6(2z y) 7x 8(3p q) 9m

THOUGHTS INTO WORDS 88. State in your own words the associative property for addition of integers.

89. State in your own words the distributive property for multiplication over addition.

Answers to the Example Practice Skills 1. 8 2. 3700 3. 18 4. 120 9. 38 10. 13 11. 2

5. 0

6. 270

7. (a) 2a 8 (b) 2x 4y

8. (a) 3a 34 (b) 10b 8

Answers to the Concept Quiz 1. True

1.4

2. False

3. True

4. False 5. True

6. True

7. True

8. False

9. True

10. False

Prime and Composite Numbers OBJECTIVES 1

Understand Divisibility

2

Identify Prime or Composite Numbers

3

Factor a Whole Number into a Product of Prime Numbers

4

Find the Greatest Common Factor

5

Find the Least Common Multiple

1

Understand Divisibility

Occasionally terms in mathematics have a special meaning in the discussion of a particular topic. Such is the case with the term “divides” as it is used in this section. We say that 6 divides 18 because 6 times the whole number 3 produces 18, but 6 does not divide 19 because there is no whole number such that 6 times the number produces 19. Likewise 5 divides 35 because 5 times the whole number 7 produces 35,

34

Chapter 1


but 5 does not divide 42 because there is no whole number such that 5 times the number produces 42. We can use this general definition:

Definition 1.1 Given that a and b are whole numbers, with a not equal to zero, a divides b if and only if there exists a whole number k such that a k b.

Remark: Note the use of the variables a, b, and k in the statement of a general definition. Also note that the definition merely generalizes the concept of divides, which we introduced in the paragraph preceding the definition. The following statements further clarify Definition 1.1. Pay special attention to the italicized words because they indicate some of the terminology used for this topic. 1.

8 divides 56, because 8 7 56.

2.

7 does not divide 38, because there is no whole number k such that 7 k 38.

3.

3 is a factor of 27, because 3 9 27.

4.

4 is not a factor of 38, because there is no whole number k such that 4 k 38.

5.

35 is a multiple of 5, because 5 7 35.

6.

29 is not a multiple of 7, because there is no whole number k such that 7 k 29.

The factor terminology is used extensively. We say that 7 and 8 are factors of 56 because 7 8 56; 4 and 14 are also factors of 56 because 4 14 56. The factors of a number are also the divisors of the number.

2

Identify Prime or Composite Numbers

Now consider two special kinds of whole numbers called “prime numbers” and “composite numbers” according to the following definition.

Definition 1.2

Prime Numbers vs. Composite Numbers

A prime number is a whole number, greater than 1, that has no factors (divisors) other than itself and 1. Whole numbers, greater than 1, that are not prime numbers are called composite numbers. The prime numbers less than 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47. Note that each number has no factors other than itself and 1. An interesting point is that the set of prime numbers is an infinite set; that is, the prime numbers go on forever, and there is no largest prime number.

3 Factor a Whole Number into a Product of Prime Numbers We can express every composite number as the indicated product of prime numbers. Let’s consider the first several composite numbers, 4, 6, 8, 9, 10, and 12, and write them as an indicated product of prime numbers. Note the use of exponents to represent repeated multiplications. 4 2 # 2 22 62#3 8 2 # 2 # 2 23 2 # # # # 12 2 2 3 2 3 10 2 5

9 3 # 3 32

The indicated product of prime numbers is sometimes called the prime factored form of the number.

1.4


35

We can use various procedures to find the prime factors of a given composite number. For our purposes, the simplest technique is to factor the composite number into any two easily recognized factors and then to continue to factor each of these until we obtain only prime factors. The order in which we write the prime factors is not important, because the commutative property of multiplication can be used to reorder the factors. The order in which we write the prime factored form is smallest to largest. This will aid in comparing prime factored forms when we are looking for greatest common factors and least common multiples. Consider this example:

EXAMPLE 1

Factor these composite numbers into an indicated product of prime numbers: (a) 18

(b) 27

(c) 24

(d) 150

Solution (a) 18 2 # 9 2 # 3 # 3 2 # 32 (c) 24 4 # 6 2 # 2 # 2 # 3 23 # 3

(b) 27 3 # 9 3 # 3 # 3 33 (d) 150 10 # 15 2 # 5 # 3 # 5 2 # 3 # 52

Remark: It does not matter which two factors we chose first. For instance, we could have factored 18 by starting with the factors 3 and 6. Doing so would give us 18 3 # 6 3 # 2 # 3 2 # 32. Either way, 18 contains two prime factors of 3 and one prime factor of 2.

▼ PRACTICE YOUR SKILL Factor these composite numbers into an indicated product of prime numbers: (a) 36

(b) 42 ■

Familiarity with a few basic divisibility rules will be helpful for determining the prime factors of some numbers. For example, if you can quickly recognize that 51 is divisible by 3, then you can divide 51 by 3 to find another factor of 17. Because 3 and 17 are both prime numbers, we have 51 3 17. The divisibility rules for 2, 3, 5, and 9 are shown here.

Rule for 2 A whole number is divisible by 2 if and only if the units digit of its base-10 numeral is divisible by 2. (In other words, the units digit must be 0, 2, 4, 6, or 8.) EXAMPLES 68 is divisible by 2, because 8 is divisible by 2. 57 is not divisible by 2, because 7 is not divisible by 2.

Rule for 3 A whole number is divisible by 3 if and only if the sum of the digits of its base-10 numeral is divisible by 3. EXAMPLES 51 is divisible by 3, because 5 1 6 and 6 is divisible by 3. 144 is divisible by 3, because 1 4 4 9 and 9 is divisible by 3. 133 is not divisible by 3, because 1 3 3 7 and 7 is not divisible by 3.

Rule for 5 A whole number is divisible by 5 if and only if the units digit of its base-10 numeral is divisible by 5. (In other words, the units digit must be 0 or 5.) EXAMPLES 115 is divisible by 5, because 5 is divisible by 5. 172 is not divisible by 5, because 2 is not divisible by 5.

36

Chapter 1


Rule for 9 A whole number is divisible by 9 if and only if the sum of the digits of its base-10 numeral is divisible by 9. EXAMPLES 765 is divisible by 9, because 7 6 5 18 and 18 is divisible by 9. 147 is not divisible by 9, because 1 4 7 12 and 12 is not divisible by 9.

4 Find the Greatest Common Factor We can use the prime factorization form of two composite numbers to conveniently find their greatest common factor (GCF). Consider this example: 42 2 3 7 70 2 5 7 Note that 2 is a factor of both, as is 7. Therefore, 14 (the product of 2 and 7) is the greatest common factor of 42 and 70. In other words, 14 is the largest whole number that divides both 42 and 70. The following examples show the process of finding the greatest common factor of two or more numbers.

EXAMPLE 2

Find the greatest common factor of 42 and 105.

Solution 42 6 # 7 2 # 3 # 7 105 3 # 35 3 # 5 # 7 Because 3 and 7 are common to both, the greatest common factor of 42 and 105 is 3 # 7 21.

▼ PRACTICE YOUR SKILL Find the greatest common factor of 42 and 60.

■

When exponents are used to write the prime factored form, the technique for finding the greatest common factor is slightly different. We actually make use of the exponents to find the number of times that factor is common. Consider this example of finding the greatest common factor of 36 and 120. First we write each number in prime factored form. 36 4 # 9 2 # 2 # 3 # 3 22 # 32 120 4 # 30 2 # 2 # 2 # 15 2 # 2 # 2 # 3 # 5 23 # 3 # 5 We determine that 2 and 3 are common factors. To find the greatest common factor, take the common factors to their lowest exponent in the factorizations, and find their product. Therefore, the product of 22 times 3 gives us the greatest common factor, which is 12. The following example should clarify the process of finding the greatest common factor when exponents are present in the prime factored form.

EXAMPLE 3


Solution 80 8 # 10 2 # 4 # 2 # 5 2 # 2 # 2 # 2 # 5 24 # 5 120 10 # 12 2 # 5 # 3 # 4 2 # 5 # 3 # 2 # 2 23 # 3 # 5 The common factors are 2 and 5, so take 2 to its lowest exponent, which is 3, times 5. Therefore, the product of 23 times 5 gives us the greatest common factor, which is 40.

1.4


37


EXAMPLE 4

■


Solution 24 2 # 2 # 2 # 3 23 # 3

35 5 # 7

Because there are no common prime factors, the greatest common factor is 1.


■

The concept of “greatest common factor” can be extended to more than two numbers, as the next example demonstrates.

EXAMPLE 5

Find the greatest common factor of 24, 28, and 120.

Solution 24 4 # 6 2 # 2 # 2 # 3 23 # 3 28 4 # 7 2 # 2 # 7 22 # 7

120 10 # 12 2 # 5 # 2 # 6 2 # 5 # 2 # 2 # 3 23 # 3 # 5 The only factor that is common to all three numbers is 2. To find the greatest common factor take 2 to its lowest exponent. Therefore 22 4 is the greatest common factor.

▼ PRACTICE YOUR SKILL Find the greatest common factor of 36, 72, and 90.

■

5 Find the Least Common Multiple We stated that 35 is a multiple of 5 because 5 # 7 35. We can produce a set of all whole number multiples of 5 by multiplying 5 times each successive whole number. The following list shows the set of multiples of 5. Whole number multiples of 5:

0, 5, 10, 15, 20, 25, 30, 35, 40, 45, . . .

In a like manner, we can produce the set of all whole number multiples of 4. The following list shows the set of multiples of 4. Whole number multiples of 4:

0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, . . .

Inspection of the two lists shows us that the numbers 0, 20, and 40 are common to both lists; in this case, these numbers would be referred to as common multiples. The least common multiple (LCM) is the smallest common nonzero multiple of two or more numbers. Therefore the least common multiple of 5 and 4 is 20. Stated another way, 20 is the smallest nonzero whole number that is divisible by both 5 and 4. Let’s find the least common multiple of 6 and 8. Start by listing the whole number multiples of 6 and 8. Whole number multiples of 6:

0, 6, 12, 18, 24, 30, 36, 42, 48, . . .

Whole number multiples of 8:

0, 8, 16, 24, 32, 40, 48, . . .

By inspection we can determine that the least common multiple is 24.

38

Chapter 1


Finding the least common multiple by inspection can be cumbersome. Another method to determine the least common multiple is to use the prime factored forms of numbers. The following steps show a systematic technique for finding the least common multiple. In the solutions to the following examples, these steps will be demonstrated.

Step 1 Express each number in prime factored form. Step 2 The least common multiple contains each different prime factor as many times as the maximum times it appears in any one factorization. If the prime factored form is written using exponents, then we want each different prime factor raised to its greatest exponent.

EXAMPLE 6

Find the least common multiple of 24 and 36.

Solution Let’s first express each number in its prime factored form. 24 4 # 6 2 # 2 # 2 # 3 23 # 3 36 6 # 6 2 # 3 # 2 # 3 22 # 32 The different prime factors are 2 and 3. So for the least common multiple we need the product of 2 (with its greatest exponent) and 3 (with its greatest exponent). The least common multiple is 23 # 32 8 # 9 72. In other words, 72 is the smallest nonzero number that is divisible by both 24 and 36.

▼ PRACTICE YOUR SKILL Find the least common multiple of 12 and 30.

EXAMPLE 7

■


Solution Let’s first express each number in its prime factored form. 48 4 # 12 2 # 2 # 2 # 6 2 # 2 # 2 # 2 # 3 24 # 3 84 4 # 21 2 # 2 # 3 # 7 22 # 3 # 7 The different prime factors are 2, 3, and 7. So for the least common multiple we need the product of 2 (with its greatest exponent), 3, and 7. The least common multiple is 24 # 3 # 7 16 # 3 # 7 336. In other words, 336 is the smallest nonzero number that is divisible by both 48 and 84.

▼ PRACTICE YOUR SKILL Find the least common multiple of 20 and 75.

EXAMPLE 8

Find the least common multiple of 12, 18, and 28.

Solution Let’s first express each number in its prime factored form. 12 3 # 4 3 # 2 # 2 22 # 3 18 2 # 9 2 # 3 # 3 2 # 32 28 4 # 7 2 # 2 # 7 22 # 7

■

1.4


39

The different prime factors are 2, 3, and 7. So for the least common multiple we need the product of 2 (with its greatest exponent), 3 (with its greatest exponent), and 7. The least common multiple is 22 # 32 # 7 4 # 9 # 7 252. In other words, 252 is the smallest nonzero number that is divisible by 12, 18, and 28.

▼ PRACTICE YOUR SKILL ■

Find the least common multiple of 12, 18, and 30.

EXAMPLE 9


Solution Let’s first express each number in its prime factored form. 8 2 # 4 2 # 2 # 2 23 9 3 # 3 32 The different prime factors are 2 and 3. So for the least common multiple we need the product of 2 (with its greatest exponent) and 3 (with its greatest exponent). The least common multiple is 23 # 32 8 # 9 72. In other words 72 is the smallest nonzero number that is divisible by both 8 and 9.



CONCEPT QUIZ

For Problems 1–5, answer true or false. 1. 2. 3. 4. 5.

Every even whole number greater than 2 is a composite number. Two is the only even prime number. One is a prime number. The prime factored form of 24 is 2 2 6. Some whole numbers are both prime and composite numbers.

Problem Set 1.4 1 Understand Divisibility For Problems 1–20, classify each statement as true or false.

15. 9 is a factor of 119 17. 3 is a prime factor of 57

1. 8 divides 56

2. 9 divides 54

18. 7 is a prime factor of 91

3. 6 does not divide 54

4. 7 does not divide 42


5. 96 is a multiple of 8

6. 78 is a multiple of 6


7. 54 is not a multiple of 4

16. 8 is a factor of 98

2 Identify Prime or Composite Numbers

8. 64 is not a multiple of 6 9. 144 is divisible by 4

10. 261 is divisible by 9

For Problems 21–30, classify each number as prime or composite.



21. 53

22. 57

23. 59



24. 61

25. 91

26. 81

40

Chapter 1

27. 89


28. 97

29. 111

4 Find the Greatest Common Factor

30. 101

For Problems 53 – 64, find the greatest common factor of the given numbers. 53. 12 and 16

54. 30 and 36

55. 56 and 64

56. 72 and 96

For Problems 31– 42, factor each composite number into a product of prime numbers. For example, 18 2 3 3.

57. 63 and 81

58. 60 and 72

59. 84 and 96

60. 48 and 52

31. 26

32. 16

33. 36

61. 36, 72, and 90

62. 27, 54, and 63

34. 80

35. 49

36. 92

63. 48, 60, and 84

64. 32, 80, and 96

37. 56

38. 144

39. 120

40. 84

41. 135

42. 98

3 Factor a Whole Number into a Product of Prime Numbers

5 Find the Least Common Multiple For Problems 65 –76, find the least common multiple of the given numbers.

For Problems 43 –52, use the divisibility rules to help determine the prime factorization of each number. 43. 118

44. 76

45. 201

46. 123

47. 85

48. 115

49. 117

50. 441

51. 129

52. 153

65. 6 and 8

66. 8 and 12

67. 12 and 16

68. 9 and 12

69. 28 and 35

70. 42 and 66

71. 49 and 56

72. 18 and 24

73. 8, 12, and 28

74. 6, 10, and 12

75. 9, 15, and 18

76. 8, 14, and 24

THOUGHTS INTO WORDS 77. How would you explain the concepts “greatest common factor” and “least common multiple” to a friend who missed class during that discussion?

78. Is it always true that the greatest common factor of two numbers is less than the least common multiple of those same two numbers? Explain your answer.

FURTHER INVESTIGATIONS 79. The numbers 0, 2, 4, 6, 8, and so on are multiples of 2. They are also called even numbers. Why is 2 the only even prime number?

83. What is the greatest common factor of x and y if x and y are nonzero whole numbers and y is a multiple of x? Explain your answer.

80. Find the smallest nonzero whole number that is divisible by 2, 3, 4, 5, 6, 7, and 8.

84. What is the least common multiple of x and y if they are both prime numbers and x does not equal y? Explain your answer.

81. Find the smallest whole number greater than 1 that produces a remainder of 1 when divided by 2, 3, 4, 5, or 6. 82. What is the greatest common factor of x and y if x and y are both prime numbers and x does not equal y? Explain your answer.

85. What is the least common multiple of x and y if the greatest common factor of x and y is 1? Explain your answer.

Answers to the Example Practice Skills 1. (a) 22 32 (b) 2 3 7

2. 6

3. 9

4. 1

5. 18


2. True 3. False

4. False 5. False

6. 60

7. 300

8. 180

9. 120

1.5

1.5

Rational Numbers

41

Rational Numbers OBJECTIVES 1

Reduce Fractions

2

Multiply Rational Numbers

3

Divide Rational Numbers

4

Add and Subtract Rational Numbers

5

Simplify Numerical Expressions

6

Solve Application Problems Involving Rational Numbers

1 Reduce Fractions Fractions and rational numbers are encountered throughout the study of algebra. The a form is called a fraction or sometimes a common fraction. The numbers we often b call fractions are rational numbers. The precise definition of rational numbers follows.

Definition 1.3

Rational Numbers

a Any number that can be written in the form , where a and b are integers, and b b is not zero, is a rational number.

Here are some examples of rational numbers: 7 9

1 2

15 7

3 4

5 7

11 13

All integers are rational numbers, because every integer can be expressed as the indicated quotient of two integers. For example, 6

6 12 18 , etc. 1 2 3

27 0

27 54 81 , etc. 1 2 3

0 0 0 , etc. 1 2 3

Our work in Section 1.2 with the division of negative integers helps with the next three examples: 4

4 8 12 , etc. 1 2 3

6

6 12 18 , etc. 1 2 3

10

10 10 20 , etc. 1 1 2

42

Chapter 1


Observe the following general property:

Property 1.2 a a a b b b

and

a a b b

2 2 2 , for example, as or . (However, 3 3 3 we seldom express rational numbers with negative denominators.) We define multiplication of rational numbers in common fractional form as follows: Therefore, we can write the rational number

Definition 1.4

Multiplication of Rational Numbers

If a, b, c, and d are integers, and b and d do not equal to zero, then a b

#

a#c c # d b d

To multiply rational numbers in common fractional form, we simply multiply numerators and multiply denominators. Because the numerators and denominators are integers, our previous agreements pertaining to multiplication of integers hold for the rationals. That is, the product of two positive rational numbers or of two negative rational numbers is a positive rational number. The product of a positive rational number and a negative rational number (in either order) is a negative rational number. Furthermore, we see from the definition that the commutative and associative properties hold for the multiplication of rational numbers. We are free to rearrange and regroup factors as we do with integers. The following examples illustrate the definition for multiplying rational numbers: 1#2 2 1 # 2 # 3 5 3 5 15

3 # 5 3#5 15 # 4 7 4 7 28

2 # 7 2 # 7 14 # 3 9 3 9 27

1 # 9 1#9 9 5 11 5(11) 55

or

or

3 # 7 3 # 7 3 # 7 21 # 4 13 4 13 4 13 52

3 # 5 3#5 15 # 1 5 3 5 3 15

14 27

9 55 or

21 52

The last example is a very special case. If the product of two numbers is 1, the numbers are said to be reciprocals of each other. Using Definition 1.4 and applying the multiplication property of one, the a#k fraction # , where b and k are nonzero integers, simplifies as shown: b k a#k a k a a # #1 # b k b k b b This result is stated as Property 1.3: Fundamental Principle of Fractions.

1.5

Property 1.3

Rational Numbers

43

Fundamental Principle of Fractions

If b and k are nonzero integers, and a is any integer, then a#k a b#k b When we work with fractions, we often use Property 1.3, which provides the basis for creating equivalent fractions. In the following examples, we will use this property to reduce fractions to lowest terms or express fractions in simplest or reduced form.

EXAMPLE 1

Reduce

12 to lowest terms. 18

Solution 2 2#6 2 6 2 12 # # #1 18 3 6 3 6 3 3

▼ PRACTICE YOUR SKILL

EXAMPLE 2

Reduce


Change

14 to simplest form. 35

■

Solution 14 2#7 2 # 35 5 7 5

Divide a common factor of 7 out of both the numerator and denominator

▼ PRACTICE YOUR SKILL Reduce

EXAMPLE 3


Reduce

■

72 . 90

Solution

2#2#2#3#3 72 4 # # # 90 2 3 3 5 5

Use the prime factored forms of the numerator and denominator to help recognize common factors

▼ PRACTICE YOUR SKILL Reduce

60 . 144

■

2 Multiply Rational Numbers We are now ready to use Definition 1.4 to multiply fractions with the understanding that the final answer should be expressed in reduced form. Study the following examples carefully, because different methods are used to handle the problems.

44

Chapter 1


EXAMPLE 4

Multiply

7 9

#

5 . 14

Solution 7#5 7#5 5 7 # 5 # # # # 9 14 9 14 3 3 2 7 18

▼ PRACTICE YOUR SKILL Multiply

EXAMPLE 5

5 # 3 . 7 10

Find the product of

■

8 18 and . 9 24

Solution 1

2

8 # 18 2 9 24 3 1

Divide a common factor of 8 out of 8 and 24 and a common factor of 9 out of 9 and 18

3

▼ PRACTICE YOUR SKILL Find the product of

EXAMPLE 6

5 12 and . 6 25

■

6 14 Multiply a b a b. 8 32

Solution 14 6 # 14 6 21 a b a b # 8 32 8 32 64 3

7

4

16

Divide a common factor of 2 out of 6 and 8 and a common factor of 2 out of 14 and 32

▼ PRACTICE YOUR SKILL Multiply a

EXAMPLE 7

10 6 b a b . 18 14

■

14 9 Multiply a b a b . 4 15

Solution 9 14 3#3#2#7 21 a b a b # # # 4 15 2 2 3 5 10

Immediately we recognize that a negative times a negative is positive

▼ PRACTICE YOUR SKILL Multiply a

10 42 b a b . 3 18

■

1.5

Rational Numbers

45

3 Divide Rational Numbers The following example motivates a definition for division of rational numbers in fractional form. 3 3 3 3 3 3 3 a ba b a ba b 4 2 4 2 4 4 2 3 3 9 ± ≤ ± ≤ a ba b 2 2 3 1 4 2 8 3 2 a ba b 3 3 2 3 2

Notice that this is a form of 1 and

3 2 is the reciprocal of 2 3

3 3 2 3 divided by is equivalent to times . The following definition 4 3 4 2 for division should seem reasonable: In other words,

Definition 1.5

Division of Rational Numbers

If b, c, and d are nonzero integers, and a is any integer, then c a a b d b

#

d c

c a c a d by , we multiply times the reciprocal of , which is . The c b d b d following example demonstrates the important steps of a division problem. Note that to divide

EXAMPLE 8

Find the quotient. (a)

2 1 3 2

(b)

5 3 6 4

(c)

9 3 12 6

(d)

6 2 7

Solution (a)

1 2 2 4 2 # 3 2 3 1 3

(b)

3 5 4 5#4 5#2#2 10 5 # # # # 6 4 6 3 6 3 2 3 3 9 3

1

3 9 # 6 3 9 (c) 12 6 12 3 2 2

1 3

(d)

6 1 6 1 3 6 2 # # 7 7 2 7 2 7 1

▼ PRACTICE YOUR SKILL Find the quotient. (a)

1 2 5 15

3 (b) 4 8

■

46

Chapter 1


4 Add and Subtract Rational Numbers Suppose that it is one-fifth of a mile between your dorm and the student center and two-fifths of a mile between the student center and the library, along a straight line as indicated in Figure 1.5. The total distance between your dorm and the 1 2 3 library is three-fifths of a mile, and we write . 5 5 5

2 mile 5

1 mile 5 Dorm

Student Center

Library

Figure 1.5

A pizza is cut into seven equal pieces and you eat two of the pieces (see 7 Figure 1.6). How much of the pizza remains? We represent the whole pizza by 7 2 5 7 and then conclude that of the pizza remains. 7 7 7

Figure 1.6

These examples motivate the following definition for addition and subtraca tion of rational numbers in form. b

Definition 1.6 Addition and Subtraction of Rational Numbers If a, b, and c are integers, and b is not zero, then a c ac b b b

Addition

a c ac b b b

Subtraction

We say that rational numbers with common denominators can be added or subtracted by adding or subtracting the numerators and placing the results over the common denominator. Consider these examples:

1.5

Rational Numbers

47

2 32 5 3 7 7 7 7 7 2 72 5 8 8 8 8 1 21 3 1 2 6 6 6 6 2

We agree to reduce the final answer

3 5 35 2 2 11 11 11 11 11 How do we add or subtract if the fractions do not have a common denominaa#k a tor? We use the fundamental principle of fractions, # , and obtain equivalent b b k fractions that have a common denominator. Equivalent fractions are fractions that name the same number. Consider the following example, which shows the details.

EXAMPLE 9

Add

1 1 . 2 3

Solution 1#3 3 1 # 2 2 3 6

1 3 and are equivalent fractions that name the same number 2 6

1 1#2 2 # 3 3 2 6

2 1 and are equivalent fractions that name the same number 3 6

1 3 2 32 5 1 2 3 6 6 6 6

▼ PRACTICE YOUR SKILL Add

1 2 . 7 3

■

Note that in Example 9 we chose 6 as our common denominator, and 6 is the least common multiple of the original denominators 2 and 3. (Recall that the least common multiple is the smallest nonzero whole number divisible by the given numbers.) In general, we use the least common multiple of the denominators of the fractions to be added or subtracted as a least common denominator (LCD). Recall from Section 1.4 that the least common multiple is found either by inspection or by using the prime factorization forms of the numbers. Let’s consider some examples involving these procedures.

EXAMPLE 10

Add

2 1 . 4 5

Solution By inspection we see that the LCD is 20. Thus both fractions can be changed to equivalent fractions that have a denominator of 20. 1 2 1#5 2#4 5 8 13 # # 4 5 4 5 5 4 20 20 20 Use of fundamental principle of fractions

48

Chapter 1



1 2 . 5 3

■

If the LCD is not obvious by inspection, then we can use the technique from Section 1.4 to find the least common multiple. We proceed as follows:

Step 1 Express each denominator as a product of prime factors. Step 2 The LCD contains each different prime factor as many times as the most times it appears in any one of the factorizations from step 1.

EXAMPLE 11

Add

7 5 . 18 24

Solution If we cannot find the LCD by inspection, then we can use the prime factorization forms. 18 2 # 3 # 3

  24 2 # 2 # 2 # 3 

LCD 2 # 2 # 2 # 3 # 3 72

5 7 5#4 7#3 20 21 41 # # 18 24 18 4 24 3 72 72 72


EXAMPLE 12

13 5 . 60 18

Subtract

■

3 8 . 14 35

Solution 14 2 # 7   35 5 # 7 

LCD 2 # 5 # 7 70

3 8 3#5 8#2 15 16 1 # 14 35 14 5 35 # 2 70 70 70

or

1 70

▼ PRACTICE YOUR SKILL Subtract

EXAMPLE 13

Add

3 8 . 15 10

■

3 5 . 8 14

Solution 82#2#2   14 2 # 7 

LCD 2 # 2 # 2 # 7 56

5 3 5 # 7 3#4 35 12 23 8 14 8#7 14 # 4 56 56 56

or

23 56

1.5

Rational Numbers

49


EXAMPLE 14

7 3 a b. 10 12

Add 3

■

2 . 5

Solution 3

3 # 5 2 15 2 15 2 13 2 # 5 1 5 5 5 5 5 5

or

13 5


5 122 . 8

■

5 Simplify Numerical Expressions Let’s now consider simplifying numerical expressions that contain rational numbers. As with integers, we first do the multiplications and divisions and then perform the additions and subtractions. In these next examples only the major steps are shown, so be sure that you can fill in all of the details.

EXAMPLE 15

Simplify

2 3 4 3

#

1 3 5 2

#

1 . 5

Solution 3 2 3 1 1 3 2 1 # # 4 3 5 2 5 4 5 10

Perform the multiplications

8 2 15 8 2 21 15 20 20 20 20 20


EXAMPLE 16

Simplify

1 4 1 3 3 # # . 14 4 7 2 7

Simplify

3 8 1 1 5 a b a b . 5 5 2 3 12

Change to equivalent fractions and combine the numerators

■

Solution 3 8 1 1 5 3 5 1 1 5 a b a b # a b a b 5 5 2 3 12 5 8 2 3 12

3 1 5 8 6 12

Change division to multiply by the reciprocal

50

Chapter 1


9 4 10 24 24 24

9 142 10

Change to equivalent fractions

24 15 5 24 8

Reduce!


3 2 1 1 5 a ba b . 11 11 3 4 6

■

The distributive property, a(b c) ab ac, holds true for rational numbers and (as with integers) can be used to facilitate manipulation.

EXAMPLE 17

Simplify 12 a

1 1 b. 3 4

Solution For help in this situation, let’s change the form by applying the distributive property. 12 a

1 1 1 1 b 12 a b 12 a b 3 4 3 4 43 7

▼ PRACTICE YOUR SKILL Simplify 20 a

EXAMPLE 18

Simplify

1 1 b. 2 5

■

1 5 1 a b. 8 2 3

Solution In this case it may be easier not to apply the distributive property but to work with the expression in its given form. 1 5 3 2 5 1 a b a b 8 2 3 8 6 6

5 5 a b 8 6

25 48

▼ PRACTICE YOUR SKILL 3 1 1 Simplify a b . 8 3 5

■

1.5

Rational Numbers

51

Examples 13 and 14 emphasize a point made in Section 1.4. Think first and decide whether or not you can use the properties to make the manipulations easier.

6 Solve Application Problems Involving Rational Numbers EXAMPLE 19

Apply Your Skill

Debra Feinman /istockphoto.com

Frank has purchased 50 candy bars to make s’mores for the Boy Scout troop. If he 2 uses of a candy bar for each s’more, how many s’mores will he be able to make? 3

Solution 2 To find how many s’mores can be made, we need to divide 50 by . 3 25

3 50 # 3 75 2 75 50 50 # 3 2 1 2 1 1

So Frank can make 75 s’mores.


3 of a foot of cable is needed 4 for each installation, how many installations can be done with the spool of coaxial cable? ■ A spool of coaxial cable contains 120 feet of cable. If

Yellow Dog Productions/Getty Images

EXAMPLE 20

Apply Your Skill Brian brought 5 cups of flour with him on a camping trip. He wants to make biscuits 3 3 and cake for tonight’s supper. It takes of a cup of flour for the biscuits and 2 4 4 cups of flour for the cake. How much flour will be left over for the rest of his camping trip?

Solution Let’s do this problem in two steps. First, add the amounts of flour needed for the biscuits and cake. 3 3 11 14 7 3 2 4 4 4 4 4 2 Then to find the amount of flour left over, we will subtract 5

10 7 3 1 7 1 2 2 2 2 2

1 So 1 cups of flour remain. 2

7 from 5. 2

52

Chapter 1


▼ PRACTICE YOUR SKILL Victor is constructing an outdoor deck. From a board 12 feet in length he needs to cut 1 3 off a piece 1 feet in length for the upright post and another piece 6 feet in length 2 8 ■

for the header. What is the length of the piece remaining?

CONCEPT QUIZ

For Problems 1–10, answer True or False. 1. 6 is a rational number. 5 5 2. 3 3 3. If the product of two rational numbers is 1, then the numbers are said to be reciprocals. 3 7 4. The reciprocal of is . 7 3 10 5. is reduced to lowest terms. 12 6. To add rational numbers with common denominators, add the numerators and place the result over the common denominator. 7. When adding or subtracting fractions, the least common multiple of the denominators can always be used as a common denominator. 1 3 8. To subtract from , we need to find equivalent fractions with a common 5 8 denominator. 5 2 9. To multiply and , we need to find equivalent fractions with a common 7 3 denominator. 3 1 10. When adding and , either 20, 40, or 60 can be used as a common denomina4 5 tor, but 20 is the least common denominator.

Problem Set 1.5 1 Reduce Fractions

2 Multiply Rational Numbers

For Problems 1–12, reduce each fraction to lowest terms. 8 12

2.

4.

18 32

7.

10.

1.

16 24

12 16

3.

5.

15 9

6.

48 36

8 48

8.

3 15

9.

9 51

11.

54 56

12.

For Problems 13 –20, multiply and express answers in reduced form. 13.

3 # 5 4 7

14.

4 # 3 5 11

16.

4 # 3 9 2

17.

6 # 26 13 9

27 36

18.

3 # 14 4 12

19. a

24 80

21 6 20. a b a b 7 24

8 10 b a b 10 32

15.

3 # 12 8 15

1.5

3 Divide Rational Numbers For Problems 21–28, divide and express answers in reduced form. 5 7 5 2 3 11 21. 22. 23. 9 9 7 5 6 13 1 5 4 6

24.

3 7 11 11

25.

26.

14 7 8 16

27. 9

57. 4

59.

3 7

58. 2

3 6 4

60.

5 6

5 7 8

For Problems 61–76, simplify each numerical expression, and express answers in reduced form.

1 3

61.

1 3 5 1 4 8 12 24

62.

2 1 5 3 4 3 6 12

63.

5 2 3 1 2 # # 6 3 4 4 5

64.

1 2 1 1 2 # # 3 2 5 3 5

65.

5 8 2 6 3 # 6 # # 4 9 6 10 3 8 2 3 1 2 3 # 5 # # 5 7 3 5 7 5

4 Add and Subtract Rational Numbers For Problems 29 – 60, add or subtract as indicated, and express answers in lowest terms. 29.

3 2 7 7

30.

5 3 11 11

66.

31.

2 7 9 9

32.

11 6 13 13

67. 4

33.

3 9 4 4

34.

7 5 6 6

69.

10 5 14 10 4 5 12 6 8 21

35.

11 3 12 12

36.

13 7 16 16

70.

6 8 # 6 5 3 4 5 12 9 12

37.

1 5 8 8

38.

5 2 9 9

71. 24 a

1 3 b 4 6

39.

11 5 24 24

40.

13 7 36 36

72. 18 a

41.

1 1 3 5

42.

1 1 6 8

1 2 b 3 9

73. 64 a

43.

15 3 16 8

44.

1 13 12 6

3 5 1 1 b 16 8 4 2

7 8 10 15

74. 48 a

46.

1 3 5 b 12 6 8

11 5 47. 24 32

53

5 Simplify Numerical Expressions

1 28. 10 4

45.

Rational Numbers

5 7 12 8

8 5 48. 18 27

49.

5 13 18 24

50.

7 1 24 36

51.

5 2 8 3

52.

3 5 4 6

53.

7 2 13 39

54.

13 3 11 33

55.

3 1 14 21

56.

14 3 20 25

75.

2 # 3 6 3 5

68. 3

1 # 1 2 2 3

Don’t forget the distributive property!

7 2 1 a b 13 3 6

76.

5 1 1 a b 9 2 4

6 Solve Application Problems Involving Rational Numbers 3 of all of the accounts within 4 the ABC Advertising Agency. Maria is personally re1 sponsible for of all accounts in her department. For 3 what portion of all of the accounts at ABC is Maria personally responsible?

77. Maria’s department has

54

Chapter 1


1 78. Pablo has a board that is 4 feet long, and he wants to cut 2 it into three pieces all of the same length (see Figure 1.7). Find the length of each of the three pieces.

85. Beth wants to make three sofa pillows for her new sofa. After consulting the chart provided by the fabric shop, she decides to make a 12 round pillow, an 18 square pillow, and a 12 16 rectangular pillow. According to the chart, how much fabric will Beth need to purchase? Fabric Shop Chart

10 round 4

1 ft 2

12 round 12 square

Figure 1.7 3 cup of sugar. How 4 much sugar is needed to make 3 cakes?

18 square

79. A recipe for birthday cake calls for

80. Caleb left an estate valued at $750,000. His will states that three-fourths of the estate is to be divided equally among his three children. How much should each receive? 1 81. One of Arlene’s recipes calls for 3 cups of milk. If she 2 wants to make one-half of the recipe, how much milk should she use? 2 82. The total length of the four sides of a square is 8 3 yards. How long is each side of the square? 1 83. If it takes 3 yards of material to make drapes for 4 1 window, how much material is needed for 5 windows? 84. Vinay has a board that is 6 a piece 2

1 feet long. If he cuts off 2

3 feet long, how long is the remaining piece 4

of board?

12 16 rectangular

3 8 1 2 5 8 3 4 7 8

yard yard yard yard yard

2 1 of his estate to the Boy Scouts, to 5 4 the local cancer fund, and the rest to his church. What fractional part of the estate does the church receive?

86. Blake Scott leaves

87. A triangular plot of ground measures 14

1 yards by 2

1 5 yards by 9 yards. How many yards of fencing is 3 6 needed to enclose the plot?

12

1 miles, then 2 1 3 walks for of a mile, and finally jogs for another 1 4 4 miles. Find the total distance that Lian covers.

88. For her exercise program, Lian jogs for 2

1 89. Mindy takes a daily walk of 2 miles. One day a thun2 3 derstorm forced her to stop her walk after of a mile. By 4 how much was her walk shortened that day?

THOUGHTS INTO WORDS 90. State in your own words the property

a a a b b b

91. Find the mistake in this simplification process: 1 2 3 1 1 1 1 1 a ba b 3 3 # 2 # 2 3 4 2 2 2 3 3 How would you correct the error? 92. Give a step-by-step description of how to add the ra5 3 tional numbers and . 8 18 93. The will of a deceased collector of antique automobiles specified that his cars be left to his three children. Half

1 1 to his daughter, and to 3 9 his younger son. At the time of his death, 17 cars were in the collection. The administrator of his estate borrowed a car to make 18. Then he distributed the cars as follows: were to go to his elder son,

Elder son:

1 1182 9 2

Daughter:

1 1182 6 3

Younger son:

1 1182 2 9

This totaled 17 cars, so he then returned the borrowed car. Where is the error in this problem?

1.6

Decimals and Real Numbers

55

FURTHER INVESTIGATIONS 94. The division problem 35 7 can be interpreted as “how many 7s are there in 35?” Likewise, a division 1 problem such as 3 can be interpreted as “how many 2 halves are there in 3?” Use this how-many interpretation to do each division problem.

1 3 4 2 1 3 c. 2 4 2 1 e. 3 4

7 8 7 8 3 4

b. 1

a.

8 7 3 f. 5 d.

a. 4

1 2

b. 3

1 4

c. 5

1 8

d. 6

1 7

a.

99 117

b.

175 225

7 1 8 8

c.

111 123

d.

234 270

e.

270 495

f.

324 459

g.

91 143

h.

187 221

11 15

11.

e.

5 1 6 6

f.

96. Reduce each fraction to lowest terms. Don’t forget that we presented some divisibility rules in Section 1.4.

95. Estimation is important in mathematics. In each of the following, estimate whether the answer is greater than 1 or less than 1 by using the how-many idea from Problem 94.

Answers to the Example Practice Skills 1.

7 8

13.

2. 17 60

5 7

3.

14.

11 8

5 12

3 14

4. 15.

1 7

2 5

5. 16.

9 4

6.

5 21

17. 14

7.

18.

70 9

1 5

8. (a)

3 3 (b) 2 32

9.

17 21

10.

89 180

12.

7 30

1 19. 160 installations 20. 4 feet 8


1.6

2. True 3. True

4. False

5. False 6. True

7. True

8. True

9. False

10. True

Decimals and Real Numbers OBJECTIVES 1

Classify Real Numbers

2

Expand the Properties of Numbers to Include Real Numbers

3

Add, Subtract, Multiply, and Divide Decimals

4

Combine Similar Terms with Decimal Coefficients

5

Evaluate Algebraic Expressions

6

Solve Application Problems that Involve Rational Numbers in Decimal Form

1 Classify Real Numbers We classify decimals—also called decimal fractions—as terminating, repeating, or nonrepeating. Here are examples of these classifications:

Terminating decimals

Repeating decimals

Nonrepeating decimals

0.3 0.26 0.347 0.9865

0.333333 . . . 0.5466666 . . . 0.14141414 . . . 0.237237237 . . .

0.5918654279 . . . 0.26224222722229 . . . 0.145117211193111148 . . . 0.645751311 . . .

56

Chapter 1


A repeating decimal has a block of digits that repeats indefinitely. This repeating block of digits may contain any number of digits, and it may or may not begin repeating immediately after the decimal point. Technically, a terminating decimal can be thought of as repeating zeros after the last digit. For example, 0.3 0.30 0.300 0.3000, and so on. In Section 1.5 we defined a rational number to be any number that can be a written in the form , where a and b are integers and b is not zero. A rational numb ber can also be defined as any number that has a terminating or repeating decimal representation. Thus we can express rational numbers in either common-fraction form or decimal-fraction form, as the next examples illustrate. A repeating decimal can be written by using a bar over the digits that repeat, for example, 0.14.

Terminating decimals

Repeating decimals

3 0.75 4

1 0.3333 . . . 3

1 0.125 8

2 0.66666 . . . 3

5 0.3125 16

1 0.166666 . . . 6

7 0.28 25

1 0.08333 . . . 12

2 0.4 5

14 0.14141414 . . . 99

The nonrepeating decimals are called irrational numbers, and they do appear in forms other than decimal form. For example, 12, 13, and p are irrational numbers. An approximate decimal representation for each of these follows. 22 1.414213562373 . . . 23 1.73205080756887 . . . p 3.14159265358979 . . .

  

We will do more work with irrational numbers in Chapter 7. The rational numbers together with the irrationals form the set of real numbers. This tree diagram of the real number system is helpful for summarizing some basic ideas. Real numbers

Rational

Irrational

Integers

0

Nonintegers

Any real number can be traced down through the diagram as follows: 5 is real, rational, an integer, and positive 4 is real, rational, an integer, and negative 3 is real, rational, a noninteger, and positive 4

1.6


57

0.23 is real, rational, a noninteger, and positive 0.161616 . . . is real, rational, a noninteger, and negative 17 is real, irrational, and positive 12 is real, irrational, and negative In Section 1.2, we associated the set of integers with evenly spaced points on a line as indicated in Figure 1.8. This idea of associating numbers with points on a −4 −3 −2 −1

0

1

2

3

4

Figure 1.8

line can be extended so that there is a one-to-one correspondence between points on a line and the entire set of real numbers (as shown in Figure 1.9). That is to say, to each real number there corresponds one and only one point on the line, and to each point on the line there corresponds one and only one real number. The line is often referred to as the real number line, and the number associated with each point on the line is called the coordinate of the point. −π

− 2

1 2

−1 2

−5 −4 −3 −2 −1

0

π

2 1

2

3

4

5

Figure 1.9

2 Expand the Properties of Numbers to Include Real Numbers The properties of integers that we discussed in Section 1.4 are true for all real numbers. We restate them here for your convenience. The multiplicative inverse property has been added to the list. A discussion follows.

Commutative Property of Addition If a and b are real numbers, then a b b a.

Commutative Property of Multiplication If a and b are real numbers, then ab ba.

Associative Property of Addition If a, b, and c are real numbers, then (a b) c a (b c).

Associative Property of Multiplication If a, b, and c are real numbers then (ab)c a(bc).

Identity Property of Addition If a is any real number, then a 0 0 a a.

58

Chapter 1


Identity Property of Multiplication If a is any real number, then a(1) 1(a) a.

Additive Inverse Property For every real number a, there exists an integer a such that a (a) (a) a 0.

Multiplication Property of Zero If a is any real number, then a(0) (0)(a) 0.

Multiplicative Property of Negative One If a is any real number, then (a)(1) (1)(a) a.

Multiplicative Inverse Property 1 For every nonzero real number a, there exists a real number , such that a 1 1 a a b 1a2 1. a a

Distributive Property If a, b, and c are real numbers, then a(b c) ab ac.

1 is called the multiplicative inverse of a or the reciprocal of a. a 1 1 1 For example, the reciprocal of 2 is and 2 a b 12 2 1. Likewise, the recipro2 2 2 1 1 1 cal of is . Therefore, 2 and are said to be reciprocals (or multiplicative inverses) 2 1 2 2 2 5 5 2 of each other. Also, and are multiplicative inverses and a b a b 1. Because 5 2 5 2 division by zero is undefined, zero does not have a reciprocal. The number

3 Add, Subtract, Multiply, and Divide Decimals The basic operations with decimals may be related to the corresponding operations 3 4 7 with common fractions. For example, 0.3 0.4 0.7 because , and 10 10 10 13 37 24 0.37 0.24 0.13 because . In general, to add or subtract deci100 100 100 mals, we add or subtract the hundredths, the tenths, the ones, the tens, and so on. To keep place values aligned, we line up the decimal points.

1.6

EXAMPLE 1


59

Find the sum. (a) 3.42 18.2 0.08

(b) 6.123 0.005 42.1 8.43

Solution Line up the numbers in a vertical format, keeping the decimal points aligned. (a) 3.42 18.2 0.08 21.70

(b) 6.123 0.005 42.1 8.43 56.658

▼ PRACTICE YOUR SKILL Find the sum for each of the following. (a) 17.3 21.05 0.4

EXAMPLE 2

(b) 12.45 1.8 0.04 1.261

■

Find the difference for each of the following. (a) 6.84 1.3

(b) 72.85 2.412

Solution Line up the numbers in a vertical format, keeping the decimal points aligned. Add a zero so that each number has the same number of places after the decimal point. (a) 6.84 1.30

(b) 72.850 2.412

5.54

70.438

▼ PRACTICE YOUR SKILL Find the difference for each of the following. (a) 13.6 10.35

(b) 8.427 0.32

■

We can use examples such as the following to help formulate a general rule for multiplying decimals. Because

7 10

#

3 21 , then (0.7)(0.3) 0.21. 10 100

Because

9 10

#

23 207 , then (0.9)(0.23) 0.207. 100 1000

Because

11 100

#

13 143 , then (0.11)(0.13) 0.0143. 100 10,000

In general, to multiply decimals, we (1) multiply the numbers and ignore the decimal points, and then (2) insert the decimal point in the product so that the number of digits to the right of the decimal point in the product is equal to the sum of the numbers of digits to the right of the decimal point in each factor.

60

Chapter 1


EXAMPLE 3

(0.7)

(0.3)

0.21

One digit to right

One digit to right

Two digits to right

(0.9)

(0.23)

0.207

One digit to right

Two digits to right

Three digits to right

(0.11)

(0.13)

0.0143

Two digits to right

Two digits to right

Four digits to right

Multiply using a vertical format. (a) 41.2(0.13)

(b) 0.021(0.03)

Solution Stack the numbers, align them on the right side, and multiply as usual. Add together the number of digits to the right of the decimal point in each number; that sum determines the placement of the decimal point in the product. In other words, if there is a total of three digits, insert the decimal point so that there are three digits to the right of it. (a) 41.2 0.13 1236 412 5.356

One digit to right Two digits to right

(b)

0.021 0.03 0.00063

Three digits to right Two digits to right Five digits to right

Three digits to right

Note that in part b, we actually multiplied 3 21 and then inserted three 0s to the left so that there would be five digits to the right of the decimal point.

▼ PRACTICE YOUR SKILL Multiply each of the following. (a) 28.4(1.32)

(b) 0.017(0.02)

■

Once again let’s look at some links between common fractions and decimals. 3 0.3 6 6 # 1 3 2 , we have 2冄0.6 . Because 10 10 2 10 3 0.03 39 39 # 1 3 13 , we have 13冄0.39 . Because 100 100 13 100 17 0.17 85 85 # 1 17 5 , we have 5冄0.85 . Because 100 100 5 100

In general, to divide a decimal by a nonzero whole number, we (1) place the decimal point in the quotient directly above the decimal point in the dividend Quotient Divisor冄Dividend

Format for long division

and then (2) divide as with whole numbers, except that in the division process, 0s are placed in the quotient immediately to the right of the decimal point in order to show the correct place value.

1.6

0.121 4冄 0.484

0.24 32冄7.68 64 1 28 1 28

0.019 12冄0.228 12 108 108


61

Zero needed to show the correct place value

Don’t forget that you can check division by multiplication. For example, because (12)(0.019) 0.228, we know that our last division example is correct. We can easily handle problems involving division by a decimal by changing to an equivalent problem that has a whole-number divisor. Consider the following examples, in which we have changed the original division problem to fractional form to show the reasoning involved in the procedure.

0.6冄 0.24

0.24 0.24 10 2.4 a ba b 0.6 0.6 10 6

0.4 6冄2.4

0.156 0.156 100 15.6 a ba b 0.12 0.12 100 12

0.12冄 0.156

1.3 12冄15.6 12 36 36

1.3冄 0.026

EXAMPLE 4

0.026 10 0.26 0.026 a ba b 1.3 1.3 10 13

0.02 13冄0.26 26

Divide using a long division format. (a) 1.176 0.21

(b) 0.148 3.7

Solution Move the decimal point in the divisor to the end of the number. Then move the decimal point in the dividend the same number of places. The decimal point for the quotient will then line up with the decimal point in the dividend as shown below. (a)

5.6 0xx21.冄1xx17.6 1 05 12 6 12 6

(b)

0.04 3xz7.冄0zx1.48 1 48

The arrows indicate that both the divisor and dividend were multiplied by 100, which changes the divisor to a whole number

The divisor and dividend were multiplied by 10

▼ PRACTICE YOUR SKILL Divide using a long division format. (a) 29.16 1.2

(b) 20.528 0.04

■

Our agreements for operating with positive and negative integers extend to all real numbers. For example, the product of two negative real numbers is a positive real number. Make sure that you agree with the following results. (You may need to do some work on scratch paper, because the steps are not shown.)

62

Chapter 1


0.24 (0.18) 0.06

(0.4)(0.8) 0.32

7.2 5.1 2.1

(0.5)(0.13) 0.065

0.6 (0.8) 1.4

(1.4) (0.2) 7

2.4 6.1 3.7

(0.18) (0.3) 0.6

0.31 (0.52) 0.83

(0.24) (4) 0.06

(0.2)(0.3) 0.06

4 Combine Similar Terms with Decimal Coefficients Numerical and algebraic expressions may contain the decimal form as well as the fractional form of rational numbers. For numerical expressions, we continue to follow the rule of doing the multiplications and divisions first and then the additions and subtractions, unless parentheses indicate otherwise. You have already combined similar terms when the coefficients were in fractional form. The next example shows how to combine similar terms for algebraic expressions when the coefficients are in decimal form.

EXAMPLE 5

Simplify 1.4x 0.03y 0.6x 21.5y by combining similar terms.

Solution Use the commutative property to rearrange the terms. Then combine the like terms by applying the distributive property. 1.4x 0.03y 0.6x 21.5y 1.4x 0.03y 0.6x 21.5y 1.4x 0.6x 0.03y 21.5y (1.4 0.6)x (0.03 21.5)y 2x 21.53y

▼ PRACTICE YOUR SKILL Simplify 5.3a 3.06b 0.08b 0.01a.

■

5 Evaluate Algebraic Expressions The following examples illustrate a variety of situations that involve both the decimal and the fractional forms of rational numbers.

EXAMPLE 6

5 3 1 Evaluate a b for a and b 1. 5 7 2

Solution 3 1 3 5 1 a b a b 112 5 7 5 2 7

3 1 2 7

21 2 14 14

23 14

for a

5 and b 1 2

1.6


63

▼ PRACTICE YOUR SKILL 1 3 2 Evaluate x y for x and y 2. 3 4 7

EXAMPLE 7

■

2 1 3 1 Evaluate x x x for x . 2 3 5 4

Solution First, let’s combine similar terms by using the distributive property. 2 1 1 2 1 1 x x x a bx 2 3 5 2 3 5 a

15 20 6 bx 30 30 30

29 x 30

Now we can evaluate. 29 3 29 x a b 30 30 4

when x

3 4

1

29 29 3 a b 30 4 40 10

▼ PRACTICE YOUR SKILL 3 1 6 1 Evaluate x x x for x . 4 5 3 7

EXAMPLE 8

■

Evaluate 2x 3y for x 1.6 and y 2.7.

Solution 2x 3y 211.62 312.72

when x 1.6 and y 2.7

3.2 8.1 11.3

▼ PRACTICE YOUR SKILL Evaluate 2x 4y for x 2.1 and y 0.07.

EXAMPLE 9

■

Evaluate 0.9x 0.7x 0.4x 1.3x for x 0.2.

Solution First, let’s combine similar terms by using the distributive property. 0.9x 0.7x 0.4x 1.3x (0.9 0.7 0.4 1.3)x 2.5x Now we can evaluate. 2.5x (2.5)(0.2)

for x 0.2

0.5

▼ PRACTICE YOUR SKILL Evaluate 5.6x 0.8x 1.8x 0.4 for x 0.3.

■

64

Chapter 1


6 Solve Application Problems That Involve Rational Numbers in Decimal Form EXAMPLE 10

Apply Your Skill A layout artist is putting together a group of images. She has four images with widths of 1.35 centimeters, 2.6 centimeters, 5.45 centimeters, and 3.2 centimeters, respectively. If the images are set side by side, what will be their combined width?

Solution To find the combined width, we need to add the four widths. 1 1

1.35 2.6 5.45 3.2 12.60 So the combined width would be 12.6 centimeters.

▼ PRACTICE YOUR SKILL A pharmacist is mixing a liquid preparation to fill a prescription. The mixture requires 4.6 ounces of magnesium citrate solution, 1.25 ounces of sodium chloride solution, 8 ounces of water, and 0.02 ounces of vanilla flavoring. How many ounces are in the final mixture? ■

CONCEPT QUIZ

For Problems 1–10, answer true or false. 1. A rational number can be defined as any number that has a terminating or repeating decimal representation. 2. A repeating decimal has a block of digits that repeat only once. 3. Every irrational number is also classified as a real number. 4. The rational numbers along with the irrational numbers form the set of natural numbers. 5. 0.141414 . . . is a rational number. 6. 15 is real, irrational, and negative. 7. 0.35 is real, rational, integer, and positive. 8. The reciprocal of c, where c 0, is also the multiplicative inverse of c. 9. Any number multiplied by its multiplicative inverse gives a result of 0. 10. Zero does not have a multiplicative inverse.

Problem Set 1.6 1 Classify Real Numbers For Problems 1– 8, classify the real numbers by tracing down the diagram on page 56.

2 Expand the Properties of Numbers to Include Real Numbers For Problems 9 –18, match the statement with the name of the property.

1. 2

2. 1/3

3. 15

4. 0.09090909. . .

5. 0.16

6. 13

10. 8(0) (0)(8) 0

7. 8/7

8. 0.125

11. 6 15 15 6

9. 17 # 42 # 25 7 # 14 # 252

1.6


12. 215 82 2 # 5 122182

51. 0.6(4.1) 0.7(3.2)

13. 4 (3 9) [4 (3)] + 9

53. 7(0.6) 0.9 3(0.4) 0.4

14. (7)(1) (1)(7) 7

54. 5(0.9) 0.6 4.1(6) 0.9

1 1 15. 6 a b 162 1 6 6

55. (0.96) (0.8) 6(1.4) 5.2

65

52. 0.5(74) 0.9(87)

56. (2.98) 0.4 5(2.3) 1.6

16. 12 (12) (12) 12 0

57. 5(2.3) 1.2 7.36 0.8 0.2

17. 3 0 0 (3) 3

58. 0.9(12) 0.4 1.36 17 9.2

18. (5)(1) (1)(5) 5 A. Commutative Property of Addition B. Associative Property of Addition C. Associative Property of Multiplication D. Identity Property of Addition E. Identity Property of Multiplication F. Additive Inverse Property G. Multiplication Property of Zero H. Multiplicative Property of Negative One I. Multiplicative Inverse Property J. Distributive Property

4 Combine Similar Terms For Problems 59 –70, simplify each algebraic expression by combining similar terms. 59. x 0.4x 1.8x 60. 2x 1.7x 4.6x 61. 5.4n 0.8n 1.6n 62. 6.2n 7.8n 1.3n 63. 3t 4.2t 0.9t 0.2t 64. 7.4t 3.9t 0.6t 4.7t 65. 3.6x 7.4y 9.4x 10.2y 66. 5.7x 9.4y 6.2x 4.4y

3 Add, Subtract, Multiply, and Divide Decimals For Problems 19 – 44, perform each indicated operation.

67. 0.3(x 4) 0.4(x 6) 0.6x 68. 0.7(x 7) 0.9(x 2) 0.5x 69. 6(x 1.1) 5(x 2.3) 4(x 1.8) 70. 4(x 0.7) 9(x 0.2) 3(x 0.6)

19. 0.37 0.25

20. 7.2 4.9

21. 2.93 1.48

22. 14.36 5.89

23. (4.7) 1.4

24. (14.1) 9.5

25. 3.8 11.3

26. 2.5 14.8

27. 6.6 (1.2)

28. 18.3 (7.4)

29. 11.5 (10.6)

30. 14.6 (8.3)

31. (0.4)(2.9)

32. (0.3)(3.6)

33. (0.8)(0.34)

34. (0.7)(0.67)

35. (9)(2.7)

36. (8)(7.6)

37. (0.7)(64)

38. (0.9)(56)

1 3 1 for x , y , and z 6 4 3 2 3 1 72. 2x y 3z for x , y , and z 5 4 2 3 2 7 5 y for y 73. y y 5 3 15 2

39. 1.56 1.3

40. 7.14 2.1

74.

41. 5.92 (0.8)

42. 2.94 0.6

43. 0.266 (0.7)

44. 0.126 (0.9)

5 Evaluate Algebraic Expressions For Problems 71– 84, evaluate each algebraic expression for the given values of the variables. Don’t forget that for some problems, it might be helpful to combine similar terms first and then to evaluate. 71. x 2y 3z

1 2 3 7 x x x for x 2 3 4 8 75. x 2y 4z for x 1.7, y 2.3, and z 3.6 76. 2x y 5z for x 2.9, y 7.4, and z 6.7

For Problems 45 –58, simplify each numerical expression.

77. 5x 7y

for x 7.8 and y 8.4

45. 16.5 18.7 9.4

78. 8x 9y

for x 4.3 and y 5.2

46. 17.7 21.2 14.6

47. 0.34 0.21 0.74 0.19

79. 0.7x 0.6y

for x 2 and y 6

48. 5.2 6.8 4.7 3.9 1.3

80. 0.8x 2.1y

for x 5 and y 9

49. 0.76(0.2 0.8)

81. 1.2x 2.3x 1.4x 7.6x

50. 9.8(1.8 0.8)

for x 2.5

66

Chapter 1


82. 3.4x 1.9x 5.2x

for x 0.3

83. 3a 1 7a 2

for a 0.9

84. 5x 2 6x 4

line to fill the tank. How many miles per gallon did she get on that tank of gas? 89. The total length of the four sides of a square is 18.8 centimeters. How long is each side of the square?

for x 1.1

6 Solve Application Problems That Involve Rational Numbers in Decimal Form 85. Tanya bought 400 shares of one stock at $14.35 per share and 250 shares of another stock at $16.68 per share. How much did she pay for the 650 shares? 86. On a trip Brent bought these amounts of gasoline: 9.7 gallons, 12.3 gallons, 14.6 gallons, 12.2 gallons, 13.8 gallons, and 15.5 gallons. How many gallons of gasoline did he purchase on the trip? 87. Kathrin has a piece of copper tubing that is 76.4 centimeters long. She needs to cut it into four pieces all of equal length. Find the length of each piece. 88. On a trip Biance filled the gas tank and noted that the odometer read 24,876.2 miles. After the next filling the odometer read 25,170.5 miles. It took 13.5 gallons of gaso-

90. When the market opened on Monday morning, Garth bought some shares of a stock at $13.25 per share. The daily changes in the market for that stock for the week were 0.75, 1.50, 2.25, 0.25, and 0.50. What was the value of one share of that stock when the market closed on Friday afternoon? 91. Victoria bought two pounds of Gala apples at $1.79 per pound and three pounds of Fuji apples at $0.99 per pound. How much did she spend for the apples? 92. In 2003 the average speed of the winner of the Daytona 500 was 133.87 miles per hour. In 2000 the average speed of the winner was 155.669 miles per hour. How much faster was the average speed of the winner in 2000 compared to the winner in 2003? 93. Use a calculator to check your answers for Problems 45 –58.

THOUGHTS INTO WORDS 94. At this time how would you describe the difference between arithmetic and algebra?

96. Do you think that 2 12 is a rational or an irrational number? Defend your answer.

95. How have the properties of the real numbers been used thus far in your study of arithmetic and algebra?

FURTHER INVESTIGATIONS 97. Without doing the actual dividing, defend the state1 ment, “ produces a repeating decimal.” [Hint: Think 7 about the possible remainders when dividing by 7.]

99. a. How can we tell that decimal? b. How can we tell that

98. Express each of the following in repeating decimal form. a.

1 7

b.

2 7

c.

4 9

d.

5 6

e.

3 11

f.

1 12

5 will produce a terminating 16 7 will not produce a terminat15

ing decimal? c. Determine which of the following will produce a 7 11 5 7 11 13 17 11 terminating decimal: , , , , , , , , 8 16 12 24 75 32 40 30 9 3 , . 20 64

Answers to the Example Practice Skills 1. (a) 38.75 (b) 15.551 2. (a) 3.25 (b) 8.107 3. (a) 37.488 (b) 0.00034 81 11 5. 5.29a 3.14b 6. 7. 8. 3.92 9. 1.78 10. 13.87 ounces 14 70

4. (a) 24.3 (b) 513.2


2. False

3. True 4. False 5. True

6. True

7. False

8. True

9. False 10. True


SUMMARY

EXAMPLE

List the elements of a set. (Sec. 1.1, Obj. 1, p. 2)

Elements of a set can be shown by a word description or a list.

List the set of even numbers greater than 4.

CHAPTER REVIEW PROBLEMS Problems 1– 4

Solution

A {6, 8, 10, 12, . . .} Determine if sets are equal. (Sec. 1.1, Obj. 1, p. 2)

Equal sets have the same members.

Is A {1, 3, 5, 7} equal to B {1, 5}?

Problems 5 – 8

Solution

AB Classify numbers in the real number system. (Sec. 1.6, Obj. 1, p. 55)

Any number that has a terminating or repeating decimal representation is a rational number. Any number that has a nonterminating or nonrepeating decimal representation is an irrational number. The rational numbers together with the irrational numbers form the set of real numbers.

Add integers. (Sec. 1.2, Obj. 2, p. 13)

The number line is a convenient visual aid for interpreting the addition of integers. See page 14 for the formal rules of addition.

Subtract integers. (Sec. 1.2, Obj. 3, p. 15)

Subtraction is defined in terms of addition. To subtract a signed number you can add its opposite.

3 Classify 1, 27, and . 4

Problems 9 –14

Solution

1 is a real number, a rational number, an integer, and negative. 27 is a real number, an irrational number, and positive. 3 is a real number, a rational 4 number, noninteger, and positive. Find the sums: 24 1132; 12 162; 13 152.

Problems 15 –24

Solution

24 1132 37 12 162 18 13 152 8 Find the differences: 35 142; 12 132 .

Problems 25 –34

Solution

35 (4) 35 (4) 39 12 (3) 12 (3) 9 Multiply and divide integers. (Sec. 1.2, Obj. 4, p. 16; Sec. 1.2, Obj. 5, p. 18)

The product (or quotient) of two positive or two negative integers is positive. The product (or quotient) of a positive and a negative integer is negative.

Solve application problems involving integers. (Sec. 1.2, Obj. 8, p. 21)

Positive and negative integers can be used to represent many real world problems.

Perform the indicated operation: (4)(5); (2)(8); (15)(3).

Problems 35 – 40

Solution

(4)(5) 20 (2)(8) 16 (15) (3) 5 A stock selling for $38.70 went up $3.52 one day, and the next day it fell $1.82. What is the price of the stock after these two days?

Problems 41– 44

Solution

38.70 3.52 (1.82) 40.40 The price is $40.40.

(continued)

67

68

Chapter 1



OBJECTIVE

SUMMARY

EXAMPLE

Know the properties of integers. (Sec. 1.3, Obj. 1, p. 24)

The commutative property deals with the order of numbers.

State the property demonstrated by 15 1102 10 15.

The associative property deals with the grouping of numbers.

Solution

See pages 25 and 26 for a full listing of properties. Identify numbers as prime or composite. (Sec. 1.4, Obj. 2, p. 34)

Factor a number into a product of primes. (Sec. 1.4, Obj. 3, p. 34)

A prime number is a number greater than 1 that only has itself and 1 as factors. Any number that is not prime is called a composite number. Start by picking any two factors of the number. If those numbers are not prime, then continue by picking factors of those numbers until every factor is a prime number.

Problems 45 – 46

Commutative property for addition

List the prime numbers less than 50.

Problems 47–52

Solution

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 Prime factor 120.

Problems 53 –58

Solution

120 12 # 10 3#4#2#5 3#2#2#2#5 23 # 3 # 5

Find the greatest common factor. (Sec. 1.4, Obj. 4, p. 36)

The greatest common factor (GCF) of two numbers is the largest divisor of both numbers. Prime factor each number, and select each common factor the least number of times it appears in the factorizations.

Find the greatest common factor (GCF) of 60 and 140.

Problems 59 – 60

Solution

60 22 # 3 # 5 210 2 # 3 # 5 # 7 GCF 2 # 3 # 5 30

Find the least common multiple. (Sec. 1.4, Obj. 5, p. 37)

The least common multiple (LCM) is the smallest nonzero common multiple of each number. Prime factor each number, and select each factor the most number of times it appears in the factorizations.

Find the least common multiple (LCM) of 12 and 54.

Problems 61– 62

Solution

12 22 # 3 54 2 # 33 LCM 22 # 33 108

Multiply and divide fractions. (Sec. 1.5, Obj. 2, p. 43; Sec. 1.5, Obj. 3, p. 45)

To multiply rational numbers in fractional form, multiply the numerators and multiply the denominators. To divide rational numbers in fractional form, change the problem to multiplying by the reciprocal of the divisor. Always express the result in reduced form.

Divide a

4 2 b a b. 15 3

Problems 63 –72

Solution

a

2 4 2 b a b a b 15 3 15

#

3 a b 4

6 6#1 1 # 60 6 10 10

(continued)

Chapter 1

OBJECTIVE

SUMMARY

Add and subtract fractions. (Sec. 1.5, Obj. 4, p. 46)

When the fractions have a common denominator, add (or subtract) the numerators. If the fractions do not have a common denominator, then use the fundamental principle a a#k of fractions, # , to obtain b b k equivalent fractions that have a common denominator.

Add or subtract decimals. (Sec. 1.6, Obj. 3, p. 58)

Multiply decimals. (Sec. 1.6, Obj. 3, p. 58)

To add or subtract decimals, write the numbers in a column so that the decimal points line up. Then add or subtract the numbers. It may be necessary to fill in zeros as placeholders.

1. Multiply the numbers and ignore the decimal points. 2. Find the sum of the number of digits to the right of the decimal point in each factor. 3. Insert the decimal point in the product that corresponds to the sum found in step 2 above. It may be necessary to insert zeros as placeholders.

Divide decimals. (Sec. 1.6, Obj. 3, p. 58)

Simplify numerical expressions involving exponents. (Sec. 1.1, Obj. 2, p. 3)

To divide a decimal number by a whole number, ignore the decimal points and divide the numbers. Place the decimal point in the quotient directly above the decimal point in the dividend. It may be necessary to put zeros in the quotient to the right of the decimal point. To divide by a decimal number, change to an equivalent problem that has a whole number divisor. Expressions of the form bn are read “b to the nth power”: b is the base, and n is the exponent. Expressions of the form bn tell us that the base, b, is used as a factor n times.

EXAMPLE Add

1 7 . 12 15

Summary

69

CHAPTER REVIEW PROBLEMS Problems 73 – 82

Solution

7 1 7#5 1#4 # 12 15 12 5 15 # 4 35 4 60 60 39 60 13 20

Add 23.48 + 1.6 + 0.075.

Problems 83 – 88

Solution

23.48 1.6 + 0.075 25.155 Multiply 24.1(0.12).

Problems 89 –94

Solution

24.1 one digit to the right 0.12 two digits to the right 482 241 2.892 three digits to the right

Divide 1.2冄50.52.

Problems 95 –100

Solution

42.1 1.2冄 50.52 48 25 24 12 12 2 4 Simplify (a) (3)4 and (b) a b . 3

Problems 101–104

Solution

(a) (3)4 (3)(3)(3)(3) 81 4 2 2 2 2 (b) a b # 3 3 3 9

(continued)

70

Chapter 1


OBJECTIVE

SUMMARY

Use the order of operations to simplify numerical expressions. (Sec. 1.1, Obj. 2, p. 3)

1. Perform the operations inside grouping symbols and above and below each fraction bar. 2. Evaluate all exponential expressions. 3. Perform all multiplications in order from left to right. 4. Perform all additions and subtractions in order from left to right.

Combine similar terms. (Sec. 1.3, Obj. 3, p. 28; Sec. 1.6, Obj. 4, p. 62)

Use the distributive property to combine similar terms.


EXAMPLE Simplify 6 12 3 1 # 2.

Problems 105 –114

Solution

6 12 3 1 # 2 6 4 2 10 2 8

Simplify 3x 12y 5x 8y.

Problems 115 –124

Solution

3x 12y 5x 8y 3x 5x 12y 8y (3 5)x (12 8)y 2x 20

Evaluate algebraic expressions. (Sec. 1.1, Obj. 3, p. 7; Sec. 1.2, Obj. 7, p. 20; Sec. 1.3, Obj. 3, p. 24; Sec. 1.6, Obj. 5, p. 62)

First simplify the algebraic expression. Then replace the variable with the given value. It is a good practice to use parentheses when substituting the value. Be sure to follow the order of operations as you simplify the resulting numerical expression.

Evaluate 2x 31x 5y2 for x 7 and y 2.

Problems 125 –132

Solution

2x 31x 5y2 2x 3x 15y 5x 15y 5(7) 15(2) when x 7, y 2 35 30 65

Solve application problems involving rational numbers in fraction or decimal form. (Sec. 1.5, Obj. 6, p. 51; Sec. 1.6, Obj. 6, p. 64)

Rational numbers are used to solve many real world problems.

for x 7 and x 2

To obtain a custom hair color for a client, Marti is mixing 3 cup of brown color with 8 1 cup of blonde color. How 4 many cups of color are being used for the custom color? Solution

To solve, add

1 3 . 8 4

1 3 2 5 3 8 4 8 8 8 So

5 cup of color is being used. 8

Problems 133 –136

Chapter 1

Review Problem Set

71

Chapter 1 Review Problem Set For Problems 1– 4, given the word description of the set, write its list description as a set. 1. A {whole numbers less than 5} 2. B {odd whole numbers less than 11} 3. C {prime numbers less than 11} 4. D {even whole numbers} For Problems 5 – 8, insert the symbol for “equal” or “not equal” to make each of the following a true statement. 5. {1, 2, 3, 4, 5} 6. {1, 3, 5, . . .}

{ 5, 3, 1}

7. {8, 6, 4, 2} 8. {1, 3, 5}

{. . . , 5, 4 , 3, 2, 1}

{2, 4, 6, 8} {1, 1, 3, 3, 5}

For Problems 9 –14, classify the real numbers by tracing down the chart on page 56. 9. 8 10. 0

1937. Find the difference between the record high and low temperatures. 42. In North America the highest elevation, which is the top of Mt. McKinley, Alaska, is 20,320 feet above sea level. The lowest elevation in North America, which is the base of Death Valley, California, is 282 feet below sea level. Find the absolute value of the difference in elevation between Mt. McKinley and Death Valley. 43. As a running back in a football game, Marquette carried the ball 7 times. On two plays he gained 6 yards each play; on another play he lost 4 yards; on the next three plays he gained 8 yards per play; and on the last play he lost 1 yard. Write a numerical expression that gives Marquette’s overall yardage for the game, and simplify the expression to give the total yardage. 44. Shelley started the month with $3278 in her checking account. During the month she deposited $175 each week for 4 weeks but had debit charges of $50, $189, $160, $20, and $115. What is the balance in her checking account after these deposits and debits? 45. State the property of integers demonstrated by (4 6) 8 (6 4) 8.

11. 0.45 7 12. 5 13. 22

46. State the property of integers demonstrated by 4(3 7) 4(3) (4)(7). For Problems 47–52, classify each number as prime or composite.

3 14. 216

For Problems 15 –24, add the integers.

47. 73

48. 87

15. 7 (10)

16. (12) (13)

49. 63

50. 81

17. (18) 5

18. 17 (29)

51. 49

52. 91

19. (12) (10)

20. (4) 10

21. 25 (50)

22. (21) (18)

For Problems 53 –58, express each number as the product of prime factors.

23. (12) 36

24. (18) 8

53. 24

54. 63

For Problems 25 –34, subtract the integers.

55. 120

56. 135

25. 8 (13)

26. 6 (9)

57. 84

58. 64

27. 12 (11)

28. 17 (19)

59. Find the greatest common factor of 36 and 54.

29. 35 (10)

30. 18 (7)

60. Find the greatest common factor of 48, 60, and 84.

31. 32 (24)

32. 16 (4)

61. Find the least common multiple of 18 and 20.

33. 24 (58)

34. 10 (55)

62. Find the least common multiple of 12, 20, and 30.

For Problems 35 – 40, perform the indicated multiplication or division.

For Problems 63 –72, perform the indicated operations and express answers in reduced form.

35. (13)(12)

36. (14)(18)

37. (72) (12)

38. 117 (9)

2 3 63. a b a b 5 4

5 1 64. a b a b 8 2

39. (12)(4)

40. (56) (8)

3 7 65. a b a b 4 8

1 3 66. a b a b 8 2

67. a

68. a

41. A record high temperature of 125°F occurred in Laughlin, Nevada, on June 29, 1994. A record low temperature of 50°F occurred in San Jacinto, Nevada, on January 8,

12 2 b a b 5 15

3 18 b a b 5 2

72

Chapter 1 12 4 b a b 15 3

70. a

10 3 b a b 3 2

4 4 72. a b a b 5 7

69. a 71. a

Basic Concepts of Arithmetic and Algebra 12 14 ba b 7 36

For Problems 73 – 82, perform the indicated operations and express answers in reduced form. 73.

3 5 8 8

74.

75.

13 15 32 32

76.

11 1 12 3 5 7 79. 12 9 77.

81.

3 5 2 9

11 1 7 7

9 7 24 24

7 1 2 10 5 17 80. 18 8 78.

82.

2 7 3 8

For Problems 83 – 88, perform the indicated operations. 83. 13.7 63.48 0.34

84. 3.56 2.8 25.04

85. 16.4 12.32

86. 34.5 8.45

87. 0.016 0.8 0.04

88. 4.08 1.8 0.007

For Problems 89 – 94, perform the indicated operations.

For Problems 115 –124, simplify each algebraic expression by combining like terms. 115. 8x 5y 13x y 116. 5xy 9xy xy y 117. 51x 42 31x 92 118. 31x 22 41x 62 119. 2x 31x 42 2x 120. 1a 12 31a 22 4a 1 121.

2 1 3 3 x y x y 8 5 4 10

122.

1 1 2 ax yb x 3 4 6

123. 1.41a b2 0.81a b2 124. 0.24ab 0.73bc 0.82ab 0.37bc For Problems 125 –132, evaluate each algebraic expression for the given values of the variables. 125. 5a 6b 7a 2b

for a 1 and b 5

126. 2xy 6 5xy 8

for x 1 and y 2

89. (24.6)(1.2)

90. (6.42)(1.8)

127. 71x 62 91x 12

91. (5.07)(0.03)

92. (0.045)(0.2)

128. 31x 42 21x 82

93. (0.084)(1.2)

94. (0.2)(0.2)

96. 0.9492 0.6

97. 37.92 12

98. 111.72 21

99. 240 2.4

100. 864 1.2

for x 7

129.

5x 2y 2x 7

for x 6 and y 4

130.

3x 4y 3x

for x 4 and y 6

For Problems 95 –100, perform the indicated operations. 95. 32.43 2.3

for x 2

131. a3 b2

for a

1 1 and b 2 3

For Problems 101–104, evaluate each numerical expression.

132. 2x2 3y2 for x 5 and y 2

101. 26

133. A caterer received $840 for catering a birthday party. If her chef is paid one-fifth of that amount, how much will the chef receive?

103. 13 22

102. 132 4

2

104. 10.42 2

For Problems 105 –114, simplify each numerical expression by following the order of operations. 105. 6 4 # 2 12 2 106. 8 4162 4 2 107. 36 24 12 3 # 2 108. 16 24 182 3 # 2 109.

4 1 2 1 # 5 5 3 4

110.

1 2 1 2 # 1 # 3 4 2 3 4

111. 21 3 2114 72 4

112. 15 3 142 72 # 34

3#69#2 113. 63

4 # 12 6 # 8 114. 24

7 yards of leather to reupholster automobile 8 3 seats. If a seat requires 1 yards of leather, how much 4 leather will be left after Julio reupholsters 3 seats?

134. Julio has 5

135. Kyle took his race car engine to a machine shop to have the cylinder head bored out. The machine shop measured and found that the diameter of the cylinder is 3.257 inches. If the diameter of the cylinder is increased by 0.12 inches, how large will the diameter of the cylinder be? 136. When Marci was planning her trip to Europe, each U.S. dollar was worth 0.66 euros. At this exchange rate, how many euros will Marci have if she exchanges $450?

Chapter 1 Test For Problems 1–5, simplify each of the numerical expressions. 1. 5(7) (3)(8)

1.

2. 4 20 2 # 5

2.

3. 6(2 5) 8

3.

4. (8 3)

4.

5. (14)(4) 4 (6)

5.

6. It was reported on the 5 o’clock weather show that the current temperature was 7 F. The temperature was forecast to drop 13 degrees by 6:00 A.M. If the forecast is correct, what will be the temperature at 6:00 A.M.?

6.

7. State the property of integers demonstrated by (5 7) (8) (7 5) (8)

7.

2

For Problems 8 –11, simplify by combining similar terms. 8. 7x 9y y 2x 2y 7x

8.

9. 2(x 4) 5(x 7) 6(x 1)

9.

10. 3a 7b 4b a

10.

11. 5x2 3x 4 10x 9

11.

2

12. Express

2

2

2

42 in reduced form. 54

12.

13. Classify 79 as prime or composite.

13.

14. Express 360 as a product of prime numbers.

14.

15. Find the greatest common factor of 36, 60, and 84.

15.

16. Find the least common multiple of 9 and 24.

16.

For Problems 17–20, perform the indicated operations, and express your answer in reduced form. 17.

15 5 12 8

17.

18.

3 1 4 15 10 6

18.

1 3 5 2 19. a b 3 2 4 6

19.

20. 1.4(2.3) (0.8)2

20.

For Problems 21–23, evaluate each algebraic expression for the given values of the variables. 21. 3x 2y xy

for x 0.5 and y 0.9

21.

22. a ab b

for a 4 and b 3

22.

2

2

23. 3(x 2) 5(x 4) 6(x 1)

for x 3

23.

73

74

Chapter 1


24.

24. Joni, a pediatric nurse, is mixing a solution of medicine for a patient. Each dose 1 1 is a mixture of teaspoon of cough syrup, teaspoon of antihistamine, and 8 2 1 teaspoon of decongestant. How many teaspoons of medicine will there be for 4 16 doses?

25.

25. On a recent fishing trip, Hector purchased gasoline daily from the marina. His purchases for the four days were 13.2 gallons, 18.4 gallons, 12.8 gallons, and 16.3 gallons. Find the cost of Hector’s gasoline purchases if the marina charged $3.30 for a gallon of gasoline.

Equations, Inequalities, and Problem Solving

2 2.1 Solving First-Degree Equations 2.2 Translating from English to Algebra 2.3 Equations and Problem Solving 2.4 More on Solving Equations and Problem Solving

Radius Images/Jupiter Images

2.5 Equations Involving Parentheses and Fractional Forms 2.6 Inequalities

95 82 93 84 s 90 can be used 5 to determine that Ashley needs a 96 or higher on her fifth exam to average 90 or higher for the five exams if she got 95, 82, 93, and 84 on her first four exams.

2.7 Inequalities, Compound Inequalities, and Problem Solving

■ The inequality

C

arlos paid $645 to have his air conditioner repaired. Included in the bill were $360 for parts and a charge for 3 hours of labor. How much was Carlos charged for each hour of labor? If we let h represent the hourly charge, then the equation 360 3h 645 can be used to determine that the per-hour charge is $95. Tracy received a cell phone bill for $136.74. Included in the $136.74 were a monthly-plan charge of $39.99 and a charge for 215 extra minutes. How much is Tracy being charged for each extra minute? If we let c represent the charge per minute, then the equation 39.99 215c 136.74 can be used to determine that the charge for each extra minute is $0.45. Chris had scores of 93, 86, and 89 on her first three algebra tests. What score must she get on the fourth test to average 90 or higher for the four tests? If we let x represent Chris’s fourth test grade, then we can use the inequality 93 86 89 x 90 to determine that Chris needs to score 92 or higher. 4 Throughout this book we follow a common theme: Develop some new skills, use the skills to help solve equations and inequalities, and finally, use the equations and inequalities to solve applied problems. In this chapter we want to use the skills we developed in the first chapter to solve equations and inequalities and to expand our work with applied problems. Video tutorials for all section learning objectives are available in a variety of delivery modes.

75

I N T E R N E T

P R O J E C T

Some of the application problems in this chapter concern supplementary and complementary angles. Do an Internet search to find a site that has an interactive demonstration of supplementary or complementary angles. Such a site might allow you to change the size of one of the supplementary angles and see how the other supplementary angle changes. Also search for the definitions of acute angles and obtuse angles. Can both (of two) supplementary angles be obtuse angles? Can both (of two) supplementary angles be acute angles?

2.1

Solving First-Degree Equations OBJECTIVES 1

Solve Equations Using the Addition-Subtraction Property of Equality

2

Solve Equations Using the Multiplication-Division Property of Equality

1 Solve Equations Using the Addition-Subtraction Property of Equality These are examples of numerical statements: 347

523

7 1 12

The first two are true statements, and the third is a false statement. When you use x as a variable, statements like x34

2x 1 7

x2 4

are called algebraic equations in x. We call a number a a solution or root of an equation if a true numerical statement is formed when we substitute a for x. (We also say that a satisfies the equation.) For example, 1 is a solution of x 3 4 because substituting 1 for x produces the true numerical statement 1 3 4. We call the set of all solutions of an equation its solution set. Thus the solution set of x 3 4 is {1}. Likewise, the solution set of 2x 1 7 is {4}, and the solution set of x2 4 is {2, 2}. Solving an equation refers to the process of determining the solution set. Remember that a set that consists of no elements is called the empty or null set and is denoted by . Thus we say that the solution set of x x 1 is ; that is, there are no real numbers that satisfy x x 1. In this chapter we will consider techniques for solving first-degree equations of one variable. This means that the equations contain only one variable, and this variable has an exponent of 1. Here are some examples of first-degree equations of one variable: 3x 4 7

0.8w 7.1 5.2w 4.8

1 y29 2

7x 2x 1 4x 1

Equivalent equations are equations that have the same solution set. For example, 5x 4 3x 8 2x 12 x6 are all equivalent equations; this can be verified by showing that 6 is the solution for all three equations.

76

2.1

Solving First-Degree Equations

77

As we work with equations, we can use the following properties of equality:

Property 2.1

Properties of Equality

For all real numbers, a, b, and c, 1. a a

Reflexive property

2. if a b, then b a

Symmetric property

3. if a b and b c, then a c

Transitive property

4. if a b, then a may be replaced by b, or b may be replaced by a, in any statement, without changing the meaning of the statement. Substitution property The general procedure for solving an equation is to continue replacing the given equation with equivalent but simpler equations until we obtain an equation of the form variable constant or constant variable. Thus in the preceding example, 5x 4 3x 8 was simplified to 2x 12, which was further simplified to x 6, from which the solution of 6 is obvious. The exact procedure for simplifying equations is our next concern. Two properties of equality play an important role in the process of solving equations. The first of these is the addition-subtraction property of equality, which we state as follows:

Property 2.2

Addition-Subtraction Property of Equality

For all real numbers a, b, and c, 1. a b if and only if a c b c 2. a b if and only if a c b c Property 2.2 states that any number can be added to or subtracted from both sides of an equation, and the result is an equivalent equation. Consider the use of this property in the next four examples.

EXAMPLE 1

Solve x 8 3.

Solution x83 x8838

Add 8 to both sides

x 11 The solution set is {11}.

▼ PRACTICE YOUR SKILL Solve x 11 4.

■

Remark: It is true that a simple equation like Example 1 can be solved by inspection. That is to say, we could think, “some number minus 8 produces 3,” and obviously, the number is 11. However, as the equations become more complex, the technique of solving by inspection becomes ineffective. This is why it is necessary to develop more formal techniques for solving equations. Therefore we will begin developing such techniques even with very simple equations.

78

Chapter 2


EXAMPLE 2

Solve x 14 8.

Solution x 14 8 x 14 14 8 14

Subtract 14 from both sides


▼ PRACTICE YOUR SKILL Solve y 6 2.

EXAMPLE 3

Solve n

■

1 1 . 3 4

Solution n n

1 1 3 4

1 1 1 1 3 3 4 3 n

4 3 12 12

n

7 12

The solution set is e

Add

1 to both sides 3

7 f. 12

▼ PRACTICE YOUR SKILL Solve x

EXAMPLE 4

1 1 . 5 2

■

Solve 0.72 y 0.35.

Solution 0.72 y 0.35 0.72 0.35 y 0.35 0.35

Subtract 0.35 from both sides

0.37 y The solution set is {0.37}.

▼ PRACTICE YOUR SKILL Solve 0.08 x 0.45.

■

Note in Example 4 that the final equation is 0.37 y instead of y 0.37. Technically, the symmetric property of equality (if a b, then b a) would permit us to change from 0.37 y to y 0.37, but such a change is not necessary to determine that the solution is 0.37. You should also realize that you could apply the symmetric

2.1


79

property to the original equation. Thus 0.72 y 0.35 becomes y 0.35 0.72, and subtracting 0.35 from both sides would produce y 0.37. We should make at this time one other comment that pertains to Property 2.2. Because subtracting a number is equivalent to adding its opposite, Property 2.2 could be stated only in terms of addition. Thus to solve an equation such as Example 4, we could add 0.35 to both sides rather than subtracting 0.35 from both sides.

2 Solve Equations Using the Multiplication-Division Property of Equality The other important property for solving equations is the multiplication-division property of equality.

Property 2.3

Multiplication-Division Property of Equality

For all real numbers, a, b, and c, where c 0, 1. a b if and only if ac bc 2. a b if and only if

b a c c

Property 2.3 states that we get an equivalent equation whenever both sides of a given equation are multiplied or divided by the same nonzero real number. The following examples illustrate the use of this property.

EXAMPLE 5

3 Solve x 6. 4

Solution 3 x6 4 4 3 4 a xb 162 3 4 3

Multiply both sides by

4 4 3 because a b a b 1 3 3 4

x8 The solution set is {8}.

▼ PRACTICE YOUR SKILL 2 Solve x 12. 3

EXAMPLE 6

■

Solve 5x 27.

Solution 5x 27 27 5x 5 5 x

27 5

Divide both sides by 5 27 2 could be expressed as 5 or 5.4 5 5


27 f. 5

80

Chapter 2


▼ PRACTICE YOUR SKILL Solve 3x 49.

EXAMPLE 7

■

2 1 Solve p . 3 2

Solution 1 2 p 3 2 3 2 3 1 a b a pb a b a b 2 3 2 2 p

Multiply both sides by because a b a b 1 3 2

2 3

3 2

3 4

3 The solution set is e f. 4

▼ PRACTICE YOUR SKILL 1 3 Solve y . 5 4

EXAMPLE 8

Solve 26

■

x . 2

Solution 26

x 2

x 21262 2 a b 2

Multiply both sides by 2

52 x The solution set is {52}.

▼ PRACTICE YOUR SKILL Solve

x 3 . 4 2

■

Look back at Examples 5 – 8, and you will notice that whenever the coefficient was an integer we divided both sides of the equation by the coefficient of the variable— otherwise, we used the multiplication part of Property 2.3. Technically, because dividing by a number is equivalent to multiplying by its reciprocal, Property 2.3 could be stated only in terms of multiplication. Thus to solve an equation such as 1 5x 27, we could multiply both sides by instead of dividing both sides by 5. 5

2.1


81

Solve 0.2n 15.

EXAMPLE 9

Solution 0.2n 15 0.2n 15 0.2 0.2

Divide both sides by 0.2

n 75 The solution set is {75}.

▼ PRACTICE YOUR SKILL Solve 0.3y 18.6.

CONCEPT QUIZ

■

For Problems 1–10, answer true or false. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Equivalent equations have the same solution set. x2 9 is a first-degree equation. The set of all solutions is called a solution set. If the solution set is the null set, then the equation has at least one solution. Solving an equation refers to obtaining any other equivalent equation. If 5 is a solution, then a true numerical statement is formed when 5 is substituted for the variable in the equation. Any number can be subtracted from both sides of an equation, and the result is an equivalent equation. Any number can divide both sides of an equation to obtain an equivalent equation. By the reflexive property, if y 2 then 2 y. By the transitive property, if x y and y 4, then x 4.

Problem Set 2.1 1 Solve Equations Using the AdditionSubtraction Property of Equality

21. b 0.19 0.46

22. b 0.27 0.74

23. n 1.7 5.2

24. n 3.6 7.3

Use the properties of equality to help solve each equation. 1. x 9 17

2. x 7 21

3. x 11 5

4. x 13 2

5. 7 x 2

6. 12 x 4

7. 8 n 14

8. 6 n 19

9. 21 y 34

2 Solve Equations Using the MultiplicationDivision Property of Equality 25. 7x 56

26. 9x 108

27. 6x 102

28. 5x 90

10. 17 y 26

29. 5x 37

30. 7x 62

11. x 17 31

12. x 22 14

31. 18 6n

32. 52 13n

13. 14 x 9

14. 17 x 28

33. 26 4n

34. 56 6n

15. 26 n 19

16. 34 n 15

35.

t 16 9

36.

t 8 12

37.

n 3 8

38.

n 5 9

17. y

19. x

2 3 3 4

18. y

3 1 5 3

20. x

2 1 5 6 5 2 8 5

39. x 15

40. x 17

82

41.

Chapter 2


3 x 18 4

42.

2 43. n 14 5

2 x 32 3

55.

3 44. n 33 8

5 7 x 12 6

5 57. x 1 7

56.

3 7 x 24 8

58.

11 x 1 12

45.

2 1 n 3 5

46.

3 1 n 4 8

59. 4n

1 3

60. 6n

47.

5 3 n 6 4

48.

3 6 n 7 8

61. 8n

6 5

62. 12n

49.

3x 3 10 20

50.

5x 5 12 36

63. 1.2x 0.36

64. 2.5x 17.5

65. 30.6 3.4n

66. 2.1 4.2n

67. 3.4x 17

68. 4.2x 50.4

y 1 51. 2 6

y 1 52. 4 9

4 9 53. x 3 8

10 6 54. x 5 14

3 4 8 3

THOUGHTS INTO WORDS 70. Are the equations 6 3x 1 and 1 3x 6 equivalent equations? Defend your answer.

69. Describe the difference between a numerical statement and an algebraic equation.

Answers to the Example Practice Skills 1. {7}

2. {4} 3. e

7 f 10

4. {0.37} 5. {18}

6. e

49 f 3

7. e

5 f 12

6. True

7. True

8. {6} 9. {62}


2.2

2. False

3. True 4. False 5. False

8. False 9. False

10. True

Translating from English to Algebra OBJECTIVES 1

Translate Algebraic Expressions into English Phrases

2

Translate English Phrases into Algebraic Expressions

3

Write Algebraic Expressions That Change the Units of Measure

1 Translate Algebraic Expressions into English Phrases In order to use the tools of algebra for solving problems, we must be able to translate back and forth between the English language and the language of algebra. In this section we will translate algebraic expressions into English phrases (word phrases) and English phrases into algebraic expressions. Let’s begin by considering the following translations from algebraic expressions to word phrases.

2.2

Translating from English to Algebra

Algebraic expression xy xy yx xy x y 3x x2 y2 2xy 21x y2 x3

83

Word phrase The sum of x and y The difference of x and y The difference of y and x The product of x and y The quotient of x and y The product of 3 and x The sum of x squared and y squared The product of 2, x, and y Two times the quantity x plus y Three less than x

2 Translate English Phrases into Algebraic Expressions Now let’s consider the reverse process: translating from word phrases to algebraic expressions. Part of the difficulty in translating from English to algebra is that different word phrases translate into the same algebraic expression. Thus we need to become familiar with different ways of saying the same thing, especially when referring to the four fundamental operations. The next examples should acquaint you with some of the phrases used in the basic operations. The sum of x and 4 x plus 4 Ex increased by 4 U Four added to x Four more than x

x4

The difference of n and 5 n minus 5 n less 5 Gn decreased by 5 W Subtract 5 from n Five less than n Five subtracted from n

n5

The product of 4 and y C Four times y S y multiplied by 4 The quotient n and 6 C n divided by 6 S Six divided into n

4y

n 6

Often a word phrase indicates more than one operation. Furthermore, the standard vocabulary of sum, difference, product, and quotient may be replaced by other terminology. Study the following translations very carefully. Also remember that the commutative property holds for addition and multiplication but not for subtraction and division. Therefore the phrase “x plus y” can be written as x y or y x. However the phrase “x minus y” means that y must be subtracted from x, and the phrase is written as x y. So be very careful of phrases that involve subtraction or division.

84

Chapter 2


Word phrase

Algebraic expression

The sum of two times x and three times y The sum of the squares of a and b

2x 3y

Five times x divided by y Two more than the square of x Three less than the cube of b Five less than the product of x and y Nine minus the product of x and y Four times the sum of x and 2 Six times the quantity w minus 4

a 2 b2 5x y x2 2 b3 3 xy 5 9 xy 41x 22 61w 42

Suppose you are told that the sum of two numbers is 12, and one of the numbers is 8. What is the other number? The other number is 12 8, which equals 4. Now suppose you are told that the product of two numbers is 56, and one of the numbers is 7. What is the other number? The other number is 56 7, which equals 8. The following examples illustrate the use of these addition-subtraction and multiplicationdivision relationships.

EXAMPLE 1

The sum of two numbers is 83, and one of the numbers is x. What is the other number?

Solution Using the addition-subtraction relationship, we can represent the other number by 83 x.

▼ PRACTICE YOUR SKILL The sum of two numbers is 56, and one of the numbers is y. Write an expression for the other number. ■

EXAMPLE 2

The difference of two numbers is 14. The smaller number is n. What is the larger number?

Solution Because the smaller number plus the difference must equal the larger number, we can represent the larger number by n 14.

▼ PRACTICE YOUR SKILL The difference of two numbers is 20. The smaller number is x. Write an expression for the larger number. ■

EXAMPLE 3

The product of two numbers is 39, and one of the numbers is y. Represent the other number.

Solution Using the multiplication-division relationship, we can represent the other number 39 by . y

2.2


85

▼ PRACTICE YOUR SKILL The product of two numbers is 45, and one of the numbers is x. Write the other number as an algebraic expression. ■ The English statement may not contain key words such as sum, difference, product, or quotient. Instead the statement may describe a physical situation; from this description you need to deduce the operations involved. We now make some suggestions for handling such situations.

EXAMPLE 4

Arlene can type 70 words per minute. How many words can she type in m minutes?

Solution In 10 minutes she would type 70(10) 700 words. In 50 minutes she would type 70(50) 3500 words. Thus in m minutes she would type 70m words.

▼ PRACTICE YOUR SKILL A fax machine can transmit 12 pages in one minute. Represent the number of pages that can be sent in y minutes. ■ Note the use of some specific examples [70(10) 700 and 70(50) 3500] to help formulate the general expression. This technique of first formulating some specific examples and then generalizing can be very effective.

EXAMPLE 5

Lynn has n nickels and d dimes. Express, in cents, this amount of money.

Solution Three nickels and 8 dimes would be 5(3) 10(8) 95 cents. Thus n nickels and d dimes would be 5n 10d cents.

▼ PRACTICE YOUR SKILL Ramon has q quarters and d dimes. Express, in cents, the amount of money that Ramon has. ■

EXAMPLE 6

A cruise ship travels at the rate of r miles per hour. How far will it travel in 8 hours?

Solution Suppose that a cruise ship travels at 50 miles per hour. Using the formula distance equals rate times time, we find that it would travel 50 # 8 400 miles. Therefore, at r miles per hour, it would travel r # 8 miles. We usually write the expression r # 8 as 8r.

▼ PRACTICE YOUR SKILL An electric automobile travels at p miles per hour. How far will it travel in 4 hours? ■

EXAMPLE 7

At a local restaurant the cost for a dinner to serve 8 people is d dollars. What is the cost per person for the dinner?

Solution The price per person is figured by dividing the total cost by the number of people. d Therefore, we represent the price per person by . 8

86

Chapter 2


▼ PRACTICE YOUR SKILL The cost of an 8-day European cruise is e euros. Express the cost per day in euros.

■

3 Write Algebraic Expressions That Change the Units of Measure The English statement to be translated into algebra may contain some geometric ideas. For example, suppose that we want to express in inches the length of a line segment that is f feet long. Because 1 foot 12 inches, we can represent f feet by 12 times f, written as 12f inches. Table 2.1 lists some of the basic relationships pertaining to linear measurements in the English and metric systems. (Additional listings of both systems are located inside the back cover of this book.)

Table 2.1

EXAMPLE 8

English system

Metric system

12 inches 1 foot 3 feet 36 inches 1 yard 5280 feet 1760 yards 1 mile

1 kilometer 1000 meters 1 hectometer 100 meters 1 dekameter 10 meters 1 decimeter 0.1 meter 1 centimeter 0.01 meter 1 millimeter 0.001 meter

The distance between two cities is k kilometers. Express this distance in meters.

Solution Because 1 kilometer equals 1000 meters, we need to multiply k by 1000. Therefore, 1000k represents the distance in meters.

▼ PRACTICE YOUR SKILL Two bolts on a machine are x centimeters apart. Express this distance in millimeters. ■

EXAMPLE 9

The length of a sofa is i inches. Express that length in yards.

Solution i To change from inches to yards, we must divide by 36. Therefore represents in 36 yards the length of the sofa.

▼ PRACTICE YOUR SKILL The height of a plasma television is p inches. Express the height in feet.

EXAMPLE 10

■

The width of a rectangle is w centimeters, and the length is 5 centimeters less than twice the width. What is the length of the rectangle? What is the perimeter of the rectangle?

Solution We can represent the length of the rectangle by 2w 5. Now we can sketch a rectangle as in Figure 2.1 and record the given information. The perimeter of a rectangle is the

2.2


87

sum of the lengths of the four sides. Therefore, the perimeter, in centimeters, is given by 2w 2(2w 5), which can be written as 2w 4w 10 and then simplified to 6w 10. 2w − 5 w

Figure 2.1

▼ PRACTICE YOUR SKILL The width of a rectangle is x inches, and the length is 3 inches more than twice the width. Write an expression to represent the length of the rectangle. Find the perimeter of the rectangle. ■

EXAMPLE 11

The length of a side of a square is x feet. Express the length of a side in inches. What is the area of the square in square inches?

Solution Because 1 foot equals 12 inches, we need to multiply x by 12. Therefore, 12x represents the length of a side in inches. The area of a square is the length of a side squared. So the area in square inches is given by (12x)2 (12x)(12x) 12 # 12 # x # x 144x2.

▼ PRACTICE YOUR SKILL The length of a side of a square is s yards. Express the length of a side in feet. What is the area of the square in square feet? ■

CONCEPT QUIZ

For Problems 1–10, match the English phrase with its algebraic expression. 1. The product of x and y 2. Two less than x 3. x subtracted from 2 4. 5. 6. 7. 8. 9. 10.

The difference of x and y The quotient of x and y The sum of x and y Two times the sum of x and y Two times x plus y x squared minus y Two more than x

A. x y B. x y x C. y D. x 2 E. xy F. x2 y G. 2(x y) H. 2 x I. x 2 J. 2x y

Problem Set 2.2 1 Translate Algebraic Expressions into English Phrases For Problems 1–12, write a word phrase for each of the algebraic expressions. For example, lw can be expressed as “the product of l and w.” 1. a b 3.

1 Bh 3

2. x y 4.

1 bh 2

5. 21l w2

7.

A w

9.

ab 2

11. 3y 2

6. pr2

8.

10.

C p ab 4

12. 31x y2

88

Chapter 2


2 Translate English Phrases into Algebraic Expressions For Problems 13 –36, translate each word phrase into an algebraic expression. For example, “the sum of x and 14” translates into x 14. 13. The sum of l and w 14. The difference of x and y 15. The product of a and b 1 16. The product of , B, and h 3 17. The quotient of d and t 18. r divided into d 19. The product of l, w, and h 20. The product of p and the square of r 21. x subtracted from y 22. The difference of x and y 23. Two larger than the product of x and y 24. Six plus the cube of x 25. Seven minus the square of y 26. The quantity, x minus 2, cubed 27. The quantity, x minus y, divided by 4 28. Eight less than x 29. Ten less x 30. Nine times the quantity, n minus 4 31. Ten times the quantity, n plus 2 32. The sum of four times x and five times y 33. Seven subtracted from the product of x and y 34. Three times the sum of n and 2 35. Twelve less than the product of x and y 36. Twelve less the product of x and y For Problems 37–72, answer the question with an algebraic expression.

41. Janet is y years old. How old will she be in 10 years? 42. Hector is y years old. How old was he 5 years ago? 43. Debra is x years old, and her mother is 3 years less than twice as old as Debra. How old is Debra’s mother? 44. Jack is x years old, and Dudley is 1 year more than three times as old as Jack. How old is Dudley? 45. Donna has d dimes and q quarters in her bank. How much money, in cents, does she have? 46. Andy has c cents that is all in dimes. How many dimes does he have? 47. A car travels d miles in t hours. What is the rate of the car? 48. If g gallons of gasoline cost d dollars, what is the price per gallon? 49. If p pounds of candy cost d dollars, what is the price per pound? 50. A bottling plant can fill x bottles of orange juice in a minute. How many bottles of orange juice can be filled in an hour? 51. Larry’s annual salary is d dollars. What is his monthly salary? 52. Nancy’s monthly salary is d dollars. What is her annual salary? 53. If n represents a whole number, what represents the next larger whole number? 54. If n represents an even number, what represents the next larger even number? 55. If n represents an odd number, what represents the next larger odd number? 56. Maria is y years old, and her sister is twice as old. What represents the sum of their ages? 57. Willie is y years old, and his father is 2 years less than twice Willie’s age. What represents the sum of their ages? 58. Harriet has p pennies, n nickels, and d dimes. How much money, in cents, does she have?

3 Write Algebraic Expressions That Change the Units of Measure

37. The sum of two numbers is 35, and one of the numbers is n. What is the other number?

59. The perimeter of a rectangle is y yards and f feet. What is the perimeter expressed in inches?

38. The sum of two numbers is 100, and one of the numbers is x. What is the other number?

60. The perimeter of a triangle is m meters and c centimeters. What is the perimeter in centimeters?

39. The difference of two numbers is 45, and the smaller number is n. What is the other number?

61. A rectangular plot of ground is f feet long. What is its length in yards?

40. The product of two numbers is 25, and one of the numbers is x. What is the other number?

62. The height of a telephone pole is f feet. What is the height in yards?

2.3

Equations and Problem Solving

89

68. The first side of a triangle is y yards long. The second side is 3 yards shorter than the first side. The third side is three times as long as the second side. What is the perimeter of the triangle in feet?

63. The width of a rectangle is w feet, and its length is three times the width. What is the perimeter of the rectangle in feet? 64. The width of a rectangle is w feet, and its length is 1 foot more than twice its width. What is the perimeter of the rectangle in feet?

69. The width of a rectangle is w yards, and the length is twice the width. What is the area of the rectangle in square yards?

65. The length of a rectangle is l inches and its width is 2 inches less than one-half of its length. What is the perimeter of the rectangle in inches?

70. The width of a rectangle is w yards, and the length is 4 yards more than the width. What is the area of the rectangle in square yards?

66. The length of a rectangle is l inches, and its width is 3 inches more than one-third of its length. What is the perimeter of the rectangle in inches?

71. The length of a side of a square is s yards. What is the area of the square in square feet?

67. The first side of a triangle is f feet long. The second side is 2 feet longer than the first side. The third side is twice as long as the second side. What is the perimeter of the triangle in inches?

72. The length of a side of a square is y centimeters. What is the area of the square in square millimeters?

THOUGHTS INTO WORDS 73. What does the phrase “translating from English to algebra” mean to you?

74. Your friend is having trouble with Problems 61 and 62. For example, for Problem 61 she doesn’t know if the f answer should be 3f or . What can you do to help her? 3

Answers to the Example Practice Skills 1. 56 y

2. x 20

3.

45 x

4. 12y

5. 25q 10d

e 8

6. 4p

7.

9. F

10. I

8. 10x

9.

p 12

10. 2x 3; 6x 6

11. 9s2

Answers to the Concept Quiz 1. E

2.3

2. D 3. H

4. A

5. C 6. B

7. G 8. J

Equations and Problem Solving OBJECTIVES 1

Solve Equations Using Both the Addition-Subtraction and the Multiplication-Division Properties

2

Solve Word Problems

1 Solve Equations Using Both the AdditionSubtraction and the Multiplication-Division Properties We often need to use more than one property of equality to help find the solution of an equation. Consider the next examples.

90

Chapter 2


EXAMPLE 1

Solve 3x 1 7.

Solution 3x 1 7 3x 1 1 7 1


3x 6 3x 6 3 3

Divide both sides by 3

x2 We can check the potential solution by substituting it into the original equation to see whether we get a true numerical statement.

✔ Check 3x 1 7 3122 1 ⱨ 7 61ⱨ7 77

Now we know that the solution set is 526 .

▼ PRACTICE YOUR SKILL Solve 4x 3 17.

EXAMPLE 2

■

Solve 5x 6 14.

Solution 5x 6 14 5x 6 6 14 6

Add 6 to both sides

5x 20 5x 20 5 5


x4

✔ Check 5x 6 14 5142 6 ⱨ 14 20 6 ⱨ 14 14 14

The solution set is 546 .


■

2.3

EXAMPLE 3


91

Solve 4 3a 22.

Solution 4 3a 22 4 3a 4 22 4


3a 18 3a 18 3 3


a 6

✔ Check 4 3a 22 4 3162 ⱨ 22 4 18 ⱨ 22 22 22

The solution set is 566.

▼ PRACTICE YOUR SKILL Solve 1 5y 36.

■

Note that in Examples 1, 2, and 3, we first used the addition-subtraction property and then used the multiplication-division property. In general, this sequence of steps provides the easiest format for solving such equations. Perhaps you should convince yourself of that fact by doing Example 1 again, this time using the multiplication-division property first and then the addition-subtraction property.

EXAMPLE 4

Solve 19 2n 4.

Solution 19 2n 4 19 4 2n 4 4


15 2n 15 2n 2 2


15 n 2

✔ Check 19 2n 4 19 ⱨ 2 a

15 b4 2

19 ⱨ 15 4 19 19 The solution set is e

15 f. 2

92

Chapter 2


▼ PRACTICE YOUR SKILL Solve 24 4a 3.

■

2 Solve Word Problems In the previous section we translated English phrases into algebraic expressions. We are now ready to extend that idea to the translation of English sentences into algebraic equations. Such translations enable us to use the concepts of algebra to solve word problems. Let’s consider some examples.

EXAMPLE 5

Apply Your Skill A certain number added to 17 yields a sum of 29. What is the number?

Solution Let n represent the number to be found. The sentence “a certain number added to 17 yields a sum of 29” translates to the algebraic equation 17 n 29. To solve this equation, we use these steps: 17 n 29 17 n 17 29 17 n 12 The solution is 12, which is the number asked for in the problem.

▼ PRACTICE YOUR SKILL A certain number added to 21 gives a sum of 78. What is the number?

■

We often refer to the statement “let n represent the number to be found” as declaring the variable. We need to choose a letter to use as a variable and indicate what it represents for a specific problem — this may seem like an insignificant idea, but as the problems become more complex, the process of declaring the variable becomes even more important. We could solve a problem such as Example 5 without setting up an algebraic equation; however, as problems increase in difficulty, the translation from English to algebra becomes critical. Therefore, even with these relatively simple problems, we need to concentrate on the translation process.

EXAMPLE 6

Apply Your Skill Six years ago Bill was 13 years old. How old is he now?

Solution Let y represent Bill’s age now; therefore, y 6 represents his age six years ago. Thus y 6 13 y 6 6 13 6 y 19 Bill is presently 19 years old.

▼ PRACTICE YOUR SKILL Five years ago Dahlia was 19 years old. How old is she now?

■

2.3

EXAMPLE 7


93

Apply Your Skill

Comma Image/Jupiter Images

Betty worked 8 hours Saturday and earned $66. How much did she earn per hour?

Solution A Let x represent the amount Betty earned per hour. The number of hours worked times the wage per hour yields the total earnings. Thus 8x 66 8x 66 8 8 x 8.25 Betty earned $8.25 per hour.

Solution B Let y represent the amount Betty earned per hour. The wage per hour equals the total wage divided by the number of hours. Thus y

66 8

y 8.25 Betty earned $8.25 per hour. Sometimes we can use more than one equation to solve a problem. In Solution A we set up the equation in terms of multiplication, whereas in Solution B we were thinking in terms of division.

▼ PRACTICE YOUR SKILL Damien talked for 34 minutes on his cell phone and was charged $2.72. How much did he pay per minute for his phone call? ■

EXAMPLE 8

Apply Your Skill If 2 is subtracted from five times a certain number, the result is 28. Find the number.

Solution Let n represent the number to be found. Translating the first sentence in the problem into an algebraic equation, we obtain 5n 2 28 To solve this equation we proceed as follows: 5n 2 2 28 2 5n 30 5n 30 5 5 n6 The number to be found is 6.

94

Chapter 2


▼ PRACTICE YOUR SKILL If 6 is subtracted from four times a certain number, the result is 30. Find the number. ■

EXAMPLE 9

Apply Your Skill

Carol Wiley/Lonely Planet Images

The cost of a five-day vacation cruise package was $534. This cost included $339 for the cruise and an amount for 2 nights of lodging on shore. Find the cost per night of the on-shore lodging.

Solution Let n represent the cost for one night of lodging; then 2n represents the total cost of lodging. Thus the total cost for the cruise and lodging is $534. We can proceed as follows: Cost of cruise Cost of lodging $534

339

2n

534

To solve this equation we proceed as follows: 339 2n 534 2n 195 195 2n 2 2 n 97.50 The cost of lodging per night is $97.50.

▼ PRACTICE YOUR SKILL The outlet store is selling a high-definition DVD player and 5 DVD movies as a package for $439.00. Find the cost for each DVD movie if the DVD player cost $349.00. ■

CONCEPT QUIZ

For Problems 1–5, answer true or false. 1. Only one property of equality is necessary to solve any equation. 2. Substituting the solution into the original equation to obtain a true numerical statement can be used to check potential solutions. 3. The statement “let x represent the number” is referred to as checking the variable. 4. Sometimes there can be two approaches to solving a word problem. 1 5. To solve the equation, x 2 7, you could begin by either adding 2 to both 3 sides of the equation or by multiplying both sides of the equation by 3. For Problems 6 –10, match the English sentence with its algebraic equation. 6. 7. 8. 9. 10.

Three added to a number is 24. The product of 3 and a number is 24. Three less than a number is 24. The quotient of a number and three is 24. A number subtracted from 3 is 24.

3x 24 3 x 24 x 3 24 x 3 24 x E. 24 3

A. B. C. D.

2.3


95

Problem Set 2.3 1 Solve Equations Using Both AdditionSubtraction and Multiplication-Division Properties For Problems 1– 40, solve each equation. 1. 2x 5 13

2. 3x 4 19

3. 5x 2 32

4. 7x 3 24

5. 3x 1 23

6. 2x 5 21

7. 4n 3 41

8. 5n 6 19

9. 6y 1 16

10. 4y 3 14

11. 2x 3 22

12. 3x 1 21

13. 10 3t 8

14. 17 2t 5

15. 5x 14 9

16. 4x 17 9

17. 18 n 23

18. 17 n 29

19. 3x 2 20

20. 6x 1 43

21. 7 4x 29

22. 9 6x 23

23. 16 2 9a

24. 18 10 7a

25. 7x 3 7

26. 9x 5 18

27. 17 2x 19

28. 18 3x 24

29. 16 4x 9

30. 14 6x 7

31. 12t 4 88

32. 16t 3 67

33. 14y 15 33

34. 12y 13 15

35. 32 16n 8

36. 41 12n 19

37. 17x 41 37

38. 19y 53 47

39. 29 7 15x

40. 49 5 14x

2 Solve Word Problems For each of the following problems, (a) choose a variable and indicate what it represents in the problem, (b) set up an equation that represents the situation being described, and (c) solve the equation. 41. Twelve added to a certain number is 21. What is the number? 42. A certain number added to 14 is 25. Find the number. 43. Nine subtracted from a certain number is 13. Find the number. 44. A certain number subtracted from 32 is 15. What is the number?

45. Suppose that two items cost $43. If one of the items costs $25, what is the cost of the other item? 46. Eight years ago Rosa was 22 years old. Find Rosa’s present age. 47. Six years from now, Nora will be 41 years old. What is her present age? 48. Chris bought eight pizzas for a total of $83.60. What was the price per pizza? 49. Chad worked 6 hours Saturday for a total of $43.50. How much per hour did he earn? 50. Jill worked 8 hours Saturday at $7.50 per hour. How much did she earn? 51. If 6 is added to three times a certain number, the result is 24. Find the number. 52. If 2 is subtracted from five times a certain number, the result is 38. Find the number. 53. Nineteen is 4 larger than three times a certain number. Find the number. 54. If nine times a certain number is subtracted from 7, the result is 52. Find the number. 55. Forty-nine is equal to 6 less than five times a certain number. Find the number. 56. Seventy-one is equal to 2 more than three times a certain number. Find the number. 57. If 1 is subtracted from six times a certain number, the result is 47. Find the number. 58. Five less than four times a number equals 31. Find the number. 59. If eight times a certain number is subtracted from 27, the result is 3. Find the number. 60. Twenty is 22 less than six times a certain number. Find the number. 61. A jeweler has priced a diamond ring at $550 (see Figure 2.2). This price represents $50 less than twice the cost of the ring to the jeweler. Find the cost of the ring to the jeweler. 62. Todd is on a 1750-calorie-perday diet plan. This plan permits 650 calories less than twice the number of calories permitted by Lerae’s diet plan. How many calories are permitted by Lerae’s plan?

$5

50.

Figure 2.2

96

Chapter 2

Equations, Inequalities, and Problem Solving less than three times the speakers of the English language. By this estimate how many million speakers of the English language were there in the year 2005?

63. The length of a rectangular floor is 18 meters (see Figure 2.3). This represents 2 meters less than five times the width of the floor. Find the width of the floor.

66. A bill from a limousine company was $510. This included $150 for the service and $80 for each hour of use. Find the number of hours that the limousine was used.

18 meters

67. Robin paid $454 for a car DVD system. This included $379 for the DVD player and $60 an hour for installation. Find the number of hours it took to install the DVD system.

64. An executive is earning $85,000 per year. This represents $15,000 less than twice her salary 4 years ago. Find her salary 4 years ago.

68. Tracy received a cell phone bill for $136.74. Included in the $136.74 were a charge of $39.99 for the monthly plan and a charge for 215 extra minutes. How much is Tracy being charged for each extra minute?

Figure 2.3

65. In the year 2005, it was estimated that there were 874 million speakers of Mandarin Chinese. This was 149 million

THOUGHTS INTO WORDS 19 x 14

69. Give a step-by-step description of how you would solve the equation 17 3x 2.

19 x 19 14 19 x 5

70. What does the phrase “declare a variable” mean when it refers to solving a word problem?

The solution set is {5}.

71. Suppose that you are helping a friend with his homework and he solves the equation 19 14 x like this:

Does he have a correct solution set? What would you tell him about his method of solving the equation?

19 14 x 19 x 14 x x

Answers to the Example Practice Skills 1. {5} 2. {5}

3. {7} 4. e

27 f 4

5. 57

6. 24 years old 7. $0.08

8. 9

9. $18


2.4

2. True

3. False 4. True 5. True

6. C

7. A

8. D 9. E

10. B

More on Solving Equations and Problem Solving OBJECTIVES 1

Solve Equations by First Simplifying One Side of the Equation

2

Solve Equations of the Form ax b cx d

3

Solve Word Problems That Involve Consecutive Numbers and Geometry

1 Solve Equations by First Simplifying One Side of the Equation As equations become more complex, we will need additional steps to solve them, so it is important that we organize our work carefully to minimize the chances for error.

2.4

More on Solving Equations and Problem Solving

97

Let’s begin this section with some suggestions for solving equations, and then we will illustrate a solution format that is effective. We can summarize the process of solving first-degree equations of one variable as follows: Step 1 Step 2

Step 3

Simplify both sides of the equation as much as possible. Use the addition-subtraction property of equality to isolate a term that contains the variable on one side of the equation and a constant on the other. Use the multiplication-division property of equality to make the coefficient of the variable 1.

The following examples illustrate this step-by-step process for solving equations. Study them carefully and be sure that you understand each step.

EXAMPLE 1

Solve 5y 4 3y 12.

Solution 5y 4 3y 12 8y 4 12 8y 4 4 12 4

Combine similar terms on the left side Add 4 to both sides

8y 16 8y 16 8 8


y2 The solution set is 526. You can do the check alone now!

▼ PRACTICE YOUR SKILL Solve 2x 7 5x 14.

■

2 Solve Equations of the Form ax b cx d EXAMPLE 2

Solve 7x 2 3x 9.

Solution Note that both sides of the equation are in simplified form; thus we can begin by using the subtraction property of equality. 7x 2 3x 9 7x 2 3x 3x 9 3x

Subtract 3x from both sides

4x 2 9 4x 2 2 9 2

Add 2 to both sides

4x 11 4x 11 4 4 x The solution set is e

11 4

11 f. 4


98

Chapter 2


▼ PRACTICE YOUR SKILL Solve x 5 6x 45.

EXAMPLE 3

■

Solve 5n 12 9n 16.

Solution 5n 12 9n 16 5n 12 9n 9n 16 9n

Subtract 9n from both sides

4n 12 16 4n 12 12 16 12


4n 28 4n 28 4 4


n7


▼ PRACTICE YOUR SKILL Solve 2a 24 6a 44.

■

3 Solve Word Problems That Involve Consecutive Numbers and Geometry As we expand our skill in solving equations, we also expand our ability to solve word problems. No one definite procedure will ensure success at solving word problems, but the following suggestions can be helpful.

Suggestions for Solving Word Problems 1. Read the problem carefully and make sure that you understand the meanings of all the words. Be especially alert for any technical terms in the statement of the problem. 2. Read the problem a second time (perhaps even a third time) to get an overview of the situation described and to determine the known facts as well as what is to be found. 3. Sketch any figure, diagram, or chart that might be helpful in analyzing the problem. 4. Choose a meaningful variable to represent an unknown quantity in the problem (perhaps t if time is an unknown quantity); represent any other unknowns in terms of that variable. 5. Look for a guideline that you can use to set up an equation. A guideline might be a formula such as distance equals rate times time, or a statement of a relationship, such as the sum of the two numbers is 28. A guideline may also be indicated by a figure or diagram that you sketch for a particular problem. 6. Form an equation that contains the variable and that translates the conditions of the guideline from English to algebra. 7. Solve the equation and use the solution to determine all facts requested in the problem. 8. Check all answers against the original statement of the problem.

2.4


99

If you decide not to check an answer, at least use the reasonableness of answer idea as a partial check. That is, ask yourself, “Is this answer reasonable?” For example, if the problem involves two investments that total $10,000, then an answer of $12,000 for one investment is certainly not reasonable. Now let’s consider some problems and use these suggestions.

EXAMPLE 4

Apply Your Skill Find two consecutive even numbers whose sum is 74.

Solution To solve this problem, we must know the meaning of the technical phrase “two consecutive even numbers.” Two consecutive even numbers are two even numbers that have one and only one whole number between them. For example, 2 and 4 are consecutive even numbers. Now we can proceed as follows: Let n represent the first even number; then n 2 represents the next even number. Because their sum is 74, we can set up and solve the following equation: n 1n 22 74 2n 2 74 2n 2 2 74 2 2n 72 2n 72 2 2 n 36 If n 36, then n 2 38; thus the numbers are 36 and 38.

✔ Check To check your answers for Example 4, determine whether they satisfy the conditions stated in the original problem. Because 36 and 38 are two consecutive even numbers, and 36 38 74 (their sum is 74), we know that the answers are correct.

▼ PRACTICE YOUR SKILL Find two consecutive even numbers whose sum is 126.

■

Suggestion 5 in our list of problem-solving suggestions was to “look for a guideline that can be used to set up an equation.” The guideline may not be explicitly stated in the problem but may instead be implied by the nature of the problem. Consider the next example.

Phil Degginger/Digital Railroad

EXAMPLE 5

Apply Your Skill Barry sells bicycles on a salary-plus-commission basis. He receives a monthly salary of $300 and a commission of $15 for each bicycle that he sells. How many bicycles must he sell in a month to have a total monthly salary of $750?

Solution Let b represent the number of bicycles to be sold in a month. Then 15b represents Barry’s commission for those bicycles. The guideline “fixed salary plus commission equals total monthly salary” generates the following equation: Fixed salary Commission Total monthly salary

$300

15b

$750

100

Chapter 2


Let’s solve this equation. 300 15b 300 750 300 15b 450 450 15b 15 15 b 30 He must sell 30 bicycles per month. (Does this number check?)

▼ PRACTICE YOUR SKILL A pizza delivery driver gets paid $50.00 a night plus $0.75 for each delivery. How many deliveries must the driver make to earn $77.00 for the night? ■ Sometimes the guideline for setting up an equation to solve a problem is based on a geometric relationship. Several basic geometric relationships pertain to angle measure. Let’s state three of these relationships and then consider some problems.

EXAMPLE 6

1.

Two angles that together measure 90° (the symbol ° indicates degrees) are called complementary angles.

2.

Two angles that together measure 180° are called supplementary angles.

3.

The sum of the measures of the three angles of a triangle is 180°.

Apply Your Skill One of two complementary angles is 14° larger than the other. Find the measure of each of the angles.

Solution If we let a represent the measure of the smaller angle, then a 14 represents the measure of the larger angle. Because they are complementary angles, their sum is 90° and we can proceed as follows: a a 14 90 2a 14 90 2a 14 14 90 14 2a 76 2a 76 2 2 a 38 If a 38, then a 14 52, and the angles have measures of 38° and 52°.

▼ PRACTICE YOUR SKILL One of two supplementary angles is 56° smaller than the other angle. Find the measure of each angle. ■

2.4

EXAMPLE 7


101

Apply Your Skill Find the measures of the three angles of a triangle if the second is three times the first and the third is twice the second.

Solution If we let a represent the measure of the smallest angle, then 3a and 2(3a) represent the measures of the other two angles. Therefore we can set up and solve the following equation: a 3a 213a2 180 a 3a 6a 180 10a 180 10a 180 10 10 a 18 If a 18, then 3a 54 and 2(3a) 108, so the angles have measures of 18°, 54°, and 108°.

▼ PRACTICE YOUR SKILL The measure of the third angle of a triangle is three times as large as the first angle. The second angle is 5° more than the third angle. Find the measure of all three angles. ■

CONCEPT QUIZ

For Problems 1– 8, answer true or false. 1. 2. 3. 4. 5. 6. 7. 8.

If n represents a whole number, then n 1 would represent the next consecutive whole number. If n represents an odd whole number, then n 1 would represent the next consecutive odd whole number. If n represents an even whole number, then n 2 would represent the next consecutive even whole number. The sum of the measures of two complementary angles is 90 . The sum of the measures of two supplementary angles is 360 . The measure of the three angles in a triangle is 120 . When checking word problems it is sufficient to check the solution in the equation. For a word problem, the reasonableness of an answer is appropriate as a partial check.

Problem Set 2.4 1 Solve Equations by First Simplifying One Side of the Equation For Problems 1–32, solve each equation. 1. 2x 7 3x 32

5. 3y 1 2y 3 4 6. y 3 2y 4 6 7. 5n 2 8n 31

2. 3x 9 4x 30

8. 6n 1 10n 51

3. 7x 4 3x 36

9. 2n 1 3n n 4 7

4. 8x 3 2x 45

10. n 7 2n 5n 3 6

102

Chapter 2


2 Solve Equations of the Form ax b cx d 11. 3x 4 2x 5 12. 5x 2 4x 6 13. 5x 7 6x 9 14. 7x 3 8x 13 15. 6x 1 3x 8 16. 4x 10 x 17 17. 7y 3 5y 10 18. 8y 4 5y 4 19. 8n 2 11n 7 20. 7n 10 9n 13 21. 2x 7 3x 10 22. 4x 6 5x 9 23. 3x 5 5x 8 24. 4x 7 6x 4 25. 7 6x 9 9x 26. 10 7x 14 12x 27. 2x 1 x 3x 5 28. 3x 4 4x 5 5x 3x 29. 5n 4 n 3n 6 n 30. 4x 3 2x 8x 3 x 31. 7 2n 6n 7n 5n 12 32. 3n 6 5n 7n 8n 9

3 Solve Word Problems That Involve Consecutive Numbers and Geometry Solve each of the following problems by setting up and solving an algebraic equation. 33. The sum of a number plus four times the number is 85. What is the number? 34. A number subtracted from three times the number yields 68. Find the number. 35. Find two consecutive odd numbers whose sum is 72.

40. One more than five times a certain number is equal to 11 less than nine times the number. What is the number? 41. The sum of a number and five times the number equals 18 less than three times the number. Find the number. 42. One of two supplementary angles is five times as large as the other. Find the measure of each angle. 43. One of two complementary angles is 6 less than twice the other angle. Find the measure of each angle. 44. If two angles are complementary, and the difference of their measures is 62°, find the measure of each angle. 45. If two angles are supplementary, and the larger angle is 20° less than three times the smaller angle, find the measure of each angle. 46. Find the measures of the three angles of a triangle if the largest is 14° less than three times the smallest, and the other angle is 4° more than the smallest. 47. One of the angles of a triangle has a measure of 40°. Find the measures of the other two angles if the difference of their measures is 10°. 48. Jesstan worked as a telemarketer on a salary-pluscommission basis. He was paid a salary of $300 a week and $12 commission for each sale. If his earnings for the week were $960, how many sales did he make? 49. Marci sold an antique vase in an online auction for $69.00. This was $15 less than twice what she paid for it. What price did she pay for the vase? 50. A set of wheels sold in an online auction for $560. This was $35 more than three times the opening bid. How much was the opening bid? 51. Suppose that Bob is paid two times his normal hourly rate for each hour worked in excess of 40 hours in a week. Last week he earned $504 for 48 hours of work. What is his hourly wage? 52. Last week on an algebra test, the highest grade was 9 points less than three times the lowest grade. The sum of the two grades was 135. Find the lowest and highest grades on the test. 53. At a university-sponsored concert, there were three times as many women as men. A total of 600 people attended the concert. How many men and how many women attended? 54. Suppose that a triangular lot is enclosed with 135 yards of fencing (see Figure 2.4). The longest side of the lot is 5 yards more than twice the length of the shortest side. The other side is 10 yards longer than the shortest side. Find the lengths of the three sides of the lot.

36. Find two consecutive even numbers whose sum is 94. 37. Find three consecutive even numbers whose sum is 114. 38. Find three consecutive odd numbers whose sum is 159. 39. Two more than three times a certain number is the same as 4 less than seven times the number. Find the number.

Figure 2.4

2.5

Equations Involving Parentheses and Fractional Forms

55. The textbook for a biology class cost $15 more than twice the cost of a used textbook for college algebra. If the cost of the two books together is $129, find the cost of the biology book. 56. A nutrition plan counts grams of fat, carbohydrates, and fiber. The grams of carbohydrates are to be 15 more than twice the grams of fat. The grams of fiber are to be 3 less than the grams of fat. If the grams of carbohydrate, fat,

103

and fiber must total 48 grams for a dinner meal, how many grams of each would be in the meal? 57. At a local restaurant, $275 in tips is to be shared between the server, bartender, and busboy. The server gets $25 more than three times the amount the busboy receives. The bartender gets $50 more than the amount the busboy receives. How much will the server receive?

THOUGHTS INTO WORDS 58. Give a step-by-step description of how you would solve the equation 3x 4 5x 2. 59. Suppose your friend solved the problem “find two consecutive odd integers whose sum is 28” like this:

1 will check in the equation. Where 2 has she gone wrong, and how would you help her? She claims that 13

x x 1 28 2x 27 x

27 1 13 2 2

FURTHER INVESTIGATIONS d. 5x 3 6x 7 x

60. Solve each of these equations.

e. 7x 4 x 4 8x

a. 7x 3 4x 3

f. 3x 2 5x 7x 2 5x

b. x 4 3x 2x 7

g. 6x 8 6x 4 h. 8x 9 8x 5

c. 3x 9 2x 5x 9

Answers to the Example Practice Skills 1. {7} 2. {8}

3. {5} 4. 62, 64

5. 36

6. 62°, 118°

7. 25°, 80°, 75°


2.5

2. False

3. True

4. True 5. False

6. False

7. False 8. True

Equations Involving Parentheses and Fractional Forms OBJECTIVES 1

Solve Equations That Involve the Use of the Distributive Property

2

Solve Equations That Involve Fractional Forms

3

Solve Equations That Are Contradictions or Identities

4

Solve Word Problems

1 Solve Equations That Involve the Use of the Distributive Property We will use the distributive property frequently in this section as we expand our techniques for solving equations. Recall that in symbolic form, the distributive property

104

Chapter 2


states that a(b c) ab ac. The following examples illustrate the use of this property to remove parentheses. Pay special attention to the last two examples, which involve a negative number in front of the parentheses. 31x 22

3#x3#2

3x 6

51y 32

5#y5#3

5y 15

214x 72

214x2 2172

8x 14

1121n2 112 142

11n 42

162 1x2 162 122

61x 22

[a(b c) ab ac]

n 4 6x 12

Do this step mentally!

It is often necessary to solve equations in which the variable is part of an expression enclosed in parentheses. The distributive property is used to remove the parentheses, and then we proceed in the usual way. Consider the following examples. (Note that when solving an equation, we are beginning to show only the major steps.)

EXAMPLE 1

Solve 41x 32 21x 62 .

Solution 41x 32 21x 62 4x 12 2x 12

Applied distributive property on each side

2x 12 12

Subtracted 2x from both sides

2x 24

Subtracted 12 from both sides

x 12

Divided both sides by 2


▼ PRACTICE YOUR SKILL Solve 31x 42 51x 22 .

■

It may be necessary to use the distributive property to remove more than one set of parentheses and then to combine similar terms. Consider the next two examples.

EXAMPLE 2

Solve 61x 72 21x 42 13.

Solution 61x 72 21x 42 13 6x 42 2x 8 13 4x 34 13

Distributive property Combined similar terms

4x 47

Added 34 to both sides

47 4


x The solution set is e

Be careful with this sign!

47 f. 4

▼ PRACTICE YOUR SKILL Solve 21x 42 71x 22 3 .

■

2.5


105

2 Solve Equations That Involve Fractional Forms 2 3 2 by adding to both 3 4 3 sides. If an equation contains several fractions, then it is usually easier to clear the equation of all fractions by multiplying both sides by the least common denominator of all the denominators. Perhaps several examples will clarify this idea.

In a previous section we solved equations such as x

EXAMPLE 3

2 5 1 Solve x . 2 3 6

Solution 2 5 1 x 2 3 6 2 5 1 6a x b 6a b 2 3 6 2 5 1 6 a xb 6 a b 6 a b 2 3 6

6 is the LCD of 2, 3, and 6 Distributive property Note how the equation has been cleared of all fractions

3x 4 5 3x 1 x

1 3

1 The solution set is e f . 3


EXAMPLE 4

Solve

4 2 7 x . 5 3 15

Solve

1 3 5n . 6 4 8

■

Solution 5n 1 3 6 4 8 24 a 24 a

1 3 5n b 24 a b 6 4 8

5n 1 3 b 24 a b 24 a b 6 4 8

Remember

5 5n n 6 6

24 is the LCD of 6, 4, and 8 Distributive property

20n 6 9 20n 15 n

3 15 20 4



7y 5 3 . 12 6 8

■

106

Chapter 2


We use many of the ideas presented in this section to help solve the equations in the next two examples. Study the solutions carefully and be sure that you can supply reasons for each step.

EXAMPLE 5

Solve

x3 x4 3 . 2 5 10

Solution x4 3 x3 2 5 10 10 a 10 a

x4 3 x3 b 10 a b 2 5 10

x4 3 x3 b 10 a b 10 a b 2 5 10

10 is the LCD of 2, 5, and 10


51x 32 21x 42 3 5x 15 2x 8 3 7x 23 3 7x 20 x The solution set is e

20 7

20 f. 7


EXAMPLE 6

Solve

x4 x2 7 . 3 5 10

Solve

x1 x2 2 . 4 6 3

■

Solution x1 x2 2 4 6 3 12 a 12 a

x1 x2 2 b 12 a b 4 6 3

x1 x2 2 b 12 a b 12 a b 4 6 3

12 is the LCD of 4, 6, and 3


31x 12 21x 22 8 3x 3 2x 4 8 x18 x7 The solution set is 576 .

Be careful with this sign!

2.5


107


x1 x4 5 . 8 3 4

■

3 Solve Equations That Are Contradictions or Identities All of the equations we have solved thus far are conditional equations. For instance, the equation 3x 12 is a true statement under the condition that x 4. Now we will consider two other types of equations— contradictions and identities. When the equation is not true under any condition, then the equation is called a contradiction. The solution set for a contradiction is the empty or null set and is denoted by . When an equation is true for any permissible value of the variable for which the equation is defined, the equation is called an identity, and the solution set for an identity is the set of all real numbers for which the equation is defined. We will denote the set of all real numbers as {All reals}. The following examples show the solutions for these types of equations.

EXAMPLE 7

Solve 4x 5 212x 82 .

Solution 4x 5 212x 82 4x 5 4x 16 5 16

Distributive property Subtracted 4x from both sides

The result is a false statement. Therefore the equation is a contradiction. There is no value of x that will make the equation a true statement, and hence the solution set is the empty set, .


EXAMPLE 8

■

Solve 5(x 3) 2x 4 7x 11.

Solution 51x 32 2x 4 7x 11


5x 15 2x 4 7x 11 7x 11 7x 11 11 11

Combined similar terms Subtracted 7x from both sides

The last step gives an equation with no variable terms, but the equation is a true statement. This equation is an identity, and any real number is a solution. The solution set would be written as {All reals}.

▼ PRACTICE YOUR SKILL Solve 3(x 1) 2x 7 5x + 4.

■

4 Solve Word Problems We are now ready to solve some word problems using equations of the different types presented in this section.

108

Chapter 2


EXAMPLE 9

Apply Your Skill Loretta has 19 coins (quarters and nickels) that amount to $2.35. How many coins of each kind does she have?

Solution Let q represent the number of quarters. Then 19 q represents the number of nickels. We can use the following guideline to help set up an equation: Value of quarters in cents Value of nickels in cents Total value in cents

25q

5(19 q)

235

We can solve the equation in this way: 25q 95 5q 235 20q 95 235 20q 140 q7 If q 7, then 19 q 12, so she has 7 quarters and 12 nickels.

▼ PRACTICE YOUR SKILL Michaela has 20 coins (quarters and dimes) that amount to $3.95. How many coins of each kind does she have? ■

EXAMPLE 10

Apply Your Skill Find a number such that 4 less than two-thirds the number is equal to one-sixth the number.

Solution Let n represent the number. Then

2 n 4 represents 4 less than two-thirds the 3

1 number, and n represents one-sixth the number. 6 2 1 n4 n 3 6 2 1 6 a n 4b 6 a nb 3 6 4n 24 n 3n 24 0 3n 24 n8 The number is 8.

▼ PRACTICE YOUR SKILL Find a number such that 12 less than three-fourths of the number is equal to one-half of the number. ■

2.5

EXAMPLE 11


109

Apply Your Skill

Burke/ Triolo Productions/Jupiter Images

1 Lance is paid 1 times his normal hourly rate for each hour he works in excess of 2 40 hours in a week. Last week he worked 50 hours and earned $462. What is his normal hourly rate?

Solution 1 3 x represents 1 times his normal 2 2 hourly rate. We can use the following guideline to help set up the equation: Let x represent his normal hourly rate. Then

Regular wages for first 40 hours Wages for 10 hours of overtime Total wages

40x

3 10 a xb 2

462

We get 40x 15x 462 55x 462 x 8.40 His normal hourly rate is $8.40.


1 Ramon is paid 1 times his normal hourly rate for each hour he works in excess of 2 40 hours a week. Last week he worked 52 hours and earned $696. Find his normal hourly rate. ■

EXAMPLE 12

Apply Your Skill Find three consecutive whole numbers such that the sum of the first plus twice the second plus three times the third is 134.

Solution Let n represent the first whole number. Then n 1 represents the second whole number, and n 2 represents the third whole number. We have n 21n 12 31n 22 134 n 2n 2 3n 6 134 6n 8 134 6n 126 n 21 The numbers are 21, 22, and 23.

▼ PRACTICE YOUR SKILL Find three consecutive whole numbers such that the sum of three times the first number plus twice the second number plus the third number is 76. ■ Keep in mind that the problem-solving suggestions we offered in Section 2.4 simply outline a general algebraic approach to solving problems. You will add to this

110

Chapter 2


list throughout this course and in any subsequent mathematics courses that you take. Furthermore, you will be able to pick up additional problem-solving ideas from your instructor and from fellow classmates as you discuss problems in class. Always be on the alert for any ideas that might help you become a better problem solver.

CONCEPT QUIZ

For Problems 1–10, answer true or false. 1. To solve an equation of the form a(x b) 14, the associative property would be applied to remove the parentheses. 2. Multiplying both sides of an equation by the common denominator of all fractions in the equation clears the equation of all fractions. 3. If Jack has 15 coins (dimes and quarters), and x represents the number of dimes, then x 15 represents the number of quarters. 4. The equation 3(x 1) 3x 3 has an infinite number of solutions. 5. The equation 2x 0 has no solution. 6. The equation 4x 5 4x 3 has no solution. 7. The solution set for an equation that is a contradiction is the null set. 8. For a conditional equation, the solution set is the set of all real numbers. 9. When an equation is true for any permissible value of the variable, then the equation is called an identity. 10. When an equation is true for only certain values of the variable, then the equation is called a contradiction.

Problem Set 2.5 1 Solve Equations That Involve the Use of the Distributive Property For Problems 1– 64, solve each equation. 1. 71x 22 21

2. 41x 42 24

3. 51x 32 35

4. 61x 22 18

5. 31x 52 12

6. 51x 62 15

7. 41n 62 5

8. 31n 42 7

9. 61n 72 8

10. 81n 32 12

11. 10 51t 82

12. 16 41t 72

24. 41x 12 51x 22 31x 82 25. 1x 22 21x 32 21x 72 26. 21x 62 313x 22 31x 42 27. 1a 12 13a 22 6 21a 12 28. 312a 12 215a 12 413a 42 29. 31x 12 21x 32 41x 22 101x 42 30. 314x 12 212x 12 21x 12 1 31. 3 71x 12 9 612x 12 32. 8 512x 12 2 61x 32

13. 51x 42 41x 62 14. 41x 72 51x 22 15. 81x 12 91x 22 16. 71t 52 51t 32

2 Solve Equations That Involve Fractional Forms 33.

3 2 5 x 4 3 6

34.

1 4 5 x 2 3 6

35.

5 1 9 x 6 4 4

36.

3 1 7 x 8 6 12

37.

1 3 3 x 2 5 4

38.

1 2 5 x 4 5 6

39.

n 5n 1 3 6 8

40.

n 3n 5 6 8 12

41.

5y 2y 3 6 5 3

42.

y 3y 1 7 2 4

17. 21x 62 1x 92 18. 1x 72 21x 102 19. 31t 42 21t 42 9 20. 51t 42 31t 22 12 21. 31n 102 51n 122 86 22. 41n 92 71n 82 83 23. 31x 12 412x 12 512x 32

2.5

111

43.

h h 1 6 8

45.

x2 x3 13 3 4 3

46.

x1 x2 39 4 5 20

47.

x1 x4 13 5 6 15

70. The sum of two numbers is 53. If three times the smaller is 1 less than the larger, find the numbers.

48.

x1 x3 4 7 5 5

71. Find a number such that 20 more than one-third of the number equals three-fourths of the number.

49.

x8 x 10 3 2 7 4

72. The sum of three-eighths of a number and five-sixths of the same number is 29. Find the number.

50.

x7 x9 5 3 6 9

51.

x2 x1 1 8 4

73. The difference of two numbers is 6. One-half of the larger number is 5 larger than one-third of the smaller. Find the numbers.

52.

x2 x4 3 2 4

44.

h h 1 4 3


x1 x3 53. 2 4 6 54.

x3 x6 1 5 2

3 Solve Equations That Are Contradictions or Identities 55. 61x 42 312x 52 56. 216t 12 413t 12 57. 81t 52 412t 102 58. 61t 52 213t 152 59. 512x 12 12x 42 412x 32 21 60. 3x 21x 62 5x 12 61. 21x 42 13x 22 2 15x 22 62. 31x 42 2x 5x 8 63. 4x 6x 21x 82 64. 7x 1 4x 31x 52

4 Solve Word Problems Solve each of the following problems by setting up and solving an appropriate algebraic equation. 65. Find two consecutive whole numbers such that the smaller plus four times the larger equals 39. 66. Find two consecutive whole numbers such that the smaller subtracted from five times the larger equals 57.

67. Find three consecutive whole numbers such that twice the sum of the two smallest numbers is 10 more than three times the largest number. 68. Find four consecutive whole numbers such that the sum of the first three equals the fourth number. 69. The sum of two numbers is 17. If twice the smaller is 1 more than the larger, find the numbers.

74. The difference of two numbers is 16. Three-fourths of the larger number is 14 larger than one-half of the smaller number. Find the numbers. 75. Suppose that a board 20 feet long is cut into two pieces. Four times the length of the shorter piece is 4 feet less than three times the length of the longer piece. Find the length of each piece. 76. Ellen is paid “time and a half” for each hour over 40 hours worked in a week. Last week she worked 44 hours and earned $391. What is her normal hourly rate? 77. Lucy has 35 coins consisting of nickels and quarters amounting to $5.75. How many coins of each kind does she have? 78. Suppose that Julian has 44 coins consisting of pennies and nickels. If the number of nickels is 2 more than twice the number of pennies, find the number of coins of each kind. 79. Max has a collection of 210 coins consisting of nickels, dimes, and quarters. He has twice as many dimes as nickels and 10 more quarters than dimes. How many coins of each kind does he have? 80. Ginny has a collection of 425 coins consisting of pennies, nickels, and dimes. She has 50 more nickels than pennies and 25 more dimes than nickels. How many coins of each kind does she have? 81. Maida has 18 coins consisting of dimes and quarters amounting to $3.30. How many coins of each kind does she have? 82. Ike has some nickels and dimes amounting to $2.90. The number of dimes is 1 less than twice the number of nickels. How many coins of each kind does he have? 83. Mario has a collection of 22 specimens in his aquarium consisting of crabs, fish, and plants. There are three times as many fish as crabs. There are two more plants than crabs. How many specimens of each kind are in the collection?

112

Chapter 2


84. Tickets for a concert were priced at $8 for students and $10 for nonstudents (see Figure 2.5). There were 1500 tickets sold for a total of $12,500. How many student tickets were sold?

N NT NO UDE ST

STUDENT

C

L

25 at Se

O

50

O

w Ro

C

$8.00

OOL

T

UNES

T

Row 03 Seat 10

U N

86. The sum of the measure of an angle and three times its complement is 202°. Find the measure of the angle. 87. In triangle ABC, the measure of angle A is 2° less than onefifth the measure of angle C. The measure of angle B is 5° less than one-half the measure of angle C. Find the measures of the three angles of the triangle. 88. If one-fourth of the complement of an angle plus one-fifth of the supplement of the angle equals 36°, find the measure of the angle.

S

w Ro

89. The supplement of an angle is 10° less than three times its complement. Find the size of the angle.

.00

E

0 $1

Row 03 Seat 10

85. The supplement of an angle is 30° more than twice its complement. Find the measure of the angle.

50 at

Se

90. In triangle ABC, the measure of angle C is eight times the measure of angle A, and the measure of angle B is 10° more than the measure of angle C. Find the measure of each angle of the triangle.

25

Figure 2.5

THOUGHTS INTO WORDS 91. Discuss how you would solve the equation

3(x 2) 5(x 3) 4(x 9) 92. Why must potential answers to word problems be checked back in the original statement of the problem? 93. Consider these two solutions: 31x 22 9

31x 42 7

31x 22

31x 42

7 3

x23

x4

7 3

x1

x

3

9 3

3

Are both of these solutions correct? How effective is the approach? 94. Make up an equation whose solution set is the null set. Explain why the solution set is null. 95. Make up an equation whose solution set is the set of all real numbers. Explain why the solution set is all real numbers.

19 3

FURTHER INVESTIGATIONS 96. Solve each of the following equations. a. 21x 12 2x 2

f.

x1 x 11 2 5 5

b. 31x 42 3x 4

g. 41x 22 21x 32 21x 62

c. 51x 12 5x 5

h. 51x 32 31x 52 21x 152

d.

x3 43 3

x2 x2 e. 1 3 3

i. 71x 12 41x 22 151x 12 97. Find three consecutive integers such that the sum of the smallest integer and the largest integer is equal to twice the middle integer.

2.6

Inequalities

113

Answers to the Example Practice Skills 1 1. e f 4

1 2. {5} 3. e f 4

4. e

29 f 14

9. 13 quarters and 7 dimes 10. 48

5. e

49 f 16

11. $12.00

1 6. e f 5

7. 8. {All reals}

12. 12, 13, and 14


2.6

2. True 3. False 4. True 5. False

6. True

7. True

8. False 9. True

10. False

Inequalities OBJECTIVES 1

Show the Solution Set of an Inequality by Using Set-Builder Notation and by Graphing

2

Solve Inequalities

Just as we use the symbol to represent is equal to, we also use the symbols and to represent is less than and is greater than, respectively. The following are examples of statements of inequality. Note that the first four are true statements, and the last two are false. 64 7

True Statement

8 2 14

True Statement

4#8 4#6

True Statement

5#2 5#7

True Statement

5 8 19

False Statement

92 3

False Statement

Algebraic inequalities contain one or more variables. Here are some examples of algebraic inequalities: x3 4 2x 1 6 x2 2x 1 0 2x 3y 7 7ab 9 An algebraic inequality such as x 1 2 is neither true nor false as it stands; it is called an open sentence. Each time a number is substituted for x, the algebraic inequality x 1 2 becomes a numerical statement that is either true or false. For example, if x 0, then x 1 2 becomes 0 1 2, which is false. If x 2, then x 1 2 becomes 2 1 2, which is true. Solving an inequality refers to the process of finding the numbers that make an algebraic inequality a true numerical statement. We say that such numbers, which are called the solutions of the inequality, satisfy the inequality. The set of all solutions of an inequality is called its solution set.

114

Chapter 2


1 Show the Solution Set of an Inequality by Using Set-Builder Notation and by Graphing We often state solution sets for inequalities with set builder notation. For example, the solution set for x 1 2 is the set of real numbers greater than 1, expressed as 5x 0x 16. The set builder notation 5x 0x 16 is read “the set of all x such that x is greater than 1.” We sometimes graph solution sets for inequalities on a number line; the solution set for 5x 0x 16 is pictured in Figure 2.6.

−4 −3 −2 −1

0

1

2

3

4

Figure 2.6

The open circle at 1 indicates that 1 is not a solution, and the red part of the line to the right of 1 indicates that all real numbers greater than 1 are solutions. We refer to the red portion of the number line as the graph of the solution set 5x 0 x 16 . The solution set for x 2 5 is the set of real numbers greater than or equal to 3; it is written as 5x 0x 36 in set-builder notation. The graph of the solution set 5x 0 x 36 is shown in Figure 2.7. The solid dot at 3 indicates that 3 is a solution, and the red part of the line to the right of 3 indicates that all real numbers greater than 3 are solutions.

−3 −2 −1

0

1

2

3

4

5

Figure 2.7

The examples in the table (Figure 2.8) below contain some simple algebraic inequalities, their solution sets, and graphs of the solution sets. Look them over very carefully to be sure you understand the symbols.

Algebraic inequality

Solution set

x 2

5x 0x 26

x 1

5x 0x 16

3 x

5x 0x 36

x1 ( is read “greater than or equal to”) x2 ( is read “less than or equal to”) 1x

Figure 2.8

5x 0x 16

5x 0 x 26 5x 0x 16

Graph of solution set 54321 0 1 2 3 4 5 54321 0 1 2 3 4 5 54321 0 1 2 3 4 5 54321 0 1 2 3 4 5

54321 0 1 2 3 4 5

54321 0 1 2 3 4 5

2.6

Inequalities

115

2 Solve Inequalities The general process for solving inequalities closely parallels that for solving equations. We continue to replace the given inequality with equivalent but simpler inequalities. For example, 2x 1 9

(1)

2x 8

(2)

x 4

(3)

are all equivalent inequalities; that is, they have the same solutions. Thus to solve (1) we can find the solutions of (3), which are obviously all numbers greater than 4. The exact procedure for simplifying inequalities is based primarily on two properties. The first of these is the addition-subtraction property of inequality.

Property 2.4

Addition-Subtraction Property of Inequality

For all real numbers a, b, and c, 1. a b if and only if a c b c. 2. a b if and only if a c b c.

Property 2.4 states that any number can be added to or subtracted from both sides of an inequality, and the result is an equivalent inequality. The property is stated in terms of , but analogous properties exist for , , and . Consider the use of this property in the next three examples.

EXAMPLE 1

Solve x 3 1 and graph the solutions.

Solution x 3 1 x 3 3 1 3

Add 3 to both sides

x 2

The solution set is 5x 0x 26 , and it can be graphed as shown in Figure 2.9.

−4 −3 −2 −1

0

1

2

3

4

Figure 2.9

▼ PRACTICE YOUR SKILL Solve x 5 2 and graph the solutions.

EXAMPLE 2

Solve x 4 5 and graph the solutions.

Solution x45 x4454 x1


■

116

Chapter 2


The solution set is 5x 0 x 16 , and it can be graphed as shown in Figure 2.10. −4 −3 −2 −1

0

1

2

3

4

Figure 2.10

▼ PRACTICE YOUR SKILL Solve x 3 1 and graph the solutions.

EXAMPLE 3

■

Solve 5 6 x and graph the solutions.

Solution 5 6x 56 6x6


1 x

Because 1 x is equivalent to x 1, the solution set is 5x 0x 16. It can be graphed as shown in Figure 2.11. −4 −3 −2 −1

0

1

2

3

4

Figure 2.11

▼ PRACTICE YOUR SKILL Solve 8 5 x and graph the solutions.

■

Now let’s look at some numerical examples to see what happens when both sides of an inequality are multiplied or divided by some number. 4 3

5142 5132

20 15

2 3

4122 4132

8 12

6 4

6 4

2 2

3 2

8 2

8 2

4 4

2

1 2

Note that multiplying or dividing both sides of an inequality by a positive number produces an inequality of the same sense. This means that if the original inequality is greater than, then the new inequality is greater than, and if the original is less than, then the resulting inequality is less than. Now note what happens when we multiply or divide both sides by a negative number: 3 5

2(3) 2(5)

6 10

4 1

5(4) 5(1)

20 5

14 2

14 2 2 2

7 1

3 6

3 6 3 3

1 2

Multiplying or dividing both sides of an inequality by a negative number reverses the sense of the inequality. Property 2.5 summarizes these ideas.

2.6

Property 2.5

Inequalities

117

Multiplication-Division Property of Inequality

(a) For all real numbers, a, b, and c, with c 0, 1. a b if and only if ac bc 2. a b if and only if

a b

c c

(b) For all real numbers, a, b, and c, with c 0, 1. a b if and only if ac bc 2. a b if and only if

b a c c

Similar properties hold when each inequality is reversed or when is replaced with , and when is replaced with . For example, if a b and c 0, then ac bc a b and . c c Observe the use of Property 2.5 in the next three examples.

EXAMPLE 4

Solve 2x 4.

Solution 2x 4 2x 4

2 2


x 2

The solution set is 5x 0x 26 .


EXAMPLE 5

■

1 3 Solve x . 4 5

Solution 3 1 x 4 5 4 3 4 1 a xb a b 3 4 3 5 x


4 3

4 15

The solution set is e x 0 x

4 f. 15

▼ PRACTICE YOUR SKILL 2 4 Solve x . 3 7

■

118

Chapter 2


EXAMPLE 6

Solve 3x 9.

Solution 3x 9 3x 9 3 3

Divide both sides by 3, which reverses the inequality

x 3

The solution set is 5x 0x 36.


■

As we mentioned earlier, many of the same techniques used to solve equations may be used to solve inequalities. However, we must be extremely careful when we apply Property 2.5. Study the following examples and note the similarities between solving equations and solving inequalities.

EXAMPLE 7

Solve 4x 3 9.

Solution 4x 3 9 4x 3 3 9 3

Add 3 to both sides

4x 12 4x 12

4 4


x 3

The solution set is 5x 0 x 36.


EXAMPLE 8

■

Solve 3n 5 11.

Solution 3n 5 11 3n 5 5 11 5


3n 6 3n 6

3 3


n 2

The solution set is 5n 0 n 26.

▼ PRACTICE YOUR SKILL Solve 4y 5 17.

■

2.6

Inequalities

119

Checking the solutions for an inequality presents a problem. Obviously we cannot check all of the infinitely many solutions for a particular inequality. However, by checking at least one solution, especially when the multiplication-division property is used, we might catch the common mistake of forgetting to reverse the sense of the inequality. In Example 8 we are claiming that all numbers greater than 2 will satisfy the original inequality. Let’s check one such number in the original inequality, say, 1. 3n 5 11 ?

3112 5 11 ?

3 5 11 8 11 Thus 1 satisfies the original inequality. If we had forgotten to reverse the sense of the inequality when we divided both sides by 3, our answer would have been n 2, and the check would have detected the error.

CONCEPT QUIZ

For Problems 1–10, answer true or false. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Numerical statements of inequality are always true. The algebraic statement x 4 6 is called an open sentence. The algebraic inequality 2x 10 has one solution. The algebraic inequality x 3 has an infinite number of solutions. The set-builder notation {x0 x 5} is read “the set of variables that are particular to x 5.” When graphing the solution set of an inequality, a solid dot is used to include the endpoint. The properties for solving inequalities are the same as the properties for solving equations. The solution set of the inequality x 6 includes the number 6. When multiplying both sides of an inequality by a negative number, the sense of the inequality stays the same. When adding a negative number to both sides of an inequality, the sense of the inequality stays the same.

Problem Set 2.6 1 Show the Solution Set of an Inequality by Using Set-Builder Notation and by Graphing For Problems 1–12, state the solution set and graph it on a number line. 1. x 2

2. x 4

3. x 3

4. x 0

5. 2 x 7. 2 x

9. x 1 11. 2 x

10. x 2 12. 1 x

2 Solve Inequalities For Problems 13 –22, solve the inequality and graph the solutions. 13. x 4 7

14. x 1 1

6. 3 x

15. x 6 5

16. x 4 5

8. 1 x

17. x 1 3

18. x 2 5

120

Chapter 2


19. x 1 0

20. x 2 0

39. 3x 2 17

40. 2x 5 19

21. x 4 4

22. x 2 2

41. 4x 3 21

42. 5x 2 28

43. 2x 1 41

44. 3x 1 35

For Problems 23 – 60, solve the inequality. 23. x 6 14

24. x 7 15

45. 6x 2 18

46. 8x 3 25

25. x 4 13

26. x 3 12

47. 3 4x 2

48. 7 6x 3

27. 4x 36

28. 3x 51

49. 2 3x 1

50. 6 2x 4

29. 6x 20

30. 8x 28

51. 38 9t 2

52. 36 7t 1

31. 5x 40

32. 4x 24

53. 5x 4 3x 24

54. 7x 8 5x 38

33. 7n 56

34. 9n 63

55. 4x 2 6x 1

56. 6x 3 8x 3

35. 48 14n

36. 36 8n

57. 5 3t 4 7t

58. 6 4t 7t 10

37. 16 9 n

38. 19 27 n

59. x 4 3x 5

60. 3 x 3x 10

THOUGHTS INTO WORDS 61. Do the greater-than and less-than relations possess the symmetric property? Explain your answer. 62. Is the solution set for x 3 the same as that for 3 x? Explain your answer.

63. How would you convince someone that it is necessary to reverse the sense of the inequality when multiplying both sides of an inequality by a negative number?

FURTHER INVESTIGATIONS Solve each of the following inequalities.

68. 3x 4 3x 6

64. x 3 x 4

69. 2x 7 2x 1

65. x 4 x 6

70. 5 4x 1 4x

66. 2x 4 2x 7

71. 7 5x 2 5x

67. 5x 2 5x 7

Answers to the Example Practice Skills 1. 5x 0 x 36 3. 5x 0 x 36 8. 5y 0 y 36

2. 5x 0 x 26

−4 −3 −2−1 0 1 2 3 4

−2 −1

0

1

2

3

4

4. 5x 0 x 36

−3 −2 −1 0

5. e x 0 x

6 f 7

1

2

3 4

6. 5x 0 x 26

7. 5x 0 x 46


2. True

3. False 4. True 5. False

6. True

7. False 8. True

9. False

10. True

2.7

2.7

Inequalities, Compound Inequalities, and Problem Solving

121

Inequalities, Compound Inequalities, and Problem Solving OBJECTIVES 1

Solve Inequalities

2

Solve Inequalities That Involve the Use of the Distributive Property

3

Solve Inequalities That Involve Fractional Forms

4

Solve Compound Inequalities

5

Solve Word Problems That Translate into Inequality Statements

1 Solve Inequalities Let’s begin this section by solving four inequalities with the same basic steps we used with equations. Again, be careful when applying the multiplication-division property of inequality.

EXAMPLE 1

Solve 5x 8 3x 10.

Solution 5x 8 3x 10 5x 8 3x 3x 10 3x


2x 8 10 2x 8 8 10 8


2x 18 2x 18 2 2


x 9

The solution set is 5x 0x 96.


■

2 Solve Inequalities That Involve the Use of the Distributive Property EXAMPLE 2

Solve 41x 32 31x 42 21x 12 .

Solution 41x 32 31x 42 21x 12 4x 12 3x 12 2x 2 7x 2x 2 7x 2x 2x 2 2x

Distributive property Combine similar terms Subtract 2x from both sides

122

Chapter 2


5x 2 5x 2 5 5 x


2 5

2 The solution set is e x 0x f . 5

▼ PRACTICE YOUR SKILL Solve 21x 12 41x 22 31x 62 .

■

3 Solve Inequalities That Involve Fractional Forms EXAMPLE 3

3 1 3 Solve n n . 2 6 4

Solution 3 1 3 n n 2 6 4 1 3 3 12 a n nb 12 a b 2 6 4 3 1 3 12 a nb 12 a nb 12 a b 2 6 4

Multiply both sides by 12, the LCD of all denominators Distributive property

18n 2n 9 16n 9 16n 9

16 16 n The solution set is e n 0n


9 16

9 f. 16

▼ PRACTICE YOUR SKILL 4 5 7 Solve y y . 3 6 6

■

In Example 3 we are claiming that all numbers greater than the original inequality. Let’s check one number, say, 0. 3 1 3 n n 2 6 4 3 1 ? 3 102 102 2 6 4 0

3 4

9 will satisfy 16

2.7


123

Therefore, 0 satisfies the original inequality. If we had forgotten to reverse the inequality sign when we divided both sides by 16, then our answer would have been 9 n , and the check would have detected the error. 16

EXAMPLE 4

Solve

x1 3 x2 . 5 10 10

Solution x1 3 x2 5 10 10 10 a 10 a

x1 3 x2 b 10 a b 5 10 10

x2 x1 3 b 10 a b 10 a b 5 10 10 21x 22 11x 12 3

Multiply both sides by 10 Distributive property Don’t forget to distribute the 1

2x 4 x 1 3 x33 The solution set is 5x冟x 06 .


x0


x4 x2 5 . 3 4 12

■

4 Solve Compound Inequalities The words “and” and “or” are used in mathematics to form compound statements. We use “and” and “or” to join two inequalities to form a compound inequality. Consider the compound inequality x 2

and

x 5

For the solution set, we must find values of x that make both inequalities true statements. The solution set of a compound inequality formed by the word “and” is the intersection of the solution sets of the two inequalities. The intersection of two sets, denoted by , contains the elements that are common to both sets. For example, if A 51, 2, 3, 4, 5, 66 and B 50, 2, 4, 6, 8, 106, then A B 52, 4, 66. So to find the solution set of the compound inequality x 2 and x 5, we find the solution set for each inequality and then determine the solutions that are common to both solution sets.

EXAMPLE 5

Graph the solution set for the compound inequality x 2 and x 5, and write the solution set in set-builder notation.

Solution x 2

2 1 0 1 2 3 4 5 6 7

x 5 x 2

2 1 0 1 2 3 4 5 6 7

and

x 5

2 1 0 1 2 3 4 5 6 7 Figure 2.12

(a) (b) (c)

124

Chapter 2


Thus all numbers greater than 2 and less than 5 are included in the solution set 5x 0 2 x 56 , and the graph is shown in Figure 2.12(c).

▼ PRACTICE YOUR SKILL Graph the solution set for the compound inequality x 0 and x 5 and write the solution set in set-builder notation. ■

EXAMPLE 6

Graph the solution set for the compound inequality x 1 and x 4 and write the solution in set-builder notation.

Solution x1 x4 x1

and

x4

2 1

0

1

2

3

4

5

2 1

0

1

2

3

4

5

2 1

0

1

2

3

4

5

(a) (b) (c)

Figure 2.13

The intersection of the two solution sets is x 1. The solution set 5x 0x 16 contains all the numbers that are less than or equal to 1, and the graph is shown in Figure 2.13(c).

▼ PRACTICE YOUR SKILL Graph the solution set for the compound inequality x 3 and x 5 and write the solution set in set-builder notation. ■

The solution set of a compound inequality formed by the word “or” is the union of the solution sets of the two inequalities. The union of two sets, denoted by , contains all the elements in both sets. For example, if A 50, 1, 26 and B 51, 2, 3, 46, then A B 50, 1, 2, 3, 46. Note that even though 1 and 2 are in both set A and set B, there is no need to write them twice in A B. To find the solution set of the compound inequality x 1

or

x 3

we find the solution set for each inequality and then take all the values that satisfy either inequality or both.

EXAMPLE 7

Graph the solution set for x 1 or x 3 and write the solution in set-builder notation.

Solution x 1

−2 −1

0

1

2

3

4

5

x 3

−2 −1

0

1

2

3

4

5

−2 −1

0

1

2

3

4

5

x 1

or x 3

Figure 2.14

(a) (b) (c)

2.7


125

Thus all numbers greater than 1 are included in the solution set 5x 0 x 16, and the graph is shown in Figure 2.14(c).

▼ PRACTICE YOUR SKILL Graph the solution set for the compound inequality x 2 or x 4 and write the solution set in set-builder notation. ■

EXAMPLE 8

Graph the solution set for x 0 or x 2, and write the solution in set-builder notation.

Solution x0

(a) 4 3 2 1 0 1 2 3 4 5

x2

(b) 4 3 2 1 0 1 2 3 4 5

x0

or x 2

(c) 4 3 2 1 0 1 2 3 4 5 Figure 2.15

Thus all numbers less than or equal to 0 and all numbers greater than or equal to 2 are included in the solution set 5x 0x 0 or x 26, and the graph is shown in Figure 2.15(c).

▼ PRACTICE YOUR SKILL Graph the solution set for the compound inequality x 1 or x 1 and write the solution set in set-builder notation. ■

5 Solve Word Problems That Translate into Inequality Statements Let’s consider some word problems that translate into inequality statements. The suggestions for solving word problems in Section 2.4 apply here, except that the situations described in these problems will translate into inequalities instead of equations.

Banana Stock /Photolibrary

EXAMPLE 9

Apply Your Skill Ashley had scores of 95, 82, 93, and 84 on her first four exams of the semester. What score must she get on the fifth exam to have an average of 90 or higher for the five exams?

Solution Let s represent the score needed on the fifth exam. Because we find the average by adding all five scores and dividing by 5 (the number of exams), we can solve this inequality: 95 82 93 84 s 90 5

126

Chapter 2


We use the following steps: 354 s 90 5 5a

354 s b 51902 5

Simplify numerator of left side


354 s 450 354 s 354 450 354


s 96 She must receive a score of 96 or higher on the fifth exam.

▼ PRACTICE YOUR SKILL Felix had scores of 88, 95, and 90 on the first three exams in biology. What score must he get on the fourth exam to have an average of 92 or higher for the four exams? ■

EXAMPLE 10

Apply Your Skill

Nam Y. Huh /AP Wide World Photos

The Cubs have won 40 baseball games and have lost 62 games. They have 60 more games to play. To win more than 50% of all their games, how many of the remaining 60 games must they win?

Solution Let w represent the number of games the Cubs must win out of the 60 games remaining. Because they are playing a total of 40 62 60 162 games, to win more than 50% of their games, they will have to win more than 81 games. Thus we have the inequality w 40 81 Solving this yields w 41 The Cubs need to win at least 42 of the remaining 60 games.

▼ PRACTICE YOUR SKILL The Red Sox want to win 70% or more of the 160 games they will play this season. So far this season, they have won 32 games. How many of the remaining games must they win to achieve their goal? ■

CONCEPT QUIZ

For Problems 1–5, answer true or false. 1. The solution set of a compound inequality formed by the word “and” is an intersection of the solution sets of the two inequalities. 2. The solution set of a compound inequality formed by the words “and” or “or” is a union of the solution sets of the two inequalities. 3. The intersection of two sets contains the elements that are common to both sets. 4. The union of two sets contains all the elements in both sets. 5. The intersection of set A and set B is denoted by A B.

2.7


127

For Problems 6 –10, match the compound statement with the graph of its solution set (Figure 2.16). 6. x 4 or x 1

A.

7. x 4 and x 1

B.

8. x 4 or x 1

C.

9. x 4 and x 1

D.

10. x 4 or x 1

E.

−2 −1

0

1

2

3

4

5

−2 −1

0

1

2

3

4

5

−2 −1

0

1

2

3

4

5

−2 −1

0

1

2

3

4

5

−2 −1

0

1

2

3

4

5

Figure 2.16

Problem Set 2.7 1 Solve Inequalities For Problems 1–50, solve each inequality. 1. 3x 4 x 8 2. 5x 3 3x 11 3. 7x 2 3x 6 4. 8x 1 4x 21 5. 6x 7 3x 3 6. 7x 5 4x 12 7. 5n 2 6n 9 8. 4n 3 5n 6

19. 31y 12 12 20. 21y 42 18 21. 21x 62 17 22. 31x 52 14 23. 31x 22 21x 12 24. 51x 32 41x 22 25. 41x 32 61x 52 26. 61x 12 81x 52 27. 31x 42 21x 32 24

9. 2t 9 4t 13

28. 21x 12 31x 22 12

10. 6t 14 8t 16

29. 51n 12 31n 12 9

11. 3x 4 2x 7

30. 41n 52 21n 12 13

12. x 2 3x 7 13. 4x 6 2x 1 14. 6x 8 4x 5

2 Solve Inequalities That Involve the Use of the Distributive Property

3 Solve Inequalities That Involve Fractional Forms 31.

2 1 n n 7 2 3

32.

1 3 n n1 4 6

33.

5 3 3 n n 4 6 8

34.

2 1 1 n n 3 2 4

15. 51x 22 30 16. 41x 12 16 17. 21n 32 9 18. 31n 22 7

128

Chapter 2

35.

3x 2 x 5 3 10

36.

5x 3 7x 4 8 12


5 Solve Word Problems That Translate into Inequality Statements Solve each of the following problems by setting up and solving an appropriate inequality.

37. n 3.4 0.15n

67. Five more than three times a number is greater than 26. Find the numbers that satisfy this relationship.

38. x 2.1 0.3x 39. 0.09t 0.11t 2002 77

68. Fourteen increased by twice a number is less than or equal to three times the number. Find the numbers that satisfy this relationship.

40. 0.07t 0.081t 1002 38 41. 0.06x 0.081250 x2 19 42. 0.08x 0.0912x2 130 43.

x1 x3 1

2 5 10

70. One side of a triangle is three times as long as another side. The third side is 15 centimeters long. If the perimeter of the triangle is to be no greater than 75 centimeters, find the largest lengths that the other two sides can be.

x3 x5 1 44. 4 7 28 45.

69. Suppose that the perimeter of a rectangle is to be no greater than 70 inches and that the length of the rectangle must be 20 inches. Find the largest possible value for the width of the rectangle.

x2 x1 2 6 5

46.

x6 x2

1 8 7

71. Sue bowled 132 and 160 in her first two games. What must she bowl in the third game to have an average of at least 150 for the three games?

47.

n3 n7

3 3 2

72. Mike has scores of 87, 81, and 74 on his first three algebra tests. What score must he get on the fourth test to have an average of 85 or higher for the four tests?

48.

n4 n2 4 4 3

49.

x3 x2 9 7 4 14

50.

x2 7 x1 5 6 15

73. This semester Lance has scores of 96, 90, and 94 on his first three algebra exams. What must he average on the last two exams to have an average greater than 92 for all five exams? 74. The Mets have won 45 baseball games and lost 55 games. They have 62 more games to play. To win more than 50% of all their games, how many of the remaining 62 games must they win?

4 Solve Compound Inequalities For Problems 51– 66, graph the solutions for each compound inequality. 51. x 1 and x 2

52. x 1 and x 4

53. x 2 or x 1

54. x 0 or x 3

55. x 2 and x 2

56. x 1 and x 3

57. x 1 and x 2

58. x 2 and x 3

59. x 4 or x 0

60. x 2 or x 4

61. x 3 and x 1

62. x 3 and x 6

63. x 0 or x 2

64. x 2 or x 1

65. x 4 or x 3

66. x 1 or x 2

75. An Internet business has costs of $4000 plus $32 per sale. The business receives revenue of $48 per sale. How many sales would ensure that the revenues exceed the costs? 76. The average height of the two forwards and the center of a basketball team is 6 feet, 8 inches. What must the average height of the two guards be for the team average to be at least 6 feet, 4 inches? 77. Scott shot rounds of 82, 84, 78, and 79 on the first four days of the golf tournament. What must he shoot on the fifth day of the tournament to average 80 or less for the five days? 78. Sydney earns $2300 a month. To qualify for a mortgage, her monthly payments must be less than 35% of her monthly income. Her monthly payments must be less than what amount to qualify for the mortgage?

2.7


129

THOUGHTS INTO WORDS 79. Give a step-by-step description of how you would solve the inequality 3x 2 4(x 6).

80. Find the solution set for each of the following compound statements, and in each case explain your reasoning. a. x 2 and 5 4

b. x 2 or 5 4

c. x 2 and 4 10

d. x 2 or 4 10

Answers to the Example Practice Skills 1. 5x冟x 16

2. 5x冟x 86

5. 5x冟0 x 56

7 3. e y冟y f 3

4. 5x冟x 56

− 3 − 2− 1 0 1 2 3 4 5 6

6. 5x冟x 36

−2 −1 0

7. 5x冟x 26

−2 − 1

8. 5x冟x 1 or x 16

0

1 2 1

2

3 4 3

4

5 5

−4 −3 −2 −1 0 1 2 3 4

9. A score of 95 or higher 10. Need to win at least 80 games


2. False

3. True

4. True 5. True 6. B

7. E

8. A 9. D

10. C


SUMMARY

EXAMPLE

Solve equations using the AdditionSubtraction Property of Equality. (Sec. 2.1, Obj. 1, p. 76)

Any number can be added to or subtracted from both sides of an equation.

Solve x 14 30.

Solve equations using the MultiplicationDivision Property of Equality. (Sec. 2.1, Obj. 2, p. 79)

An equivalent equation is obtained whenever both sides of an equation are multiplied or divided by a nonzero real number.

Solve equations using both the AdditionSubtraction Property of Equality and the MultiplicationDivision Property of Equality. (Sec. 2.3, Obj. 1, p. 89)

To solve most equations, both properties must be applied.

Solve equations that involve the use of the distributive property. (Sec. 2.5, Obj. 1, p. 103)

To solve equations in which the variable is part of an expression enclosed in parentheses, the distributive property is used. The distributive property removes the parentheses and the resulting equation is solved in the usual way.

Solve equations that involve fractional forms. (Sec. 2.5, Obj. 2, p. 105)

When an equation contains several fractions, it is usually best to start by clearing the equation of all fractions. The fractions can be cleared by multiplying both sides of the equation by the least common denominator of all the denominators.

130


Solution

x 14 14 30 14 x 16 The solution set is 5166 . Solve

2 y 24. 3

Solution

3 3 2 a yb 1242 2 3 2 y 36 The solution set is 5366 . Solve 5x 12 10. Solution

Problems 1– 6

5x 12 12 10 12 5x 22 22 5x 5 5 22 x 5 22 The solution set is e f . 5 Solve 41x 12 21x 32 . Solution

Problems 7–13

4x 4 2x 6 4x 4 4 2x 6 4 4x 2x 10 4x 2x 2x 2x 10 2x 10 10 2x 2 2 x5 The solution set is 556 . Solve

2n 5 3n . 4 7 14

Problems 14 –20

Solution

28 a

3n 2n 5 b 28 a b 4 7 14 3n 2n 5 28 a b 28 a b 28 a b 4 7 14 21n 8n 10 29n 10 29n 10 29 29

(continued)

Chapter 2

OBJECTIVE

SUMMARY

10 29

The solution set is e Solve equations that are contradictions or identities. (Sec. 2.5, Obj. 3, p. 107)

When an equation is not true for any value of x, then the equation is called “a contradiction.”

131


EXAMPLE n

Summary

10 f. 29

Solve 21x 42 2x 5.

Problems 21–26

Solution

2x 8 2x 5 2x 2x 8 2x 2x 5 85 This is a false statement so there is no solution. The solution set is .

When an equation is true for any permissible value of x, then the equation is called “an identity.”

Solve 4x 8 212x 42 . Solution

4x 8 4x 8 4x 4x 8 4x 4x 8 8 8 This is a true statement so any value of x is a solution. The solution set is {All reals}.

Show the solution set of an inequality in setbuilder notation and by graphing. (Sec. 2.6, Obj. 1, p. 113)

The solution set of x 2 5 is all numbers greater than 3. The solution set in set-builder notation is 5x冟x 36 and is read “the set of all x such that x is greater than 3.” A number line is used to graph the solution. An open circle on the number line means that number is not included in the solution set. A solid dot on the number line indicates that the number is included in the solution set.

Write the solution set of x 2 in set-builder notation and graph the solution.

Problems 27–30

Solution

5x冟x 26 −3 −2 −1 0

1 2

3 4

Write the solution set of x 1 in set-builder notation and graph the solution. Solution

5x冟x 16 −3 −2 −1

Solve inequalities. (Sec. 2.6, Obj. 2, p. 115; Sec. 2.7, Obj. 1, p. 121)

Properties for solving inequalities are similar to the properties for solving equations— except for properties that involve multiplying or dividing by a negative number. When multiplying or dividing both sides of an inequality by a negative number, you must reverse the inequality symbol.

0

1

2

3

4

Solve 2x 8 14. Solution

2x 8 8 14 8 2x 6 6 2x 2 2 x 3 The solution set is 5x冟x 36 .

Problems 31–36

(continued)

132

Chapter 2


OBJECTIVE

SUMMARY

EXAMPLE

Solve inequalities that involve the use of the distributive property. (Sec. 2.7, Obj. 2, p. 121)

To solve inequalities when the variable is part of an expression enclosed in parentheses, the distributive property is used. The distributive property removes the parentheses and the resulting inequality is solved in the usual way.

Solve 15 21x 12 5.

Solve inequalities that involve fractional forms. (Sec. 2.7, Obj. 3, p. 122)

When an inequality contains several fractions, it is usually best to clear the inequality of all fractions. The fractions can be cleared by multiplying both sides of the inequality by the least common denominator of all the denominators.

Solve compound inequalities formed by the word “and.” (Sec. 2.7, Obj. 4, p. 123)

The solution set of a compound inequality formed by the word “and” is the intersection of the solution sets of the two inequalities. To solve inequalities involving “and,” we must satisfy all of the conditions. Thus the compound inequality x 1 and x 3 is satisfied by all numbers between 1 and 3.

Solve compound inequalities formed by the word “or.” (Sec. 2.7, Obj. 4, p. 123)

The solution set of a compound inequality formed by the word “or” is the union of the solution sets of the two inequalities. To solve inequalities involving “or,” we must satisfy one or more of the conditions. Thus the compound inequality x 1 or x 5 is satisfied by all numbers less than 5.

Solution


15 2x 2 5 15 2x 3 15 3 2x 3 3 18 2x 2x 18

2 2 9 x The solution set is 5x冟x 96 . Solve

3y y1 4. 5 3

Problems 43 – 46

Solution

y1 3y b 15142 5 3 y1 3y b 60 15 a b 15 a 5 3 313y2 51y 12 60 9y 5y 5 60 4y 5 60 4y 65 4y 65 4 4 65 y 4 65 The solution set is e y冟y f. 4 15 a

Solve the compound inequality x 4 and x 2 and graph the solution set.

Problems 47–50

Solution

All of the conditions must be satisfied. Thus the compound inequality x 4 and x 2 is satisfied by all numbers greater than 2. The solution set is 5x冟x 26 . Solve the compound inequality x 1 or x 2 and graph the solution set.

Problems 51–54

Solution

One or more of the conditions must be satisfied. Thus the compound inequality x 1 or x 2 is satisfied by all real numbers. The solution set is {All reals}.

(continued)

Chapter 2

Review Problem Set

OBJECTIVE

SUMMARY

EXAMPLE

Translate English phrases into algebraic expressions. (Sec. 2.2, Obj. 2, p. 83)

To translate English phrases into algebraic expressions, you should know the algebraic vocabulary for “addition,” “subtraction,” “multiplication,” and “division.”

Translate the phrase “four less than a number” into an algebraic expression.

Keep these suggestions in mind as you solve word problems: 1. Read the problem carefully. 2. Sketch any figure or diagram that might be helpful. 3. Choose a meaningful variable. 4. Look for a guideline. 5. Form an equation. 6. Solve the equation. 7. Check your answer.

The difference of two numbers is 14. If 35 is the larger number, find the smaller number.

Follow the same steps as in the previous box; but for step 5, substitute Form an inequality, and for step 6, substitute Solve the inequality.

Martin must average at least 240 points for a series of three bowling games to get into the playoffs. If he has bowled games of 220 points and 210 points, what must his score be on the third game to get into the playoffs?

Solve word problems. (Sec. 2.3, Obj. 2, p. 92; Sec. 2.4, Obj. 3, p. 98; Sec. 2.5, Obj. 4, p. 107)

Solve word problems involving inequalities. (Sec. 2.7, Obj. 5, p. 125)

Problems 55 – 64

Solution

Let n represent the number. The algebraic expression is n 4. Problems 65 –74

Solution

Let n represent the smaller number. Guideline

Larger number smaller number 14 35 n 14 n 21 n 21 The smaller number is 21.

Solution

Let p represent the points for the third game. Guideline

Average 240 220 210 p 240 3 430 p 720 p 290 Martin must bowl 290 points or more.

Chapter 2 Review Problem Set For Problems 1–26, solve each of the equations.


3. 7 4x 10

1. 9x 2 29

4. 6y 5 4y 13

2. 3 4y 1

5. 4n 3 7n 9

Problems 75 –77

133

134

Chapter 2


6. 3x 4x 2 7x 14 9x

35. 16 8 2y 3y

7. 71y 42 41y 32

36. 24 5x 4 7x

8. 21x 12 51x 32 111x 22

37. 61x 32 41x 132

9. 31x 62 5x 3

38. 21x 32 31x 62 14

10. 21x 42 31x 82

39. 31n 42 51n 22 3n

11. 51n 12 41n 22 31n 12 3n 5

40. 41n 22 1n 12 41n 62

12. 1t 32 12t 12 31t 52 21t 12

41. 12 41x 12 2

13. 312t 42 213t 12 214t 32 1t 12

42. 36 31x 22 1

14.

2 1 7 n n 5 2 10

43.

2n n 3 5 4 10

15.

3n 5n 1 4 7 14

44.

n4 n3 7

5 6 15

16.

x3 x5 11 6 8 12

45.

3 2 n6 n4 4 3

17.

n1 3 n 2 4 8

46.

1 1 3 n n4 n2 2 3 5

18.

x4 x3 9 8

19.

x1 x2 3 4

20.

3x 2 2x 1 3 2

For Problems 47–54, graph the solution set for each of the compound inequalities. 47. x 3 and x 2 48. x 2 and x 5 49. x 6 and x 3

21. 4x 3 6x 7 2x

50. x 1 and x 0

22. 5x 8 2x 3x 8

51. x 1 or x 4

23. 2(x 5) 10 2x

52. x 2 or x 0

24. (x 3) 6x 5x 7

53. x 3 or x 4

25. x 2 3x 2(x 1)

54. x 1 or x 4

26. 3x (2x 5) x For Problems 27–30, write the solution set in set-builder notation and graph the solution.

For Problems 55 – 64, translate each word phrase into an algebraic expression. 55. Five less than n

27. x 1

56. Five less n

28. x 2

57. Ten times the quantity, x minus 2

29. x 0

58. Ten times x minus 2

30. x 3

59. x minus 3

For Problems 31– 46, solve each inequality.

60. d divided by r

31. 3x 2 10

61. x squared plus 9

32. 2x 5 3

62. x plus 9, the quantity squared

33. 2x 9 x 4

63. The sum of the cubes of x and y

34. 3x 1 5x 10

64. Four less than the product of x and y

Chapter 2 For Problems 65 –76, set up an equation or an inequality and solve the problem. 65. Three-fourths of a number equals 18. Find the number. 66. Nineteen is 2 less than three times a certain number. Find the number. 67. The difference of two numbers is 21. If 12 is the smaller number, find the other number. 68. One subtracted from nine times a certain number is the same as 15 added to seven times the number. Find the number. 69. The sum of two numbers is 40. Six times the smaller number equals four times the larger. Find the numbers. 70. Find a number such that 2 less than two-thirds of the number is 1 more than one-half of the number. 71. Miriam has 30 coins consisting of nickels and dimes amounting to $2.60. How many coins of each kind does she have?

Review Problem Set

135

72. Suppose that Russ has a bunch of nickels, dimes, and quarters amounting to $15.40. The number of dimes is 1 more than three times the number of nickels, and the number of quarters is twice the number of dimes. How many coins of each kind does he have? 73. The supplement of an angle is 14° more than three times the complement of the angle. Find the measure of the angle. 74. Pam rented a car from a rental agency that charges $25 a day and $0.20 per mile. She kept the car for 3 days and her bill was $215. How many miles did she drive during that 3-day period? 75. Monica had scores of 83, 89, 78, and 86 on her first four exams. What score must she get on the fifth exam so that her average for all five exams is 85 or higher? 76. Ameya’s average score for her first three psychology exams was 84. What must she get on the fourth exam so that her average for the four exams is 85 or higher?

Chapter 2 Test For Problems 1–12, solve each of the equations. 1.

1. 7x 3 11

2.

2. 7 3x 2

3.

3. 4n 3 2n 15

4.

4. 3n 5 8n 20

5.

5. 41x 22 51x 92

6.

6. 91x 42 61x 32

7.

7. 51y 22 21y 12 31y 62

8.

8.

3 2 1 x 5 3 2

9.

9.

x2 x3 4 6

10.

10.

x2 x1 2 3 2

11.

11.

x3 x1 13 6 8 24

12.

12. 51n 22 31n 72 For Problems 13 –18, solve each of the inequalities.

13.

13. 3x 2 13

14.

14. 2x 5 3

15.

15. 31x 12 51x 32

16.

16. 4 71x 12 3

17.

17. 21x 12 51x 22 51x 32

18.

18.

3 1 n2 n1 2 4

For Problems 19 and 20, graph the solution set for each compound inequality. 19.

19. x 2 and x 4

20.

20. x 1 or x 3 For Problems 21–25, set up an equation or an inequality and solve each problem.

21.

21. A car repair bill without the tax was $441. This included $153 for parts and 4 hours of labor. Find the hourly rate that was charged for labor.

22.

22. Suppose that a triangular plot of ground is enclosed by 70 meters of fencing. The longest side of the lot is two times the length of the shortest side, and the third side is 10 meters longer than the shortest side. Find the length of each side of the plot.

23.

23. Tina had scores of 86, 88, 89, and 91 on her first four history exams. What score must she get on the fifth exam to have an average of 90 or higher for the five exams?

136

Chapter 2

24. Sean has 103 coins consisting of nickels, dimes, and quarters. The number of dimes is 1 less than twice the number of nickels, and the number of quarters is 2 more than three times the number of nickels. How many coins of each kind does he have?

24.

25. In triangle ABC, the measure of angle C is one-half the measure of angle A, and the measure of angle B is 30 more than the measure of angle A. Find the measure of each angle of the triangle.

25.

Test

137

Chapters 1–2 Cumulative Practice Test For Problems 1– 4, simplify each numerical expression. 1. 3(4) 2 (3)(6) 1 2. (2)7

5. 4x 2y xy 1 2 x y 5 3

18. 3(2x 8) (x 4) 5x 12

19. 2(x 1) 3(x 6)

1 1 and y 2 6

20. 2 (x 1) 4

for x

for x 0.1 and y 0.2

8. Find the greatest common factor of 48, 60, and 96. 9. Find the least common multiple of 9 and 12. 3 3 5 3 10. Simplify x y x y by combining similar 8 7 12 4 terms. For Problems 11–14, simplify each expression by applying the distributive property and combining like terms. 11. (x 2) 6(x 4) 2(x 7) 12. 5(2x 1) 3(4x 7) 3 1 12x 82 120x 42 2 4

14. 3(x 5) (2x 7)

138

For Problems 19 and 20, solve the inequality.

for x 2 and y 3

7. 0.2(x y) 0.3(x y)

13.

x2 x1 2 4 3

17. 2(2x 1) 3(x 3) 4(x 7)

1 1 2 3 2 4

For Problems 5 –7, evaluate each algebraic expression for the given values of the variables.

6.

15. 3(x 4) 4(x 1) 16.

3. 6.2 7.1 3.4 1.9 4.

For Problems 15 –18, solve each equation.

21. Express 300 as the product of prime factors. 22. Graph the solutions for the compound inequality x 1 and x 3. For Problems 23 –25, use an equation or inequality to help solve each problem. 23. On Friday and Saturday nights, the police made a total of 42 arrests at a DUI checkpoint. On Saturday night they made 6 more than three times the arrests of Friday night. Find the number of arrests for each night. 24. For a wedding reception, the caterer charges a $125 fee plus $35 per person for dinner. If Peter and Rynette must keep the cost of the caterer to less than $2500, how many people can attend the reception? 25. Find two consecutive odd numbers whereby the smaller plus five times the larger equals 76.

Ratio, Percent, Formulas, and Problem Solving

3 3.1 Ratio, Proportion, and Percent 3.2 More on Percents and Problem Solving 3.3 Formulas: Geometric and Others 3.4 Problem Solving

Mel Yates/Cultura/Photolibrary

3.5 More about Problem Solving

■ The equation s 3 0.6s can be used to determine how much the owner of a pizza parlor must charge for a pizza if it costs $3 to make, and he wants to make a profit of 60% based on the selling price.

K

irk starts jogging at the rate of 5 miles per hour. One-half hour later, Consuela starts jogging on the same route at 7 miles per hour. How long will it take Consuela to catch Kirk? If we let t represent the time that Consuela jogs, then 1 1 represents Kirk’s time. We can use the equation 7t 5 a t b to determine that 2 2 1 Consuela should catch Kirk in 1 hours. 4

t

We used the formula distance equals rate times time, which is usually expressed

as d rt, to set up the equation 7t 5 a t b . Throughout this chapter we will use a 1 2

variety of formulas in a problem-solving setting to connect algebraic and geometric concepts.


139

I N T E R N E T

P R O J E C T

In this chapter there are suggestions for solving problems. Do an Internet search on Polya problem solving. What are Polya’s four principles of problem solving? How do Polya’s four principles align with the suggestions given in this chapter for solving word problems?

3.1

Ratio, Proportion, and Percent OBJECTIVES 1

Solve Proportions

2

Solve Word Problems Using Proportions

3

Use a Proportion to Convert a Fraction to a Percent

4

Solve Basic Percent Problems

1 Solve Proportions B

A

Ratio In Figure 3.1, as gear A revolves four times, gear B will revolve three times. We say that the gear ratio of A to B is 4 to 3, or the gear ratio of B to A is 3 to 4. Mathematically, a ratio is the comparison of two numbers by division. We can write the gear ratio of A to B in these equivalent expressions: 4 to 3

4:3

4 3

We express ratios as fractions in reduced form. For example, if there are 7500 women 3 7500 and 5000 men at a certain university, then the ratio of women to men is . 5000 2 Figure 3.1

Proportion A statement of equality between two ratios is called a proportion. For example, 2 8 3 12 8 2 is a proportion that states that the ratios and are equal. In general, if we 3 12 have a proportion c a b d

b 0 and d 0

and we multiply both sides of the equation by the common denominator, bd, we obtain a c 1bd2 a b 1bd2 a b b d ad bc Let’s state this as a property of proportions. a c b d

if and only if ad bc, where b 0 and d 0

The products ad and bc are commonly called cross products. Thus according to the property, cross products in a proportion are equal. 140

3.1

EXAMPLE 1

Solve

Ratio, Proportion, and Percent

141

3 x . 20 4

Solution x 3 20 4 4x 60

Cross products are equal

x 15



EXAMPLE 2

Solve

5 x . 18 6

Solve

x2 x3 . 5 4

■

Solution x3 x2 5 4 41x 32 51x 22 4x 12 5x 10 12 x 10 22 x


Cross products are equal Distributive property Subtracted 4x from both sides Subtracted 10 from both sides


x4 x2 . 3 5

■

If a variable appears in one or both of the denominators, then certain restrictions must be imposed to avoid division by zero, as the next example illustrates.

EXAMPLE 3

Solve

4 7 . a2 a3

Solution 7 4 a2 a3

a 2 and a 3

71a 32 41a 22


7a 21 4a 8


3a 21 8

Subtracted 4a from both sides

3a 29 a

Subtracted 21 from both sides

29 3



29 f. 3

142

Chapter 3



EXAMPLE 4

Solve

8 3 . a1 a2

Solve

x x 3 . 4 5

■

Solution This is not a proportion. This example demonstrates the importance of thinking first before pushing the pencil. Because the equation is not in the form of a proportion, we need to revert to a previous technique for solving it. Let’s multiply both sides by 20, the least common denominator, to clear the equation of all fractions. x x 3 4 5 20 a

x x 3b 20 a b 4 5

x x 20 a b 20132 20 a b 4 5



5x 60 4x x 60 0 x 60

Subtracted 4x from both sides Subtracted 60 from both sides



x x 1 . 4 2

■

2 Solve Word Problems Using Proportions Some word problems can be conveniently set up and solved using the concepts of ratio and proportion. Consider the next examples.

EXAMPLE 5

Apply Your Skill On the map in Figure 3.2, 1 inch represents 20 miles. If Newton and Descartes are 1 6 inches apart on the map, find the number of miles between the cities. 2

Solution Let m represent the number of miles between the two cities. Now let’s set up a proportion where one ratio compares distances in inches on the map, and the other ratio compares corresponding distances in miles on land. 20 1 m 1 6 2

3.1


143

To solve this equation, we equate the cross products. Newton

1 m112 a 6 b 1202 2 Kenmore

m a

13 b 1202 130 2

East Islip

6

The distance between the two cities is 130 miles.

1 inches 2


Islip

Windham

Descartes

The plot plan for a new subdivision is scaled so that 1 inch represents 12 feet. Two 1 driveways on the plot plan are 3 inches apart. What is the distance in feet between 4 the two driveways? ■

Figure 3.2

EXAMPLE 6

Apply Your Skill A sum of $1750 is to be divided between two people in the ratio of 3 to 4. How much does each person receive?

Solution Let d represent the amount of money to be received by one person. Then 1750 d represents the amount for the other person. We set up this proportion: d 3 1750 d 4

4d 311750 d 2

4d 5250 3d 7d 5250 d 750 If d 750, then 1750 d 1000; therefore, one person receives $750, and the other person receives $1000.

▼ PRACTICE YOUR SKILL A banker is told to deposit a $1500.00 paycheck into two accounts in the ratio of 2:3. How much does each account receive? ■

3 Use a Proportion to Convert a Fraction to a Percent The word percent means per one hundred, and we use the symbol % to express it. 7 For example, we write 7 percent as 7%, which means , or 0.07. In other words, 100 percent is a special kind of ratio—a ratio in which the denominator is always 100. Proportions provide a convenient basis for changing common fractions to percents. Consider the following examples.

144

Chapter 3


EXAMPLE 7

Express

7 as a percent. 20

Solution We are asking, “What number compares to 100 as 7 compares to 20?” Therefore, if we let n represent that number, we can set up the proportion like this: n 7 100 20 20n 700


n 35 Thus

35 7 35% . 20 100


EXAMPLE 8

Express

13 as a percent. 25

Express

5 as a percent. 6

■

Solution n 5 100 6 6n 500


500 250 1 n 83 6 3 3 Therefore

1 5 83 % . 6 3

▼ PRACTICE YOUR SKILL Express

7 as a percent. 12

■

4 Solve Basic Percent Problems What is 8% of 35? Fifteen percent of what number is 24? Twenty-one is what percent of 70? These are the three basic types of percent problems. Each of these problems can be solved easily by translating it into, and solving, a simple algebraic equation.

EXAMPLE 9

Apply Your Skill What is 8% of 35?

Solution Let n represent the number to be found. The word “is” refers to equality, and the word “of” implies multiplication. Thus the question translates into n (8%)(35)

3.1


145

which can be solved as follows: n 10.0821352 2.8 Therefore 2.8 is 8% of 35.


What is 12% of 58?

EXAMPLE 10

Apply Your Skill Fifteen percent of what number is 24?

Solution Let n represent the number to be found. 115% 21n2 1242 0.15n 24 15n 2400

Multiplied both sides by 100

n 160 Therefore 15% of 160 is 24.

▼ PRACTICE YOUR SKILL Eight percent of what number is 56?

EXAMPLE 11

■

Apply Your Skill Twenty-one is what percent of 70?

Solution Let r represent the percent to be found. 21 r 1702 21 r 70 3 r 10

Reduce!

30 r 100

Changed

3 30 to 10 100

30% r Therefore 21 is 30% of 70.

▼ PRACTICE YOUR SKILL Thirty-two is what percent of 80?

■

146

Chapter 3


EXAMPLE 12

Apply Your Skill Seventy-two is what percent of 60?

Solution Let r represent the percent to be found. 72 r 1602 72 r 60 6 r 5 120 r 100 120% r

Changed

6 120 to 5 100

Therefore, 72 is 120% of 60.


Sixty is what percent of 40?

Get into the habit of checking answers for reasonableness. We also suggest that you alert yourself to a potential computational error by estimating the answer before you actually do the problem. For example, prior to doing Example 12, you may have estimated: “Because 72 is larger than 60, the answer has to be greater than 100%. Furthermore 1.5 (or 150%) times 60 equals 90.” Therefore, you can estimate the answer to be somewhere between 100% and 150%. That may seem rather broad, but such an estimate will often detect a computational error.

CONCEPT QUIZ

For Problems 1– 8, answer true or false. 1. A ratio is the comparison of two numbers by division. 2. The ratio of 7 to 3 can be written 3:7. 3. A proportion is a statement of equality between two ratios. y x 4. For the proportion , the cross product would be 5x 3y. 3 5 w w 5. The algebraic statement 1 is a proportion. 2 5 6. The word “percent” means parts per one thousand. 7. One hundred twenty percent of 30 is 24. 8. If the cross product of a proportion is wx yz, then

y x . z w

Problem Set 3.1 1 Solve Proportions

x 5 3 2 x2 x4 7. 4 3 x1 x2 9. 6 4 h h 11. 1 2 3 5.

For Problems 1–32, solve each equation. Note: not all equations are proportions. 1.

x 3 6 2

2.

5 x 9 3

3.

n 5 12 24

4.

n 7 8 16

4 x 7 3 x9 x6 8. 7 8 x6 x2 10. 6 8 h h 12. 2 5 4 6.

3.1

13.

x1 x2 4 3 2

14.

x2 x3 4 5 6

15.

4 3 x2 x7

16.

9 8 x1 x5

17.

1 5 x7 x1

18.

3 2 x 10 x6

19.

3 2 2x 1 3x 2

20.

1 2 4x 3 5x 3

21.

n1 8 n 7

22.

23.

x1 3 1 2 4

24. 2

25. 3

x4 3 5 2


147

10 gallons water

5 n 6 n1 x3 5 4 6

26.

x5 5 2 3 9

27.

n 1 150 n 2

28.

n 3 200 n 5

29.

300 n 3 n 2

30.

80 n 7 n 9

31.

1 2 5x 1 3x 7

32.

3 4 2x 5 x3

Figure 3.3

At this rate, how many pounds of salt are needed for 25 gallons of water? 40. A home valued at $150,000 is assessed $2700 in real estate taxes. At the same rate, what would the taxes be on a home valued at $175,000? 41. If 20 pounds of fertilizer will cover 1500 square feet of lawn, how many pounds are needed for 2500 square feet?

For Problems 33 –50, solve each problem by using a proportion.

42. It was reported that a flu epidemic is affecting 6 out of every 10 college students in a certain university. At this rate, how many students will be affected if there are 15,000 students in the university?

33. A house plan has a scale in which 1 inch represents 6 feet. Find the dimensions of a rectangular room that mea1 1 sures 2 inches by 3 inches on the house plan. 2 4

43. A preelection poll indicated that three out of every seven eligible voters were going to vote in an upcoming election. At this rate, how many people are expected to vote in a city of 210,000 eligible voters?

34. On a certain map, 1 inch represents 15 miles. If two cities are 7 inches apart on the map, find the number of miles between the cities.

44. A board 28 feet long is cut into two pieces, and the lengths of the two pieces are in the ratio of 2 to 5. Find the lengths of the two pieces.

35. Suppose that a car can travel 264 miles using 12 gallons of gasoline. How far will it go on 15 gallons?

45. In a nutrition plan, the ratio of calories to grams of carbohydrates is 16 to 1. According to this ratio, how many grams of carbohydrates would be in a plan that has 2200 calories?

2 Solve Word Problems Using Proportions

36. Jesse used 10 gallons of gasoline to drive 170 miles. How much gasoline will he need to travel 238 miles? 37. If the ratio of the length of a rectangle to its width is

5 , and the width is 24 centimeters, find its length. 2 38. If the ratio of the width of a rectangle to its length is

4 , and the length is 45 centimeters, find the width. 5 39. A saltwater solution is made by dissolving 3 pounds of salt in 10 gallons of water (see Figure 3.3).

46. The ratio of male students to female students at a certain university is 5 to 4. If there is a total of 6975 students, find the number of male students and the number of female students. 47. An investment of $500 earns $45 in a year. At the same rate, how much additional money must be invested to raise the earnings to $72 per year? 48. A sum of $1250 is to be divided between two people in the ratio of 2 to 3. How much does each person receive?

148

Chapter 3


49. An inheritance of $180,000 is to be divided between a child and the local cancer fund in the ratio of 5 to 1. How much money will the child receive? 50. Fuel for a two-stroke motor consists of oil and gasoline mixed in the ratio of 1 part oil to 50 parts gasoline. How many ounces of oil should be added to 6 gallons of gasoline? Round your answer to the nearest ounce. A gallon of gasoline is equivalent to 128 ounces.

51. 54. 57. 60.

52. 55. 58. 61.

53. 56. 59. 62.

63. What is 7% of 38? 64. What is 35% of 52? 65. 15% of what number is 6.3?

67. 76 is what percent of 95?

For Problems 51– 62, use proportions to change each common fraction to a percent. 17 20 1 6 1 16 12 5

For Problems 63 –74, answer the question by setting up and solving an appropriate equation.

66. 55% of what number is 38.5?

3 Use a Proportion to Convert a Fraction to a Percent

11 20 7 25 3 8 5 4

4 Solve Basic Percent Problems

68. 72 is what percent of 120? 69. What is 120% of 50?

3 5 5 7 3 2 13 6

70. What is 160% of 70? 71. 46 is what percent of 40? 72. 26 is what percent of 20? 73. 160% of what number is 144? 74. 220% of what number is 66?

THOUGHTS INTO WORDS 75. Explain the difference between a ratio and a proportion. 76. What is wrong with this solution?

6a

77. Estimate an answer for each of the following problems, and explain how you arrived at your estimate. Then work out the problem to see how well you estimated.

x x 4 2 6

a.

x 4b 21x2 2

b.

3x 24 2x x 24

c.

Explain how it should be solved. d.

The ratio of female students to male students at a small private college is 5 to 3. If there is a total of 1096 students, find the number of male students. If 15 pounds of fertilizer will cover 1200 square feet of lawn, how many pounds are needed for 3000 square feet? An investment of $5000 earns $300 interest in a year. At the same rate, how much money must be invested to earn $450? If the ratio of the length of a rectangle to its width is 5 to 3, and the length is 70 centimeters, find its width.

FURTHER INVESTIGATIONS Solve each of the following equations. Don’t forget that division by zero is undefined.

78.

3 6 x2 2x 4

79.

8 4 2x 1 x3

80.

5 10 x3 x6

81.

6 5 x1 x1

82.

x2 x 1 2 2

83.

3 x3 1 x x

Answers to the Example Practice Skills 1. {15} 11. 40%

2. {1} 3. e

13 f 5

4. {4} 5. 39 feet

6. $600, $900

7. 52%

1 8. 58 % 3

12. 150%


2. False

3. True 4. True 5. False 6. False

7. False 8. True

9. 6.96

10. 700

3.2

3.2

More on Percents and Problem Solving

149

More on Percents and Problem Solving OBJECTIVES 1

Solve Equations Involving Decimals

2

Solve Word Problems Involving Discount, Selling Price, Cost, or Profit

3

Solve Simple Interest Problems

1 Solve Equations Involving Decimals We can solve the equation x 0.35 0.72 by subtracting 0.35 from both sides of the equation. Another technique for solving equations that contain decimals is to clear the equation of all decimals by multiplying both sides by an appropriate power of 10. The following examples demonstrate the use of that strategy in a variety of situations.

EXAMPLE 1

Solve 0.5x 14.

Solution 0.5x 14 5x 140 x 28

Multiplied both sides by 10 Divided both sides by 5


▼ PRACTICE YOUR SKILL Solve 0.4x 36.

EXAMPLE 2

■

Solve x 0.07x 0.13.

Solution x 0.07x 0.13 1001x 0.07x2 10010.132 1001x2 10010.07x2 10010.132

Multiply both sides by 100 Distributive property

100x 7x 13 107x 13 x The solution set is e

13 107

13 f. 107

▼ PRACTICE YOUR SKILL Solve 3x 0.2x 128.

■

150

Chapter 3


EXAMPLE 3

Solve 0.08y 0.09y 3.4.

Solution 0.08y 0.09y 3.4 8y 9y 340 17y 340 y 20



▼ PRACTICE YOUR SKILL Solve 0.04a 0.07a 16.5.

EXAMPLE 4

■

Solve 0.10t 560 0.121t 10002 .

Solution 0.10t 560 0.121t 10002 10t 56,000 121t 10002


10t 56,000 12t 12,000


22t 44,000 t 2000


▼ PRACTICE YOUR SKILL Solve 0.08y 0.15(y 200) 99.

■

2 Solve Word Problems Involving Discount, Selling Price, Cost, or Profit We can solve many consumer problems with an equation approach. For example, here is a general guideline regarding discount sales: Original selling price Discount Discount sale price Let’s work some examples using our algebraic techniques along with this basic guideline.

Brand X Pictures/Jupiter Images

EXAMPLE 5

Apply Your Skill Amy bought a dress at a 30% discount sale for $35. What was the original price of the dress?

Solution Let p represent the original price of the dress. We can use the basic discount guideline to set up an algebraic equation: Original selling price Discount Discount sale price

(100%)( p)

(30%)(p)

$35

3.2


151

Solving this equation, we get 1100% 21p2 130% 2 1p2 35 170% 2 1p2 35 0.7p 35 7p 350 p 50 The original price of the dress was $50.

▼ PRACTICE YOUR SKILL Shawna bought a notebook computer at a 20% off sales event for $640. What was the original price of the notebook computer? ■ Don’t forget that if an item is on sale for 30% off, then you are going to pay 100% 30% 70% of the original price. Thus at a 30% off sales event, you can buy a $50 dress for 170%21$502 $35. (Note that we just checked our answer for Example 5.)

EXAMPLE 6

Apply Your Skill Find the cost of a $60 pair of jogging shoes on sale for 20% off.

Doug Cannell/istockphoto.com

Solution Let x represent the discount sale price. Because the shoes are on sale for 20% off, we must pay 80% of the original price. x 180%2 1602

10.82 1602 48

The sale price is $48.

▼ PRACTICE YOUR SKILL Find the cost of a $140 pair of sunglasses on sale for 30% off.

■

Here is another equation that is useful for solving consumer problems: Selling price Cost Profit Profit (also called markup, markon, margin, and margin of profit) may be stated in different ways: as a percent of the selling price, as a percent of the cost, or simply in terms of dollars and cents. Let’s consider some problems where the profit is either a percent of the selling price or a percent of the cost.

EXAMPLE 7

Apply Your Skill

John A. Rizzo/Photolibrary

A retailer has some shirts that cost him $20 each. He wants to sell them at a profit of 60% of the cost. What should the selling price be on the shirts?

Solution Let s represent the selling price. The basic relationship selling price equals cost plus profit can be used as a guideline: Selling price Cost Profit (% of cost)

s

$20

(60%)(20)

152

Chapter 3


Solving this equation, we obtain s 20 160% 21202 20 10.621202 20 12 32 The selling price should be $32.

▼ PRACTICE YOUR SKILL A tire dealer has some alloy wheels that cost him $450. The dealer wants to sell the wheels at a profit of 80% of the cost. What selling price should he mark on the wheels? ■

Frances M. Roberts/Digital Railroad

EXAMPLE 8

Apply Your Skill Kathrin bought a painting for $120 and later decided to resell it. She made a profit of 40% of the selling price. How much money did she receive for the painting?

Solution We can use the same basic relationship as a guideline, except this time the profit is a percent of the selling price. Let s represent the selling price. Selling price Cost Profit (% of selling price)

120

s

(40%)(s)

We can solve this equation: s 120 140% 21s2 s 120 0.4s 0.6s 120 s

Subtracted 0.4s from both sides

120 200 0.6

She received $200 for the painting.

▼ PRACTICE YOUR SKILL Carlos bought an autographed NASCAR photo for $80. He will sell the photo only if he makes a profit of 60% of the selling price. What will the selling price have to be for Carlos to sell the photo? ■

3 Solve Simple Interest Problems We can also translate certain types of investment problems into algebraic equations. In some of these problems, we use the simple interest formula i Prt, where i represents the amount of interest earned by investing P dollars at a yearly rate of r percent for t years.

3.2

EXAMPLE 9


153

Apply Your Skill John invested $9300 for 2 years and received $1395 in interest. Find the annual interest rate John received on his investment.

Solution i Prt 1395 9300r122 1395 18600r 1395 r 18600 0.075 r The annual interest rate is 7.5%.

▼ PRACTICE YOUR SKILL After three years, Maribel received $1620 in interest on her $10,000 investment. Find the annual interest rate Maribel received on her investment. ■

EXAMPLE 10

Apply Your Skill How much principal must be invested to receive $1500 in interest when the investment is made for 3 years at an annual interest rate of 6.25%?

Solution i Prt

1500 P10.06252 132 1500 P10.18752 1500 P 0.1875 8000 P The principal must be $8000.

▼ PRACTICE YOUR SKILL How much principal must be invested for 2 years to receive $715 in interest when the annual interest rate is 6.5%? ■

EXAMPLE 11

Apply Your Skill How much monthly interest will be charged on a credit card bill with a balance of $754 when the credit card company charges an 18% annual interest rate?

Solution i Prt i 75410.182 a

1 b 12

Remember, 1 month is

i 11.31 The interest charge would be $11.31.

1 of a year 12

154

Chapter 3


▼ PRACTICE YOUR SKILL How much monthly interest will be charged on a student loan with a balance of $14,000 when the loan company charges 6% annual interest rate? ■

CONCEPT QUIZ

For Problems 1–5, answer true or false. 1. To clear the decimals from the equation, 0.5x 1.24 0.07x 1.8, you would multiply both sides of the equation by 10. 2. If an item is on sale for 35% off, then you are going to pay 65% of the original price. 3. Profit is always a percent of the selling price. 4. In the formula, i Prt, the r represents the interest return. 5. The basic relationship, selling price equals cost plus profit, can be used whether the profit is based on selling price or cost.

Problem Set 3.2 1 Solve Equations Involving Decimals For Problems 1–22, solve each equation. 1. x 0.36 0.75

2. x 0.15 0.42

3. x 7.6 14.2

4. x 11.8 17.1

5. 0.62 y 0.14

6. 7.4 y 2.2

7. 0.7t 56

8. 1.3t 39

9. x 3.36 0.12x

10. x 5.3 0.06x

11. s 35 0.3s

12. s 40 0.5s

13. s 42 0.4s

14. s 24 0.6s

15. 0.07x 0.081x 6002 78 16. 0.06x 0.091x 2002 63 17. 0.09x 0.112x2 130.5 18. 0.11x 0.1213x2 188 19. 0.08x 0.111500 x2 50.5 20. 0.07x 0.0912000 x2 164 21. 0.09x 550 0.1115400 x2 22. 0.08x 580 0.116000 x2

2 Solve Word Problems Involving Discount, Selling Price, Cost, or Profit For Problems 23 –38, set up an equation and solve each problem.

25. Find the cost of a $4800 widescreen television that is on sale for 25% off. 26. Byron purchased a computer monitor at a 10% discount sale for $121.50. What was the original price of the monitor? 27. Suppose that Jack bought a $32 putter on sale for 35% off. How much did he pay for the putter? 28. Mindy bought a 13-inch portable color TV for 20% off the list price. The list price was $149.95. What did she pay for the TV? 29. Pierre paid $126 for a coat that was listed for $180. What rate of discount did he receive? 30. Phoebe paid $32 for a pair of sandals that were listed for $40. What rate of discount did she receive? 31. A retailer has some toe rings that cost him $5 each. He wants to sell them at a profit of 70% of the cost. What should be the selling price of the toe rings? 32. A retailer has some video games that cost her $25 each. She wants to sell them at a profit of 80% of the cost. What price should she charge for the games? 33. The owner of a pizza parlor wants to make a profit of 55% of the cost for each pizza sold. If it costs $8 to make a pizza, at what price should it be sold? 34. Produce in a food market usually has a high markup because of loss due to spoilage. If a head of lettuce costs a retailer $0.40, at what price should it be sold to realize a profit of 130% of the cost?

23. Tom bought an electric drill at a 30% discount sale for $35. What was the original price of the drill?

35. Jewelry has a very high markup rate. If a ring costs a jeweler $400, what should its price be to yield a profit of 60% of the selling price?

24. Magda bought a dress for $140, which represents a 20% discount off the original price. What was the original price of the dress?

36. If a box of candy costs a retailer $2.50, and he wants to make a profit of 50% based on the selling price, what price should he charge for the candy?

3.2 37. If the cost of a pair of shoes is $32 for a retailer, and he sells them for $44.80, what is his rate of profit based on the cost? 38. A retailer has some candle sets that cost her $24. If she sells them for $31.20, find her rate of profit based on the cost.

3 Solve Simple Interest Problems


155

41. How much principal, invested at 8% annual interest for 3 years, is needed to earn $1000? 42. How long will $2400 need to be invested at a 5.5% annual interest rate to earn $330? 43. What will be the interest earned on a $5000 certificate of deposit invested at 6.8% annual interest for 10 years?

For Problems 39 – 46, use the formula i Prt to reach a solution.

44. One month a credit card company charged $38.15 in interest on a balance of $2725. What annual interest rate is the credit card company charging?

39. Find the annual interest rate if $560 in interest is earned when $3500 is invested for 2 years.

45. How much is a month’s interest on a mortgage balance of $95,000 at an 8% annual interest rate?

40. How much interest will be charged on a student loan if $8000 is borrowed for 9 months at a 19.2% annual interest rate?

46. For how many years must $2000 be invested at a 5.4% annual interest rate to earn $162?

THOUGHTS INTO WORDS 48. From a consumer’s standpoint, would you prefer that a retailer figure his profit on the basis of cost or the selling price of an item? Explain your answer.

47. What is wrong with this solution? 1.2x 2 3.8 1011.2x2 2 1013.82 12x 2 38 12x 36 x3 How should it be solved?

FURTHER INVESTIGATIONS 49. A retailer buys an item for $40, resells it for $50, and claims that he is making only a 20% profit. Is his claim correct? 50. A store has a special discount sale of 40% off on all items. It also advertises an additional 10% off on items bought in quantities of a dozen or more. How much will it cost to buy a dozen of some particular item that regularly sells for $5 each? (Be careful: a 40% discount followed by a 10% discount is not equal to a 50% discount.) 51. Is a 10% discount followed by a 40% discount the same as a 40% discount followed by a 10% discount? Justify your answer. 52. Some people use the following formula to determine the selling price of an item when the profit is based on a percent of the selling price.

Selling price

Cost 100% Percent of profit

Show how this formula is developed. For Problems 53 – 60, solve each equation, and express the solution in decimal form. Your calculator might be helpful. 53. 2.4x 5.7 9.6 54. 3.2x 1.6 5.8 55. 0.08x 0.091800 x2 68.5 56. 0.10x 0.121720 x2 80 57. 7x 0.39 0.03

58. 9x 0.37 0.35

59. 0.21t 1.62 3.4

60. 0.41t 3.82 2.2


2. {40}

3. {150}

4. {300}

5. $800

6. $98



5. True

7. $810

8. $200

9. 5.4%

10. $5500

11. $70

156

Chapter 3

3.3


Formulas: Geometric and Others OBJECTIVES 1

Evaluate Formulas

2

Apply Geometric Formulas

3

Solve Formulas for a Specific Variable

4

Solve an Equation for a Specific Variable

1 Evaluate Formulas To find the distance traveled in 3 hours at a rate of 50 miles per hour, we multiply the rate by the time. Thus the distance is 50(3) 150 miles. We usually state the rule distance equals rate times time as the formula d rt. Formulas are simply rules that we state in symbolic language and express as equations. Thus the formula d rt is an equation that involves three variables: d, r, and t. As we work with formulas, it is often necessary to solve for a specific variable when we know numerical values for the remaining variables. Consider the next examples.

EXAMPLE 1

Solve d rt for r if d 330 and t 6.

Solution Substitute 330 for d and 6 for t in the given formula: 330 r162 Solving this equation yields 330 6r 55 r

▼ PRACTICE YOUR SKILL Solve d rt for t if d 150 and r 60.

EXAMPLE 2

■

5 1F 322 for F if C 10. (This formula expresses the relationship be9 tween the Fahrenheit and Celsius temperature scales.) Solve C

Solution Substitute 10 for C to obtain 10

5 1F 322 9

We can solve this equation: 9 9 5 1102 a b 1F 322 5 5 9 18 F 32 50 F


9 5

3.3

Formulas: Geometric and Others

157

▼ PRACTICE YOUR SKILL 1 Solve A h1B1 182 for B1 if A 150 and h 10. 2

■

Apply Geometric Formulas

2

There are several formulas in geometry that we use quite often. We review them below; we will use them periodically throughout the remainder of the text. These formulas (along with some others) and Figures 3.4 through 3.14 are also shown in the inside front cover of this text. Rectangle

Triangle

w

h

l

A P l w

b A = 1 bh 2 A area b base h altitude (height)

P = 2l + 2w

A=lw

area perimeter length width

Figure 3.5

Figure 3.4

Parallelogram

Trapezoid b1 h

h

b

b2 A = bh

A = 1 h(b1 + b2 ) 2 A area b1, b2 bases h altitude

A area b base h altitude (height) Figure 3.7

Figure 3.6

Prism Circle

Sphere

r

r h base

A = πr2

C = 2πr

A area C circumference r radius Figure 3.8

V = 4 π r 3 S = 4π r 2 3 S surface area V volume r radius Figure 3.9

V = Bh V volume B area of base h altitude (height) Figure 3.10

158

Chapter 3

Ratio, Percent, Formulas, and Problem Solving Rectangular Prism

Pyramid l h h

w V = lwh V S w l h

base

S = 2hw + 2 hl + 2lw

volume total surface area width length altitude (height)

V = 1 Bh 3 V volume B area of base h altitude (height) Figure 3.12

Figure 3.11

Right Circular Cylinder

Right Circular Cone

r

s

h h r V = πr 2h V S r h

S = 2 πr 2 + 2 π r h

volume total surface area radius altitude ( height)

Figure 3.14

Figure 3.13

Christina Richards/Shutterstock

EXAMPLE 3

V = 1 πr 2h S = πr 2 + πrs 3 V volume S total surface area r radius h altitude (height) s slant height

Apply Your Skill Mark wants to construct a garden plot in the shape of a triangle. If the length of the altitude is 18 feet, what must the length of the base of the triangle be if he wants the garden to be 720 square feet in area?

Solution We substitute 18 for h and 720 for A in the formula to find the area of a triangle. 1 A bh 2 720

1 b1182 2

720 9b 80 b The length of the base should be 80 feet.

3.3


159

▼ PRACTICE YOUR SKILL Find the height of a triangle whose area is 52 square feet and whose base is 16 feet in length. ■

EXAMPLE 4

Apply Your Skill

Lisa F. Young/Shutterstock

A contractor needs to determine the area of a foyer in a church that is in the shape of a trapezoid. The trapezoid has an altitude of length 25 feet, the length of one base is 30 feet, and the length of the other base is 42 feet. Find the area of the foyer.

Solution We substitute 25 for h, 30 for b1, and 42 for b2 in the formula for finding the area of a trapezoid. 1 A h1b1 b2 2 2 A

1 1252 130 422 2

A

1 1252 1722 2

A 900 The area of the foyer is 900 square feet.

▼ PRACTICE YOUR SKILL John is building a deck that is in the shape of a trapezoid. Find the area of the deck if the trapezoid has a height whose length is 10 feet and bases with lengths of 12 feet and 16 feet, respectively. ■

EXAMPLE 5

Find the total surface area of a right circular cylinder that has a radius of 10 inches and a height of 14 inches.

Solution 10 inches

Let’s sketch a right circular cylinder and record the given information as in Figure 3.15. We substitute 10 for r and 14 for h in the formula for finding the total surface area of a right circular cylinder. 14 inches

S 2pr2 2prh

2p1102 2 2p1102 1142 200p 280p

Figure 3.15

480p The total surface area is 480p square inches.

▼ PRACTICE YOUR SKILL Find the total surface area of a right circular cylinder that has a radius of 15 feet and a height of 24 feet. ■ In Example 5 we used Figure 3.15 to record the given information; you can also consult Figure 3.13 for the geometric formula. Now let’s consider an example where a figure is very useful in the analysis of the problem.

160

Chapter 3


EXAMPLE 6

A sidewalk 3 feet wide surrounds a rectangular plot of ground that measures 75 feet by 100 feet. Find the area of the sidewalk.

Solution Let’s make a sketch and record the given information as in Figure 3.16. 3 feet

3 feet

75 feet

100 feet

3 feet

Figure 3.16

We can find the area of the sidewalk by subtracting the area of the rectangular plot from the area of the plot plus the sidewalk (the large dashed rectangle). The width of the large rectangle is 75 3 3 81 feet, and its length is 100 3 3 106 feet, so A 1812 11062 175211002 8586 7500 1086 The area of the sidewalk is 1086 square feet.

▼ PRACTICE YOUR SKILL A deck 4 feet wide surrounds a rectangular swimming pool that measures 20 feet by 30 feet. Find the area of the deck. ■

3 Solve Formulas for a Specific Variable Sometimes it is convenient to change a formula’s form by using the properties of equality. For example, we can change the formula d rt as follows: d rt rt d r r

Divide both sides by r

d t r We say that the formula d rt “has been solved for the variable t.” The formula can also be solved for r. d rt rt d t t d r t

Divide both sides by t

3.3

EXAMPLE 7


161

Solve C 2pr for r.

Solution C 2pr C 2pr 2p 2p

Divide both sides by 2p

C r 2p

▼ PRACTICE YOUR SKILL Solve i Prt for P.

EXAMPLE 8

■

1 Solve V Bh for h. 3

Solution 1 V Bh 3 1 31V2 3 a Bhb 3


3V Bh 3V Bh B B

Divide both sides by B

3V h B

▼ PRACTICE YOUR SKILL 1 Solve A bh for b. 2

EXAMPLE 9

■

Solve P 2l 2w for w.

Solution P 2l 2w P 2l 2l 2w 2l

Subtract 2l from both sides

P 2l 2w 2w P 2l 2 2


P 2l w 2

▼ PRACTICE YOUR SKILL Solve S 2pr 2 2prh for h.

■

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Chapter 3


4 Solve an Equation for a Specific Variable In Chapter 4 you will be working with equations that contain two variables. At times you will need to solve for one variable in terms of the other variable—that is, to change the form of the equation as we have been doing with formulas. The next examples illustrate how we can use the properties of equality for such situations.

EXAMPLE 10

Solve 3x y 4 for x .

Solution 3x y 4 3x y y 4 y

Subtract y from both sides

3x 4 y 4y 3x 3 3 x


4y 3

▼ PRACTICE YOUR SKILL Solve 2x y 3 for x.

EXAMPLE 11

■

Solve 4x 5y 7 for y.

Solution 4x 5y 7 4x 5y 4x 7 4x


5y 7 4x 5y 7 4x 5 5


y

7 4x 1 a b 5 1

Multiply numerator and denominator of fraction on the right by 1

y

4x 7 5

We commonly do this so that the denominator is positive

▼ PRACTICE YOUR SKILL Solve 3x 8y 5 for y.

EXAMPLE 12

■

Solve y mx b for m.

Solution y mx b y b mx b b y b mx

Subtract b from both sides

3.3

yb mx x x yb m x


163

Divide both sides by x

▼ PRACTICE YOUR SKILL Solve y ax b for a.

CONCEPT QUIZ

■

For Problems 1–10, match the correct formula for each. 1. Area of a rectangle

A. A pr 2

2. Circumference of a circle

B. V lwh

3. Volume of a rectangular prism

C. P 2l 2w

4. Area of a triangle

4 D. V pr3 3

5. Area of a circle

E. A lw

6. Volume of a right circular cylinder

F.

7. Perimeter of a rectangle

G. A

8. Volume of a sphere 9. Area of a parallelogram 10. Area of a trapezoid

A bh

1 h1b1 b2 2 2 1 H. A bh 2 I.

C 2pr

J.

V pr 2h

Problem Set 3.3 1 Evaluate Formulas For Problems 1–10, solve for the specified variable using the given facts. 1. Solve d rt

for t if d 336 and r 48.

2. Solve d rt

for r if d 486 and t 9.

3. Solve i Prt

for P if i 200, r 0.08, and t 5.

4. Solve i Prt for t if i 880, P 2750, and r 0.04. 5. Solve F

9 C 32 5

6. Solve C

5 1F 322 9

1 7. Solve V Bh 3 1 8. Solve V Bh 3

for C if F 68. for F if C 15.

for B if V 112 and h 7. for h if V 216 and B 54.

9. Solve A P Prt for t if A 5080, P 4000, and r 0.03. 10. Solve A P Prt for P if A 1032, r 0.06, and t 12.

2 Apply Geometric Formulas For Problems 11–32, use the geometric formulas given in this section to help find solutions. 11. Find the perimeter of a rectangle that is 14 centimeters long and 9 centimeters wide. 12. If the perimeter of a rectangle is 80 centimeters, and its length is 24 centimeters, find its width. 13. If the perimeter of a rectangle is 108 inches, and its length 1 is 3 feet, find its width in inches. 4 14. How many yards of fencing would it take to enclose a rectangular plot of ground that is 69 feet long and 42 feet wide?

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15. A dirt path 4 feet wide surrounds a rectangular garden that is 38 feet long and 17 feet wide. Find the area of the dirt path. 16. Find the area of a cement walk 3 feet wide that surrounds a rectangular plot of ground 86 feet long and 42 feet wide. 17. Suppose that paint costs $6.00 per liter and that 1 liter will cover 9 square meters of surface. We are going to paint (on one side only) 50 rectangular pieces of wood of the same size, which have a length of 60 centimeters and a width of 30 centimeters. What will be the total cost of the paint? 18. A lawn is in the shape of a triangle with one side 130 feet long and the altitude to that side 60 feet long. Will one sack of fertilizer, which covers 4000 square feet, be enough to fertilize the lawn? 19. Find the length of an altitude of a trapezoid with bases of 8 inches and 20 inches and an area of 98 square inches. 20. A flower garden is in the shape of a trapezoid with bases of 6 yards and 10 yards. The distance between the bases is 4 yards. Find the area of the garden. 21. The diameter of a metal washer is 4 centimeters, and the diameter of the hole is 2 centimeters (see Figure 3.17). How many square centimeters of metal are there in 50 washers? Express the answer in terms of p.

4 cm 2 cm

34 feet

Figure 3.18 27. Find the volume and total surface area of a right circular cylinder that has a radius of 8 feet and a height of 18 feet. Express answers in terms of p. 28. Find the total surface area and volume of a sphere that has a diameter of 12 centimeters. Express the answers in terms of p. 29. If the volume of a right circular cone is 324p cubic inches, and a radius of the base is 9 inches long, find the height of the cone. 30. Find the volume and total surface area of a tin can if the radius of the base is 3 centimeters, and the height of the can is 10 centimeters. Express answers in terms of p. 31. If the total surface area of a right circular cone is 65p square feet, and a radius of the base is 5 feet long, find the slant height of the cone. 32. If the total surface area of a right circular cylinder is 104p square meters, and a radius of the base is 4 meters long, find the height of the cylinder.

3 Solve Formulas for a Specific Variable Figure 3.17

22. Find the area of a circular plot of ground that has a 1 radius of length 14 meters. Use 3 as an approximation 7 for p.

For Problems 33 – 44, solve each formula for the indicated variable. (Before doing these problems, cover the right-hand column and see how many of the formulas you recognize!) 33. V Bh for h

Volume of a prism

34. A lw for l

Area of a rectangle

23. Find the area of a circular region that has a diameter of 1 yard. Express the answer in terms of p.

1 35. V Bh for B 3

24. Find the area of a circular region if the circumference is 12p units. Express the answer in terms of p.

36. A

25. Find the total surface area and the volume of a sphere that has a radius 9 inches long. Express the answers in terms of p.

37. P 2l 2w for w Perimeter of a rectangle

26. A circular pool is 34 feet in diameter and has a flagstone walk around it that is 3 feet wide (see Figure 3.18). Find the area of the walk. Express the answer in terms of p.

39. V

1 bh 2

Volume of a pyramid

for h

Area of a triangle

38. V pr 2h for h

Volume of a cylinder

1 2 pr h for h 3

40. i Prt for t

Volume of a cone

Simple interest formula

3.3

41. F

9 C 32 for C Celsius to Fahrenheit 5

42. A P Prt for t Simple interest formula 43. A 2pr 2 2prh for h 5 44. C 1F 322 9

for F

51. y 3x 4 for x 52. y 7x 10 for x 53.

y3 x2 4 6

for y

54.

y5 x1 3 2

for y

Fahrenheit to Celsius

For Problems 45 – 60, solve each equation for the indicated variable. 45. 3x 7y 9 for x 46. 5x 2y 12 for x

55. ax by c 0 for y 56. ax by c for y 57.

y4 x6 2 5

for x

58.

y4 x3 6 8

for x

47. 9x 6y 13 for y 48. 3x 5y 19 for y 49. 2x 11y 14 for x

165

50. x 14y 17 for x

Surface area of a cylinder

4 Solve an Equation for a Specific Variable


59. m

yb x

for y

60. y mx b for x

THOUGHTS INTO WORDS 61. Suppose that both the length and width of a rectangle are doubled. How does this affect the perimeter of the rectangle? Defend your answer.

63. To estimate the change from a Fahrenheit temperature reading to a Celsius reading, some people subtract 32 and then divide by 2. How good is this estimate?

62. Suppose that the length of the radius of a circle is doubled. How does this affect the area of the circle? Defend your answer.

FURTHER INVESTIGATIONS For each of the following problems, use 3.14 as an approximation for p. Your calculator should be helpful with these problems.

66. Find, to the nearest square inch, the area of each of these pizzas: 10-inch diameter, 12-inch diameter, and 14-inch diameter.

64. Find the area of a circular plot of ground that has a radius 16.3 meters long. Express your answer to the nearest tenth of a square meter.

67. Find, to the nearest square centimeter, the total surface area of the tin can in Figure 3.20.

65. Find, to the nearest tenth of a square centimeter, the area of the colored ring in Figure 3.19.

3 centimeters

7 centimeters

3 centimeters

Figure 3.19

10 centimeters

Figure 3.20

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Chapter 3

Ratio, Percent, Formulas, and Problem Solving 70. Find, to the nearest cubic meter, the volume of the figure shown in Figure 3.23.

68. Find, to the nearest square centimeter, the total surface area of a baseball that has a radius of 4 centimeters (see Figure 3.21).

8 meters 4 centimeters

20 meters

Figure 3.21 69. Find, to the nearest cubic inch, the volume of a softball that has a diameter of 5 inches (see Figure 3.22).

12 meters Figure 3.23

5-inch diameter

Figure 3.22

Answers to the Example Practice Skills

i 2. 12 3. 6.5 feet 4. 140 square feet 5. 1170p square feet 6. 464 square feet 7. P rt yb 3 y 3x 5 2A S 2pr2 8. b 9. h 10. x 11. y 12. a x h 2pr 2 8 1. 2.5

Answers to the Concept Quiz 1. E

3.4

2. I

3. B 4. H 5. A

6. J

7. C

8. D

9. F

10. G

Problem Solving OBJECTIVES 1

Solve Equations

2

Solve Word Problems Involving Simple Interest

3

Solve Word Problems Involving Geometry

4

Solve Word Problems Involving Motion

1 Solve Equations Before we delve into word problems, let’s take a look at solving equations. The next example shows specific types of equations that will often show up in word problems.

3.4

EXAMPLE 1

Problem Solving

Solve: 3 (a) y y 2 48 4 (b) n (n 1) (2n 3) 124 (c)

2 3 x 1x 32 6 7 7

Solution (a)

3 y y 2 48 4 3 4 y y 2 48 4 4

Change 1 to

4 4

7 y 2 48 4 7 y 50 4

Add 2 to each side

4 4 7 a yb 1502 7 4 7 y

200 7


200 f. 7

(b) n (n 1) (2n 3) 124 n n 1 2n 3 124 4n 4 124 4n 128 n 32

Add 4 to each side Divide both sides by 4

The solution set is {32}. (c)

3 2 x 1x 32 6 7 7 3 2 7 c x 1x 32 d 7162 7 7 3 2 7 a xb 7 c 1x 32 d 7162 7 7



2 3x 7 a b 1x 32 42 7 3x 21x 32 42 3x 2x 6 42 5x 6 42 5x 36 x The solution set is e

36 f. 5

36 5

2 7 Distributive property Multiply 7 times

167

168

Chapter 3


▼ PRACTICE YOUR SKILL Solve each of the following equations: (a)

2 b b 4 39 5

(c)

2 1 c 1c 42 15 3 3

(b) n (n 1) (n 2) 153 ■

Let’s restate the suggestions for solving word problems that we offered in Section 2.4.

Suggestions for Solving Word Problems 1. Read the problem carefully, and make sure that you understand the meanings of all the words. Be especially alert for any technical terms used in the statement of the problem. 2. Read the problem a second time (perhaps even a third time) to get an overview of the situation being described and to determine the known facts as well as what is to be found. 3. Sketch any figure, diagram, or chart that might be helpful in analyzing the problem. 4. Choose a meaningful variable to represent an unknown quantity in the problem (perhaps t if time is the unknown quantity); represent any other unknowns in terms of that variable. 5. Look for a guideline that can be used to set up an equation. A guideline might be a formula, such as selling price equals cost plus profit, or a relationship, such as interest earned from a 9% investment plus interest earned from a 10% investment equals total amount of interest earned. A guideline may also be illustrated by a figure or diagram that you sketch for a particular problem. 6. Form an equation that contains the variable that translates the conditions of the guideline from English into algebra. 7. Solve the equation, and use the solution to determine all the facts requested in the problem. 8. Check all answers back in the original statement of the problem.

We emphasize the importance of suggestion 5. Determining the guideline to follow when setting up the equation is key to analyzing the problem. Sometimes the guideline is a formula—such as one of the formulas we presented in the previous section and accompanying problem set. Let’s consider a problem of that type.

2 Solve Word Problems Involving Simple Interest EXAMPLE 2

Apply Your Skill How long will it take $500 to double itself if it is invested at 8% simple interest?

Solution Let’s use the basic simple interest formula, i Prt, where i represents interest, P is the principal (money invested), r is the rate (percent), and t is the time in years.

3.4

Problem Solving

169

For $500 to “double itself ” means that we want the original $500 to earn another $500 in interest. Thus using i Prt as a guideline, we can proceed as follows: i Prt

500 50018% 2 1t2 Now let’s solve this equation: 500 50010.082 1t2 1 0.08t 100 8t 100 t 8 1 12 t 2 1 It will take 12 years. 2

▼ PRACTICE YOUR SKILL How long will it take $3000 to earn $525 interest if it is invested at 10% simple interest? ■

3 Solve Word Problems Involving Geometry If the problem involves a geometric formula, then a sketch of the figure is helpful for recording the given information and analyzing the problem. Try it in the next example.

2007 Scott Boehm /Getty Images

EXAMPLE 3

Apply Your Skill The length of a football field is 40 feet more than twice its width, and the perimeter of the field is 1040 feet. Find the length and width of the field.

Solution Because the length is stated in terms of the width, we can let w represent the width, and then 2w 40 represents the length (see Figure 3.24 to see how we labeled the diagram). A guideline for this problem is the perimeter formula P 2l 2w. Thus we use the following equation to set up and solve the problem: P 2l 2w

1040 212w 402 2w 1040 4w 80 2w 1040 6w 80 960 6w 160 w If w 160, then 2w 40 211602 40 360. Thus the football field is 360 feet long and 160 feet wide.

170

Chapter 3


10 20 30 40 50 40 30 20 10

w

10 20 30 40 50 40 30 20 10

2w + 40 Figure 3.24

▼ PRACTICE YOUR SKILL The length of a rectangular computer display screen is twice its width, and the perimeter is 48 inches. Find the length and width of the screen. ■

4 Solve Word Problems Involving Motion Sometimes the formulas we use when we are analyzing a problem are different from those we use as a guideline for setting up the equation. For example, uniform-motion problems involve the formula d rt, but the main guideline for setting up an equation for such problems is usually a statement about either times, rates, or distances. Let’s consider an example.

Roderick Chen/Jupiter Images

EXAMPLE 4

Apply Your Skill Pablo leaves city A on a moped and travels toward city B at 18 miles per hour. At the same time, Cindy leaves city B on a moped and travels toward city A at 23 miles per hour. The distance between the two cities is 123 miles. How long will it take before Pablo and Cindy meet on their mopeds?

Solution First, sketch a diagram as in Figure 3.25. Then let t represent the time that Pablo travels and also the time that Cindy travels.

Pablo traveling at 18 mph

Cindy traveling at 23 mph

A

B

total of 123 miles Figure 3.25 Distance Pablo travels Distance Cindy travels Total distance

18t

Solving this equation yields 18t 23t 123 41t 123 t3 They both travel for 3 hours.

23t

123

3.4

Problem Solving

171

▼ PRACTICE YOUR SKILL Justin leaves the park and walks toward home at 5 miles per hour. At the same time, Justin’s sister Hope leaves home and heads to the park walking at 2 miles per hour. The distance between the park and home is 3.5 miles. How long will it take for Justin and Hope to meet? ■ Some people find it helpful to use a chart to organize the known and unknown facts in a uniform-motion problem. We will illustrate with the next example.

EXAMPLE 5

Apply Your Skill

Stockbyte/Getty Images

A car leaves a town traveling at 60 kilometers per hour. How long will it take a second car traveling at 75 kilometers per hour to catch the first car if the second car leaves 1 hour later and travels the same route?

Solution Let t represent the time of the second car. Then t 1 represents the time of the first car, because it travels 1 hour longer. We can now record the information of the problem in a chart.

First car Second car

Rate

Time

Distance (d rt)

60 75

t1 t

601t 12 75t

Because the second car is to overtake the first car, the distances must be equal. Distance of second car

75t

Distance of first car

60(t 1)

We can solve this equation: 75t 75t 15t t

601t 12 60t 60 60 4

The second car should overtake the first car in 4 hours. (Check the answer!)

▼ PRACTICE YOUR SKILL A student left his home to drive to the university traveling at 60 miles per hour. Fifteen minutes after he left, his mother discovered that he forgot his laptop computer; she left the house traveling at 70 miles per hour to catch up with him. How long will it take her to catch up with him? ■ We offer one bit of advice at this time. Don’t become discouraged if solving word problems is giving you trouble. Problem solving is not a skill that you can develop overnight. It takes time, patience, hard work, and an open mind. Keep giving it your best shot, and gradually you’ll become more confident in your approach to such problems. Furthermore, we realize that some (perhaps many) of these problems may not seem “practical” to you; however, keep in mind that the real goal here is to develop your skill in applying problem-solving techniques. Finding and using a guideline, sketching a figure to record information and help in the analysis, estimating an answer before attempting to solve the problem, and using a chart to record information are important skills to develop.

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Chapter 3


CONCEPT QUIZ

Arrange the following steps in the correct order to solve word problems. A. Declare a variable and represent any other unknown quantities in terms of that variable. B. Check the answer back into the original statement of the problem. C. Write an equation for the problem, and remember to look for a formula or guideline that could be used to write the equation. D. Read the problem carefully, and be sure that you understand all the terms in the stated problem. E. Sketch a diagram or figure that helps you analyze the problem. F. Solve the equation, and figure out the answer being asked for in the problem.

Problem Set 3.4 1 Solve Equations For Problems 1–12, solve each equation. These equations are of the types you will be using in Problems 13 – 40.

15. How long will it take $8000 to triple itself if it is invested at 6% simple interest? 16. How many years will it take $500 to earn $750 in interest if it is invested at 6% simple interest?

1. 95010.122 t 950 2. 120010.092 t 1200 1 3. l l 1 19 4 2 4. l l 1 41 3 5. 50010.082 t 1000 6. 80010.112 t 1600 7. s 12s 12 13s 42 37 8. s 13s 22 14s 42 42 9.

5 5 r 1r 62 135 2 2

10.

10 10 r 1r 32 90 3 3

2 11. 24 at b 18t 8 3 12. 16t 8 a

9 tb 60 2

For Problems 13 – 40, set up an equation, and solve each of the problems. Keep in mind the suggestions we offered in this section.

2 Solve Word Problems Involving Simple Interest 13. How long will it take $4000 to double itself if it is invested at 8% simple interest? 14. How many years will it take $4000 to double itself if it is invested at 5% simple interest?

3 Solve Word Problems Involving Geometry 17. The length of a rectangle is three times its width. If the perimeter of the rectangle is 112 inches, find its length and width. 18. The width of a rectangle is one-half of its length. If the perimeter of the rectangle is 54 feet, find its length and width. 19. Suppose that the length of a rectangle is 2 centimeters less than three times its width. The perimeter of the rectangle is 92 centimeters. Find the length and width of the rectangle. 20. Suppose that the length of a certain rectangle is 1 meter more than five times its width. The perimeter of the rectangle is 98 meters. Find the length and width of the rectangle. 21. The width of a rectangle is 3 inches less than one-half of its length. If the perimeter of the rectangle is 42 inches, find the area of the rectangle. 22. The width of a rectangle is 1 foot more than one-third of its length. If the perimeter of the rectangle is 74 feet, find the area of the rectangle. 23. The perimeter of a triangle is 100 feet. The longest side is 3 feet less than twice the shortest side, and the third side is 7 feet longer than the shortest side. Find the lengths of the sides of the triangle. 24. A triangular plot of ground has a perimeter of 54 yards. The longest side is twice the shortest side, and the third side is 2 yards longer than the shortest side. Find the lengths of the sides of the triangle.

3.4 25. The second side of a triangle is 1 centimeter more than three times the first side. The third side is 2 centimeters longer than the second side. If the perimeter is 46 centimeters, find the length of each side of the triangle. 26. The second side of a triangle is 3 meters less than twice the first side. The third side is 4 meters longer than the second side. If the perimeter is 58 meters, find the length of each side of the triangle. 27. The perimeter of an equilateral triangle is 4 centimeters more than the perimeter of a square, and the length of a side of the triangle is 4 centimeters more than the length of a side of the square. Find the length of a side of the equilateral triangle. (An equilateral triangle has three sides of the same length.) 28. Suppose that a square and an equilateral triangle have the same perimeter. Each side of the equilateral triangle is 6 centimeters longer than each side of the square. Find the length of each side of the square. (An equilateral triangle has three sides of the same length.) 29. Suppose that the length of a radius of a circle is the same as the length of a side of a square. If the circumference of the circle is 15.96 centimeters greater than the perimeter of the square, find the length of a radius of the circle. (Use 3.14 as an approximation for p.) 30. The circumference of a circle is 2.24 centimeters more than six times the length of a radius. Find the radius of the circle. (Use 3.14 as an approximation for p.)

4 Solve Word Problems Involving Motion 31. Sandy leaves a town traveling in her car at a rate of 45 miles per hour. One hour later, Monica leaves the same town traveling the same route at a rate of 50 miles per hour. How long will it take Monica to overtake Sandy? 32. Two cars start from the same place traveling in opposite directions. One car travels 4 miles per hour faster than the other car. Find their speeds if after 5 hours they are 520 miles apart. 33. The distance between city A and city B is 325 miles. A freight train leaves city A and travels toward city B at 40 miles per hour. At the same time, a passenger train leaves city B and travels toward city A at 90 miles per hour. How long will it take the two trains to meet? 34. Kirk starts jogging at 5 miles per hour. Half an hour later Consuela starts jogging on the same route at

Problem Solving

173

7 miles per hour. How long will it take Consuela to catch Kirk? 35. A car leaves a town at 40 miles per hour. Two hours later, a second car leaves the town traveling the same route and overtakes the first car in 5 hours and 20 minutes. How fast was the second car traveling? 36. Two airplanes leave St. Louis at the same time and fly in opposite directions. If one travels at 500 kilometers per hour and the other at 600 kilometers per hour, how long will it take for them to be 1925 kilometers apart (see Figure 3.26)? 1925 kilometers

500 kph

600 kph St. Louis Airport

Figure 3.26 37. Two trains leave from the same station at the same time, one traveling east and the other traveling west. 1 At the end of 9 hours they are 1292 miles apart. If the 2 rate of the train traveling east is 8 miles per hour faster than the other train, find their rates. 38. Dawn started on a 58-mile trip on her moped at 20 miles per hour. After a while the motor “kicked out,” and she pedaled the remainder of the trip at 12 miles 1 per hour. The entire trip took 3 hours. How far had 2 Dawn traveled when the motor on the moped quit running? 39. Jeff leaves home and rides his bicycle out into the country for 3 hours. On his return trip along the same route, it takes him three-quarters of an hour longer. If his rate on the return trip was 2 miles per hour slower than his rate out into the country, find the total round-trip distance. 1 40. In 1 hours more time, Rita, riding her bicycle at 12 miles 4 per hour, rode 2 miles farther than Sonya, who was riding her bicycle at 16 miles per hour. How long did each girl ride?

THOUGHTS INTO WORDS 41. Suppose that your friend analyzes Problem 31 as follows: “Sandy has traveled 45 miles before Monica starts. Because Monica travels 5 miles per hour faster than Sandy, 45 it will take her 9 hours to catch Sandy.” How would 5 you react to this analysis of the problem?

42. Summarize the new ideas about problem solving that you have acquired thus far in this course.

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Chapter 3


Answers to the Example Practice Skills 1. (a) {25} (b) {50} (c) e

37 f 3

2. 1.75 years 3. Width is 8 inches; length is 16 inches 4.

1 hour 5. 1.5 hours 2

Answers to the Concept Quiz D E A C F B

3.5

More about Problem Solving OBJECTIVES 1

Solve Word Problems Involving Mixture

2

Solve Word Problems Involving Interest

3

Solve Word Problems Involving Age Let’s begin this section with an important but often overlooked facet of problem solving: the importance of looking back over your solution and considering some of the following questions: 1. 2. 3. 4. 5.

Is your answer to the problem a reasonable answer? Does it agree with the answer you estimated before doing the problem? Have you checked your answer by substituting it back into the conditions stated in the problem? Do you see another plan that you could use to solve the problem? Is there another guideline that could be used? Is this problem closely related to another problem that you have previously solved? Have you tucked away for future reference the technique you used to solve this problem?

Looking back over the solution of a newly solved problem can often lay important groundwork for solving similar problems in the future.

1 Solve Word Problems Involving Mixture Now let’s consider three problems that we often refer to as mixture problems. No basic formula applies for all of these problems, but the suggestion that you think in terms of a pure substance is often helpful in setting up a guideline. For example, a phrase such as “a 30% solution of acid” means that 30% of the solution is acid and the remaining 70% is water.

EXAMPLE 1

Apply Your Skill How many milliliters of pure acid must be added to 150 milliliters of a 30% solution of acid to obtain a 40% solution of acid? (See Figure 3.27.)

Solution To solve this problem, we can use the following guideline: Amount of pure acid in original solution

Amount of pure acid to be added

Amount of pure acid in final solution

3.5

More about Problem Solving

175

To apply this guideline, it is necessary to multiply 150 ml by 30% to determine the amount of pure acid in the solution. We also need to recognize that adding “pure acid” is equivalent to a 100% acid solution. Let p represent the amount of pure acid to be added. Then the amount of the final solution is 150 milliliters plus the amount of pure acid added. Let’s organize the information that we have in a table and use the formula, amount of solution percent amount of pure acid. 150 milliliters 30% solution

Figure 3.27

Amount of solution Percent Amount of pure acid Amount of solution

Percent (in decimal form)

Amount of pure acid

150 p 150 p

0.30 1.00 0.40

0.30(150) 1.00p 0.40(150 p)

Original solution Pure acid Final solution

Note that the last column of the table yields the expressions for the guideline. Using the guideline and the last column of the table we can write the equation: 0.30(150) 1.00p 0.40(150 p) Now let’s solve the equation to determine the amount of pure acid to be added. 0.30(150) 1.00p 0.40(150 p) 45 p 60 0.40p 0.60p 15 p

15 25 0.60

We must add 25 milliliters of pure acid. (You should check this answer.)

▼ PRACTICE YOUR SKILL A new preschool regulation requires that juice mixtures be at least 20% pure juice. How many quarts of pure orange juice must be added to 36 quarts of a 10% juice mixture to obtain a 20% juice mixture? ■

EXAMPLE 2

Apply Your Skill Suppose that you have a supply of a 30% solution of alcohol and a 70% solution of alcohol. How many quarts of each should be mixed to produce a 20-quart solution that is 40% alcohol?

Solution We can use a guideline similar to the one in Example 1:

Pure alcohol in 30% solution



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Chapter 3


Let x represent the amount of 30% solution. Then 20 x represents the amount of 70% solution. Now, let’s organize the information that we have in a table and use the formula, amount of solution percent amount of pure acid.

Amount of solution Percent Amount of pure acid

30% Solution 70% Solution 40% Solution

Amount of solution

Percent (in decimal form)

Amount of pure alcohol

x 20 x 20

0.30 0.70 0.40

0.30x 0.70(20 x) 0.40(20)

The last column of the table yields the expressions for the guideline. Using the guideline and the last column of the table we can write the equation: 0.30x 0.70(20 x) 0.40(20) Solving this equation to determine the amount of the solutions to be mixed, we obtain 0.30x 0.70120 x2 0.401202 0.30x 0.70120 x2 8 30x 70120 x2 800

Multiplied both sides of the equation by 100

30x 1400 70x 800 40x 600 x 15

Therefore, 20 x 5. We should mix 15 quarts of the 30% solution with 5 quarts of the 70% solution.

▼ PRACTICE YOUR SKILL An oil company obtains fuel from two different refineries. One refinery supplies fuel that is 10% ethanol, and the other refinery supplies fuel that is 20% ethanol. How many gallons of each should be mixed to produce 6000 gallons of fuel that is 16% ethanol? ■

2 Solve Word Problems Involving Interest EXAMPLE 3

Apply Your Skill A woman has a total of $5000 to invest. She invests part of it at 4%, and the remainder at 6%. Her total yearly interest from the two investments is $260. How much did she invest at each rate?

Solution Let x represent the amount invested at 6%. Then 5000 x represents the amount invested at 4%. Use the following guideline: Interest earned from 6% investment

(6%)(x)

Interest earned from 4% investment

(4%)($5000 x)

Total interest earned

$260

3.5


177

Solving this equation yields 16%21x2 14%215000 x2 260 0.06x 0.0415000 x2 260 6x 415000 x2 26,000 6x 20,000 4x 26,000 2x 20,000 26,000 2x 6000 x 3000 Therefore, 5000 x 2000. She invested $3000 at 6% and $2000 at 4%.

▼ PRACTICE YOUR SKILL A sum of $8000 is invested in two different accounts. The total yearly interest from the two investments is $524. How much was invested in each account if the first account paid 7% interest, and the second account paid 5% interest? ■

EXAMPLE 4

Apply Your Skill An investor invests a certain amount of money at 3%. Then he finds a better deal and invests $5000 more than that amount at 5%. His yearly income from the two investments is $650. How much did he invest at each rate?

Solution Let x represent the amount invested at 3%. Then x 5000 represents the amount invested at 5%. 13% 21x2 15% 21x 50002 650 0.03x 0.051x 50002 650 3x 51x 50002 65,000 3x 5x 25,000 65,000 8x 25,000 65,000 8x 40,000 x 5000 Therefore, x 5000 10,000. He invested $5000 at 3% and $10,000 at 5%.

▼ PRACTICE YOUR SKILL Mrs. Stewart invested a certain amount of money at 6%. Then she found a second investment that paid 9%. Because that second investment carried more risk, she invested $2000 less than the amount she invested at 6%. Her yearly income from the two investments is $870. Find the amount invested in each investment. ■

178

Chapter 3


3 Solve Word Problems Involving Age The key to solving the problem in the next example is to represent the various unknown quantities in terms of one variable.

EXAMPLE 5

Apply Your Skill Jody is 6 years younger than her sister Cathy, and in 7 years Jody will be three-fourths as old as Cathy is at that time. Find their present ages.

Solution By letting c represent Cathy’s present age, we can represent all of the unknown quantities like this: c:

Cathy’s present age

c 6:

Jody’s present age

c 7:

Cathy’s age in 7 years

c 6 7 or c 1:

Jody’s age in 7 years

The statement “in 7 years Jody will be three-fourths as old as Cathy is at that time” serves as the guideline, so we can set up and solve the following equation: c1

3 1c 72 4

4c 4 31c 72 4c 4 3c 21 c 17 Therefore, Cathy’s present age is 17, and Jody’s present age is 17 6 11.

▼ PRACTICE YOUR SKILL Jennifer is 3 years older than her cousin Kyle. In 1 year, Kyle’s age will be two-thirds of Jennifer’s age at that time. Find their present ages. ■

CONCEPT QUIZ

For Problems 1–7, answer true or false. 1. The phrase “a 40% solution of alcohol” means that 40% of the amount of the solution is alcohol. 2. The amount of pure acid in 300ml of a 30% solution is 100ml. 3. If we want to produce 10 quarts by mixing solution A and solution B, the amount of solution A needed could be represented by x, and the amount of solution B would then be represented by 10 x. d 4. The formula d rt is equivalent to r . t r 5. The formula d rt is equivalent to t . d 6. If y represents John’s current age, then his age 4 years ago would be represented by y 4. 7. If Shane’s current age is represented by x, then his age in 10 years would be represented by 10x.

3.5


179

Problem Set 3.5 Set up an equation and solve each of the following problems.

1 Solve Word Problems Involving Mixture 1. How many milliliters of pure acid must be added to 100 milliliters of a 10% acid solution to obtain a 20% solution? 2. How many liters of pure alcohol must be added to 20 liters of a 40% solution to obtain a 60% solution? 3. How many centiliters of distilled water must be added to 10 centiliters of a 50% acid solution to obtain a 20% acid solution?

14. Sally invested a certain sum of money at 3%, twice that sum at 4%, and three times that sum at 6%. Her total yearly interest from all three investments was $145. How much did she invest at each rate? 15. If $2000 is invested at 4% interest, how much money must be invested at 7% interest so that the total return for both investments averages 6%? 16. Fawn invested a certain amount of money at 3% interest and $1250 more than that amount at 5%. Her total yearly interest was $134.50. How much did she invest at each rate?

4. How many milliliters of distilled water must be added to 50 milliliters of a 40% acid solution to reduce it to a 10% acid solution?

17. A sum of $2300 is invested, part of it at 6% interest and the remainder at 8%. If the interest earned by the 8% investment is $100 more than the interest earned by the 6% investment, find the amount invested at each rate.

5. Suppose that we want to mix some 30% alcohol solution with some 50% alcohol solution to obtain 10 quarts of a 35% solution. How many quarts of each kind should we use?

18. If $3000 is invested at 9% interest, how much money must be invested at 12% so that the total return for both investments averages 11%?

6. We have a 20% alcohol solution and a 50% solution. How many pints must be used from each to obtain 8 pints of a 30% solution?

19. How can $5400 be invested, part of it at 8% and the remainder at 10%, so that the two investments will produce the same amount of interest?

7. How many gallons of a 15% salt solution must be mixed with 8 gallons of a 20% salt solution to obtain a 17% salt solution?

20. A sum of $6000 is invested, part of it at 5% interest and the remainder at 7%. If the interest earned by the 5% investment is $160 less than the interest earned by the 7% investment, find the amount invested at each rate.

8. How many liters of a 10% salt solution must be mixed with 15 liters of a 40% salt solution to obtain a 20% salt solution? 9. Thirty ounces of a punch containing 10% grapefruit juice is added to 50 ounces of a punch containing 20% grapefruit juice. Find the percent of grapefruit juice in the resulting mixture. 10. Suppose that 20 gallons of a 20% salt solution are mixed with 30 gallons of a 25% salt solution. What is the percent of salt in the resulting solution?

2 Solve Word Problems Involving Interest 11. Suppose that Cory invested a certain amount of money at 3% interest and $750 more than that amount at 5%. His total yearly interest was $157.50. How much did he invest at each rate? 12. Nina received an inheritance of $12,000 from her grandmother. She invested part of it at 6% interest and the remainder at 8%. If the total yearly interest from both investments was $860, how much did she invest at each rate? 13. Udit received $1200 from his parents as a graduation present. He invested part of it at 4% interest and the remainder at 6%. If the total yearly interest amounted to $62, how much did he invest at each rate?

3 Solve Word Problems Involving Age 21. Pam is half as old as her brother Bill. Six years ago, Bill was four times older than Pam. How old are Pam and Bill now? 22. Christopher is twice as old as his brother Chance. In 10 years, Chance’s age will be three-fourths of Christopher’s age. How old are Christopher and Chance now? 23. Suppose that the sum of the present ages of Tom and his father is 100 years. Ten years ago, Tom’s father was three times as old as Tom was at that time. Find their present ages. 24. The sum of the present ages of Beth and her aunt Greta is 40 years. In 10 years, Greta’s age will be twice Beth’s age. Find their present ages. 25. Dylan’s present age is three times Tobin’s present age. Four years ago, the sum of their ages was 52 years. Find their present ages. 26. Scott’s present age is two-thirds of Lisa’s present age. In 5 years, the sum of their ages will be 45 years. Find their present ages.

180

Chapter 3


27. Mario is 7 years older than his little sister Maria. In 8 years, Mario’s age will be 4 years less than twice Maria’s age at that time. Find their present ages.

29. Julio’s present age is one-half of his mother’s present age. Seven years ago, the sum of their ages was 82 years. How old was Julio’s mother when he was born?

28. Kim’s age is 4 years more than twice Marty’s age. Four years ago the difference in their ages was 17 years. Find their present ages.

30. At the present time Jason’s mother is three times as old as Jason. In 10 years, Jason’s mother will be twice as old as Jason. How old was Jason’s mother when he was born?

Answers to the Example Practice Skills 1. 4.5 quarts 2. 2400 gallons of 10% ethanol and 3600 gallons of 20% ethanol 3. $6200 at 7% and $1800 at 5% 4. $7000 at 6% and $5000 at 9% 5. Jennifer is 8 years old and Kyle is 5 years old


2. False

3. True 4. True 5. False 6. True

7. False


SUMMARY

Solve proportions. (Sec. 3.1, Obj. 1, p. 140)

A statement of equality between two ratios is a proportion. In a proportion, the cross products are equal; that is to say: a c If , then ad bc, b d when b 0 and d 0.

Solve word problems using proportions. (Sec. 3.1, Obj. 2, p. 142)

A variety of word problems can be set up and solved using proportions.

EXAMPLE Solve

x x1 . 3 2

CHAPTER REVIEW PROBLEMS Problems 1–10

Solution

x x1 , then 3 2 21x 12 3x 2x 2 3x 2 x The solution set is {2}. If

The scale on a blueprint shows that 1 inch represents 8 feet. If the length of the house on the blueprint is 4.5 inches, what is the length of the actual house?

Problems 11–18

Solution

Set up the proportion 1 4.5 and solve for x. x 8 x 4.5(8) 36 The house is 36 feet long. Use a proportion to convert a fraction to a percent. (Sec. 3.1, Obj. 3, p. 143)

The concept of percent means per one hundred and is therefore a ratio that has a denominator of 100. To convert a fraction to a percent, set up a proportion and solve.

Solve basic percent problems. (Sec. 3.1, Obj. 4, p. 144)

There are three basic types of percent problems. Each type can be solved by translating the problem into a simple algebraic equation and then solving.

Convert

7 to a percent. 40

Problems 19 –26

Solution

x 7 40 100 40x 711002 x 17.5 7 Therefore 17.5% . 40 1. What is 35% of 400?

Problems 27–32

Solution

n 0.35(400) 140 Therefore 140 is 35% of 400. 2. Twelve percent of what number is 30? Solution

0.12x 30 30 x 250 0.12 Therefore 12% of 250 is 30. 3. Fifteen is what percent of 80? Solution

15 x1802 15 0.1875 x 80 Therefore, 15 is 18.75% of 80.

(continued)

181

182

Chapter 3


OBJECTIVE

SUMMARY

EXAMPLE


Solve equations involving decimal numbers. (Sec. 3.2, Obj. 1, p. 149)

One technique for solving equations that contain decimal numbers is to clear the equation of all the decimals by multiplying both sides of the equation by an appropriate power of 10.

Solve 0.3x 0.12x 2.1

Problems 33 –38

Solve word problems involving discount, selling price, cost, or profit. (Sec. 3.2, Obj. 2, p. 150)

Many consumer problems involve the concept of percent and can be solved with an equation approach. The basic relationships selling price equals cost plus profit and original selling price minus discount equals the discount sale price are frequently used to solve problems.

Solve simple interest problems. (Sec. 3.2, Obj. 3, p. 152)

Evaluate formulas. (Sec. 3.3, Obj. 1, p. 156)

The simple interest formula, i Prt where i represents the amount of interest earned by investing P dollars at a yearly interest rate of r percent for t years, is used to solve certain types of investment problems.

Substitute the given values into the formula and solve for the remaining variable.

Solution

Multiply both sides of the equation by 100. 10010.3x 0.12x2 10012.12 30x 12x 210 42x 210 x 5 The solution set is {5}. A computer game store paid $20 each for some games. What should the selling price be to make a profit of 80% of the cost?

Problems 39 – 46

Solution

Let s represent the selling price. s 20 180%21202 s 20 0.801202 s 36 Therefore the selling price should be $36. For how many years must $2000 be invested at 8% interest to earn $240?

Problems 47–50

Solution

i Prt 240 0.08120002 1t2 240 160t 1.5 t The $2000 must be invested for 1.5 years at 8% interest. Solve d rt for r if d 140 and t 2.5.

Problems 51–54

Solution

d rt 140 r12.52 56 r Apply geometric formulas. (Sec. 3.3, Obj. 2, p. 157)

Geometry formulas are used to solve problems involving geometric figures. The formulas we use quite often are presented in Section 3.3.

Find the width of a rectangle whose length is 8 inches and whose perimeter is 24.

Problems 55 –58

Solution

Use the formula P 2l 2w. P 2l 2w 24 2182 2w 24 16 2w 8 2w 4w Therefore the width is 4 inches.

(continued)

Chapter 3

Summary

183

OBJECTIVE

SUMMARY

EXAMPLE


Solve formulas for a specific variable. (Sec. 3.3, Obj. 3, p. 160)

We can solve a formula such as P 2l 2w for l or for w by applying the properties of equality.

Solve P 2l 2w for l.

Problems 59 – 62

Solve an equation for a specific variable. (Sec. 3.3, Obj. 4, p. 162)

In future chapters, there will be equations with two variables—for example, x and y. At times you will need to solve for one variable in terms of the other variable.


Solve word problems. (Sec. 3.4, Obj. 2, p. 168; Sec. 3.4, Obj. 3, p. 169; Sec. 3.4, Obj. 4, p. 170)

Use these suggestions to help you solve word problems: 1. Read the problem carefully. 2. Sketch a figure, diagram, or chart to help organize the facts. 3. Choose a meaningful variable. Be sure to write down what the variable represents. 4. Look for a guideline. This is often the key component when solving a problem. Many times formulas are used as guidelines. 5. Use the guideline to set up an equation. 6. Solve the equation and answer the question in the problem. 7. Check your answer in the original statement of the problem.

Solution

P 2l 2w P 2w 2l P 2w l 2 Problems 63 – 66

Solution

2x 3y 12 3y 2x 12 2x 12 y 3 Mary starts jogging at 4 miles per hour. One-half hour later, Tom starts jogging on the same route at 6 miles per hour. How long will it take Tom to catch up with Mary? Solution

Let x represent Tom’s time. Then because Mary left one-half 1 hour before Tom, x repre2 sents the time she is jogging. One guideline is that the distance traveled by each person is the same. The distance can be found by applying the formula d rt. Mary’s distance Tom’s distance 1 4 ax b 6x 2 4x 2 6x 2 2x 1x Therefore, it takes Tom 1 hour to catch up with Mary.

Problems 67– 88

184

Chapter 3 Ratio, Percent, Formulas, and Problem Solving

Chapter 3 Review Problem Set For Problems 1–10, solve each proportion. 1.

3.

5.

x 5 8 2

2.

3 9 2 x

4.

x1 6 3 5

6.

x1 x6 7. 3 2 9.

7 3 2x 2 x4

7 x 6 2 5 10 x 4 5 x2 4 8

x2 x3 8. 6 4 10.

2 6 x2 3x 4

For Problems 11–18, set up a proportion and solve. 11. A blueprint has a scale in which 1 inch represents 4 feet. Find the dimensions of a rectangular room that measures 3.5 inches by 4.25 inches on the blueprint. 12. On a certain map, 1 inch represents 20 miles. If two cities are 6.5 inches apart on the map, find the number of miles between the cities. 13. If a car travels 200 miles using 10 gallons of gasoline, how far will it travel on 15 gallons of gasoline?

23.

7 4

24.

3 16

25.

2 3

26.

7 3

For Problems 27–32, set up an equation and solve. 27. Eighteen is what percent of 30? 28. Twenty-four is what percent of 150? 29. Fifteen percent of a number is 6. Find the number. 30. Ten percent of a number is 56. Find the number. 31. What is 120% of 500? 32. What is 42% of 20? For Problems 33 –38, solve each equation. 33. 0.5x 0.7x 6 34. 0.02x 1.8x 6.37 35. 0.07t 0.12(t 3) 0.59 36. 0.1x 0.12(1700 x) 188 37. x 0.25x 20 38. 0.2(x 3) 14

14. A home valued at $150,000 is assessed $3000 in real estate taxes. At the same rate, how much are the taxes on a home with a value of $200,000? 15. If 20 pounds of fertilizer will cover 1400 square feet of grass, how many pounds of fertilizer are needed for 1750 square feet of grass?

For Problems 39 – 46, set up an equation and solve. 39. Louise bought a pair of jeans for $36.40, which represents a 30% discount of the original price. Find the original price of the jeans. 40. Find the cost of a $75 shirt that is on sale for 20% off.

16. A board 24 feet long is cut into two pieces whose lengths are in the ratio of 2 to 3. Find the lengths of the two pieces. 17. A sum of $750 is to be divided between two people in the ratio of 3 to 2. How much does each person receive? 18. The ratio of female students to male students at a certain junior college is 5 to 3. If there is a total of 4400 students, find the number of female students and the number of male students. For Problems 19 –26, use proportions to change each common fraction to a percent. 19.

13 20

7 21. 8

20.

8 25

1 22. 8

41. Ely bought a putter for $72 that was originally listed for $120. What rate of discount did he receive? 42. Dominic bought a computer monitor for $162.50 that was originally listed for $250. What rate of discount did he receive? 43. A retailer has some candles that cost him $5 each. He wants to sell them at a profit of 60% of the cost. What should he charge for the candles? 44. A retailer has some shoes that cost her $25 a pair. She wants to sell them at a profit of 45% of the cost. What should she charge for the shoes? 45. A bookstore manager buys some textbooks for $88 each. He wants to sell them at a profit of 20% based on the selling price. What should he charge for the books? 46. A supermarket manager buys some apples at $0.80 per pound. He wants to sell them at a profit of 50% based on the selling price. What should he charge for the apples?

Chapter 3 For Problems 47–50, use the formula i Prt. 47. How much interest will be charged on a 4-year student loan of $8,500 at an 8% annual interest rate? 48. How long will $2500 need to be invested at a 6% annual interest rate to earn $600? 49. How much principal, invested at 7% annual interest rate for 3 years, is needed to earn $630? 50. Find the annual interest rate if $820 in interest is earned when $2,500 is invested for 4 years. 51. Solve P 2l 2w 52. Solve F

for w if P 50 and l 19.

9 C 32 5

for C if F 77.

53. Solve A P Prt 54. Solve d rt

for r if A 450, P 400, and t 2.

for t if d 150 and r 60.

55. Find the area of a trapezoid with one base 8 inches long, the other base 14 inches long, and the altitude between the two bases 7 inches. 56. If the area of a triangle is 27 square centimeters, and the length of one side is 9 centimeters, find the length of the altitude to that side. 57. How many yards of fencing are required to enclose a rectangular field that is 120 yards long and 57 yards wide? 58. If the total surface area of a right circular cylinder is 152p square feet, and a radius of a base is 4 feet long, find the height of the cylinder. 59. Solve V lwh for w.

Review Problem Set

185

10 centimeters longer than twice the width. Find the dimensions of the rectangle. 69. Seventy-eight yards of fencing were purchased to enclose a rectangular garden. The length of the garden is 1 yard shorter than three times its width. Find the length and width of the garden. 70. One angle of a triangle has a measure of 47°. Of the other two angles, one of them is 3° smaller than three times the other angle. Find the measures of the two remaining angles. 71. Two airplanes leave Chicago at the same time and fly in opposite directions. If one travels at 350 miles per hour and the other at 400 miles per hour, how long will it take them to be 1125 miles apart? 72. Two cars start from the same place travelling in opposite directions. One car travels 7 miles per hour faster than the other car. If at the end of 3 hours they are 369 miles apart, find the speed of each car. 73. Connie rides out into the country on her bicycle at a rate of 10 miles per hour. An hour later Zak leaves from the same place that Connie did and rides his bicycle along the same route at 12 miles per hour. How long will it take Zak to catch Connie? 74. Two cities, A and B, are 406 miles apart. Billie starts at city A in her car traveling toward city B at 52 miles per hour. At the same time, using the same route, Zorka leaves in her car from city B traveling toward city A at 64 miles per hour. How long will it be before the cars meet? 75. How many liters of pure alcohol must be added to 10 liters of a 70% solution to obtain a 90% solution? 76. How many gallons of a 10% salt solution must be mixed with 12 gallons of a 15% salt solution to obtain a 12% salt solution?

60. Solve V pr 2h

for h.

1 61. Solve A bh 2

for b.

77. Suppose that 20 ounces of a punch containing 20% orange juice is added to 30 ounces of a punch containing 30% orange juice. Find the percent of orange juice in the resulting mixture.

1 62. Solve V Bh 3

for B.

78. How many liters of pure alcohol must be added to 10 liters of a 30% solution to obtain a 50% solution?

63. Solve 2x 5y 12 64. Solve 3x y 2

for y.

for y.

65. Solve 6x 2y 5

for x.

66. Solve 3x 4y 7

for x.

67. Suppose that the length of a certain rectangle is 5 meters longer than twice the width. The perimeter of the rectangle is 46 meters. Find the length and width of the rectangle. 68. A copper wire 110 centimeters long was bent in the shape of a rectangle. The length of the rectangle was

79. How many cups of grapefruit juice must be added to 40 cups of punch that contains 5% grapefruit juice to obtain a punch that is 10% grapefruit juice? 80. How many milliliters of pure acid must be added to 150 milliliters of a 30% solution of acid to obtain a 40% solution? 81. A sum of $2100 is invested, part of it at 3% interest and the remainder at 5%. If the interest earned by the 5% investment is $51 more than the interest from the 3% investment, find the amount invested at each rate. 82. Eva invested a certain amount of money at 5% interest, and $1500 more than that amount at 6%. Her total yearly interest income was $420. How much did she invest at each rate?

186

Chapter 3


83. A total of $4000 was invested, part of it at 8% and the remainder at 9%. If the total interest income amounted to $350, how much was invested at each rate? 84. If $500 is invested at 6% interest, how much additional money must be invested at 9% so that the total return for both investments averages 8%? 85. Shane is 8 years older than his brother Caleb. In 4 years, Caleb’s age will be two-thirds of Shane’s age. Find their present ages. 86. The sum of the present ages of Melinda and her mother is 64 years. In 8 years, Melinda will be three-fifths as old as

her mother at that time. Find the present ages of Melinda and her mother. 87. At the present time, Nikki is one-third as old as Kaitlin. In 10 years, Kaitlin’s age will be 6 years less than twice Nikki’s age at that time. Find the present age of both girls. 88. Annilee’s present age is two-thirds of Jessie’s present age. In 12 years, the sum of their ages will be 54 years. Find their present ages.

Chapter 3 Test For Problems 1–10, solve each equation. 1.

x2 x3 4 5

1.

2.

4 3 2x 1 3x 5

2.

3.

x1 x2 2 6 5

3.

4.

x8 x4 2 7 4

4.

5.

n 7 20 n 3

5.

6.

h h 1 4 6

6.

7. 0.05n 0.061400 n2 23

7.

8. s 35 0.5s

8.

9. 0.07n 45.5 0.081600 n2

9.

10. 12t 8 a

7 tb 50 2

11. Solve F

9C 160 5

12. Solve y 21x 42 13. Solve

10.

for C. for x.

y5 x3 4 9

for y.

11. 12. 13.

For Problems 14 –16, use the geometric formulas given in this chapter to help you find the solution. 14. Find the area of a circular region if the circumference is 16p centimeters. Express the answer in terms of p.

14.

15. If the perimeter of a rectangle is 100 inches and its length is 32 inches, find the area of the rectangle.

15.

16. The area of a triangular plot of ground is 133 square yards. If the length of one side of the plot is 19 yards, find the length of the altitude to that side.

16.

For Problems 17–25, set up an equation and solve. 17. Express

5 as a percent. 4

17.

18. Thirty-five percent of what number is 24.5?

18.

19. Cora bought a digital camera for $132.30, which represented a 30% discount off the original price. What was the original price of the camera?

19.

20. A retailer has some lamps that cost her $40 each. She wants to sell them at a profit of 30% of the cost. What price should she charge for the lamps?

20.

187

188

Chapter 3


21.

21. Hugh paid $48 for a pair of golf shoes that were listed for $80. What rate of discount did he receive?

22.

22. The election results in a certain precinct indicated that the ratio of female voters to male voters was 7 to 5. If a total of 1500 people voted, how many women voted?

23.

23. A car leaves a city traveling at 50 miles per hour. One hour later a second car leaves the same city traveling on the same route at 55 miles per hour. How long will it take the second car to overtake the first car?

24.

24. How many centiliters of pure acid must be added to 6 centiliters of a 50% acid solution to obtain a 70% acid solution?

25.

25. How long will it take $4000 to double if it is invested at 9% simple interest?

Chapters 1–3 Cumulative Review Problem Set For Problems 1–10, simplify each algebraic expression by combining similar terms. 1. 7x 9x 14x

For Problems 27–38, solve each equation. 27. 5x 2 22 28. 3x 4 7x 4

2. 10a 4 13a a 2

29. 3(4x 1) 6(2x 1)

3. 51x 32 71x 62

30. 21x 12 31x 22 12

4. 31x 12 412x 12 5. 3n 21n 12 513n 22 n 6. 6n 314n 22 212n 32 5 7.

1 3 2 1 x x x x 2 4 3 6

8.

4 5 1 n n nn 3 15 6

31.

1 1 1 2 x x 5 3 3 2

32.

t3 1 t2 4 3 6

33.

2n 1 n2 1 5 4

34. 0.09x 0.121500 x2 54 35. 51n 12 1n 22 312n 12

9. 0.4x 0.7x 0.8x x 10. 0.51x 22 0.41x 32 0.2x

36.

3 2 x1 x4

37. 0.2x 0.11x 42 0.7x 1 For Problems 11–20, evaluate each algebraic expression for the given values of the variables. 11. 5x 7y 2xy for x 2 and y 5 12. 2ab a 6b for a 3 and b 4 13. 31x 12 21x 62

for x 5

14. 51n 32 1n 42 n for n 7 15.

16.

3x 2y 2x 3y

39. 4x 6 3x 1 40. 3x 6 12

42. for n

2 3

for x

1 1 and y 2 4

19. 5x 7y 8x 3y for x 9 and y 8 3a b 4a 3b a 6b 4b 3a

1 1 1 2 x x 7 4 4 2

43. 0.08t 0.11300 t2 28

for a 0.2 and b 0.3

18. x2 3xy 2y2

20.

For Problems 39 – 46, solve each inequality.

41. 21n 12 31n 22 1

for x 3 and y 6

3 1 5 n n n 4 3 6

17. 2a 2 4b2

1 1 38. 1t 22 1t 42 2 at b 3 at b 2 3

for a 1 and b 3

44. 4 5x 2 3x 45.

2 1 n2 n1 3 2

46. 3 21x 12 x

For Problems 47–54, set up an equation or an inequality and solve each problem.

21. 34

22. 26

47. Erin’s salary this year is $32,000. This represents $2000 more than twice her salary 5 years ago. Find her salary 5 years ago.

23. 10.42 3

1 5 24. a b 2

48. One of two supplementary angles is 45° less than four times the other angle. Find the measure of each angle.

For Problems 21–26, evaluate each expression.

25. a

1 1 2 b 2 3

26. a

3 7 3 b 4 8

49. Jaamal has 25 coins, consisting of nickels and dimes, that amount to $2.10. How many coins of each kind does he have?

189

190

Chapter 3


50. Hana bowled 144 and 176 in her first two games. What must she bowl in the third game to have an average of at least 150 for the three games? 51. A board 30 feet long is cut into two pieces, and the lengths are in the ratio of 2 to 3. Find the lengths of the two pieces. 52. A retailer has some shoes that cost him $32 per pair. He wants to sell them at a profit of 20% of the selling price. What price should he charge for the shoes?

53. Two cars start from the same place traveling in opposite directions. One car travels 5 miles per hour faster than the other car. Find their speeds if after 6 hours they are 570 miles apart. 54. How many liters of pure alcohol must be added to 15 liters of a 20% solution to obtain a 40% solution?

Exponents and Polynomials

4 4.1 Addition and Subtraction of Polynomials 4.2 Multiplying Monomials 4.3 Multiplying Polynomials 4.4 Dividing by Monomials 4.5 Dividing by Binomials

Bruno Herdt /Photolibrary

4.6 Zero and Negative Integers as Exponents

■ The average distance between the sun and the Earth is approximately 93,000,000 miles. Using scientific notation, 93,000,000 can be written as (9.3)(107).

A

rectangular dock that measures 12 feet by 16 feet is treated with a uniform strip of nonslip coating along both sides and both ends. How wide is the strip if one-half of the dock is treated? If we let x represent the width of the strip, then we 1 can use the equation (16 2x)(12 2x) (12)(16) to determine that the width of 2 the strip is 2 feet. The equation we used to solve this problem is called a quadratic equation. Quadratic equations belong to a larger classification called polynomial equations. To solve problems involving polynomial equations, we need to develop some basic skills that pertain to polynomials. That is to say, we need to be able to add, subtract, multiply, divide, and factor polynomials. Chapters 4 and 5 will help you develop those skills as you work through problems that involve quadratic equations.


191

I N T E R N E T

P R O J E C T

For ease of operations, many of the numbers you find in science courses are written in scientific notation. In scientific notation, numbers are written in the form: a 10b

(“a times 10 to the power of b”), where the exponent b is an integer, and 1 0 a 0 10. For example, 7900 can be written as 7.9 103 and 0.00062 can be written as 6.2 104. Do an Internet search to find the mass of the Earth. Is the mass expressed in scientific notation? Engineering notation is similar to scientific notation. Do an Internet search on engineering notation and determine how it differs from scientific notation.

4.1

Addition and Subtraction of Polynomials OBJECTIVES 1

Classify Polynomials by Size and Degree

2

Add Polynomials

3

Subtract Polynomials Using a Horizontal Format

4

Subtract Polynomials Using a Vertical Format

5

Perform Operations on Polynomials Involving Both Addition and Subtraction

6

Use Polynomials to Represent the Area or Perimeter of Geometric Figures

1 Classify Polynomials by Size and Degree In earlier chapters, we called algebraic expressions such as 4x, 5y, 6ab, 7x2, and 9xy2z3 “terms.” Recall that a term is an indicated product that may contain any number of factors. The variables in a term are called “literal factors,” and the numerical factor is called the “numerical coefficient” of the term. Thus, in 6ab, a and b are literal factors and the numerical coefficient is 6. Terms that have the same literal factors are called “similar” or “like” terms. Terms that contain variables with only whole numbers as exponents are called monomials. The previously listed terms, 4x, 5y, 6ab, 7x2, and 9xy2z3, are all monomials. (We will work with some algebraic expressions later, such as 7x1y1 and 4a2b3, which are not monomials.) The degree of a monomial is the sum of the exponents of the literal factors. Here are some examples: 4xy is of degree 2. 5x is of degree 1. 14a2b is of degree 3. 17xy2z3 is of degree 6. 9y4 is of degree 4. If the monomial contains only one variable, then the exponent of the variable is the degree of the monomial. Any nonzero constant term is said to be of degree zero. A polynomial is a monomial or a finite sum (or difference) of monomials. The degree of a polynomial is the degree of the term with the highest degree in the polynomial. Some special classifications of polynomials are made according to the number of terms. We call a one-term polynomial a monomial, a two-term polynomial a 192

4.1

Addition and Subtraction of Polynomials

193

binomial, and a three-term polynomial a trinomial. The following examples illustrate some of this terminology: The polynomial 5x3y4 is a monomial of degree 7. The polynomial 4x2y 3xy is a binomial of degree 3. The polynomial 5x2 6x 4 is a trinomial of degree 2. The polynomial 9x4 7x3 6x2 x 2 is given no special name, but it is of degree 4.

2 Add Polynomials In the preceding chapters, you have worked many problems involving the addition and subtraction of polynomials. For example, simplifying 4x 2 6x 7x 2 2x to 11x 2 4x by combining similar terms can actually be considered the addition problem 14x 2 6x2 17x 2 2x2 . At this time we will simply review and expand some of those ideas.

EXAMPLE 1

Add 5x 2 7x 2 and 9x 2 12x 13.

Solution We commonly use the horizontal format for such work. Thus, 15x 2 7x 22 19x 2 12x 132 15x 2 9x 2 2 17x 12x2 12 132 14x 2 5x 11

▼ PRACTICE YOUR SKILL Add 8x 2 2x 6 and 3x 2 5x 10.

■

The commutative, associative, and distributive properties provide the basis for rearranging, regrouping, and combining similar terms.

EXAMPLE 2

Add 5x 1, 3x 4, and 9x 7.

Solution 15x 12 13x 42 19x 72 15x 3x 9x2 31 4 172 4 17x 4

▼ PRACTICE YOUR SKILL Add 6x 4, 2x 3, and 5x 8 .

EXAMPLE 3

■

Add x 2 2x 1, 2x 3 x 4, and 5x 6.

Solution 1x 2 2x 12 12x 3 x 42 15x 62

12x 3 2 1x 2 2 12x x 5x2 11 4 62 2x3 x 2 4x 9

▼ PRACTICE YOUR SKILL Add 2x2 3x 4, 5x3 7x 2, and 2x 6.

■

194

Chapter 4


3 Subtract Polynomials Using a Horizontal Format Recall from Chapter 1 that a b a (b). We define subtraction as adding the opposite. This same idea extends to polynomials in general. The opposite of a polynomial is formed by taking the opposite of each term. For example, the opposite of 12x2 7x 32 is 2x2 7x 3. Symbolically, we express this as 12x2 7x 32 2x2 7x 3 Now consider some subtraction problems.

EXAMPLE 4

Subtract 2x2 9x 3 from 5x2 7x 1.

Solution Use the horizontal format: 15x2 7x 12 12x2 9x 32 15x2 7x 12 12x2 9x 32

15x2 2x2 2 17x 9x2 11 32

3x2 16x 2

▼ PRACTICE YOUR SKILL Subtract 3x2 2x 1 from 9x2 8x 2.

EXAMPLE 5

■

Subtract 8y2 y 5 from 2y2 9.

Solution 12y2 92 18y2 y 52 12y2 92 18y2 y 52

12y 2 8y 2 2 1y2 19 52 10y 2 y 4

▼ PRACTICE YOUR SKILL Subtract 2a2 3a 1 from 6a 4 .

■

4 Subtract Polynomials Using a Vertical Format Later, when dividing polynomials, you will need to use a vertical format to subtract polynomials. Let’s consider two such examples.

EXAMPLE 6

Subtract 3x2 5x 2 from 9x2 7x 1.

Solution 9x2 7x 1

Notice which polynomial goes on the bottom and the alignment of similar terms in columns

3x2 5x 2 Now we can mentally form the opposite of the bottom polynomial and add. 9x2 77x 1 3x2 55x 2

The opposite of 3x2 5x 2 is 3x2 5x 2

6x2 12x 1

▼ PRACTICE YOUR SKILL Subtract 4x2 x 3 from 11x2 8x 2 .

■

4.1

EXAMPLE 7


195

Subtract 15y3 5y2 3 from 13y3 7y 1.

Solution 13y3 5y2 7y 1

Similar terms are arranged in columns

15y3 5y2 5y 3 2y3 5y2 7y 4

We mentally formed the opposite of the bottom polynomial and added

▼ PRACTICE YOUR SKILL Subtract 8y3 6y 4 from 5y3 2y2 8.

■

5 Perform Operations on Polynomials Involving Both Addition and Subtraction We can use the distributive property along with the properties a 1(a) and a 1(a) when adding and subtracting polynomials. The next examples illustrate this approach.

EXAMPLE 8

Perform the indicated operations. 13x 42 12x 52 17x 12

Solution 13x 42 12x 52 17x 12 113x 42 112x 52 117x 12 113x2 1142 112x2 1152 117x2 1112 3x 4 2x 5 7x 1 3x 2x 7x 4 5 1 2x 8

▼ PRACTICE YOUR SKILL Perform the indicated operations: 16x 12 13x 22 15x 82

■

Certainly we can do some of the steps mentally; Example 9 gives a possible format.

EXAMPLE 9

Perform the indicated operations. 1y2 5y 22 12y2 8y 62 14y2 2y 52

Solution 1y2 5y 22 12y2 8y 62 14y2 2y 52 y2 5y 2 2y2 8y 6 4y2 2y 5 y2 2y2 4y2 5y 8y 2y 2 6 5 5y2 5y 13

196

Chapter 4


▼ PRACTICE YOUR SKILL Perform the indicated operations: 1x2 3x 42 14x2 5x 62 12x 72

■

When we use the horizontal format, as in Examples 8 and 9, we use parentheses to indicate a quantity. In Example 8, the quantities (3x 4) and (2x 5) are to be added; from this result we are to subtract the quantity (7x 1). Brackets, [ ], are also sometimes used as grouping symbols, especially if there is a need to indicate quantities within quantities. To remove the grouping symbols, perform the indicated operations, starting with the innermost set of symbols. Let’s consider two examples of this type.

EXAMPLE 10

Perform the indicated operations. 3x 32x 13x 12 4

Solution First we need to add the quantities 2x and (3x 1). 3x 32x 13x 12 4 3x 3 2x 3x 1 4 3x 3 5x 1 4

Now we need to subtract the quantity [5x 1] from 3x. 3x 35x 14 3x 5x 1 2x 1

▼ PRACTICE YOUR SKILL Perform the indicated operations: 10y 37y 1y 42 4

EXAMPLE 11

■

Perform the indicated operations. 8 57x 3 2 1x 12 4 4x6

Solution Start with the innermost set of grouping symbols (the parentheses) and proceed as follows: 8 57x 3 2 1x 12 4 4x6 8 57x 3 x 1 4 4x6 8 17x x 1 4x2

8 110x 12 8 10x 1 10x 9

▼ PRACTICE YOUR SKILL Perform the indicated operations: 12 54y 3 6 13y 22 4 106

■

4.1


197

6 Use Polynomials to Represent the Area or Perimeter of Geometric Figures For a final example in this section, we look at polynomials in a geometric setting.

EXAMPLE 12

Suppose that a parallelogram and a rectangle have dimensions as indicated in Figure 4.1. Find a polynomial that represents the sum of the areas of the two figures. x x

x 20 Figure 4.1

Solution Using the area formulas A bh and A lw for parallelograms and rectangles, respectively, we can represent the sum of the areas of the two figures as follows: x1x2 x2

Area of the parallelogram

201x2 20x

Area of the rectangle

We can represent the total area by x2 20x.

▼ PRACTICE YOUR SKILL Suppose that a circle and a rectangle have the dimensions as indicated in the figure below. Find a polynomial that represents the sum of the areas of the two figures. 10 r

r

■

CONCEPT QUIZ

For Problems 1–5, answer true or false. 1. 2. 3. 4. 5.

The degree of the monomial 4x2y is 3. The degree of the polynomial 2x4 5x3 7x2 4x 6 is 10. A three-term polynomial is called a binomial. A polynomial is a monomial or a finite sum of monomials. Monomial terms must have whole number exponents for each variable.

Problem Set 4.1 1 Classify Polynomials by Size and Degree For Problems 1– 6, match the polynomial with its description.

2 Add Polynomials For Problems 7–22, add the polynomials.

1. 5xy

A. Monomial of degree 5

7. 2x 1 and 8x 3

2. 5xy2 3x2

B. Binomial of degree 5

9. 3x 4 and 5x 7

3. 5x2y 3xy4

C. Monomial of degree 3

11. 5y 3 and 9y 13

4. 3x5 2x3 5x 1

D. Binomial of degree 3

12. x2 2x 1 and 2x2 x 4

5. 3x2y3

E. Polynomial of degree 5

13. 2x2 7x 9 and 4x2 9x 14

6. x2 2xy y2

F.

14. 3a2 4a 7 and 3a2 7a 10

2

Trinomial of degree 2

8. 4x 6 and 7x 10 10. 3x 5 and 2x 9

198

Chapter 4


15. 5x 2, 3x 7, and 9x 10 16. x 4, 8x 9, and 7x 6

5 Perform Operations on Polynomials Involving Both Addition and Subtraction

17. 2x2 x 4, 5x2 7x 2, and 9x2 3x 6

For Problems 45 – 64, perform the indicated operations.

18. 3x2 2x 6, 6x2 7x 3, and 4x2 9

45. (5x 3) (7x 2) (3x 6)

19. 4n2 n 1 and 4n2 6n 5

46. (3x 4) (9x 1) (14x 7)

20. 5n2 7n 9 and 5n 4

47. (x 1) (2x 6) (4x 7)

21. 2x2 7x 10, 6x 2, and 9x2 5

48. (3x 6) (x 8) (7x 10)

22. 7x 11, x2 5x 9, and 4x 5

3 Subtract Polynomials Using a Horizontal Format For Problems 23 –34, subtract the polynomials using a horizontal format.

49. 1x2 7x 42 12x2 8x 92 14x2 2x 12 50. 13x2 x 62 18x2 9x 12 17x2 2x 62 51. 1x2 3x 42 12x2 x 22 14x2 7x 102 52. 13x2 22 17x2 82 19x2 2x 42 53. (3a 2b) (7a 4b) (6a 3b)

23. 7x 1 from 12x 6

54. (5a 7b) (8a 2b) (5a 6b)

24. 10x 3 from 14x 13

55. 1n 62 12n2 n 42 1n2 72

25. 5x 2 from 3x 7

56. 13n 42 1n2 9n 102 12n 42

26. 7x 2 from 2x 3 27. x 1 from 4x 6 28. 3x 2 from x 9 29. x2 7x 2 from 3x2 8x 4 30. 2x2 6x 1 from 8x2 2x 6 31. 2n2 3n 4 from 3n2 n 7 32. 3n2 7n 9 from 4n2 6n 10 33. 4x3 x2 6x 1 from 7x3 x2 6x 12 34. 4x2 6x 2 from 3x3 2x2 7x 1

57. 7x [3x (2x 1)] 58. 6x [2x (5x 2)] 59. 7n [4n (6n 1)] 60. 9n [3n (5n 4)] 61. (5a 1) [3a (4a 7)] 62. (3a 4) [7a (9a 1)] 63. 13x {5x [4x (x 6)]} 64. 10x {7x [3x (2x 3)]} 65. Subtract 5x 3 from the sum of 4x 2 and 7x 6.

4 Subtract Polynomials Using a Vertical Format For Problems 35 – 44, subtract the polynomials using a vertical format. 35. 3x 2 from 12x 4 36. 4x 6 from 7x 3 37. 5a 6 from 3a 9 38. 7a 11 from 2a 1 39. 8x2 x 6 from 6x2 x 11

66. Subtract 7x 5 from the sum of 9x 4 and 3x 2. 67. Subtract the sum of 2n 5 and n 7 from 8n 9. 68. Subtract the sum of 7n 11 and 4n 3 from 13n 4.

6 Use Polynomials to Represent the Area or Perimeter of Geometric Figures 69. Find a polynomial that represents the perimeter of the rectangle in Figure 4.2.

40. 3x2 2 from 2x2 6x 4 41. 2x3 6x2 7x 9 from 4x3 6x2 7x 14 42. 4x3 x 10 from 3x2 6 43. 2x2 6x 14 from 4x3 6x2 7x 2 44. 3x 7 from 7x3 6x2 5x 4

3x + 5

x−2

Figure 4.2 70. Find a polynomial that represents the area of the shaded region in Figure 4.3. The length of a radius of the larger

4.2

Multiplying Monomials

199

circle is r units, and the length of a radius of the smaller circle is 4 units. 2x 3x

4x

2x x

3x x

3x Figure 4.4 Figure 4.3 71. Find a polynomial that represents the sum of the areas of the rectangles and squares in Figure 4.4.

72. Find a polynomial that represents the total surface area of the rectangular solid in Figure 4.5.

2

9 x

Figure 4.5

THOUGHTS INTO WORDS 73. Explain how to subtract the polynomial

3x 6x 2 from 4x 7. 2

2

75. Is the sum of two binomials ever a trinomial? Defend your answer.

74. Is the sum of two binomials always another binomial? Defend your answer.

Answers to the Example Practice Skills 1. 11x2 3x 4 2. 13x 7 3. 5x3 2x2 2x 4. 6x2 6x 3 5. 2a2 9a 5 6. 7x2 9x 5 7. 3y3 2y2 6y 12 8. 4x 7 9. 3x 2 9 10. 2y 4 11. y 6 12. pr 2 10r


4.2

2. False


Multiplying Monomials OBJECTIVES 1

Apply the Properties of Exponents to Multiply Monomials

2

Apply the Properties of Exponents to Raise a Monomial to a Power

3

Multiply a Polynomial by a Monomial

4

Apply Multiplication of Monomials to Geometric Problems

1 Apply the Properties of Exponents to Multiply Monomials The basic idea of using exponents to indicate repeated multiplication was introduced in Section 1.1. Expanding on that idea with some properties of exponents can make

200

Chapter 4


multiplying monomials easier. These properties are the direct result of the definition of an exponent. The following examples lead to Property 4.1. x2 # x3 1x # x2 1x # x # x2 x5

a3 # a4 1a # a # a2 1a # a # a # a2 a7 b # b2 1b2 1b # b2 b3

In general, bn # bm 1b # b # b # p # b2 1b # b # b # p # b2 144424443 144424443

n factors of b

m factors of b

#b#b#p#b b144424443

(n m) factors of b

bnm

Property 4.1 If b is any real number, and n and m are positive integers, then bn # bm bnm

Property 4.1 states that when multiplying powers with the same base, add exponents.

EXAMPLE 1

Multiply the following: (a) x4 # x3

(b) a8 # a7

Solution (a) x4 # x3 x43 x7

(b) a8 # a7 a87 a15

▼ PRACTICE YOUR SKILL Multiply the following: (a) y3 # y5

(b) a2 # a

■

Consider the following example in which we use the properties of exponents to help simplify the process of multiplying monomials.

EXAMPLE 2

Multiply the following: (a) 13x3 215x4 2

(b) 14a2b3 2 16ab2 2

(c) 1xy2 17xy5 2

Solution (a) 13x3 215x4 2 3 # 5 # x3 # x4 15x7

(b) 14a2b3 2 16ab2 2 4 # 6 # a2 # a # b3 # b2 24a3b5

3 1 (d) a x2y3 b a x3y5 b 4 2

4.2


201

(c) 1xy217xy5 2 1 # 7 # x # x # y # y 5 7x 2y6 3 1 3 1 (d) a x2y3 b a x3y5 b # # x 2 # x3 # y3 # y5 4 2 4 2 3 x5y8 8

▼ PRACTICE YOUR SKILL Multiply the follwing: (a) 14x2 2 16x3 2

2 6 (b) a a5bb a a4b2 b 3 7

■

2 Apply the Properties of Exponents to Raise a Monomial to a Power Another property of exponents is demonstrated by these examples. 1x2 2 3 x2 # x2 # x2 x222 x6 1a3 2 2 a3 # a3 a33 a6

1b3 2 4 b3 # b3 # b3 # b3 b3333 b12 In general, n # n # n # p # n 1bn 2 m b144424443 b b b

m factors of bn

m of these ns

64748 p bnnn n bmn

Property 4.2 If b is any real number, and m and n are positive integers, then 1bn 2 m bmn

Property 4.2 states that when raising a power to a power, multiply exponents.

EXAMPLE 3

Raise each to the indicated power. (a) 1x 4 2 3

(b) 1a 5 2 6

Solution # (a) 1x 4 2 3 x 3 4 x 12

# (b) 1a 5 2 6 a 6 5 a 30

▼ PRACTICE YOUR SKILL Raise each to the indicated power. (a) (y3)2

(b) (m2)4

■

202

Chapter 4


The third property of exponents we use in this section raises a monomial to a power. 12x2 3 12x2 12x2 12x2 2 # 2 # 2 # x # x # x 23 # x3 13a4 2 2 13a4 213a4 2 3 # 3 # a4 # a4 132 2 1a4 2 2

12xy5 2 2 12xy5 2 12xy5 2 122 122 1x21x2 1y5 2 1y5 2 122 2 1x2 2 1y5 2 2 In general, 1ab2 n ab # ab # ab # p # ab 14444244443

n factors of ab

1a # a # a # p # a21b # b # b # p # b2 144424443 144424443

n factors of a

n factors of b

anbn

Property 4.3 If a and b are real numbers, and n is a positive integer, then 1ab2 n anbn

Property 4.3 states that when raising a monomial to a power, raise each factor to that power.

It is a simple process to raise a monomial to a power when using the properties of exponents. Study the next example.

EXAMPLE 4

Raise each to the indicated power. (a) 12x2y3 2 4

(b) 13ab5 2 3

(c) 12a4 2 5

3 2 (d) a x2y3 b 5

Solution (a) 12x2y3 2 4 122 4 1x2 2 4 1y3 2 4 16x8y12

(b) 13ab5 2 3 132 3 1a1 2 3 1b5 2 3 27a3b15 (c) 12a4 2 5 122 5 1a4 2 5 32a20 3 2 2 3 (d) a x2y3 b a b 1x2 2 3 1y3 2 3 5 5

8 6 9 xy 125

▼ PRACTICE YOUR SKILL Raise each to the indicated power. (a) 12x3 2 4

(b) 13a3b5 2 2

■

Sometimes problems involve first raising monomials to a power and then multiplying the resulting monomials, as in the following example.

4.2

EXAMPLE 5

Multiply the following: (a) 13x2 2 3 12x3 2 2


203

(b) 1x2y3 2 5 12x2y2 2

Solution (a) 13x2 2 3 12x3 2 2 132 3 1x2 2 3 122 2 1x3 2 2 12721x6 21421x6 2 108x12

(b) 1x2y3 2 5 12x2y2 2 112 5 1x2 2 5 1y3 2 5 122 2 1x2 2 2 1y2 2 1121x10 2 1y15 21421x4 21y2 2 4x14y17

▼ PRACTICE YOUR SKILL Multiply the following: (a) 15a3 2 2 12a4 2 3

(b) 13m4n2 2 3 1mn2 2

■

3 Multiply a Polynomial by a Monomial The distributive property along with the properties of exponents forms a basis for finding the product of a monomial and a polynomial. The next example illustrates these ideas.

EXAMPLE 6

Multiply the following:

(b) 15a2 2 1a3 2a 2 12

(a) 13x2 12x2 6x 12

(c) 12xy216x2y 3xy2 4y3 2

Solution (a) 13x212x2 6x 12 13x212x2 2 13x2 16x2 13x2 112

Apply distributive property

6x3 18x2 3x

(b) 15a2 2 1a3 2a2 12 15a2 2 1a3 2 15a2 2 12a2 2 15a2 2 112 5a5 10a4 5a2

(c) 12xy216x2y 3xy2 4y3 2

12xy2 16x2y2 12xy213xy2 2 12xy2 14y3 2 12x3y2 6x2y3 8xy4


(a) 14a3 21a2 3ab 5b2 2

(b) 12y2 15x2y 6xy y3 2

■

Once you feel comfortable with this process, you may want to perform most of the work mentally and then simply write down the final result. See whether you understand the following examples. 1. 3x12x 32 6x2 9x 2. 4x12x2 3x 12 8x3 12x2 4x

3. ab13a2b 2ab2 b3 2 3a3b2 2a2b3 ab4

204

Chapter 4


4 Apply Multiplication of Monomials to Geometric Problems We conclude this section by making a connection between algebra and geometry.

EXAMPLE 7

Suppose that the dimensions of a rectangular solid are represented by x, 2x, and 3x as shown in Figure 4.6. Express the volume and total surface area of the figure.

Solution Using the formula V lwh, we can express the volume of the rectangular solid as (2x)(3x)(x), which equals 6x3. The total surface area can be described as follows:

x

Area of front and back rectangles: 21x2 13x2 6x2

2x

Area of left side and right side: 212x2 1x2 4x2

3x

Area of top and bottom: 212x213x2 12x2

Figure 4.6

We can represent the total surface area by 6x2 4x2 12x2, or 22x2.

▼ PRACTICE YOUR SKILL The dimensions of some rectangular regions of a floor plan are shown in the figure below. Express the area of the floor plan. x

x

x

x 2x

CONCEPT QUIZ

4x

■

2x

For Problems 1– 6, answer true or false. 1. 2. 3. 4. 5. 6.

When multiplying factors with the same base, add the exponents. 32 # 32 94 2x 2 # 3x3 6x 6 1x 2 2 3 x 5 14x3 2 2 4x6 To simplify 13x 2y212x3y2 2 4, use the order of operations to first raise 2x3y2 to the fourth power, and then multiply the monomials.

Problem Set 4.2 1 Apply the Properties of Exponents to Multiply Monomials For Problems 1–30, multiply using the properties of exponents to help with the manipulation. 1. 15x219x2

2. 17x218x2

3. 13x2 217x2

4. 19x214x3 2

5. 13xy2 12xy2

6. 16xy2 13xy2

2

7. 12x y2 17x2

8. 15xy2 2 14y2

9. 14a2b2 2 112ab2

10. 13a3b2 113ab2 2

15. 15x2 212x213x3 2

16. 14x212x2 216x4 2

17. 14xy212x217y2 2

18. 15y2 213xy215x2 2

19. 12ab21ab213b2

20. 17ab214a21ab2

21. 16cd213c2d214d2

22. 12c3d216d3 215cd2

2 3 23. a xyb a x2y4 b 3 5

8 5 24. a xb a x2yb 6 3

25. a

9 15 26. a a3b4 b a ab2 b 5 6

7 2 8 a bb a b4 b 12 21

11. 1xy2 15x3 2

12. 17y2 21x2y2

27. 10.4x5 210.7x3 2

28. 11.2x4 210.3x2 2

13. 18ab2c2113a2c2

14. 19abc3 2114bc2 2

29. 14ab211.6a3b2

30. 16a2b211.4a2b4 2

4.2

2 Apply the Properties of Exponents to Raise a Monomial to a Power


74. 215x 3y2 51x 4y2 75. 2x1x2 3x 42 x12x2 3x 62

For Problems 31– 46, raise each monomial to the indicated power. Use the properties of exponents to help with the manipulation.

76. 3x12x2 x 52 2x1x2 4x 72

31. 12x 2

32. 13x 2

78. 233x 12x 12 4 213x 42

33. 13a2b3 2 2

34. 18a4b5 2 2

79. 413x 22 5 32x 13x 42 4

35. 13x 2

36. 12x 2

80. 512x 12 3 3x 14x 32 4

37. 14x4 2 3

38. 13x3 2 3

4 2

2 3

4 5 2 39. 19x y 2

3 2

4 3

40. 18x6y4 2 2

41. 12x2y2 4

42. 12x2y3 2 5

43. 13a3b2 2 4

44. 12a4b2 2 4

45. 1x2y2 6

46. 1x2y3 2 7

77. 3 32x 1x 22 4 41x 22

4 Apply Multiplication of Monomials to Geometric Problems 81. Express in simplified form the sum of the areas of the two rectangles shown in Figure 4.7.

4

3 For Problems 47–56, perform the indicated operations and simplify. 47. 13x2 2 12x3 2

48. 12x2 3 14x5 2

49. 13x2 3 14x2 2

50. 13xy2 2 12x2y2 4

51. 15x2y2 2 1xy2 2 3

52. 1x2y2 3 16xy2 2

53. 1a2bc3 2 3 1a3b2 2

54. 1ab2c3 2 4 1a2b2 3

55. 12x2y2 2 4 1xy3 2 3

56. 13xy2 3 1x2y3 2 4

205

x−1

x+2

Figure 4.7 82. Express in simplified form the volume and the total surface area of the rectangular solid in Figure 4.8.

x 4

3 Multiply a Polynomial by a Monomial For Problems 57–70, multiply by using the distributive property. 57. 5x13x 22

58. 7x12x 52

59. 3x2 16x 22

60. 4x2 17x 22

61. 4x17x2 42

62. 6x19x2 52

63. 2x1x2 4x 62

64. 3x12x2 x 52

65. 6a13a2 5a 72

66. 8a14a2 9a 62

67. 7xy14x2 x 52

68. 5x2y13x2 7x 92

69. xy19x2 2x 62

70. xy2 16x2 x 12

2x Figure 4.8 83. Represent the area of the shaded region in Figure 4.9. The length of a radius of the smaller circle is x, and the length of a radius of the larger circle is 2x.

Figure 4.9 84. Represent the area of the shaded region in Figure 4.10.

For Problems 71– 80, remove the parentheses by multiplying and then simplify by combining similar terms; for example, x−2

31x y2 21x 3y2 3x 3y 2x 6y 5x 9y

x 4

71. 51x 2y2 412x 3y2 3x + 2

72. 312x 5y2 214x y2 73. 41x 3y2 312x y2

Figure 4.10

206

Chapter 4


THOUGHTS INTO WORDS 85. How would you explain to someone why the product of x3 and x4 is x7 and not x12?

87. How can Figure 4.11 be used to demonstrate geometrically that x1x 22 x2 2x?

86. Suppose your friend was absent from class the day that this section was discussed. How would you help her understand why the property 1bn 2 m bmn is true?

x

x

2

Figure 4.11

FURTHER INVESTIGATIONS For Problems 88 –97, find each of the indicated products. Assume that the variables in the exponents represent positive integers; for example,

1x2n 21x4n 2 x2n4n x6n 88. 1xn 2 1x3n 2

89. 1x2n 2 1x5n 2

90. 1x2n1 21x3n2 2

91. 1x5n2 21xn1 2

92. 1x3 21x4n5 2

93. 1x6n1 21x4 2

94. 12xn 213x2n 2

95. 14x3n 215x7n 2

96. 16x2n4 215x3n4 2

97. 13x5n2 214x2n2 2

Answers to the Example Practice Skills 4 9 3 a b 3. (a) y6 (b) m8 4. (a) 16x12 (b) 9a6b10 7 6. (a) 4a5 12a4b 20a3b2 (b) 10x2y2 12xy2 2y2 7. 12x2 2. (a) 24x5 (b)

1. (a) y8 (b) a3 (b) 27m14n8

5. (a) 200a18


4.3

2. False

3. False 4. False

5. False 6. True

Multiplying Polynomials OBJECTIVES 1

Use the Distributive Property to Find the Product of Two Polynomials

2

Use the Shortcut Pattern to Find the Product of Two Binomials

3

Raise a Binomial to a Power

4

Use Special Product Patterns to Find Products

5

Apply Polynomials to Geometric Problems

1 Use the Distributive Property to Find the Product of Two Polynomials In general, to go from multiplying a monomial times a polynomial to multiplying two polynomials requires the use of the distributive property. Consider some examples.

4.3

EXAMPLE 1

Multiplying Polynomials

207

Find the product of 1x 32 and 1y 42 .

Solution 1x 32 1y 42 x1y 42 31y 42 x1y2 x142 31y2 3142 xy 4x 3y 12


Find the product of 1x 52 and 1y 22 .

■

Notice that each term of the first polynomial is multiplied times each term of the second polynomial.

EXAMPLE 2

Find the product of 1x 22 and 1y z 52 .

Solution 1x 22 1y z 52 x1y z 52 21y z 52 x1y2 x1z2 x152 21y2 21z2 2152 xy xz 5x 2y 2z 10


Find the product of 1x 42 and 1y z 32 .

■

Usually, multiplying polynomials will produce similar terms that can be combined to simplify the resulting polynomial.

EXAMPLE 3

Multiply 1x 32 1x 22 .

Solution 1x 32 1x 22 x1x 22 31x 22 x2 2x 3x 6 x2 5x 6

Combine like terms

▼ PRACTICE YOUR SKILL Multiply 1x 621x 52 .

EXAMPLE 4

■


Solution 1x 42 1x 92 x1x 92 41x 92 x2 9x 4x 36 x2 5x 36

Combine like terms

208

Chapter 4



EXAMPLE 5

■

Multiply 1x 42 1x2 3x 22 .

Solution 1x 42 1x2 3x 22 x1x2 3x 22 41x2 3x 22 x3 3x2 2x 4x2 12x 8 x3 7x2 14x 8


Multiply 1x 321x2 2x 62 .

EXAMPLE 6

■

Multiply 12x y2 13x2 2xy 4y2 2 .

Solution 12x y2 13x2 2xy 4y2 2 2x13x2 2xy 4y2 2 y13x2 2xy 4y2 2 6x3 4x2y 8xy2 3x2y 2xy2 4y3 6x3 7x2y 10xy2 4y3


Multiply 1x 2y2 15x2 3xy 4y2 2 .

■

2 Use the Shortcut Pattern to Find the Product of Two Binomials Perhaps the most frequently used type of multiplication problem is the product of two binomials. It will be a big help later if you can become proficient at multiplying binomials without showing all of the intermediate steps. This is quite easy to do if you use a three-step shortcut pattern demonstrated by the following examples.

EXAMPLE 7


Solution 1 3

1

2

3

(x + 5)(x + 7) = x 2 + 12 x + 35. 2 Figure 4.12

Step 1 Multiply x # x. Step 2 Multiply 5 # x and 7 # x and combine them. Step 3 Multiply 5 # 7.

4.3


209


EXAMPLE 8

■


Solution 1 3

1

2

3

(x − 8)(x + 3) = x 2 − 5x − 24. 2 Figure 4.13


EXAMPLE 9

■


Solution 1 3

1

2

3

(3x + 2)(2x − 5) = 6x 2 − 11 x − 10. 2 Figure 4.14

▼ PRACTICE YOUR SKILL Multiply (4x 1)(2x 3).

■

The mnemonic device FOIL is often used to remember the pattern for multiplying binomials. The letters in FOIL represent First, Outside, Inside, and Last. If you look back at Examples 7 through 9, step 1 is to find the product of the first terms in each binomial; step 2 is to find the product of the outside terms and the inside terms; and step 3 is to find the product of the last terms in each binomial. Now see whether you can use the pattern to find these products: 1x 32 1x 72 13x 12 12x 52 1x 22 1x 32 14x 52 1x 22 Your answers should be x 2 10x 21, 6x 2 17x 5, x 2 5x 6, and 4x 2 3x 10.

210

Chapter 4


Keep in mind that the shortcut pattern applies only to finding the product of two binomials. For other situations, such as finding the product of a binomial and a trinomial, we suggest showing the intermediate steps as follows: 1x 32 1x2 6x 72 x1x2 2 x16x2 x172 31x2 2 316x2 3172 x3 6x2 7x 3x2 18x 21 x3 9x2 11x 21 Perhaps you could omit the first step, and shorten the form as follows: 1x 42 1x2 5x 62 x3 5x2 6x 4x2 20x 24 x3 9x2 14x 24 Remember that you are multiplying each term of the first polynomial times each term of the second polynomial and combining similar terms.

3 Raise a Binomial to a Power Exponents are also used to indicate repeated multiplication of polynomials. For example, we can write 1x 421x 42 as 1x 42 2. Thus to square a binomial we simply write it as the product of two equal binomials and apply the shortcut pattern. 1x 42 2 1x 421x 42 x2 8x 16 1x 52 2 1x 521x 52 x2 10x 25 12x 32 2 12x 3212x 32 4x2 12x 9 When you square binomials, be careful not to forget the middle term. That is to say, 1x 32 2 x2 32; instead, 1x 32 2 1x 321x 32 x2 6x 9. The next example suggests a format to use when cubing a binomial.

EXAMPLE 10

Find the indicated product of 1x 42 3.

Solution 1x 42 3 1x 421x 421x 42 1x 421x2 8x 162

Use FOIL to multiply two of the factors

x1x2 8x 162 41x2 8x 162

Use the distributive property

x3 8x2 16x 4x2 32x 64 x3 12x2 48x 64


Find the indicated product of 1x 52 3.

■

4 Use Special Product Patterns to Find Products When we multiply binomials, some special patterns occur that you should recognize. We can use these patterns to find products and later to factor polynomials. We will state each of the patterns in general terms followed by examples to illustrate the use of each pattern.

4.3

PAT T E R N 1

1a b2 2 1a b2 1a b2 a 2 Square of the first term of the binomial

Examples


2ab

Twice the product of the two terms of the binomial

211

b2 Square of the second term of the binomial

1x 42 2 x2 8x 16 12x 3y2 2 4x2 12xy 9y2 15a 7b2 2 25a2 70ab 49b2


Use the pattern 1a b2 2 a2 2ab b2 to find 1x 82 2.

PAT T E R N 2

1a b2 2 1a b2 1a b2 a2 Square of the first term of the binomial

Examples

2ab

Twice the product of the two terms of the binomial

■


1x 82 2 x2 16x 64 13x 4y2 2 9x2 24xy 16y2 14a 9b2 2 16a2 72ab 81b2


Use the pattern 1a b2 2 a2 2ab b2 to find 1x 12 2.

PAT T E R N 3

1a b21a b2

a2

Square of the first term of the binomial

Examples

■


1x 721x 72 x2 49 12x y212x y2 4x2 y2 13a 2b213a 2b2 9a2 4b2


Use the pattern 1a b21a b2 a2 b2 to find 13x 52 13x 52 .

■

212

Chapter 4


5 Apply Polynomials to Geometric Problems As you might expect, there are geometric interpretations for many of the algebraic concepts presented in this section. We will give you the opportunity to make some of these connections between algebra and geometry in the next problem set. We conclude this section with a problem that allows us to use some algebra and geometry.

EXAMPLE 11

Apply Your Skill A rectangular piece of tin is 16 inches long and 12 inches wide as shown in Figure 4.15. From each corner a square piece x inches on a side is cut out. The flaps are then turned up to form an open box. Find polynomials that represent the volume and the exterior surface area of the box.

16 inches x x

12 inches

Figure 4.15

Solution The length of the box is 16 2x, the width is 12 2x, and the height is x. From the volume formula V lwh, the polynomial (16 2x)(12 2x)(x), which simplifies to 4x3 56x2 192x, represents the volume. The outside surface area of the box is the area of the original piece of tin minus the four corners that were cut off. Therefore, the polynomial 16(12) 4x2, or 192 4x2, represents the outside surface area of the box.

▼ PRACTICE YOUR SKILL A rectangular piece of cardboard is 10 inches long and 6 inches wide. From each corner a square piece y inches on a side is cut out. The flaps are then turned up to form an open box. Find the polynomials that represent the volume and the exterior surface area of the box. ■

CONCEPT QUIZ

For Problems 1–5, answer true or false. 1. The algebraic expression (x y)2 is called the square of a binomial. 2. The algebraic expression (x y)(x 2xy y) is called the product of two binomials. 3. The mnemonic device FOIL stands for first, outside, inside, and last. 4. 1a 22 2 a2 4 5. 1y 321y 32 y2 9

4.3


213

Problem Set 4.3 1 Use the Distributive Property to Find the Product of Two Polynomials For Problems 1–10, find the indicated products by applying the distributive property; for example, 1x 121y 52 x1y2 x152 11y2 1152 1. 1x 221y 32

For Problems 37– 80, find the indicated products by using the shortcut pattern for multiplying binomials. 37. 1x 221x 92

38. 1x 321x 82

39. 1x 621x 22

40. 1x 821x 62

41. 1x 321x 112

42. 1x 421x 102

43. 1n 421n 32

44. 1n 521n 92

45. 1n 621n 122

46. 1n 821n 132

47. 1y 321y 72

48. 1y 221y 122

49. 1y 721y 122

50. 1y 421y 132

51. 1x 521x 72

52. 1x 121x 92

53. 1x 1421x 82

54. 1x 1521x 62

55. 1a 1021a 92

56. 1a 721a 62

x2 8x 3x 24

57. 12a 121a 62

58. 13a 221a 42

x2 11x 24

59. 15x 221x 72

60. 12x 321x 82

61. 13x 7212x 12

62. 15x 6214x 32

63. 14a 3213a 42

64. 15a 4214a 52

65. 16n 5212n 32

66. 14n 3216n 72

xy 5x y 5

2. 1x 321y 62

3. 1x 421y 12

4. 1x 521y 72

5. 1x 521y 62

6. 1x 721y 92

7. 1x 221y z 12

8. 1x 421y z 42

9. 12x 3213y 12

10. 13x 2212y 52

For Problems 11–36, find the indicated products by applying the distributive property and combining similar terms. Use the following format to show your work: 1x 321x 82 x1x2 x182 31x2 3182

11. 1x 321x 72

2 Use the Shortcut Pattern to Find the Product of Two Binomials

12. 1x 421x 22

13. 1x 821x 32

14. 1x 921x 62

15. 1x 721x 12

16. 1x 1021x 82

17. 1n 421n 62

18. 1n 321n 72

67. 17x 4212x 32

68. 18x 5213x 72

19. 13n 121n 62

20. 14n 321n 62

69. 15 x219 2x2

70. 14 3x212 x2

21. 15x 2213x 72

22. 13x 4217x 12

71. 12x 3214x 52

72. 13x 1219x 22

23. 1x 321x2 4x 92

24. 1x 221x2 6x 22

73. 13x 1213x 42

74. 12x 5214x 12

25. 1x 421x2 x 62

26. 1x 521x2 2x 72

75. 18n 3219n 42

76. 16n 5219n 72

27. 1x 5212x2 3x 72

28. 1x 4213x2 4x 62

77. 13 2x219 x2

78. 15 4x214 5x2

79. 14x 3215x 22

80. 12x 7217x 32

29. 12a 1214a2 5a 92 30. 13a 2212a2 3a 52 31. 13a 521a2 a 12

3 Raise a Binomial to a Power

32. 15a 221a2 a 32

For Problems 81– 88, find the indicated products. Don’t forget that 1x 22 3 means (x 2)(x 2)(x 2).

33. 1x2 2x 32 1x2 5x 42

81. 1x 22 3

82. 1x 42 3

34. 1x2 3x 42 1x2 5x 22

83. 1x 32 3

84. 1x 12 3

35. 1x2 6x 72 1x2 3x 92

85. 12n 12 3

86. 13n 22 3

36. 1x2 5x 42 1x2 7x 82

87. 13n 22 3

88. 14n 32 3

214

Chapter 4


4 Use Special Product Patterns to Find Products

5 Apply Polynomials to Geometric Problems

For Problems 89 –118, use one of the appropriate patterns (a b)2 a2 2ab b2, (a b)2 a2 2ab b2, or (a b)(a b) a2 b2 to find the indicated products. 89. 1x 72 2

90. 1x 92 2

91. 15x 2215x 22

92. 16x 1216x 12

93. 1x 12 2

94. 1x 42 2

95. 13x 72 2

96. 12x 92 2

97. 12x 32

98. 14x 52

2

99. 12x 3y2 12x 3y2 101. 11 5n2

3

x

2

100. 13a b2 13a b2 102. 12 3n2

2

2

103. 13x 4y2 2

104. 12x 5y2 2

105. 13 4y2 2

106. 17 6y2 2

107. 11 7n211 7n2

108. 12 9n212 9n2

109. 14a 7b2

110. 16a b2

2

121. Explain how Figure 4.16 can be used to demonstrate geometrically that (x 3)(x 5) x2 8x 15.

x

5

Figure 4.16 122. Explain how Figure 4.17 can be used to demonstrate geometrically that 1x 521x 32 x2 2x 15. x−3

3

x

2

112. 1x 6y2 2

111. 1x 8y2 2 113. 15x 11y2 15x 11y2

5

114. 17x 9y2 17x 9y2 115. x18x 12 18x 12

x

116. 3x15x 72 15x 72

Figure 4.17

117. 2x14x y2 14x y2 118. 4x12 3x2 12 3x2 119. John, who has not mastered squaring a binomial, turns in the homework shown below. Identify any mistakes in John’s work, and then write your answer for the problem. 2 4) =22 x + 16 x 16 (a) 1x (x42+2

2

123. A square piece of cardboard is 14 inches long on each side. From each corner a square piece x inches on a side is cut out as shown in Figure 4.18. The flaps are then turned up to form an open box. Find polynomials that represent the volume and the exterior surface area of the box. 14 inches

2

2

2

14 inches

2 25 x=22 x –5x5x+ 25 (b) 1x (x52–2 5)

x x

6) x=22 x 12x – 12x– 36 36 62–22 (c) 1x (x 2

2

22 (3x22+22 2) = 3x 12x + 44 3x +12x (d) 13x

2

2

2 1) =22 x + 2x + 2 x 2x 2 (e) 1x (x12+2

2

2

3) x=22 x 6x + 6x+ 99 x (x 332–22 (f) 1x 2

2

120. Use the pattern a2 2ab b2 to find 130 62 2.

Figure 4.18

THOUGHTS INTO WORDS 124. Describe the process of multiplying two polynomials. 125. Illustrate as many uses of the distributive property as you can.

126. Determine the number of terms in the product of (x y z) and (a b c) without doing the multiplication. Explain how you arrived at your answer.

4.4

Dividing by Monomials

215

FURTHER INVESTIGATIONS 127. The following two patterns result from cubing binomials: 1a b2 3 a3 3a2b 3ab2 b3 1a b2 3 a3 3a2b 3ab2 b3 Use these patterns to redo Problems 81– 88. 128. Find a pattern for the expansion of 1a b2 4. Then use the pattern to expand 1x 22 4, 1x 32 4, and 12x 12 4. 129. We can use some of the product patterns to do arithmetic computations mentally. For example, let’s use the pattern 1a b2 2 a2 2ab b2 to mentally compute 312. Your thought process should be 312 130 12 2 302 2(30)(1) 12 961. Compute each of the following numbers mentally and then check your answers.

(a) 192

(b) 292

(c) 492

(d) 792

(e) 382

(f) 582

131. Every whole number with a units digit of 5 can be represented by the expression 10x 5, where x is a whole number. For example, 35 10(3) 5 and 145 10(14) 5. Now observe the following pattern for squaring such a number: 110x 52 2 100x2 100x 25 100x1x 12 25 The pattern inside the dashed box can be stated as “add 25 to the product of x, x 1, and 100.” Thus to mentally compute 352 we can figure 352 3(4)(100) 25 1225. Compute each of the following numbers mentally and then check your answers.

(a) 212

(b) 412

(c) 712

(a) 152

(b) 252

(c) 452

(d) 322

(e) 522

(f) 822

(d) 552

(e) 652

(f) 752

(g) 852

(h) 952

(i) 1052

130. Use the pattern 1a b2 2 a2 2ab b2 to compute each of the following numbers mentally and then check your answers.

Answers to the Example Practice Skills 1. 5. 10. 11.

xy 2x 5y 10 2. xy xz 3x 4y 4z 12 3. x2 11x 30 4. x2 6x 16 x3 5x2 12x 18 6. 5x3 7x2y 2xy2 8y3 7. x2 7x 6 8. x2 4x 21 9. 8x2 10x 3 x3 15x2 75x 125 Pattern 1. x2 16x 64 Pattern 2. x2 2x 1 Pattern 3. 9x2 25 V 4y3 32y 2 60y cubic inches; A 60 4y2 square inches


4.4

2. False

3. True

4. False 5. False

Dividing by Monomials OBJECTIVES 1

Apply the Properties of Exponents to Divide Monomials

2

Divide Polynomials by Monomials

1 Apply the Properties of Exponents to Divide Monomials To develop an effective process for dividing by a monomial we must rely on yet another property of exponents. This property is also a direct consequence of the definition of exponents and is illustrated by the following examples.

216

Chapter 4


x5 x#x#x#x#x x3 2 x#x x a#a#a#a a4 a 3 a#a#a a y7 3

y

y#y#y#y#y#y#y y4 y#y#y

x#x#x#x x4 # # # 1 4 x x x x x y3 3

y

y#y#y 1 y#y#y

Property 4.4 If b is any nonzero real number, and n and m are positive integers, then 1.

bn bnm bm

2.

bn 1 bm

when n m

when n m

(The situation n m will be discussed in a later section.)

Applying Property 4.4 to the previous examples yields these results: x5 x52 x3 x2 a4 a43 a1 a3 y7 y3

Usually written as a

y73 y4

x4 1 x4 y3 y3

1

Property 4.4 along with our knowledge of dividing integers provides the basis for dividing a monomial by another monomial. Consider the next example.

EXAMPLE 1

Divide the monomials. (a)

16x5 2x3

(b)

81a12 9a4

Solution (a)

16x5 8x53 8x2 2x3

(b)

81a12 9a124 9a8 9a4

(c)

45x4 9x4

(d)

54x3y7 6xy5

4.4

(c) (d)

45x4 5 9x4 54x3y7 6xy5

Dividing by Monomials

217

x4 1 x4

9x31y75 9x2y2

▼ PRACTICE YOUR SKILL Divide the monomials. 32b8 4b2

(a)

(b)

6a4b7 6ab2

■

2 Divide Polynomials by Monomials a b ab ; this property serves as the basis for dividing a polynomial c c c by a monomial. Consider these examples. Recall that

25x3 10x2 25x3 10x2 5x2 2x 5x 5x 5x 35x8 28x6 35x8 28x6 5x5 4x3 3 3 7x 7x 7x3

ab a b c c c

To divide a polynomial by a monomial, we simply divide each term of the polynomial by the monomial. Here are some additional examples.

EXAMPLE 2

Perform the division. (a)

12x3y2 14x2y5 2xy

(b)

48ab5 64a2b 16ab

(c)

33x6 24x5 18x4 3x

Solution (a)

12x3y2 14x2y5 12x3y2 14x2y5 6x2y 7xy4 2xy 2xy 2xy

(b)

48ab5 64a2b 48ab5 64a2b 3b4 4a 16ab 16ab 16ab

(c)

33x6 24x5 18x4 33x6 24x5 18x4 3x 3x 3x 3x 11x5 8x4 6x3

▼ PRACTICE YOUR SKILL Perform the division. (a)

21m4n5 14m2n3 7m2n

(b)

12a5 18a4 36a3 6a3

■

As with many skills, once you feel comfortable with the process, you may want to perform some of the steps mentally. Your work could take on the following format. 24x4y5 56x3y9 8x2y3

3x2y2 7xy6

13a2b 12ab2 13a 12b ab

218

Chapter 4


CONCEPT QUIZ

For Problems 1–5, answer true or false. 1. When dividing factors that have the same base, add the exponents. 10a 6 8a4 2. 2a2 y8 3. 4 y 2 y 6x5 3x 2x4 4. 3x x3 5. 3 0 x

Problem Set 4.4 32.

27x7 36x5 45x3 3x

33.

24n8 48n5 78n3 6n3

34.

56n9 84n6 91n2 7n2

84x 7x5

65x2y3 9. 5xy

35.

60a7 96a3 12a

91a4b6 11. 13a3b4

72a5b4 12. 12ab2

37.

32x6y2 15. x

39.

48a2b2 60a3b4 6ab

41.

12a2b2c2 52a2b3c5 4a2bc

56a b c 4abc

42.

48a3b2c 72a2b4c5 12ab2c

90x y z

43.

9x2y3 12x3y4 xy

45.

42x6 70x4 98x2 14x2

For Problems 25 –50, perform each division of polynomials by monomials.

46.

48x8 80x6 96x4 16x4

8x4 12x5 25. 2x2

12x3 16x6 26. 4x

47.

15a3b 35a2b 65ab2 5ab

1 Apply the Properties of Exponents to Divide Monomials For Problems 1–24, divide the monomials. 10

1.

12

x x2

2.

x x5

5.

16n 2n2

5

4.

10.

13.

16.

19.

8x 4x3

4x 2x

6.

54n 6n4

72x 9x3 70x3y4 2

5x y 18x2y6 xy2 54x5y3 y2 ab ab 60a b c 15a2c

8

5

8.

14.

17.

20.

24x3y4 x2y2 96x5y7 12y3

18.

23.

84x4y9 14x4 2 3 5

6ab ab

21.

80xy z

2 6

3 2

22.

3. 6

3

7.

3

5xyz2

3 2 8

24.

6xy2z4

2 Divide Polynomials by Monomials

27x2y4 45xy4 9xy

3

27.

9x6 24x4 3x3

28.

35x8 45x6 5x4

48.

24a4b2 36a3b 48a2b 6ab

29.

28n5 36n2 4n2

30.

42n6 54n4 6n4

49.

xy 5x2y3 7x2y6 xy

31.

35x6 56x5 84x3 7x2

50.

9x2y3 xy 14xy4 xy

36.

38.

65a8 78a4 13a2 40x4y7 64x5y8 8x3y4

40.

45a3b4 63a2b6 9ab2

44.

15x3y 27x2y4 xy

4.5

Dividing by Binomials

219

THOUGHTS INTO WORDS 51. How would you explain to someone why the quotient of x8 and x2 is x6 and not x4?

coefficient in the denominator. What can you tell him that might help?

52. Your friend is having difficulty with problems such as 36x3y2 12x2y and where there appears to be no numerical xy xy

Answers to the Example Practice Skills 1. (a) 8b6 (b) a3b5

2. (a) 3m 2n4 2n2 (b) 2a2 3a 6


4.5

2. False 3. False

4. False 5. False

Dividing by Binomials OBJECTIVE 1

Divide Polynomials by Binomials

1 Divide Polynomials by Binomials Perhaps the easiest way to explain the process of dividing a polynomial by a binomial is to work a few examples and describe the step-by-step procedure as we go along.

EXAMPLE 1

Divide x2 5x 6 by x 2.

Solution Step 1 Use the conventional long division for- x 2冄 x2 5x 6 mat from arithmetic, and arrange both the dividend and the divisor in descending powers of the variable.

Step 2 Find the first term of the quotient by

x dividing the first term of the dividend x 2冄 x2 5x 6 by the first term of the divisor.

x2 x x

Step 3 Multiply the entire divisor by the term

x of the quotient found in step 2, and x 2冄 x2 5x 6 x1x 22 x2 2x x2 2x position this product to be subtracted from the dividend.

Step 4 Subtract. Remember to add the opposite!

x x 2冄x2 5x 6 x2 2x 3x 6

Step 5 Repeat the process beginning with

x 3 step 2; use the polynomial that resulted x 2冄 x2 5x 6 from the subtraction in step 4 as a new x2 2x dividend. 3x 6 3x 6

3x 3 x 31x 22 3x 6

220

Chapter 4


Thus 1x2 5x 62 1x 22 x 3, which can be checked by multiplying (x 2) and (x 3). 1x 22 1x 32 x2 5x 6


Divide 1x2 11x 282 by 1x 42 .

■

A division problem such as 1x2 5x 62 1x 22 can also be written as x 5x 6 . Using this format, we can express the final result for Example 1 as x2 2 x 5x 6 x 3. (Technically, the restriction x 2 should be made to x2 avoid division by zero.) In general, to check a division problem we can multiply the divisor times the quotient and add the remainder, which can be expressed as 2

Dividend (Divisor)(Quotient) Remainder Sometimes the remainder is expressed as a fractional part of the divisor. The relationship then becomes Remainder Dividend Quotient Divisor Divisor

EXAMPLE 2

Divide 2x2 3x 20 by x 4.

Solution Step 1 x 4冄 2x2 3x 20 Step 2

2x x 4冄 2x2 3x 20

2x2 2x x

Step 3

2x x 4冄 2x2 3x 20

2x1x 42 2x2 8x

2x2 8x

Step 4

2x x 4冄 2x2 3x 20 2x2 8x 5x 20

Step 5

2x 5 x 4冄 2x2 3x 20 2x2 8x 5x 20

5x 5 x 51x 42 5x 20

5x 20

✔ Check

1x 42 12x 52 2x2 3x 20

Therefore,

2x2 3x 20 2x 5. x4

4.5


221

▼ PRACTICE YOUR SKILL Divide (3x2 5x 2) by (x 2).

■

Now let’s continue to think in terms of the step-by-step division process but organize our work in the typical long division format.

EXAMPLE 3

Divide 12x2 x 6 by 3x 2.

Solution 4x 3 3x 2冄12x2 x 6 12x2 8x 9x 6 9x 6

✔ Check

13x 22 14x 32 12x2 x 6

Therefore,

12x2 x 6 4x 3. 3x 2


Divide 18x2 14x 152 by 14x 32.

■

Each of the next three examples illustrates another aspect of the division process. Study them carefully; then you should be ready to work the exercises in the next problem set.

EXAMPLE 4

Perform the division 17x2 3x 42 1x 22 .

Solution 7x 11 1 x 2冄 7x2 3x 4 7x2 14x 11x 4 11x 22 18

A remainder of 18

✔ Check Just as in arithmetic, we check by adding the remainder to the product of the divisor and quotient. 1x 22 17x 112 18 ⱨ 7x2 3x 4 7x2 3x 22 18 ⱨ 7x2 3x 4 7x2 3x 4 7x2 3x 4 Therefore,

7x2 3x 4 18 . 7x 11 x2 x2


Perform the division 14x2 6x 42 1x 32.

■

222

Chapter 4


EXAMPLE 5

Perform the division

x3 8 . x2

Solution x2 2x 4 x 2冄 x3 0x2 0x 8 x3 2x2 2x2 0x 8 2x2 4x 4x 8 4x 8

✔ Check

Notice the insertion of x2 and x terms with zero coefficients

1x 22 1x2 2x 42 ⱨ x3 8

x3 2x2 4x 2x2 4x 8 ⱨ x3 8 x3 8 x3 8 Therefore,

x3 8 x2 2x 4. x2


EXAMPLE 6


a3 27 . a3


x3 5x2 3x 4 . x2 2x

■

Solution x 3 x2 2x冄x3 5x2 3x 4 x3 2x2 3x2 3x 4 3x2 6x 9x 4

A remainder of 9x 4

We stop the division process when the degree of the remainder is less than the degree of the divisor.

✔ Check

1x2 2x2 1x 32 19x 42 ⱨ x3 5x2 3x 4

x3 3x2 2x2 6x 9x 4 ⱨ x3 5x2 3x 4 x3 5x2 3x 4 x3 5x2 3x 4 Therefore,

9x 4 x3 5x2 3x 4 x3 2 . x2 2x x 2x

▼ PRACTICE YOUR SKILL Perform the division

a3 5a2 2a 3 . a2 a

■

4.5

CONCEPT QUIZ


223

For Problems 1– 6, answer true or false. 1. A division problem written as (x2 x 6) (x 1) could also be written as x2 x 6 . x1 x2 7x 12 2. The division of x 4 could be checked by multiplying (x 4) x3 by (x 3). 3. For the division problem (2x2 5x 9) (2x 1), the remainder is 7. The 7 remainder for the division problem is sometimes expressed as . 2x 1 4. In general, to check a division problem we can multiply the divisor times the quotient and subtract the remainder. 5. If a term is inserted to act as a placeholder, then the coefficient of the term must be zero. 6. When performing division, the process ends when the degree of the remainder is less than the degree of the divisor.

Problem Set 4.5 1 Divide Polynomials by Binomials For Problems 1– 40, perform the divisions. 1. 1x2 16x 482 1x 42 2. 1x2 15x 542 1x 62 3. 1x2 5x 142 1x 72 4. 1x2 8x 652 1x 52

22. 1n3 67n 242 1n 82 23. 1x3 272 1x 32 24. 1x3 82 1x 22 25.

27x3 64 3x 4

26.

8x3 27 2x 3

27.

1 3n2 2n n2

28.

x 5 12x2 3x 2

29.

9t2 3t 4 1 3t

30.

4n2 6n 1 4 2n

31.

6n3 5n2 7n 4 2n 1

32.

21n3 23n2 9n 10 3n 2

33.

4x3 23x2 30x 32 x7

34.

5x3 12x2 13x 14 x1

5. 1x2 11x 282 1x 32 6. 1x 11x 152 1x 22 2

7. 1x2 4x 392 1x 82 8. 1x2 9x 302 1x 122 9. 15n2 n 42 1n 12 10. 17n2 61n 902 1n 102 11. 18y2 53y 192 1y 72 12. 16y2 47y 722 1y 92 13. 120x 31x 72 15x 12 2

14. 127x2 21x 202 13x 42 15. 16x2 25x 82 12x 72 16. 112x2 28x 272 16x 52

35. 1x3 2x2 3x 12 1x2 2x2

17. 12x3 x2 2x 82 1x 22

36. 1x3 6x2 2x 12 1x2 3x2

18. 13x3 7x2 26x 242 1x 42

37. 12x3 4x2 x 52 1x2 4x2

19. 15n3 11n2 15n 92 1n 32

38. 12x3 x2 3x 52 1x2 x2

20. 16n3 29n2 6n 52 1n 52

39. 1x4 162 1x 22

21. 1n3 40n 242 1n 62

40. 1x4 812 1x 32

224

Chapter 4


THOUGHTS INTO WORDS 41. Give a step-by-step description of how you would do the division problem 12x3 8x2 29x 302 1x 62 .

42. How do you know by inspection that the answer to the following division problem is incorrect?

13x3 7x2 22x 82 1x 42 3x2 5x 1

Answers to the Example Practice Skills 1. x 7

2. 3x 1

3. 2x 5

4. 4x 6

14 x3

5. a2 3a 9

6. a 4

6a 3 a2 a


4.6

2. True 3. True

4. False

5. True 6. True

Zero and Negative Integers as Exponents OBJECTIVES 1

Apply the Properties of Exponents Including Negative and Zero Exponents to Simplify Expressions

2

Write Numbers in Scientific Notation

3

Write Numbers Expressed in Scientific Notation in Standard Decimal Notation

4

Use Scientific Notation to Evaluate Numerical Expressions

1 Apply the Properties of Exponents Including Negative and Zero Exponents to Simplify Expressions Thus far in this text we have used only positive integers as exponents. The next definitions and properties serve as a basis for our work with exponents.

Definition 4.1 If n is a positive integer and b is any real number, then bn bbb p b

1 424 3 n factors of b

Property 4.5 If m and n are positive integers and a and b are real numbers, except b 0 whenever it appears in a denominator, then 1. bn # bm bnm 2. 1bn 2 m bmn

3. 1ab2 n anbn

4.6

a n an 4. a b n b b 5.

Zero and Negative Integers as Exponents

225

Part 4 has not been stated previously

bn bnm bm

When n m

bn 1 bm

When n m

Property 4.5 pertains to the use of positive integers as exponents. Zero and the negative integers can also be used as exponents. First let’s consider the use of 0 as an exponent. We want to use 0 as an exponent in such a way that the basic properties of exponents will continue to hold. Consider the example x4 # x0. If part 1 of Property 4.5 is to hold, then x4 # x0 x40 x4 Note that x0 acts like 1 because x4 # x0 x4. This suggests the following definition.

Definition 4.2 If b is a nonzero real number, then b0 1 According to Definition 4.2 the following statements are all true. 40 1

16282 0 1 4 0 a b 1 7

n0 1,

1x2y5 2 0 1,

n0 x 0 and y 0

A similar line of reasoning indicates how negative integers should be used as exponents. Consider the example x3 # x3. If part 1 of Property 4.5 is to hold, then x3 # x3 x3132 x0 1 Thus x3 must be the reciprocal of x3 because their product is 1; that is, x3

1 x3

This process suggests the following definition.

Definition 4.3 If n is a positive integer, and b is a nonzero real number, then bn

1 bn

According to Definition 4.3, the following statements are all true. x6

1 x6

226

Chapter 4


23

1 1 8 23

102

1 1 2 100 10

or

0.01

1 1 x4 1# x4 4 1 1 x x4 2 2 a b 3

1 1 9 9 1# 4 4 4 2 2 a b 9 3

2 2 3 2 a b . In other words, to raise a 3 2 fraction to a negative power, we can invert the fraction and raise it to the corresponding positive power.

Remark: Note in the last example that a b

We can verify (we will not do so in this text) that all parts of Property 4.5 hold for all integers. In fact, we can replace part 5 with this statement.

Replacement for Part 5 of Property 4.5 bn bnm for all integers n and m bm

The next example illustrates the use of this new property. In each part of the example, we simplify the original expression and use only positive exponents in the final result.

EXAMPLE 1

Simplify the expressions and use only positive exponents in the answer. x2 x5

(a)

(b)

a3 a7

(c)

y5 2

y

(d)

x6 x6

Solution (a)

1 x2 x 25 x3 3 x x5

(b)

a3 a3172 a37 a 4 a7

(c) (d)

y5 y2

y5122 y52 y3

1 y3

x6 x6162 x66 x0 1 x6

▼ PRACTICE YOUR SKILL Simplify the expressions and use only positive exponents in the answer. (a)

m4 m7

(b)

y2 y6

(c)

z3 z2

(d)

k5 k5

■

4.6


227

The properties of exponents provide a basis for simplifying certain types of numerical expressions, as the following example illustrates.

EXAMPLE 2

Simplify the expressions and use only positive exponents in the answer. (a) 24 # 26

(b) 10 5 # 106

(c)

102 102

(d) (23)2

Solution (a) 24 # 26 246 22 4 (b) 105 # 106 105162 101 (c)

1 10

or

0.1

102 102122 1022 104 10,000 102

(d) 123 2 2 23122 26 64

▼ PRACTICE YOUR SKILL Simplify the expressions and use only positive exponents in the answer. 101 (a) 35 # 38 (b) 102 # 101 (c) (d) 125 2 1 104

■

Having the use of all integers as exponents also expands the type of work that we can do with algebraic expressions. In the following example, we simplify each given expression and use only positive exponents in the final result.

EXAMPLE 3

Simplify the expressions and use only positive exponents in the answer. (a) x8 # x2

(b) a4 # a3 (c) (y3)4 6 2 12x (f) (4x2)(3x1) (g) a b 6x2

(d) (x2 y4)3

(e) a

x1 2 b y2

Solution (a) x8x2 x8122 x6 (b) a4a3 a4132 a7 (c) 1y3 2 4 y3142 y12

1 a7

1 y12

(d) 1x2y4 2 3 1x2 2 3 1y4 2 3 x6y12 (e) a

1x1 2 2 x1 2 x2 b 4 x2y4 2 2 2 y 1y 2 y

x6 y12

(f) 14x2 2 13x1 2 12x2112 12x3 (g) a

12x6 2 b 12x6122 2 2 12x4 2 2 6x2

Divide the coefficients

122 2 1x4 2 2 a

12 x3

1 x8 b 1x8 2 2 4 2

12 2 6

228

Chapter 4


▼ PRACTICE YOUR SKILL Simplify the expressions and use only positive exponents in the answer. (a) z1 # z4

(c) a

(b) 1a5b1 2 2

m2 1 b n4

(d) (2a5b2)(5a3b)

■

2 Write Numbers in Scientific Notation Many scientific applications of mathematics involve the measurement of very large and very small quantities. For example: The speed of light is approximately 29,979,200,000 centimeters per second. A light year—the distance light travels in 1 year—is approximately 5,865,696,000,000 miles. A gigahertz equals 1,000,000,000 hertz. The length of a typical virus cell equals 0.000000075 of a meter. The length of the diameter of a water molecule is 0.0000000003 of a meter. Working with numbers of this type in standard form is quite cumbersome. It is much more convenient to represent very small and very large numbers in scientific notation, sometimes called scientific form. Although negative numbers can be written in scientific form, we will restrict our discussion to positive numbers. A number is in scientific notation when it is written as the product of a number between 1 and 10 (including 1) and an integral power of 10. Symbolically, a number in scientific notation has the form 1N2110k 2 , where 1 N 10, and k is an integer. For example, 621 can be written as 16.2121102 2 , and 0.0023 can be written as 12.32 1103 2 . To switch from ordinary notation to scientific notation, you can use the following procedure.

Write the given number as the product of a number greater than or equal to 1 (and less than 10) and an integral power of 10. To determine the exponent of 10, count the number of places that the decimal point moved when going from the original number to the number greater than or equal to 1 and less than 10. This exponent is (a) negative if the original number is less than 1, (b) positive if the original number is greater than 10, and (c) zero if the original number itself is between 1 and 10.

EXAMPLE 4

Write the following numbers in scientific notation. (a) 0.000179

(b) 8175

(c) 3.14

Solution We can write (a) 0.000179 11.7921104 2 (b) 8175 18.1752110 2 3

(c) 3.14 13.142 1100 2

According to part (a) of the procedure According to part (b) According to part (c)

▼ PRACTICE YOUR SKILL Write the following numbers in scientific notation. (a) 0.00000683

(b) 5,600,000

(c) 9.54

■

4.6


229

We can express the applications given earlier in scientific notation as follows: Speed of light: 29,979,200,000 12.997922 11010 2 centimeters per second Light year: 5,865,696,000,000 15.8656962 11012 2 miles Gigahertz: 1,000,000,000 (1)(10 9) hertz Length of a virus cell: 0.000000075 (7.5)(108) meter Length of the diameter of a water molecule 0.0000000003 (3)(1010) meter

3 Write Numbers Expressed in Scientific Notation in Standard Decimal Notation To switch from scientific notation to ordinary decimal notation you can use the following procedure.

Move the decimal point the number of places indicated by the exponent of 10. The decimal point is moved to the right if the exponent is positive and to the left if it is negative.

EXAMPLE 5

Write the following numbers in standard decimal notation. (a) (4.71)(104)

(b) (1.78)(102)

Solution Thus we can write (a) 14.712 1104 2 47,100

(b) 11.782 110 2 0.0178 2

Two zeros are needed for place value purposes One zero is needed for place value purposes

▼ PRACTICE YOUR SKILL Write the following numbers in standard decimal notation. (a) (8.2)(107)

(b) (3.46)(104)

■

4 Use Scientific Notation to Evaluate Numerical Expressions The use of scientific notation along with the properties of exponents can make some arithmetic problems much easier to evaluate. The next examples illustrate this point.

EXAMPLE 6

Evaluate (4000)(0.000012).

Solution 14000210.0000122 142 1103 2 11.22 1105 2 142 11.221103 2 1105 2 14.821102 2 0.048

▼ PRACTICE YOUR SKILL Use scientific notation and properties of exponents to evaluate (50,000)(0.0000013). ■

230

Chapter 4


EXAMPLE 7

Evaluate

960,000 . 0.032

Solution 19.62 1105 2 960,000 0.032 13.22 1102 2 132 1107 2

10 5 105122 107 102

30,000,000

▼ PRACTICE YOUR SKILL Use scientific notation and properties of exponents to evaluate

EXAMPLE 8

Evaluate

16000210.000082 140,0002 10.0062

8,400 . 0.042

■

.

Solution 16000210.000082 140,0002 10.0062

1621103 2 182 1105 2

1421104 2 162 1103 2 14821102 2 12421101 2

122 1103 2

102 1021 103 101

0.002

▼ PRACTICE YOUR SKILL Use scientific notation and properties of exponents to evaluate

CONCEPT QUIZ

1900210.00062

130,0002 10.022

.

■

For Problems 1–10, answer true or false. 1. Any nonzero number raised to the zero power is equal to one. 2. The algebraic expression x2 is the reciprocal of x2 if x 0. 3. To raise a fraction to a negative exponent, we can invert the fraction and raise it to the corresponding positive exponent. 1 4. 3 y3 y 5. A number in scientific notation has the form 1N2 110 k 2 , where 1 N 10 and k is any real number. 6. A number is less than zero if the exponent is negative when the number is written in scientific notation. 1 7. 2 x2 x 102 8. 100 104 9. 13.1121102 2 311 10. 15.2421101 2 0.524

4.6


231

Problem Set 4.6 1 Apply the Properties of Exponents Including Negative and Zero Exponents to Simplify Expressions For Problems 1–30, evaluate each numerical expression. 1. 32

2. 25

3 1 5. a b 2 7.

1 24

4 9. a b 3

3. 43 3 2 6. a b 4 8.

0

4. 52

1 31

1 10. a b 2

3

2 3 11. a b 3

12. 1162 0

13. 122 2

14. 132 2

15. 132 2

16. 122 2

17.

1 3 3 a b 4

18.

1 3 4 a b 2

19. 26 # 29

20. 35 # 32

21. 36 # 33

22. 27 # 22

49.

4n1 2n3

50.

12n2 3n5

51.

24x6 8x2

52.

56x5 7x1

53.

52y2 2

13y

54.

91y3 7y3

55. 1x3 2 2

56. 1x1 2 5

57. 1x2 2 2

58. 1x3 2 1

59. 1x3y4 2 1

60. 1x4y2 2 2

61. 1x2y1 2 3

62. 1x3y4 2 2

63. 12n2 2 3

64. 13n1 2 4

65. 14n3 2 2

66. 12n2 2 3

67. 13a2 2 4

68. 15a1 2 2

69. 15x1 2 2

70. 14x2 2 2

71. 12x2y1 2 1

72. 13x2y3 2 2

73. a

x2 1 b y

74. a

75. a

a1 4 b b2

76. a

a3 3 b b2

y2 3

x

b

2

23.

102 101

24.

101 103

77. a

78. a

26.

102 102

x3 1 b y4

25.

101 102

x1 2 b y3

79. a

x2 1 b x3

80. a

x4 2 b x

81. a

2x1 3 b x2

82. a

3x2 1 b x5

83. a

18x1 2 b 9x

84. a

35x2 1 b 7x1

27. 121 # 32 2 1 29. a

1

4 b 3

2

28. 131 # 42 2 1 30. a

3 b 21

3

For Problems 31– 84, simplify each algebraic expression and express your answers using positive exponents only. 31. x6x1

32. x2x7

33. n4n2

34. n8n3

35. a2a3

36. a4a6

37. 12x3 214x2 2

38. 15x4 2 16x7 2

39. 13x6 2 19x2 2

40. 18x8 2 14x2 2

41. 15y1 2 13y2 2

42. 17y3 2 19y4 2

43. 18x4 2 112x4 2

44. 13x2 2 16x2 2

2 Write Numbers in Scientific Notation 85. The U.S. Social Security Administration pays approximately $42,000,000,000 in monthly benefits to all beneficiaries. Write this number in scientific notation. 86. In 1998 approximately 10,200,000,000 pennies were made. Write, in scientific notation, the number of pennies made. 87. The thickness of a dollar bill is 0.0043 inches. Write this number in scientific notation.

45.

x7 x3

46.

x2 x4

88. The length of an Ebola virus cell is approximately 0.0000002 meters. Write this number in scientific notation.

47.

n1 n3

48.

n2 n5

89. The diameter of Jupiter is approximately 89,000 miles. Write this number in scientific notation.

232

Chapter 4


90. Avogadro’s Number, which has a value of approximately 602,200,000,000,000,000,000,000, refers to the calculated value of the number of molecules in a gram mole of any chemical substance. Write Avogadro’s Number in scientific notation. 91. The cooling fan for a computer hard drive has a thickness of 0.025 meters. Write this number in scientific notation. 92. A sheet of 20-lb bond paper has a thickness of 0.097 millimeters. Write the thickness in scientific notation. 93. According to the 2004 annual report for Coca-Cola, the Web site CokePLAY.com was launched in Korea, and it was visited by 11,000,000 people in its first six months. Write the number of hits for the Web site in scientific notation. 94. In 2005, Wal-Mart reported that they served 138,000,000 customers worldwide each week. Write, in scientific notation, the number of customers for each week.

4 Use Scientific Notation to Evaluate Numerical Expressions For Problems 107–118, use scientific notation and the properties of exponents to evaluate each numerical expression. 107. (0.007)(120)

108. (0.0004)(13)

109. (5,000,000)(0.00009)

110. (800,000)(0.0000006)

111.

6000 0.0015

112.

480 0.012

113.

0.00086 4300

114.

0.0057 30,000

115.

0.00039 0.0013

116.

0.0000082 0.00041

120,000210.00042

118.

117.

10.0008210.072

10.006216002

10.0000421302

For Problems 119 –122, convert the numbers to scientific notation and compute the answer.

3 Write Numbers Expressed in Scientific Notation in Standard Decimal Notation For Problems 95 –106, write each number in standard decimal form; for example, 11.421103 2 1400. 95. 1821103 2

96. 162 1102 2

97. 15.2121104 2

98. 17.22 1103 2

99. 11.1421107 2

100. 15.642 1108 2

101. 1721102 2

102. 18.142 1101 2

103. 19.8721104 2

104. 14.372 1105 2

105. 18.6421106 2

106. 13.142 1107 2

119. The U.S. Social Security Administration pays approximately $42,000,000,000 in monthly benefits to all beneficiaries. If there are 48,000,000 beneficiaries, find the average dollar amount each beneficiary receives. 120. In 1998 approximately 10,200,000,000 pennies were made. If the population then was 270,000,000, find the average number of pennies produced per person. Round the answer to the nearest whole number. 121. The thickness of a dollar bill is 0.0043 inches. How tall will a stack of 1,000,000 dollars be? Express the answer to the nearest foot. 122. The diameter of Jupiter is approximately 11 times larger than the diameter of Earth. Find the diameter of Earth given that Jupiter has a diameter of approximately 89,000 miles. Express the answer to the nearest mile.

THOUGHTS INTO WORDS 123. Is the following simplification process correct?

122 2 1 a

1 1 1 1 b a b 2 4 2

124. Explain the importance of scientific notation.

1 4 1 1 a b 4

Can you suggest a better way to do the problem?

FURTHER INVESTIGATIONS 125. Use your calculator to redo Problems 1–16. Be sure that your answers are equivalent to the answers you obtained without the calculator. 126. Use your calculator to evaluate 1140,0002 2. Your answer should be displayed in scientific notation; the format of

the display depends on the particular calculator. For example, it may look like 1.96 10 or 1.96E 10 . Thus in ordinary notation the answer is 19,600,000,000. Use your calculator to evaluate each of the following expressions. Express final answers in ordinary notation.

4.6 (a) 190002 3

(b) 140002 3

(c) 1150,0002 2

(d) 1170,0002 2

(e) 10.0122 5

(f) 10.00152 4


(g) 10.0062 3

(h) 10.022 6

127. Use your calculator to check your answers to Problems 107–118.

Answers to the Example Practice Skills 1 1 1 b2 1 1 1 4 (b) y (c) (d) 1 2. (a) (b) (c) 1000 (d) 32 3. (a) (b) (c) 2 4 3 10 5 5 27 1000 m mn a z z 10b3 6 6 0 (d) 2 4. (a) (6.83)(10 ) (b) (5.6)(10 ) (c) (9.54)(10 ) 5. (a) 82,000,000 (b) 0.000346 6. 0.065 a 7. 200,000 8. 0.0009 1. (a)


2. True 3. True

4. False

233

5. False 6. False

7. True

8. True

9. False 10. True


SUMMARY

EXAMPLE

Classify polynomials by size and degree. (Sec. 4.1, Obj. 1, p. 192)

Terms with variables that contain only whole number exponents are called monomials. A polynomial is a monomial or a finite sum of monomials. The degree of a monomial is the sum of the exponents of the literal factors. The degree of a polynomial is the degree of the term with the highest degree in the polynomial. A one-term polynomial is called a monomial; a two-term polynomial is called a binomial; and a three-term polynomial is called a trinomial.

The polynomial 6x2y3 is a monomial of degree 5.

Add and subtract polynomials. (Sec. 4.1, Obj. 2, p. 193; Sec. 4.1, Obj. 3, p. 194; Sec. 4.1, Obj. 4, p. 194)

Polynomials are added by combining similar terms. A polynomial is subtracted by adding the opposite. The opposite of a polynomial is formed by taking the opposite of each term.

Add 3x2 4x 6 and 5x2 7x 8.

Perform operations on polynomials involving both addition and subtraction. (Sec. 4.1, Obj. 5, p. 195)

We can use the distributive property along with the properties, a 1(a) and a 1(a), when adding and subtracting polynomials.

Perform the indicated operation: (3y 4) (6y 5)

Apply properties of exponents to multiply monomials. (Sec. 4.2, Obj. 1, p. 199)

When multiplying powers with the same base, add the exponents. When raising a power to a power, multiply the exponents.

Find the products. (a) x5 # x2 (b) (y 3)2

Apply properties of exponents to raise a monomial to a power. (Sec. 4.2, Obj. 2, p. 201)

When raising a monomial to a power, raise each factor to that power.

Simplify (7a3b5)2.

Find the product of two polynomials. (Sec. 4.3, Obj. 1, p. 206)

The distributive property along with the properties of exponents forms a basis for multiplying polynomials.

Multiply (2x 3)(x2 4x 5).

234


The polynomial 3x2y2 7xy2 is a binomial of degree 4. The polynomial 8x 2 3x 1 is a trinomial of degree 2. The polynomial x4 6x 3 5x 2 12x 2 is a polynomial of degree 4.

Problems 7–10

Solution

(3x2 4x 6) (5x2 7x 8) (3x2 5x2) (4x 7x) (6 8) 8x2 3x 2 Problems 11–18

Solution

(3y 4) (6y 5) 1(3y 4) 1(6y 5) 1(3y) 1(4) 1(6y) 1(5) 3y 4 6y 5 3y 9 Problems 19 –22

Solution (a) x5 # x2 x52 x7 # (b) 1y3 2 2 y3 2 y6

Problems 23 –26

Solution

17a3b5 2 72 1a3 2 2 1a5 2 2 49a6b10

Problems 27–34

Solution

(2x 3)(x2 4x 5) 2x(x2 4x 5) 3(x2 4x 5) 2x3 8x2 10x 3x2 12x 15 2x3 5x2 2x 15

(continued)

Chapter 4

Summary

235

OBJECTIVE

SUMMARY

EXAMPLE


Use the shortcut pattern to find the product of two binomials. (Sec. 4.3, Obj. 2, p. 208)

The product of two binomials is perhaps the most frequently used type of multiplication problem. A threestep shortcut pattern, called FOIL, is often used to find the product of two binomials. FOIL stands for first, outside, inside, and last.

Multiply (2x 5)(x 3).

Problems 35 – 48

Raise a binomial to a power. (Sec. 4.3, Obj. 3, p. 210)

When raising a binomial to a power, rewrite the problem as a product of binomials. Then you can apply the FOIL shortcut to multiply two binomials. Then apply the distributive property to find the product.

Find the product (x 4)3.

Use a special product pattern to find products. (Sec. 4.3, Obj. 4, p. 210)

When multiplying binomials you should be able to recognize the following special patterns and use them to find the product. (a b)2 a2 2ab b2 (a b)2 a2 2ab b2 (a b)(a b) a2 b2

Find the product (5x 6)2.

Apply polynomials to geometric problems. (Sec. 4.1, Obj. 6, p. 197; Sec. 4.2, Obj. 4, p. 204; Sec. 4.3, Obj. 5, p. 212)

Polynomials can be applied to represent perimeters, areas, and volumes of geometric figures.

A piece of aluminum that is 15 inches by 20 inches has a square piece x inches on a side cut out from two corners. Find the area of the aluminum piece after the corners are removed.

Solution

(2x 5)(x 3) 2x(x) (2x)(3) (5)(x) (5)(3) 2x2 6x 5x 15 2x2 x 15 Problems 49 –50

Solution

(x 4)3 (x 4)(x 4)(x 4) (x 4)(x2 8x 16) x(x2 8x 16) 4(x2 8x 16) x3 8x2 16x 4x2 32x 64 x3 12x2 48x 64 Problems 51–56

Solution

(5x 6)2 (5x)2 2(5x)(6) (6)2 25x2 60x 36

Problems 57–58

Solution

The area before the corners are removed is found by applying the formula A lw. Therefore A 15(20) 300. Each of the two corners removed has an area of x2. Therefore 2x2 must be subtracted from the area of the aluminum piece. The area of the aluminum piece after the corners are removed is 300 2x2.

(continued)

236

Chapter 4


OBJECTIVE

SUMMARY

Apply properties of exponents to divide monomials. (Sec. 4.4, Obj. 1, p. 215)

The following properties of exponents, along with our knowledge of dividing integers, serve as a basis for dividing monomials: bn bnm when n m bm bn 1 when n m bm

Divide polynomials by monomials. (Sec. 4.4, Obj. 2, p. 217)

Dividing a polynomial by a monomial is based on the property ab a b c c c

Divide polynomials by binomials. (Sec. 4.5, Obj. 1, p. 219)

Use the conventional long division format from arithmetic. Arrange both the dividend and the divisor in descending powers of the variable. You may have to insert terms with zero coefficients if terms with some powers of the variable are missing.

Apply the properties of exponents (including negative and zero exponents) to simplify expressions. (Sec. 4.6, Obj. 1, p. 224)

By definition, if b is a nonzero real number, then b0 1. By definition, if n is a positive integer, and b is a nonzero real number, 1 then bn n . b When simplifying expressions use only positive exponents in the final result.

Write numbers expressed in scientific notation in standard decimal notation. (Sec. 4.6, Obj. 3, p. 229)

To change from scientific notation to ordinary decimal notation, move the decimal point the number of places indicated by the exponent of 10. The decimal point is moved to the right if the exponent is positive and to the left if it is negative.

EXAMPLE Divide

32a5b4 . 8a3b

CHAPTER REVIEW PROBLEMS Problems 59 – 60

Solution

32a5b4 4a2b3 8a3b

Divide

8x5 16x4 4x3 . 4x

Problems 61– 64

Solution

8x5 16x4 4x3 4x 16x4 4x3 8x5 4x 4x 4x 2x4 4x3 x2 Divide (y3 5y 2) (y 2).

Problems 65 – 68

Solution

y2 2y 1 y 2冄y3 0y2 5y 2 y3 2y2 2y2 5y 2y2 4y y 2 y 2 28n6 . Express the 4n2 answer using positive exponents only. Simplify

Problems 69 –94

Solution

28n6 7n6122 7n62 4n2 7 7n4 4 n Write (3.28)(104) in standard decimal notation.

Problems 95 –98

Solution

(3.28)(104) 0.000328 (continued)

Chapter 4

Review Problem Set


OBJECTIVE

SUMMARY

EXAMPLE

Write numbers in scientific notation. (Sec. 4.6, Obj. 2, p. 228)

To represent a number in scientific notation, express it as a product of a number between 1 and 10 (including 1) and an integral power of 10.

Write 657,000,000 in scientific notation.

Scientific notation can be used to make arithmetic problems easier to evaluate.

Evaluate (61,000)(0.000005).

Use scientific notation to evaluate numerical expressions. (Sec. 4.6, Obj. 4, p. 229)

237

Problems 99 –102

Solution

Count the number of decimal places from the existing decimal point until you have a number between 1 and 10. Becasue 657,000,000 is greater than 1, for the scientific notation, raise 10 to that number. 657,000,000 (6.57)(108)

Solution

Problems 103 –106

(61,000)(0.000005) (6.1)(104)(5)(106) (6.1)(5)(104)(106) (30.5)(102) 0.305

Chapter 4 Review Problem Set For Problems 1– 6, match each polynomial with its description.

14. 31a 12 213a 42 512a 72

1. 3xy

A. Binomial of degree 4

15. 1a 42 51a 22 713a 12

2. a2 2ab b2

B. Monomial of degree 3

16. 213n 12 412n 62 513n 42

3. 4x y 8xy

C. Monomial of degree 5

17. 31n2 2n 42 412n2 n 32

4. 2x4 x3 5x2

D. Trinomial of degree 4

18. 51n2 n 12 314n2 3n 72

5. 5a3

E. Binomial of degree 3

6. 3x2y2 4y3

F.

4

2

Trinomial of degree 2

For Problems 7–10, perform the additions and subtractions. 7. 15x 6x 42 13x 7x 22 2

2

8. 17y2 9y 32 14y2 2y 62 9. 12x2 3x 42 14x2 3x 62 13x2 2x 12 10. 13x2 2x 42 1x2 5x 62 14x2 3x 82

For Problems 11–18, remove parentheses and combine similar terms.

For Problems 19 –26, find the indicated products. 19. 15x2 217x4 2

20. 16x3 219x5 2

21. 14xy2 216x2y3 2

22. 12a3b4 213ab5 2

23. 12a2b3 2 3

24. 13xy2 2 2

25. 14xy3 2 2

26. 15a2b2 3

For Problems 27–34, find the indicated products. Be sure to simplify your answers. 27. 5x17x 32

28. 13x2 218x 12

29. 1x 2213x2 x 12

30. 1x 6212x2 5x 42

31. 1x 221x2 x 62

11. 512x 12 71x 32 213x 42

32. 12x 121x2 4x 72

12. 312x2 4x 52 513x2 4x 12

33. 1x2 x 121x2 2x 52

13. 61y2 7y 32 41y2 3y 92

34. 1n2 2n 421n2 7n 12

238

Chapter 4


For Problems 35 –50, find the indicated products. Be sure to simplify your answers.

For Problems 69 – 80, evaluate each expression. 69. 32 22

70. 13 22 2

71. 24

72. 152 0

35. (x 9)(x 8)

36. (x 4)(x 6)

37. (x 6)(x 2)

38. (x 5)(x 6)

39. (x 7)(x 2)

40. (x 10)(x 4)

73. 50

74.

41. (2x 1)(x 5)

42. (3x 2)(x 4)

43. (5x 1)(x 3)

44. (4x 3)(x 5)

3 2 75. a b 4

76.

45. (2x 1)(7x 3)

46. (4a 7)(5a 8)

47. (5n 1)(6n 5)

48. (3n 4)(4n 1)

49. 1a 52 3

50. 1a 62 3

77.

1 122 3

79. 30 22

For Problems 51–56, use one of the appropriate patterns, (a b)2 a2 2ab2 b2, (a b)2 a2 2ab b2, or (a b)(a b) a2 b2, to find the indicated products. 52. 1m 42 2

81. x5x8

53. 13a 52 2

54. 12x 72 2

83.

55. 12n 1212n 12

56. 14n 5214n 52

58. A piece of cardboard that is 12 inches by 18 inches has a square piece x inches on a side cut out from each corner. After the corners are removed, the flaps are turned up to form an open box. Find the polynomial that represents the volume of the box.

61.

36x4y5 3xy

2

18x4y3 54x6y2 6x2y2

60.

56a5b7 8a2b3

62.

30a b 39a b 3ab 5 10

78. 21 32 80. 12 32 2

85.

87.

x4 x6 18xy4 2x3y 24a5 3a1

82. 13x5 214x2 2 84.

86.

88.

x6 x4 20x2y 5x4y3 48n2 12n1

89. 1x2y2 1

90. 1a2b3 2 2

91. 12x2 1

92. 13n2 2 2

93. 12n1 2 3

94. 14ab1 213a1b2 2

For Problems 95 –98, write each expression in standard decimal form.

For Problems 59 – 68, perform the divisions. 59.

1 1 1 a b 4

For Problems 81–94, simplify each of the following, and express your answers using positive exponents only.

51. 1y 82 2

57. A piece of vinyl flooring that is 20 feet by 30 feet has a square piece x feet on a side cut out from each corner. Find the area of the vinyl flooring piece after the corners are removed.

1 32

95. 16.121102 2

96. 15.621104 2

97. 1821102 2

98. 19.221104 2

4 8

For Problems 99 –102, write each number in scientific notation. 56x4 40x3 32x2 63. 4x2 64.

12a3 15a2 3a 3a

65. 1x2 9x 12 1x 52 66. 121x2 4x 122 13x 22 67. 12x3 3x2 2x 42 1x 22 68. 16x 7x 92 12x 12 2

99. 9000 101. 0.047

100. 47 102. 0.00021

For Problems 103 –106, use scientific notation and the properties of exponents to evaluate each expression. 103. (0.00004)(12,000) 105.

0.0056 0.0000028

104. (0.0021)(2000) 106.

0.00078 39,000

Chapter 4 Test 1. Find the sum of 7x2 6x 2 and 5x2 8x 7.

1.

2. Subtract x 9x 14 from 4x 3x 6.

2.

3. Remove parentheses and combine similar terms for the expression 3(2x 1) 6(3x 2) (x 7).

3.

2

2

4. Find the product 14xy2 217x2y3 2 .

4.

5. Find the product 12x y2 13xy 2 . 2

2

3

5.

For Problems 6 –12, find the indicated products and express answers in simplest form. 6. (x 9)(x 2)

6.

7. (n 14)(n 7)

7.

8. (5a 3)(8a 7)

8.

9. 13x 7y2

2

9.

10. 1x 32 12x2 4x 72

10.

11. (9x 5y)(9x 5y)

11.

12. (3x 7)(5x 11)

12.

13. Find the indicated quotient: 14. Find the indicated quotient:

96x4y5 12x2y

.

56x2y 72xy2 . 8xy

15. Find the indicated quotient: 12x3 5x2 22x 152 12x 32 .

13. 14. 15.

16. Find the indicated quotient: 14x3 23x2 362 1x 62 .

16.

2 3 17. Evaluate a b . 3

17.

18. Evaluate 42 41 40.

18.

19. Evaluate

1 . 24

20. Find the product 16x4 214x2 2 and express the answer using a positive exponent. 21. Simplify a

8x1 1 b and express the answer using a positive exponent. 2x2

19. 20.

21.

22. Simplify 1x3y5 2 2 and express the answer using positive exponents.

22.

23. Write 0.00027 in scientific notation.

23.

24. Express 19.221106 2 in standard decimal form.

24.

25. Evaluate (0.000002)(3000).

25.

239

Chapters 1– 4 Cumulative Review Problem Set For Problems 1–10, evaluate each of the numerical expressions. 1. 5 312 72 2 3 # 5 112 3

3. 7 2

4. 4 122 3162

5. 132 4

2.

82#

2

2 1 7. a b 3

6. 25 9. a

1 2 1 b 2 3

# 5 112

8.

1 42

10. 20 21 22

2x 3y xy

for x

1 1 and y 2 3

3a 2b 4a 7b a 3a b 2b

1 3

for x 2

16. 21n2 3n 12 1n2 n 42 312n 12 for n 3 For Problems 17–29, find the indicated products. 17. 13x2y3 2 15xy4 2

18. 16ab4 2 12b3 2

19. 12x2y5 2 3

20. 3xy12x 5y2

21. 15x 2213x 12

22. 17x 1213x 42

23. 1x 2212x 32

24. 17 2y217 2y2

25. 1x 2213x x 42 2

26. 12x 521x2 x 42 28. 11 2n2 3

29. 1x 2x 62 12x 5x 62 2

2

For Problems 30 –34, perform the indicated divisions. 30.

32.

52x3y4

31.

2

13xy

126a3b5 9a2b3

56xy2 64x3y 72x4y4 8xy

33. 12x3 2x2 19x 212 1x 32 34. 13x3 17x2 6x 42 13x 12 For Problems 35 –38, simplify each expression and express your answers using positive exponents only. 35. 12x 213x 2 3

240

4

4x2 36. 2x1

0.00034 0.0000017

For Problems 42 – 49, solve each of the equations. 42. 5x 8 6x 3 43. 214x 12 5x 3 2x y y 8 2 3

46. 1.6 2.4x 5x 65

15. 1x2 2x 42 1x2 x 22 12x2 3x 12 for x 1

27. 12n 32 2

40. 10.0002210.0032 2

45. 6x 8 4x 1013x 22

for a 1 and b

14. 21x 42 312x 12 13x 22

41.

44.

3 2 1 1 12. n n n n for n 5 3 2 4 13.

38. 1xy2z1 2 2

For Problems 39 – 41, use scientific notation and the properties of exponents to help evaluate each numerical expression. 39. (0.00003)(4000)

For Problems 11–16, evaluate each algebraic expression for the given values of the variables. 11.

37. 13x1y2 2 1

47. 31x 12 21x 32 4 48.

3n 1 n2 2 5 3 15

49. 0.06x 0.0811500 x2 110 For Problems 50 –55, solve each of the inequalities. 50. 2x 7 31x 42

51. 6x 5 3x 5

52. 41x 52 213x 62 0 53. 5x 3 4x 5 54.

3x x 5x 1 4 2 6

55. 0.081700 x2 0.11x 65 For Problems 56 – 62, set up an equation and solve each problem. 56. The sum of 4 and three times a certain number is the same as the sum of the number and 10. Find the number. 57. Fifteen percent of some number is 6. Find the number. 58. Lou has 18 coins consisting of dimes and quarters. If the total value of the coins is $3.30, how many coins of each denomination does he have? 59. A sum of $1500 is invested, part of it at 8% interest and the remainder at 9%. If the total interest amounts to $128, find the amount invested at each rate. 60. How many gallons of water must be added to 15 gallons of a 12% salt solution to change it to a 10% salt solution? 61. Two airplanes leave Atlanta at the same time and fly in opposite directions. If one travels at 400 miles per hour and the other at 450 miles per hour, how long will it take them to be 2975 miles apart? 62. The length of a rectangle is 1 meter more than twice its width. If the perimeter of the rectangle is 44 meters, find the length and width.

Factoring, Solving Equations, and Problem Solving

5 5.1 Factoring by Using the Distributive Property 5.2 Factoring the Difference of Two Squares 5.3 Factoring Trinomials of the Form x2 bx c 5.4 Factoring Trinomials of the Form ax2 bx c

Photodisc/Getty Images

5.5 Factoring, Solving Equations, and Problem Solving

■ Algebraic equations can be used to solve a large variety of problems involving geometric relationships.

A

flower garden is in the shape of a right triangle with one leg 7 meters longer than the other leg and the hypotenuse 1 meter longer than the longer leg. Find the lengths of all three sides of the right triangle. A popular geometric formula, called the Pythagorean theorem, serves as a guideline for setting up an equation to solve this problem. We can use the equation x2 1x 72 2 1x 82 2 to determine that the sides of the right triangle are 5 meters, 12 meters, and 13 meters long. The distributive property has allowed us to combine similar terms and multiply polynomials. In this chapter, we will see yet another use of the distributive property as we learn how to factor polynomials. Factoring polynomials will allow us to solve other kinds of equations, which will, in turn, help us to solve a greater variety of word problems.


241

I N T E R N E T

P R O J E C T

Pythagoras is widely known for the Pythagorean theorem pertaining to right triangles. Do an Internet search to determine at least two other fields where Pythagoras made significant contributions. Pythagoras also founded a school. While conducting your search, find the name given to the students attending Pythagoras’ school and some of the school rules for students. Can you think of any modern-day schools that might have the same requirements?

5.1

Factoring by Using the Distributive Property OBJECTIVES 1

Find the Greatest Common Factor

2

Factor Out the Greatest Common Factor

3

Factor by Grouping

4

Solve Equations by Factoring

5

Solve Word Problems Using Factoring

1 Find the Greatest Common Factor In Chapter 1, we found the greatest common factor of two or more whole numbers by inspection or by using the prime factored form of the numbers. For example, by inspection we see that the greatest common factor of 8 and 12 is 4. This means that 4 is the largest whole number that is a factor of both 8 and 12. If it is difficult to determine the greatest common factor by inspection, then we can use the prime factorization technique as follows: 42 2 # 3 # 7 70 2 # 5 # 7 We see that 2 # 7 14 is the greatest common factor of 42 and 70. It is meaningful to extend the concept of greatest common factor to monomials. Consider the next example.

EXAMPLE 1

Find the greatest common factor of 8x 2 and 12x 3.

Solution 8x2 2 # 2 # 2 # x # x 12x3 2 # 2 # 3 # x # x # x Therefore, the greatest common factor is 2 # 2 # x # x 4x2.

▼ PRACTICE YOUR SKILL Find the greatest common factor of 14a2 and 7a5.

■

By the greatest common factor of two or more monomials we mean the monomial with the largest numerical coefficient and highest power of the variables that is a factor of the given monomials. 242

5.1

EXAMPLE 2

Factoring by Using the Distributive Property

243

Find the greatest common factor of 16x 2y , 24x 3y 2, and 32xy.

Solution 16x2y 2 # 2 # 2 # 2 # x # x # y 24x3y2 2 # 2 # 2 # 3 # x # x # x # y # y 32xy 2 # 2 # 2 # 2 # 2 # x # y Therefore, the greatest common factor is 2 # 2 # 2 # x # y 8xy.

▼ PRACTICE YOUR SKILL Find the greatest common factor of 18m2n4, 4m3n5, and 10m4n3.

■

2 Factor Out the Greatest Common Factor We have used the distributive property to multiply a polynomial by a monomial; for example, 3x1x 2 2 3x 2 6x Suppose we start with 3x 2 6x and want to express it in factored form. We use the distributive property in the form ab ac a(b c). 3x 2 6x 3x1x2 3x122 3x1x 22

3x is the greatest common factor of 3x2 and 6x Use the distributive property

The next four examples further illustrate this process of factoring out the greatest common monomial factor.

EXAMPLE 3

Factor 12x 3 8x 2.

Solution 12x3 8x2 4x2 13x2 4x2 122 4x2 13x 22

ab ac a(b c)

▼ PRACTICE YOUR SKILL Factor 15a2 21a6.

EXAMPLE 4

■

Factor 12x 2y 18xy 2.

Solution 12x2y 18xy2 6xy12x2 6xy13y2 6xy12x 3y2

▼ PRACTICE YOUR SKILL Factor 8m3n2 2m6n.

■

244

Chapter 5


EXAMPLE 5

Factor 24x 3 30x 4 42x 5.

Solution 24x 3 30x 4 42x 5 6x 3 142 6x 3 15x2 6x 3 17x 2 2 6x 3 14 5x 7x 2 2

▼ PRACTICE YOUR SKILL Factor 48y8 16y6 24y4.

EXAMPLE 6

■

Factor 9x 2 9x.

Solution 9x 2 9x 9x1x2 9x112 9x1x 12

▼ PRACTICE YOUR SKILL Factor 8b3 8b2.

■

We want to emphasize the point made just before Example 3. It is important to realize that we are factoring out the greatest common monomial factor. We could factor an expression such as 9x 2 9x in Example 6 as 91x 2 x 2 , 313x 2 3x 2 , 1 3x13x 3 2 , or even 118x2 18x2 , but it is the form 9x1x 1 2 that we want. We 2 can accomplish this by factoring out the greatest common monomial factor; we sometimes refer to this process as factoring completely. A polynomial with integral coefficients is in completely factored form if these conditions are met: 1.

It is expressed as a product of polynomials with integral coefficients.

2.

No polynomial, other than a monomial, within the factored form can be further factored into polynomials with integral coefficients.

Thus 91x 2 x 2 , 313x 2 3x 2 , and 3x13x 2 2 are not completely factored be1 cause they violate condition 2. The form 118x2 18x2 violates both conditions 2 1 and 2.

3 Factor by Grouping Sometimes there may be a common binomial factor rather than a common monomial factor. For example, each of the two terms of x(y 2) z(y 2) has a common binomial factor of (y 2). Thus we can factor (y 2) from each term and get x(y 2) z(y 2) (y 2)(x z) Consider a few more examples involving a common binomial factor: a1b c2 d1b c2 1b c2 1a d2

x1x 22 31x 22 1x 221x 32

x1x 52 41x 52 1x 521x 42

5.1


245

It may be that the original polynomial exhibits no apparent common monomial or binomial factor, which is the case with ab 3a bc 3c However, by factoring a from the first two terms and c from the last two terms, we see that ab 3a bc 3c a(b 3) c(b 3) Now a common binomial factor of (b 3) is obvious, and we can proceed as before: a(b 3) c(b 3) (b 3)(a c) This factoring process is called factoring by grouping. Let’s consider two more examples of factoring by grouping.

EXAMPLE 7

Factor each polynomial completely. (a) x2 x 5x 5

(b) 6x2 4x 3x 2

Solution (a) x2 x 5x 5 x1x 12 51x 12 1x 12 1x 52 (b) 6x2 4x 3x 2 2x13x 22 113x 22 13x 22 12x 12

Factor x from first two terms and 5 from last two terms Factor common binomial factor of (x 1) from both terms Factor 2x from first two terms and 1 from last two terms Factor common binomial factor of (3x 2) from both terms

▼ PRACTICE YOUR SKILL Factor each polynomial completely. (a) ab 5a 3b 15

(b) xy 2x 4y 8

■

4 Solve Equations by Factoring Suppose we are told that the product of two numbers is 0. What do we know about the numbers? Do you agree we can conclude that at least one of the numbers must be 0? The next property formalizes this idea.

Property 5.1 For all real numbers a and b, ab 0

if and only if a 0 or b 0

Property 5.1 provides us with another technique for solving equations.

246

Chapter 5


EXAMPLE 8

Solve x 2 6x 0.

Solution To solve equations by applying Property 5.1, one side of the equation must be a product, and the other side of the equation must be zero. This equation already has zero on the right-hand side of the equation, but the left-hand side of this equation is a sum. We will factor the left-hand side, x2 6x, to change the sum into a product. x 2 6x 0 x1x 62 0 x0

or

x0

or

x60

ab 0 if and only if a 0 or b 0

x 6

The solution set is {6, 0}. (Be sure to check both values in the original equation.)

▼ PRACTICE YOUR SKILL Solve y2 7y 0.

EXAMPLE 9

■

Solve x 2 12x.

Solution In order to solve this equation by Property 5.1, we will first get zero on the right-hand side of the equation by adding 12x to each side. Then we factor the expression on the left-hand side of the equation. x 2 12x x 2 12x 0

Added 12x to both sides

x1x 122 0 x0

or

x0

or

x 12 0


x 12

The solution set is {0, 12}.

▼ PRACTICE YOUR SKILL Solve a2 15a.

■

Remark: Note in Example 9 that we did not divide both sides of the original equation by x. Doing so would cause us to lose the solution of 0.

EXAMPLE 10

Solve 4x 2 3x 0.

Solution 4x2 3x 0 x14x 32 0 x0

or

4x 3 0

x0

or

4x 3

x0

or

x

The solution set is e 0,

3 f. 4

3 4


5.1


247

▼ PRACTICE YOUR SKILL Solve 5y2 2y 0.

EXAMPLE 11

■

Solve x1x 2 2 31x 2 2 0.

Solution In order to solve this equation by Property 5.1, we will factor the left-hand side of the equation. The greatest common factor of the terms is (x 2). x1x 22 31x 22 0 1x 221x 32 0

x20 x 2

x30

or or

The solution set is 53, 26 .


x 3

▼ PRACTICE YOUR SKILL Solve m(m 2) 5(m 2) 0.

■

5 Solve Word Problems Using Factoring Each time we expand our equation-solving capabilities, we gain more techniques for solving word problems. Let’s solve a geometric problem with the ideas we learned in this section.

EXAMPLE 12

Apply Your Skill The area of a square is numerically equal to twice its perimeter. Find the length of a side of the square.

Solution s

s

Sketch a square and let s represent the length of each side (see Figure 5.1). Then the area is represented by s 2 and the perimeter by 4s. Thus s 2 214s2

s

s 2 8s s Figure 5.1

s 2 8s 0 s1s 82 0 s0

or

s80

s0

or

s8

Because 0 is not a reasonable answer to the problem, the solution is 8. (Be sure to check this solution in the original statement of the example!)

▼ PRACTICE YOUR SKILL The area of a square is numerically equal to three times its perimeter. Find the length of a side of the square. ■

248

Chapter 5


CONCEPT QUIZ

For Problems 1–10, answer true or false. 1. The greatest common factor of 6x2y3 12x3y2 18x4y is 2x2y. 2. If the factored form of a polynomial can be factored further, then it has not met the conditions for being considered “factored completely.” 3. Common factors are always monomials. 4. If the product of x and y is zero, then x is zero or y is zero. 5. The factored form 3a(2a2 4) is factored completely. 6. The solutions for the equation x(x 2) 7 are 7 and 5. 7. The solution set for x2 7x is {7}. 8. The solution set for x(x 2) 3(x 2) 0 is {2, 3}. 9. The solution set for 3x x2 is {3, 0}. 10. The solution set for x(x 6) 2(x 6) is {6}.

Problem Set 5.1 1 Find the Greatest Common Factor For Problems 1–10, find the greatest common factor of the given expressions. 1. 24y and 30xy

2. 32x and 40xy

3. 60x2y and 84xy 2

4. 72x3 and 63x2

5. 42ab3 and 70a2b2

6. 48a2b2 and 96ab4

7. 6x3, 8x, and 24x2

8. 72xy, 36x2y, and 84xy 2

32. 30x2y 40xy 55y 33. 2x3 3x2 4x

34. x4 x3 x2

35. 44y5 24y3 20y2 36. 14a 18a3 26a5 37. 14a2b3 35ab2 49a3b 38. 24a3b2 36a2b4 60a4b3 39. x1y 12 z1y 12

9. 16a2b2, 40a2b3, and 56a3b4

40. a1c d2 21c d2

10. 70a 3b3, 42a2b4, and 49ab5

41. a1b 42 c1b 42

2 Factor Out the Greatest Common Factor

42. x1y 62 31y 62

For Problems 11– 46, factor each polynomial completely.

43. x1x 32 61x 32

11. 8x 12y

12. 18x 24y

44. x1x 72 91x 72

13. 14xy 21y

14. 24x 40xy

45. 2x1x 12 31x 12

15. 18x2 45x

16. 12x 28x3

46. 4x1x 82 51x 82

17. 12xy2 30x2y

18. 28x2y2 49x2y

19. 36a2b 60a3b4

20. 65ab3 45a2b2

21. 16xy3 25x2y 2

22. 12x 2y2 29x2y

For Problems 47– 60, use the process of factoring by grouping to factor each polynomial.

23. 64ab 72cd

24. 45xy 72zw

47. 5x 5y bx by

25. 9a2b4 27a2b

26. 7a3b5 42a2b6

48. 7x 7y zx zy

27. 52x4y 2 60x6y

28. 70x5y3 42x8y2

49. bx by cx cy

29. 40x2y 2 8x2y

30. 84x2y3 12xy3

50. 2x 2y ax ay

31. 12x 15xy 21x2

3 Factor by Grouping

51. ac bc a b

5.1 52. x y ax ay


249

82. x13x 22 713x 22 0

53. x 5x 12x 60 2

83. 41x 62 x1x 62 0

54. x2 3x 7x 21

84. x1x 92 21x 92

55. x 2x 8x 16 2

56. x2 4x 9x 36

5 Solve Word Problems Using Factoring

57. 2x2 x 10x 5

For Problems 85 –91, set up an equation and solve each problem.

58. 3x2 2x 18x 12

85. The square of a number equals nine times that number. Find the number.

59. 6n 3n 8n 4 2

86. Suppose that four times the square of a number equals 20 times that number. What is the number?

60. 20n2 8n 15n 6

4 Solve Equations by Factoring For Problems 61– 84, solve each equation. 61. x 8x 0

62. x 12x 0

63. x2 x 0

64. x2 7x 0

65. n2 5n

66. n2 2n

67. 2y2 3y 0

68. 4y2 7y 0

69. 7x2 3x

70. 5x2 2x

71. 3n2 15n 0

72. 6n2 24n 0

73. 4x2 6x

74. 12x2 8x

75. 7x x2 0

76. 9x x2 0

77. 13x x2

78. 15x x2

79. 5x 2x2

80. 7x 5x2

2

2

81. x1x 52 41x 52 0

87. The area of a square is numerically equal to five times its perimeter. Find the length of a side of the square. 88. The area of a square is 14 times as large as the area of a triangle. One side of the triangle is 7 inches long, and the altitude to that side is the same length as a side of the square. Find the length of a side of the square. Also find the areas of both figures, and be sure that your answer checks. 89. Suppose that the area of a circle is numerically equal to the perimeter of a square whose length of a side is the same as the length of a radius of the circle. Find the length of a side of the square. Express your answer in terms of p. 90. One side of a parallelogram, an altitude to that side, and one side of a rectangle all have the same measure. If an adjacent side of the rectangle is 20 centimeters long, and the area of the rectangle is twice the area of the parallelogram, find the areas of both figures. 91. The area of a rectangle is twice the area of a square. If the rectangle is 6 inches long, and the width of the rectangle is the same as the length of a side of the square, find the dimensions of both the rectangle and the square.

THOUGHTS INTO WORDS 92. Suppose that your friend factors 24x2y 36xy like this: 24x2y 36xy 4xy16x 92

14xy213212x 32

12xy12x 32 Is this correct? Would you suggest any changes?

93. The following solution is given for the equation x1x 102 0. x1x 102 0 x2 10x 0 x1x 102 0 x0

or

x0

or

x 10 0 x 10

The solution set is {0, 10}. Is this solution correct? Would you suggest any changes?

250

Chapter 5


FURTHER INVESTIGATIONS A P Prt

94. The total surface area of a right circular cylinder is given by the formula A 2pr 2 2prh, where r represents the radius of a base, and h represents the height of the cylinder. For computational purposes, it may be more convenient to change the form of the right side of the formula by factoring it.

P11 rt2 Use A P(1 rt) to find the total amount of money accumulated for each of the following investments. a. $100 at 8% for 2 years

A 2pr2 2prh

b. $200 at 9% for 3 years

2pr1r h2

c. $500 at 10% for 5 years

Use A 2pr1r h2 to find the total surface area of 22 each of the following cylinders. Use as an approxi7 mation for p.

d. $1000 at 10% for 10 years

a. r 7 centimeters and h 12 centimeters

For Problems 96 –99, solve each equation for the indicated variable.

b. r 14 meters and h 20 meters

96. ax bx c

c. r 3 feet and h 4 feet

97. b2x2 cx 0

d. r 5 yards and h 9 yards

98. 5ay2 by

95. The formula A P Prt yields the total amount of money accumulated (A) when P dollars are invested at r percent simple interest for t years. For computational purposes it may be convenient to change the right side of the formula by factoring.

for x for x

for y

99. y ay by c 0 for y

Answers to the Example Practice Skills 1. 7a2

2. 2m2n3

3. 3a2(5 7a4) 4. 2m3n(4n m3)

7. (a) (a 3)(b 5) (b) (x 4)(y 2) 8. {0, 7} 12. The length is 12.

5. 8y4(6y4 2y2 3) 6. 8b2(b 1) 2 9. {0, 15} 10. e , 0 f 11. {5, 2} 5


5.2

2. True


6. False 7. False

8. True

9. True

10. False

Factoring the Difference of Two Squares OBJECTIVES 1

Factor the Difference of Two Squares

2

Solve Equations by Factoring the Difference of Two Squares

3

Solve Word Problems Using Factoring

1 Factor the Difference of Two Squares In Section 4.3, we noted some special multiplication patterns. One of these patterns was 1a b 21a b 2 a 2 b 2

5.2

Factoring the Difference of Two Squares

251

Here is another version of that pattern:

Difference of Two Squares a2 b2 (a b)(a b) To apply the difference-of-two-squares pattern is a fairly simple process, as these next examples illustrate. The steps inside the box are often performed mentally. x 2 36 4x 25 2

9x 2 16y 2 64 y 2

1x 2 2 16 2 2

1x 6 2 1x 6 2

12x 2 2 15 2 2

12x 5 2 12x 5 2

18 2 1y 2

18 y 2 18 y 2

13x 2 2 14y 2 2 2

2

13x 4y 2 13x 4y 2

Because multiplication is commutative, the order of writing the factors is not important. For example, (x 6)(x 6) can also be written as (x 6)(x 6).

Remark: You must be careful not to assume an analogous factoring pattern for the sum of two squares; it does not exist. For example, x 2 4 ≠ (x 2)(x 2) because (x 2)(x 2) x 2 4x 4. We say that the sum of two squares is not factorable using integers. The phrase “using integers” is necessary because x 2 4 could be 1 written as 12x 2 82 , but such factoring is of no help. Furthermore, we do not 2 consider 11 21x 2 4 2 as factoring x 2 4. It is possible that both the technique of factoring out a common monomial factor and the difference of two squares pattern can be applied to the same polynomial. In general, it is best to look for a common monomial factor first.

EXAMPLE 1

Factor 2x 2 50.

Solution 2x2 50 21x2 252 21x 521x 52

Common factor of 2 Difference of squares

▼ PRACTICE YOUR SKILL Factor 3y2 12.

■

In Example 1, by expressing 2x 2 50 as 2(x 5)(x 5), we say that it has been factored completely. That means the factors 2, x 5, and x 5 cannot be factored any further using integers.

EXAMPLE 2

Factor completely 18y 3 8y.

Solution 18y3 8y 2y19y2 42 2y13y 2213y 22

Common factor of 2y Difference of squares

▼ PRACTICE YOUR SKILL Factor completely 32x3 2x.

■

Sometimes it is possible to apply the difference-of-two-squares pattern more than once. Consider the next example.

252

Chapter 5


EXAMPLE 3

Factor completely x 4 16.

Solution x 4 16 1x 2 42 1x 2 42

1x 2 42 1x 22 1x 22

▼ PRACTICE YOUR SKILL Factor completely a4 81.

■

The following examples should help you to summarize the factoring ideas presented thus far. 5x 2 20 51x 2 4 2

25 y 2 15 y 215 y 2

3 3x 2 311 x 2 2 311 x 2 11 x 2

36x2 49y2 16x 7y2 16x 7y2 a2 9

is not factorable using integers

9x 17y

is not factorable using integers

2 Solve Equations by Factoring the Difference of Two Squares Each time we learn a new factoring technique, we also develop more power for solving equations. Let’s consider how we can use the difference-of-squares factoring pattern to help solve certain kinds of equations.

EXAMPLE 4

Solve x 2 25.

Solution x 2 25 x 2 25 0

1x 52 1x 52 0 x50 x 5

or

x50

or

x5

Remember: ab 0 if and only if a 0 or b 0

The solution set is {5, 5}. Check these answers!

▼ PRACTICE YOUR SKILL Solve x2 64.

EXAMPLE 5

■

Solve 9x 2 25.

Solution 9x 2 25 9x 2 25 0

5.2


253

13x 52 13x 52 0 3x 5 0

or

3x 5 0

3x 5

or

3x 5

5 3

or

x

x

5 3

5 5 The solution set is e , f . 3 3

▼ PRACTICE YOUR SKILL Solve 4x2 9.

EXAMPLE 6

■

Solve 5y 2 20.

Solution 5y 2 20 5y 2 20 5 5


y2 4 y2 4 0

1y 221y 22 0 y20 y 2

or

y20

or

y2

The solution set is {2, 2}. Check it!

▼ PRACTICE YOUR SKILL Solve 3x2 75.

EXAMPLE 7

■

Solve x 3 9x 0.

Solution x 3 9x 0 x1x 2 92 0

x1x 32 1x 32 0 x0

or

x30

or

x0

or

x3

or

x30 x 3

The solution set is {3, 0, 3}.

▼ PRACTICE YOUR SKILL Solve x3 49x 0.

■

The more we know about solving equations, the more easily we can solve word problems.

254

Chapter 5


3 Solve Word Problems Using Factoring EXAMPLE 8

Apply Your Skill The combined area of two squares is 20 square centimeters. Each side of one square is twice as long as a side of the other square. Find the lengths of the sides of each square.

Solution We can sketch two squares and label the sides of the smaller square s (see Figure 5.2). Then the sides of the larger square are 2s. The sum of the areas of the two squares is 20 square centimeters, so we set up and solve the following equation: s 2 12s2 2 20 s 2 4s 2 20

s

5s 2 20

2s

s

s2 4

2s

s2 4 0

1s 22 1s 22 0 s20 s 2

Figure 5.2

or

s20

or

s2

Because s represents the length of a side of a square, we must disregard the solution 2. Thus one square has sides of length 2 centimeters, and the other square has sides of length 2(2) 4 centimeters.

▼ PRACTICE YOUR SKILL The combined area of two squares is 250 square feet. Each side of one square is three times as long as a side of the other square. Find the lengths of the sides of each square. ■

CONCEPT QUIZ

For Problems 1– 8, answer true or false. 1. A binomial that has two perfect square terms that are subtracted is called the difference of two squares. 2. The sum of two squares is factorable using integers. 3. When factoring it is usually best to look for a common factor first. 4. The polynomial 4x2 y2 factors into (2x y)(2x y). 5. The completely factored form of y4 81 is (y2 9)(y2 9). 6. The solution set for x2 16 is {4}. 7. The solution set for 5x3 5x 0 is {1, 0, 1}. 8. The solution set for x4 9x2 0 is {3, 0, 3}.

Problem Set 5.2 1 Factor the Difference of Two Squares For Problems 1–12, use the difference-of-squares pattern to factor each polynomial. 1. x2 1

2. x2 25

3. x2 100

4. x2 121

5. x2 4y2

6. x2 36y2

7. 9x2 y2

8. 49y2 64x2

9. 36a2 25b2 11. 1 4n2

10. 4a2 81b2 12. 4 9n2

5.2 For Problems 13 – 44, factor each polynomial completely. Indicate any that are not factorable using integers. Don’t forget to look for a common monomial factor first.


61. 5 45x2 0

62. 3 12x2 0

63. 4x3 400x 0

64. 2x3 98x 0

65. 64x2 81

66. 81x2 25

67. 36x3 9x

68. 64x3 4x

255

13. 5x2 20

14. 7x2 7

15. 8x2 32

16. 12x2 60

17. 2x2 18y2

18. 8x2 32y2

19. x3 25x

20. 2x3 2x

21. x2 9y2

22. 18x 42y

For Problems 69 – 80 set up an equation and solve the problem.

23. 45x2 36xy

24. 16x2 25y2

69. Forty-nine less than the square of a number equals zero. Find the number.

25. 36 4x2

26. 75 3x2

27. 4a4 16a2

28. 9a4 81a2

29. x4 81

30. 16 x4

31. x4 x2

32. x5 2x3

72. Ten times the square of a number equals 40. Find the number.

33. 3x3 48x

34. 6x3 24x

35. 5x 20x3

36. 4x 36x3

73. The sum of the areas of two squares is 234 square inches. Each side of the larger square is five times the length of a side of the smaller square. Find the length of a side of each square.

37. 4x2 64

38. 9x2 9

39. 75x3y 12xy3

40. 32x3y 18xy3

41. 16x4 81y4

42. x4 1

43. 81 x4

44. 81x4 16y4

2 Solve Equations by Factoring the Difference of Two Squares For Problems 45 – 68, solve each equation. 45. x2 9

46. x2 1

47. 4 n2

48. 144 n2

49. 9x2 16

50. 4x2 9

51. n2 121 0

52. n2 81 0

53. 25x 4

54. 49x 36

55. 3x2 75

56. 7x2 28

57. 3x3 48x 0

58. x3 x 0

59. n3 16n

60. 2n3 8n

2

2

3 Solve Word Problems Using Factoring

70. The cube of a number equals nine times the number. Find the number. 71. Suppose that five times the cube of a number equals 80 times the number. Find the number.

74. The difference of the areas of two squares is 75 square feet. Each side of the larger square is twice the length of a side of the smaller square. Find the length of a side of each square. 1 75. Suppose that the length of a certain rectangle is 2 2 times its width, and the area of that same rectangle is 160 square centimeters. Find the length and width of the rectangle. 76. Suppose that the width of a certain rectangle is threefourths of its length and that the area of this same rectangle is 108 square meters. Find the length and width of the rectangle. 77. The sum of the areas of two circles is 80p square meters. Find the length of a radius of each circle if one of them is twice as long as the other. 78. The area of a triangle is 98 square feet. If one side of the triangle and the altitude to that side are of equal length, find the length. 79. The total surface area of a right circular cylinder is 100p square centimeters. If a radius of the base and the altitude of the cylinder are the same length, find the length of a radius. 80. The total surface area of a right circular cone is 192p square feet. If the slant height of the cone is equal in length to a diameter of the base, find the length of a radius.

256

Chapter 5


THOUGHTS INTO WORDS 81. How do we know that the equation x2 1 0 has no solutions in the set of real numbers?

83. Consider the following solution: 4x2 36 0 41x2 92 0

82. Why is the following factoring process incomplete? 16x2 64 14x 82 14x 82

41x 321x 32 0

How should the factoring be done?

40

or

40

or

x30 x 3

or

x30

or

x3

The solution set is {3, 3}. Is this a correct solution? Do you have any suggestion to offer the person who worked on this problem?

FURTHER INVESTIGATIONS The following patterns can be used to factor the sum of two cubes and the difference of two cubes, respectively. a3 b3 1a b2 1a2 ab b2 2 a3 b3 1a b2 1a2 ab b2 2

Consider these examples: x3 8 1x2 3 122 3 1x 221x2 2x 42

x 1 1x2 112 1x 121x x 12 3

3

3

2

Use the sum-of-two-cubes and the difference-of-two-cubes patterns to factor each polynomial. 84. x3 1

86. n3 27

87. n3 64

88. 8x3 27y3

89. 27a3 64b3

90. 1 8x3

91. 1 27a3

92. x3 8y3

93. 8x3 y3

94. a3b3 1

95. 27x3 8y3

96. 8 n3

97. 125x3 8y3

98. 27n3 125

99. 64 x3

85. x3 8

Answers to the Example Practice Skills 1. 3(y 2)(y 2) 2. 2x(4x 1)(4x 1) 7. {7, 0, 7} 8. 5 feet and 15 feet

3. (a2 9)(a 3)(a 3)

3 3 4. {8, 8} 5. e , f 2 2

6. {5, 5}


5.3

2. False


6. False 7. True

8. True

Factoring Trinomials of the Form x2 bx c OBJECTIVES 1

Factor Trinomials of the Form x 2 bx c

2

Use Factoring of Trinomials to Solve Equations

3

Solve Word Problems Including Consecutive Number Problems

4

Use the Pythagorean Theorem to Solve Problems

1 Factor Trinomials of the Form x2 bx c One of the most common types of factoring used in algebra is to express a trinomial as the product of two binomials. In this section, we will consider trinomials where the coefficient of the squared term is 1, that is, trinomials of the form x 2 bx c.

5.3

Factoring Trinomials of the Form x2 bx c

257

Again, to develop a factoring technique we first look at some multiplication ideas. Consider the product (x r)(x s), and use the distributive property to show how each term of the resulting trinomial is formed. 1x r21x s2 x1x2 x1s2 r1x2 r1s2 14243 x2

(s r)x

rs

Note that the coefficient of the middle term is the sum of r and s and that the last term is the product of r and s. These two relationships are used in the next examples.

EXAMPLE 1

Factor x 2 7x 12.

Solution We need to fill in the blanks with two numbers whose sum is 7 and whose product is 12. x 2 7x 12 (x _____)(x _____) This can be done by setting up a table showing possible numbers.

Product

Sum

1(12) 12 2(6) 12 3(4) 12

1 12 13 268 347

The bottom line contains the numbers that we need. Thus x 2 7x 12 1x 3 2 1x 4 2

▼ PRACTICE YOUR SKILL Factor y2 9y 20.

EXAMPLE 2

■

Factor x 2 11x 24.

Solution We need two numbers whose product is 24 and whose sum is 11.

Product 1121242 1221122 132182 142162

24 24 24 24

Sum 1 1242 2 1122 3 182 4 162

25 14 11 10

The third line contains the numbers that we want. Thus x 2 11x 24 1x 3 2 1x 8 2

▼ PRACTICE YOUR SKILL Factor m2 8m 15.

■

258

Chapter 5


EXAMPLE 3

Factor x 2 3x 10.


Product 11102 11102 2152 2152

Sum 1 1102 9 1 10 9 2 152 3 2 5 3

10 10 10 10

The bottom line is the key line. Thus x 2 3x 10 1x 5 2 1x 2 2


EXAMPLE 4

■

Factor x 2 2x 8.


Product 1182 1182 2142 2142

8 8 8 8

Sum 1 182 7 1 8 7 2 142 2 2 4 2

The third line has the information we want. x 2 2x 8 1x 4 2 1x 2 2


■

The tables in the last four examples illustrate one way of organizing your thoughts for such problems. We showed complete tables; that is, for Example 4, we included the last line even though the desired numbers were obtained in the third line. If you use such tables, keep in mind that as soon as you get the desired numbers, the table need not be continued beyond that point. Furthermore, there will be times that you will be able to find the numbers without using a table. The key ideas are the product and sum relationships.

EXAMPLE 5

Factor x 2 13x 12.

Solution Product

Sum

(1)(12) 12

(1) (12) 13

5.3


259

We need not complete the table.

x 2 13x 12 1x 1 2 1x 12 2


■

In the next example, we refer to the concept of absolute value. Recall that the absolute value is the number without regard for the sign. For example, 040 4

EXAMPLE 6

and

0 4 0 4

Factor x 2 x 56.

Solution Note that the coefficient of the middle term is 1. Therefore, we are looking for two numbers whose product is 56; their sum is 1, so the absolute value of the negative number must be 1 larger than the absolute value of the positive number. The numbers are 8 and 7, and we have x 2 x 56 1x 8 2 1x 7 2

▼ PRACTICE YOUR SKILL Factor a2 a 12.

EXAMPLE 7

■

Factor x 2 10x 12.

Solution Product

Sum

11122 12 2162 12 3142 12

1 12 13 268 347

Because the table is complete and no two factors of 12 produce a sum of 10, we conclude that x 2 10x 12 is not factorable using integers.


■

In a problem such as Example 7, we need to be sure that we have tried all possibilities before we conclude that the trinomial is not factorable.

2 Use Factoring of Trinomials to Solve Equations The property ab 0 if and only if a 0 or b 0 continues to play an important role as we solve equations that involve the factoring ideas of this section. Consider the following examples.

260

Chapter 5


EXAMPLE 8

Solve x 2 8x 15 0.

Solution x 2 8x 15 0

1x 32 1x 52 0

Factor the left side

x30

x50

or

x 3

Use ab 0 if and only if a 0 or b 0

x 5

or


▼ PRACTICE YOUR SKILL Solve y2 14y 24 0.

EXAMPLE 9

■

Solve x 2 5x 6 0.

Solution x 2 5x 6 0

1x 62 1x 12 0 x60 x 6

or

x10

or

x1


▼ PRACTICE YOUR SKILL Solve a2 3a 28 0.

EXAMPLE 10

■

Solve y 2 4y 45.

Solution y 2 4y 45 y 2 4y 45 0

1y 921y 52 0 y90

or

y9

or

y50 y 5


▼ PRACTICE YOUR SKILL Solve x2 18x 40.

■

Don’t forget that we can always check to be absolutely sure of our solutions. Let’s check the solutions for Example 10. If y 9, then y2 4y 45 becomes 92 4192 ⱨ 45 81 36 ⱨ 45 45 45

5.3


261

If y 5, then y2 4y 45 becomes 152 2 4152 ⱨ 45

25 20 ⱨ 45 45 45

3 Solve Word Problems Including Consecutive Number Problems The more we know about factoring and solving equations, the more easily we can solve word problems.

EXAMPLE 11

Apply Your Skill Find two consecutive integers whose product is 72.

Solution Let n represent one integer. Then n 1 represents the next integer. n1n 12 72

The product of the two integers is 72

n2 n 72 n2 n 72 0

1n 92 1n 82 0 n90 n 9

or

n80

or

n8

If n 9, then n 1 9 1 8. If n 8, then n 1 8 1 9. Thus the consecutive integers are 9 and 8 or 8 and 9.


Find two consecutive integers whose product is 110.

EXAMPLE 12

Apply Your Skill A rectangular plot is 6 meters longer than it is wide. The area of the plot is 16 square meters. Find the length and width of the plot.

Solution We let w represent the width of the plot, and then w 6 represents the length (see Figure 5.3).

w

w+6 Figure 5.3

262

Chapter 5


Using the area formula A lw, we obtain w1w 62 16 w2 6w 16 w2 6w 16 0

1w 82 1w 22 0 w80 w 8

or

w20

or

w2

The solution 8 is not possible for the width of a rectangle, so the plot is 2 meters wide, and its length (w 6) is 8 meters.

▼ PRACTICE YOUR SKILL A rectangular plot is 2 yards longer than it is wide. The area of the plot is 80 square yards. Find the width and length of the plot. ■

4 Use the Pythagorean Theorem to Solve Problems The Pythagorean theorem, an important theorem pertaining to right triangles, can also serve as a guideline for solving certain types of problems. The Pythagorean theorem states that in any right triangle, the square of the longest side (called the hypotenuse) is equal to the sum of the squares of the other two sides (called legs); see Figure 5.4. We can use this theorem to help solve a problem. a2 + b2 = c2 c

b

a Figure 5.4

EXAMPLE 13

Apply Your Skill Suppose that the lengths of the three sides of a right triangle are consecutive whole numbers. Find the lengths of the three sides.

Solution Let s represent the length of the shortest leg. Then s 1 represents the length of the other leg, and s 2 represents the length of the hypotenuse. Using the Pythagorean theorem as a guideline, we obtain the following equation: Sum of squares of two legs Square of hypotenuse 6447448 64748

s 2 1s 1 2 2

Solving this equation yields s2 s2 2s 1 s2 4s 4 2s2 2s 1 s2 4s 4

1s 2 2 2

5.3


s 2 2s 1 4s 4

Add s2 to both sides

s 2 2s 1 4

Add 4s to both sides

s 2 2s 3 0

Add 4 to both sides

263

1s 32 1s 12 0 s30

or

s3

or

s10 s 1

The solution of 1 is not possible for the length of a side, so the shortest side (s) is of length 3. The other two sides (s 1 and s 2) have lengths of 4 and 5.

▼ PRACTICE YOUR SKILL The length of one leg of a right triangle is 7 inches more then the length of the other leg. The length of the hypotenuse is 13 inches. Find the length of the two legs. ■

CONCEPT QUIZ

For Problems 1–10, answer true or false. 1. Any trinomial of the form x2 bx c can be factored (using integers) into the product of two binomials. 2. To factor x2 4x 60 we look for two numbers whose product is 60 and whose sum is 4. 3. A trinomial of the form x2 bx c will never have a common factor other than 1. 4. If n represents an odd integer, then n 1 represents the next consecutive odd integer. 5. The Pythagorean theorem only applies to right triangles. 6. In a right triangle the longest side is called the hypotenuse. 7. The polynomial x2 25x 72 is not factorable. 8. The polynomial x2 27x 72 is not factorable. 9. The solution set of the equation x2 2x 63 0 is {9, 7}. 10. The solution set of the equation x2 5x 66 0 is {11, 6}.

Problem Set 5.3 1 Factor Trinomials of the Form x2 bx c For Problems 1–30, factor each trinomial completely. Indicate any that are not factorable using integers.

19. y2 y 72

20. y2 y 30

21. x2 21x 80

22. x2 21x 90

23. x2 6x 72

24. x2 8x 36

25. x2 10x 48

26. x2 12x 64

27. x2 3xy 10y2

28. x2 4xy 12y2

29. a2 4ab 32b2

30. a2 3ab 54b2

1. x2 10x 24

2. x2 9x 14

3. x 13x 40

4. x 11x 24

5. x2 11x 18

6. x2 5x 4

7. n 11n 28

8. n 7n 10

9. n2 6n 27

10. n2 3n 18

11. n2 6n 40

12. n2 4n 45

2 Use Factoring of Trinomials to Solve Equations

13. t2 12t 24

14. t2 20t 96

For Problems 31–50, solve each equation.

15. x2 18x 72

16. x2 14x 32

31. x2 10x 21 0

32. x2 9x 20 0

17. x2 5x 66

18. x2 11x 42

33. x2 9x 18 0

34. x2 9x 8 0

2

2

2

2

264

Chapter 5


35. x2 3x 10 0

36. x2 x 12 0

37. n2 5n 36 0

38. n2 3n 18 0

39. n2 6n 40 0

40. n2 8n 48 0

41. t2 t 56 0

42. t2 t 72 0

43. x2 16x 28 0 44. x2 18x 45 0 45. x2 11x 12

46. x2 8x 20

47. x1x 102 16

48. x1x 122 35

49. x2 2x 24 0 50. x 6x 16 0 2

3 Solve Word Problems Including Consecutive Number Problems For Problems 51– 64, set up an equation and solve each problem.

59. Suppose that the width of a certain rectangle is 3 inches less than its length. The area is numerically 6 less than twice the perimeter. Find the length and width of the rectangle. 60. The sum of the areas of a square and a rectangle is 64 square centimeters. The length of the rectangle is 4 centimeters more than a side of the square, and the width of the rectangle is 2 centimeters more than a side of the square. Find the dimensions of the rectangle and the square. 61. The perimeter of a rectangle is 30 centimeters, and the area is 54 square centimeters. Find the width and length of the rectangle. [Hint: Let w represent the width; then 15 w represents the length.] 62. The perimeter of a rectangle is 44 inches, and its area is 120 square inches. Find the width and length of the rectangle. 63. An apple orchard contains 84 trees. The number of trees per row is 5 more than the number of rows. Find the number of rows. 64. A room contains 54 chairs. The number of rows is 3 less than the number of chairs per row. Find the number of rows.

51. Find two consecutive integers whose product is 56. 52. Find two consecutive odd whole numbers whose product is 63. 53. Find two consecutive even whole numbers whose product is 168. 54. One number is 2 larger than another number. The sum of their squares is 100. Find the numbers. 55. Find four consecutive integers such that the product of the two larger integers is 22 less than twice the product of the two smaller integers. 56. Find three consecutive integers such that the product of the two smaller integers is 2 more than 10 times the largest integer. 57. One number is 3 smaller than another number. The square of the larger number is 9 larger than 10 times the smaller number. Find the numbers. 58. The area of the floor of a rectangular room is 84 square feet. The length of the room is 5 feet more than its width. Find the width and length of the room.

4 Use the Pythagorean Theorem to Solve Problems For Problems 65 – 68, set up an equation and solve for each problem. 65. Suppose that one leg of a right triangle is 7 feet shorter than the other leg. The hypotenuse is 2 feet longer than the longer leg. Find the lengths of all three sides of the right triangle. 66. Suppose that one leg of a right triangle is 7 meters longer than the other leg. The hypotenuse is 1 meter longer than the longer leg. Find the lengths of all three sides of the right triangle. 67. Suppose that the length of one leg of a right triangle is 2 inches less than the length of the other leg. If the length of the hypotenuse is 10 inches, find the length of each leg. 68. The length of one leg of a right triangle is 3 centimeters more than the length of the other leg. The length of the hypotenuse is 15 centimeters. Find the lengths of the two legs.

THOUGHTS INTO WORDS 69. What does the expression “not factorable using integers” mean to you? 70. Discuss the role that factoring plays in solving equations.

71. Give an explanation of how you would solve the equation (x 3)(x 4) 0, and also how you would solve (x 3)(x 4) 8.

5.4

Factoring Trinomials of the Form ax2 bx c

265

FURTHER INVESTIGATIONS For Problems 72 –75, factor each trinomial and assume that all variables appearing as exponents represent positive integers. 72. x2a 10xa 24

73. x2a 13xa 40

74. x2a 2xa 8

75. x2a 6xa 27

Therefore, we can solve the given equation as follows: n2 26n 168 0

1n 1221n 142 0 n 12 0

or

n 14 0

n 12 or

n 14


76. Suppose that we want to factor n2 26n 168 so that we can solve the equation n2 26n 168 0. We need to find two positive integers whose product is 168 and whose sum is 26. Because the constant term, 168, is rather large, let’s look at it in prime factored form:

Solve each of the following equations. a. n2 30n 216 0 b. n2 35n 294 0

168 2 # 2 # 2 # 3 # 7

c. n2 40n 384 0 d. n2 40n 375 0

Now we can mentally form two numbers by using all of these factors in different combinations. Using two 2s and the 3 in one number and the other 2 and the 7 in another number produces 2 # 2 # 3 12 and 2 # 7 14.

e. n2 6n 432 0 f. n2 16n 512 0

Answers to the Example Practice Skills 1. (y 5)(y 4) 2. (m 3)(m 5) 3. (y 8)(y 3) 4. (y 2)(y 6) 5. (y 1)(y 6) 6. (a 4)(a 3) 7. Not factorable 8. {2, 12} 9. {7, 4} 10. {2, 20} 11. 11, 10 or 10, 11 12. 8 yards, 10 yards 13. 5 inches and 12 inches


5.4

2. True 3. True

4. False 5. True

6. True

7. True

8. False

9. True

10. False

Factoring Trinomials of the Form ax2 bx c OBJECTIVES 1

Factor Trinomials Where the Leading Coefficient Is Not 1

2

Solve Equations That Involve Factoring

1 Factor Trinomials Where the Leading Coefficient Is Not 1 Now let’s consider factoring trinomials where the coefficient of the squared term is not 1. We first illustrate an informal trial-and-error technique that works well for certain types of trinomials. This technique relies on our knowledge of multiplication of binomials.

EXAMPLE 1

Factor 2x 2 7x 3.

Solution By looking at the first term, 2x2, and the positive signs of the other two terms, we know that the binomials are of the form (2x _____)(x _____)

266

Chapter 5


Because the factors of the constant term 3 are 1 and 3, we have only two possibilities to try: (2x 3)(x 1)

or

(2x 1)(x 3)

By checking the middle term of both of these products, we find that the second one yields the correct middle term of 7x. Therefore, 2x2 7x 3 12x 12 1x 32

▼ PRACTICE YOUR SKILL Factor 3y2 16y 5.

EXAMPLE 2

■

Factor 6x 2 17x 5.

Solution First, we note that 6x 2 can be written as 2x # 3x or 6x # x. Second, because the middle term of the trinomial is negative, and the last term is positive, we know that the binomials are of the form (2x _____)(3x _____)

or

(6x _____)(x _____)

The factors of the constant term 5 are 1 and 5, so we have the following possibilities: 12x 52 13x 12

16x 52 1x 12

12x 1213x 52 16x 121x 52

By checking the middle term for each of these products, we find that the product (2x 5)(3x 1) produces the desired term of 17x. Therefore, 6x 2 17x 5 12x 5 2 13x 1 2

▼ PRACTICE YOUR SKILL Factor 4a2 8a 3.

EXAMPLE 3

■

Factor 8x2 8x 30.

Solution First, we note that the polynomial 8x2 8x 30 has a common factor of 2. Factoring out the common factor gives us 2(4x2 4x 15). Now we need to factor 4x2 4x 15. Now we note that 4x 2 can be written as 4x # x or 2x # 2x. Also, the last term, 15, can be written as (1)(15), (1)(15), (3)(5), or (3)(5). Thus we can generate the possibilities for the binomial factors as follows:

Using 1 and 15

Using 1 and 15

(4x 15)(x 1)

(4x 1)(x 15)

(4x 1)(x 15)

(4x 15)(x 1)

(2x 1)(2x 15)

(2x 1)(2x 15)

5.4


Using 3 and 5

Using 3 and 5

(4x 3)(x 5)

(4x 3)(x 5)

(4x 5)(x 3)

(4x 5)(x 3)

✓ (2x 5)(2x 3)

(2x 5)(2x 3)

267

By checking the middle term of each of these products, we find that the product indicated with a check mark produces the desired middle term of 4x. Therefore, 8x2 8x 30 212x 5212x 32

▼ PRACTICE YOUR SKILL Factor 6y2 27y 30.

■

Let’s pause for a moment and look back over Examples 1, 2, and 3. Example 3 clearly created the most difficulty because we had to consider so many possibilities. We have suggested one possible format for considering the possibilities, but as you practice such problems, you may develop a format of your own that works better for you. Whatever format you use, the key idea is to organize your work so that you consider all possibilities. Let’s look at another example.

EXAMPLE 4

Factor 4x 2 6x 9.

Solution First, we note that 4x 2 can be written as 4x # x or 2x # 2x. Second, because the middle term is positive and the last term is positive, we know that the binomials are of the form (4x _____)(x _____)

or

(2x _____)(2x _____)

Because 9 can be written as 9 # 1 or 3 # 3, we have only the following five possibilities to try: 14x 92 1x 12

14x 32 1x 32

12x 32 12x 32

14x 121x 92

12x 1212x 92

When we try all of these possibilities, we find that none of them yields a middle term of 6x. Therefore, 4x2 6x 9 is not factorable using integers.

▼ PRACTICE YOUR SKILL Factor 6a2 3a 7.

■

Remark: Example 4 illustrates the importance of organizing your work so that you try all possibilities before you conclude that a particular trinomial is not factorable.

2 Solve Equations That Involve Factoring The ability to factor certain trinomials of the form ax2 bx c provides us with greater equation-solving capabilities. Consider the next examples.

268

Chapter 5


EXAMPLE 5

Solve 3x 2 17x 10 0.

Solution 3x 2 17x 10 0

1x 52 13x 22 0

x50

Factoring 3x2 17x 10 as (x 5)(3x 2) may require some extra work on scratch paper

3x 2 0

or


x 5

or

3x 2

x 5

or

x

2 3

2 The solution set is e5, f. Check it! 3

▼ PRACTICE YOUR SKILL Solve 2y2 7y 6 0.

EXAMPLE 6

■

Solve 24x 2 2x 15 0.


14x 32 16x 52 0 6x 5 0

4x 3 0

or

4x 3

or

6x 5

3 4

or

x

x

5 6

5 3 The solution set is e , f. 6 4

▼ PRACTICE YOUR SKILL Solve 2a2 5a 12 0.

CONCEPT QUIZ

■

For Problems 1– 8, answer true or false. 1. Any trinomial of the form ax2 bx c can be factored (using integers) into the product of two binomials. 2. To factor 2x2 x 3, we look for two numbers whose product is 3 and whose sum is 1. 3. A trinomial of the form ax2 bx c will never have a common factor other than 1. 4. The factored form (x 3)(2x 4) is factored completely. 5. The difference-of-squares polynomial 9x2 25 could be written as the trinomial 9x2 0x 25. 6. The polynomial 12x2 11x 12 is not factorable. 1 2 7. The solution set of the equation 6x2 13x 5 0 is e , f . 3 5 5 4 8. The solution set of the equation 18x2 39x 20 0 is e , f . 6 3

5.4


269

Problem Set 5.4 1 Factor Trinomials Where the Leading Coefficient Is Not 1 For Problems 1–30, factor each of the trinomials completely. Indicate any that are not factorable using integers. 1. 3x2 7x 2

2. 2x2 9x 4

3. 6x2 19x 10

4. 12x2 19x 4

5. 4x2 25x 6

6. 5x2 22x 8

7. 12x2 31x 20

8. 8x2 30x 7

2 9. 5y 33y 14

2 10. 6y 4y 16

33. 12x2 11x 2 0 34. 15x2 56x 20 0 35. 3x2 25x 8 0 36. 4x2 31x 21 0 37. 15n2 41n 14 0 38. 6n2 31n 40 0

11. 4n2 26n 48

12. 4n2 17n 15

39. 6t2 37t 35 0

13. 2x2 x 7

14. 7x2 19x 10

40. 2t2 15t 27 0

15. 18x2 45x 7

16. 10x2 x 5

17. 21x2 90x 24

18. 6x2 17x 12

19. 8x2 2x 21

20. 9x2 15x 14

21. 9t2 15t 14

22. 12t3 20t2 25t

2 23. 12y 79y 35

2 24. 9y 52y 12

25. 6n2 2n 5

26. 20n2 27n 9

27. 14x 55x 21

28. 15x 34x 15

29. 20x2 31x 12

30. 8t2 3t 4

2

41. 16y2 18y 9 0 42. 9y2 15y 14 0 43. 9x2 6x 8 0 44. 12n2 28n 5 0 45. 10x2 29x 10 0

2

46. 4x2 16x 15 0 47. 6x2 19x 10

2 Solve Equations That Involve Factoring For Problems 31–50, solve each equation.

48. 12x2 17x 6

31. 2x2 13x 6 0

49. 16x1x 12 5

32. 3x2 16x 5 0

50. 5x15x 22 8

THOUGHTS INTO WORDS 51. Explain your thought process when factoring 24x2 17x 20

53. Your friend solves the equation 8x2 32x 32 0 as follows: 8x2 32x 32 0

52. Your friend factors 8x 32x 32 as follows: 2

8x2 32x 32 14x 82 12x 42

41x 22 122 1x 22

81x 22 1x 22

Is she correct? Do you have any suggestions for her?

14x 8212x 42 0 4x 8 0

or

2x 4 0

4x 8

or

2x 4

x2

or

x2

The solution set is {2}. Is your friend correct? Do you have any changes to recommend?

270

Chapter 5


Answers to the Example Practice Skills 1. (3y 1)(y 5) 3 6. e , 4 f 2

2. (2a 1)(2a 3)

3. 3(y 2)(2y 5)

3 4. Not factorable 5. e2, f 2


5.5

2. False 3. False

4. False 5. True

6. True

7. False

8. True

Factoring, Solving Equations, and Problem Solving OBJECTIVES 1

Factor Perfect-Square Trinomials

2

Recognize the Different Types of Factoring Patterns

3

Use Factoring to Solve Equations

4

Solve Word Problems That Involve Factoring

1 Factor Perfect-Square Trinomials Before we summarize our work with factoring techniques, let’s look at two more special factoring patterns. These patterns emerge when multiplying binomials. Consider the following examples. 1x 52 2 1x 521x 52 x 2 10x 25

12x 32 2 12x 3212x 32 4x 2 12x 9

14x 72 2 14x 7214x 72 16x 2 56x 49 In general, 1a b 2 2 1a b 2 1a b 2 a 2 2ab b 2. Also, 1x 6 2 2 1x 6 21x 6 2 x 2 12x 36

13x 4 2 2 13x 4 213x 4 2 9x 2 24x 16

15x 2 2 2 15x 2 215x 2 2 25x 2 20x 4 In general, 1a b 2 2 1a b 21a b 2 a 2 2ab b 2. Thus we have the following patterns.

Perfect-Square Trinomials a 2 2ab b 2 1a b 2 2 a 2 2ab b 2 1a b 2 2

Trinomials of the form a 2 2ab b 2 or a 2 2ab b 2 are called perfect-square trinomials. They are easy to recognize because of the nature of their terms. For example, 9x 2 30x 25 is a perfect-square trinomial for these reasons:

5.5

1. 2. 3.


271

The first term is a square: 13x2 2 The last term is a square: 152 2

The middle term is twice the product of the quantities being squared in the first and last terms: 213x2 152

Likewise, 25x2 40xy 16y2 is a perfect-square trinomial for these reasons: 1. 2. 3.

The first term is a square: 15x 2 2 The last term is a square: 14y2 2

The middle term is twice the product of the quantities being squared in the first and last terms: 215x2 14y2

Once we know that we have a perfect-square trinomial, the factoring process follows immediately from the two basic patterns. 9x 2 30x 25 13x 5 2 2

25x2 40xy 16y2 15x 4y2 2 Here are some additional examples of perfect-square trinomials and their factored forms.

EXAMPLE 1

Factor the following. (a) x2 16x 64 (d) 1 6y 9y2

(b) 16x2 56x 49 (e) 4m2 4mn n2

(c) 25x2 20xy 4y2

Solution (a) x2 16x 64 1x2 2 21x2182 182 2 1x 82 2

(b) 16x2 56x 49 14x2 2 214x2 172 172 2 14x 72 2

(c) 25x2 20xy 4y2 15x2 2 215x212y2 12y2 2 15x 2y2 2 (d) 1 6y 9y2 112 2 211213y2 13y2 2 11 3y2 2

(e) 4m2 4mn n2 12m2 2 212m2 1n2 1n2 2 12m n2 2 You may want to do the middle part mentally, after you feel comfortable with the process.

▼ PRACTICE YOUR SKILL Factor the following. (a) a2 10a 25

(b) 36x2 12x 1

(c) 49m2 56mn 16n2

■

2 Recognize the Different Types of Factoring Patterns In this chapter, we have considered some basic factoring techniques one at a time, but you must be able to apply them as needed in a variety of situations. Let’s first summarize the techniques and then consider some examples. These are the techniques we have discussed in this chapter: 1.

Factoring by using the distributive property to factor out the greatest common monomial or binomial factor

2.

Factoring by grouping

3.

Factoring by applying the difference-of-squares pattern

272

Chapter 5


4.

Factoring by applying the perfect-square-trinomial pattern

5.

Factoring trinomials of the form x2 bx c into the product of two binomials

6.

Factoring trinomials of the form ax2 bx c into the product of two binomials

As a general guideline, always look for a greatest common monomial factor first, and then proceed with the other factoring techniques. In each of the following examples, we have factored completely whenever possible. Study these examples carefully and note the factoring techniques we have used.

EXAMPLE 2

Factor completely 2x2 12x 10.

Solution 2x2 12x 10 21x2 6x 52

First, factor out the common factor of 2

21x 121x 52

▼ PRACTICE YOUR SKILL Factor completely 3x2 6x 72.

EXAMPLE 3

■

Factor completely 4x2 36.

Solution 4x 2 36 41x 2 9 2

Remember that the sum of two squares is not factorable using integers unless there is a common factor

▼ PRACTICE YOUR SKILL Factor completely 4x2 100.

EXAMPLE 4

■

Factor completely 4t2 20t 25.

Solution 4t 2 20t 25 12t 5 2 2

If you fail to recognize a perfect-square trinomial, no harm is done; simply proceed to factor into the product of two binomials, and then you will recognize that the two binomials are the same


EXAMPLE 5

Factor completely x2 3x 8.

Solution x2 3x 8 is not factorable using integers. This becomes obvious from the table.

■

5.5

Product 1182 1182 2142 2142


273

Sum 1 182 7 1 8 7 2 142 2 2 4 2

8 8 8 8

No two factors of 8 produce a sum of 3.

▼ PRACTICE YOUR SKILL Factor completely x2 8x 24.

EXAMPLE 6

■

Factor completely 6y2 13y 28.

Solution 6y2 13y 28 12y 72 13y 42 . We found the binomial factors as follows: (y _____)(6y _____) or

1 # 28

or 28 # 1

(y _____)(6y _____)

2 # 14

or 14 # 2

or

4#7

or

7#4

(2y _____)(3y _____) or (2y _____)(3y _____)


EXAMPLE 7

■

Factor completely 32x2 50y2.

Solution 32x2 50y2 2116x2 25y2 2 214x 5y2 14x 5y2

First, factor out the common factor of 2 Factor the difference of squares

▼ PRACTICE YOUR SKILL Factor completely 4x2 100.

■

3 Use Factoring to Solve Equations Each time we considered a new factoring technique in this chapter, we used that technique to help solve some equations. It is important that you be able to recognize which technique works for a particular type of equation.

274

Chapter 5


EXAMPLE 8

Solve x 2 25x.

Solution x2 25x x2 25x 0 x1x 252 0 x0

or

x0

or

x 25 0 x 25


▼ PRACTICE YOUR SKILL Solve y2 4y.

EXAMPLE 9

■

Solve x 3 36x 0.

Solution x 3 36x 0 x1x 2 362 0 x1x 621x 62 0 x0

or

x0

or

x60 x 6

or

x60

or

x6

If abc 0, then a 0 or b 0 or c 0

The solution set is {6, 0, 6}. Does it check?

▼ PRACTICE YOUR SKILL Solve z3 25z 0.

EXAMPLE 10

■

Solve 10x 2 13x 3 0.


15x 12 12x 32 0 5x 1 0

or

2x 3 0

5x 1

or

2x 3

1 5

or

x

x

3 2

1 3 The solution set is e , f. Does it check? 5 2


■

5.5

EXAMPLE 11


275

Solve 4x 2 28x 49 0.

Solution 4x 2 28x 49 0 12x 72 2 0

12x 72 12x 72 0 2x 7 0

or

2x 7 0

2x 7

or

2x 7

7 2

or

x

x

7 2

7 The solution set is e f. 2


■

Pay special attention to the next example. We need to change the form of the original equation before we can apply the property ab 0 if and only if a 0 or b 0. The unique feature of this property is that an indicated product is set equal to zero.

EXAMPLE 12

Solve 1x 1 2 1x 4 2 40.

Solution 1x 121x 42 40 x 2 5x 4 40 x 5x 36 0 2

1x 92 1x 42 0 x90 x 9

Multiply the binomials Added 40 to each side

or

x40

or

x4


▼ PRACTICE YOUR SKILL Solve (x 1)(x 2) 20.

EXAMPLE 13

■

Solve 2n 2 16n 40 0.

Solution 2n2 16n 40 0 21n2 8n 202 0 n2 8n 20 0

1n 102 1n 22 0 n 10 0 n 10

Multiplied both sides by

or

n20

or

n2

The solution set is {10, 2}. Does it check?

1 2

276

Chapter 5



■

4 Solve Word Problems That Involve Factoring The preface to this book states that a common thread throughout the book is to learn a skill, to use that skill to help solve equations, and then to use equations to help solve problems. This approach should be very apparent in this chapter. Our new factoring skills have provided more ways of solving equations, which in turn gives us more power to solve word problems. We conclude the chapter by solving a few more examples.

EXAMPLE 14

Apply Your Skill Find two numbers whose product is 65 if one of the numbers is 3 more than twice the other number.

Solution Let n represent one of the numbers; then 2n 3 represents the other number. Because their product is 65, we can set up and solve the following equation: n12n 32 65 2n 3n 65 0 2

12n 1321n 52 0 2n 13 0

or

n50

2n 13

or

n5

13 2

or

n5

n

13 13 , then 2n 3 2 a b 3 10. If n 5, then 2n 3 2 2 13 2(5) 3 13. Thus the numbers are and 10, or 5 and 13. 2 If n

▼ PRACTICE YOUR SKILL Find two numbers whose product is 36 if one of the numbers is 1 more than twice the other number. ■

EXAMPLE 15

Apply Your Skill The area of a triangular sheet of paper is 14 square inches. One side of the triangle is 3 inches longer than the altitude to that side. Find the length of the one side and the length of the altitude to that side.

Solution Let h represent the altitude to the side. Then h 3 represents the length of the side of the triangle (see Figure 5.5). Because the formula for finding the area of 1 a triangle is A bh, we have 2

h

h+3 Figure 5.5

5.5


277

1 h1h 32 14 2 h1h 32 28


h2 3h 28 h2 3h 28 0

1h 72 1h 42 0 h70 h 7

or

h40

or

h4

The solution 7 is not reasonable. Thus the altitude is 4 inches, and the length of the side to which that altitude is drawn is 7 inches.

▼ PRACTICE YOUR SKILL The area of a triangular piece of glass is 30 square inches. One side of the triangle is 4 inches longer than the altitude to that side. Find the length of that side and the length of the altitude to that side. ■

CONCEPT QUIZ

For Problems 1–7, match each factoring problem with the type of pattern that would be used to factor the problem. 1. 2. 3. 4. 5. 6. 7.

x2 2xy y2 x2 y2 ax ay bx by x2 bx c ax2 bx c ax2 ax a (a b)x (a b)y

A. B. C. D. E. F. G.

Trinomial with an x-squared coefficient of one Common binomial factor Difference of two squares Common factor Factor by grouping Perfect-square trinomial Trinomial with an x-squared coefficient of not one

Problem Set 5.5 1 Factor Perfect-Square Trinomials For Problems 1–12, factor each of the perfect-square trinomials.

17. n2 7n 60

18. 4n3 100n

19. 3a2 7a 4

20. a2 7a 30

21. 8x2 72

22. 3y3 36y2 96y

23. 9x2 30x 25

24. 5x2 5x 6

1. x 4x 4

2. x 18x 81

3. x 10x 25

4. x2 24x 144

5. 9n2 12n 4

6. 25n2 30n 9

25. 15x2 65x 70

26. 4x2 20xy 25y2

7. 16a2 8a 1

8. 36a2 84a 49

27. 24x2 2x 15

28. 9x2y 27xy

2 2

9. 4 36x 81x2

2

10. 1 4x 4x2

11. 16x2 24xy 9y2

29. xy 5y 8x 40 30. xy 3y 9x 27

12. 64x2 16xy y2

31. 20x2 31xy 7y2 32. 2x2 xy 36y2

33. 24x2 18x 81

34. 30x2 55x 50

35. 12x2 6x 30

For Problems 13 – 40, factor each polynomial completely. Indicate any that are not factorable using integers.

36. 24x2 8x 32

37. 5x4 80

13. 2x2 17x 8

14. x2 19x

38. 3x5 3x

39. x2 12xy 36y2

15. 2x3 72x

16. 30x2 x 1

40. 4x2 28xy 49y2

2 Recognize the Different Types of Factoring Patterns

278

Chapter 5


3 Use Factoring to Solve Equations For Problems 41–70, solve each equation. 41. 4x2 20x 0

42. 3x2 24x 0

43. x 9x 36 0

44. x 8x 20 0

45. 2x3 8x 0

46. 4x3 36x 0

2

2

47. 6n 29n 22 0 2

72. Find two numbers whose product is 12 such that one of the numbers is four less than eight times the other number. 73. Find two numbers whose product is 1. One of the numbers is three more than twice the other number. 74. Suppose that the sum of the squares of three consecutive integers is 110. Find the integers. 75. One number is one more than twice another number. The

48. 30n2 n 1 0

sum of the squares of the two numbers is 97. Find the

49. 13n 1214n 32 0

numbers.

50. 12n 3217n 12 0

76. One number is one less than three times another number. If the product of the two numbers is 102, find the numbers.

51. 1n 221n 62 15 52. 1n 321n 72 25 53. 2x2 12x

54. 3x2 15x

55. t3 2t2 24t 0

56. 2t3 16t2 18t 0

57. 12 40x 25x2 0

77. In an office building, a room contains 54 chairs. The number of chairs per row is three less than twice the number of rows. Find the number of rows and the number of chairs per row. 78. An apple orchard contains 85 trees. The number of trees in each row is three less than four times the number of rows. Find the number of rows and the number of trees per row.

58. 12 7x 12x2 0 59. n2 28n 192 0

79. Suppose that the combined area of two squares is 360 square feet. Each side of the larger square is three times as long as a side of the smaller square. How big is each square?

60. n2 33n 270 0 61. 13n 121n 22 12 62. 12n 521n 42 1 63. x3 6x2

64. x3 4x2

65. 9x2 24x 16 0 66. 25x2 60x 36 0

80. The area of a rectangular slab of sidewalk is 45 square feet. Its length is 3 feet more than four times its width. Find the length and width of the slab. 81. The length of a rectangular sheet of paper is 1 centimeter more than twice its width, and the area of the rectangle is 55 square centimeters. Find the width and length of the rectangle.

67. x3 10x2 25x 0 68. x3 18x2 81x 0 69. 24x2 17x 20 0 70. 24x2 74x 35 0

4 Solve Word Problems That Involve Factoring For Problems 71– 86, set up an equation and solve each problem. 71. Find two numbers whose product is 15 such that one of the numbers is seven more than four times the other number.

82. Suppose that the length of a certain rectangle is three times its width. If the length is increased by 2 inches and the width increased by 1 inch, the newly formed rectangle has an area of 70 square inches. Find the width and length of the original rectangle. 83. The area of a triangle is 51 square inches. One side of the triangle is 1 inch less than three times the length of the altitude to that side. Find the length of that side and the length of the altitude to that side. 84. Suppose that a square and a rectangle have equal areas. Furthermore, suppose that the length of the rectangle is twice the length of a side of the square, and the width of the rectangle is 4 centimeters less than the length of a side of the square. Find the dimensions of both figures.

5.5 85. The sum of the areas of two circles is 180p square inches. The length of a radius of the smaller circle is 6 inches less than the length of a radius of the larger circle. Find the length of a radius of each circle.


86. The sum of the areas of two circles is 100p square centimeters. The length of a radius of the larger circle is 2 centimeters more than the length of a radius of the smaller circle. Find the length of a radius of each circle.

THOUGHTS INTO WORDS 87. When factoring polynomials, why do you think that it is best to look for a greatest common monomial factor first?

89. Explain how you would solve (x 2)(x 3) (x 2)(3x 1)

88. Explain how you would solve (4x 3)(8x 5) 0 and also how you would solve (4x 3)(8x 5) 9.

Do you see more than one approach to this problem?

Answers to the Example Practice Skills 1. (a) (a 5)2 (b) (6x 1)2 (c) (7m 4n)2 5. Not factorable 6. (3x 2)(x 4) 2 11. e f 5

2. 3(x 4)(x 6)

7. 4(x 5)(x 5)

12. {6, 3} 13. {2, 6} 14. 4, 9 or 8,

Answers to the Concept Quiz 1. F or A

2. C 3. E

4. A 5. G 6. D

7. B

279

9 2

3. 4(x2 25) 4. (3x 5)2 4 8. {0, 4} 9. {5, 0, 5} 10. e2, f 3

15. Altitude is 6 inches and side is 10 inches


SUMMARY

EXAMPLE

Find the greatest common factor. (Sec. 5.1, Obj. 1, p. 242)

To find the greatest common factor of two or more monomials, determine the greatest common factor of the numerical coefficients of the given monomials. Then for each variable that is common to all the given monomials, raise that variable to the lowest of its exponents in the given monomials.

Find the greatest common factor of 12a3b3, 18a2b2, and 54ab4.

Factor out the greatest common factor. (Sec. 5.1, Obj. 2, p. 243)

The distributive property in the form ab ac a(b c) provides the basis for factoring out a common factor.

Factor 8x5y2 20x3y4.

Factor by grouping. (Sec. 5.1, Obj. 3, p. 244)

Rewriting an expression such as ab 3a bc 3c as a(b 3) c(b 3) and then factoring out the common binomial factor of b 3 so that a(b 3) c(b 3) becomes (b 3)( a c) is called “factoring by grouping.”

Problems 1– 4

Solution

The greatest common factor of the numerical coefficient of all three terms is 6. For the variable, a, a1 is the greatest common factor of the given monomials. For the variable, b, b2 is the greatest common factor of the given monomials. So the greatest common factor is 6ab2. Problems 5 – 8

Solution

8x5y2 20x3y4 4x3y2(2x2 5y2) Factor 15xy 6x 10y2 4y.

Problems 9 –12

Solution

15xy 6x 10y2 4y 3x(5y 2) 2y(5y 2) (5y 2)(3x 2y)

Factor the difference of two squares. (Sec. 5.2, Obj. 1, p. 250)

The factoring pattern a2 b2 (a b)(a b) is called “the difference of two squares.” Be careful not to apply the pattern to the sum of two squares such as a2 b2. There is no pattern for the sum of two squares.

Factor 4x2 81y2.

Factor trinomials of the form x2 bx c. (Sec. 5.3, Obj. 1, p. 256)

For trinomials of the form x2 bx c, we want two factors of c whose sum will be equal to b.

Factor a2 2a 24.

Factor trinomials where the leading coefficient is not 1. (Sec. 5.4, Obj. 1, p. 265)

To factor trinomials of the form ax2 bx c, the process is typically trial and error. Make an organized listing of possible factor combinations. Then multiply to see if they are the correct factors. Be sure to factor out any common factors first.

280


Problems 13 –16

Solution

4x2 81y2 (2x 9y)(2x 9y)

Problems 17–20

Solution

The factors of 24 that sum to 2 are 6 and 4. a2 2a 24 (a 6)(a 4) Factor 3x2 x 14.

Problems 21–24

Solution

3x2 x 14 (3x 7)(x 2)

(continued)

OBJECTIVE

SUMMARY

EXAMPLE

Factor perfectPerfect-square trinomials are easy to rec- Factor 16x2 56x 49. square trinomials. ognize because of the nature of their (Sec. 5.5, Obj. 1, p. 270) terms. The first term and the last term will Solution 16x2 56x 49 be the squares of a quantity. The middle (4x)2 2(4x)(7) (7)2 term is twice the product of the quantities (4x 7)2 being squared in the first and last terms. Recognize the different types of factoring. (Sec. 5.5, Obj. 2, p. 271)

As a general guideline for factoring com- Factor 3x2 12xy 12y2. pletely, always look for a greatest comSolution mon factor first, and then proceed with 3x2 12xy 12y2 one or more of the following techniques. 3(x2 4xy 4y2) 1. Apply the difference-of-squares 3(x 2y)2 pattern. 2. Apply the perfect-square pattern. 3. Factor a trinomial of the form x2 bx c into the product of two binomials. 4. Factor a trinomial of the form ax2 bx c into the product of two binomials.

Use factoring to solve equations. (Sec. 5.1, Obj. 4, p. 245; Sec. 5.2, Obj. 2, p. 252; Sec. 5.3, Obj. 2, p. 259; Sec. 5.4, Obj. 2, p. 267; Sec. 5.5, Obj. 3, p. 273)

Property 5.1 states that for all real numbers a and b, ab 0 if and only if a 0 or b 0. To solve equations by applying this property, first set the equation equal to 0. Proceed by factoring the other side of the equation. Then set each factor equal to 0 and solve the equations.

Solve word problems that involve factoring. (Sec. 5.1, Obj. 5, p. 247; Sec. 5.2, Obj. 3, p. 254; Sec. 5.3, Obj. 3, p. 261; Sec. 5.3, Obj. 4, p. 262; Sec. 5.5, Obj. 4, p. 276)

Knowledge of factoring will expand the techniques available for solving word problems. This chapter introduced the Pythagorean theorem. This theorem pertains to right triangles and states that in any right triangle, the square of the longest side is equal to the sum of the squares of the other two sides. The formula for the theorem is written as a2 b2 c2, where a and b are the legs of the triangle, and c is the hypotenuse.

Solve x2 7x 8.

CHAPTER REVIEW PROBLEMS Problems 25 –28

Problems 29 –52

Problems 53 –72

Solution

x2 7x 8 x 7x 8 0 (x 8)(x 1) 0 x80 or x10 x 8 or x1 The solution set is {8, 1}. 2

Problems 73 – 84 The length of one leg of a right triangle is 1 inch longer than the length of the other leg. The length of the hypotenuse is 5 inches. Find the length of the two legs. Solution

Let x represent one leg of the triangle. Then x 1 will represent the other leg. We know that the hypotenuse is equal to 5. Apply the Pythagorean theorem. x2 (x 1)2 5 2 x x2 2x 1 25 2x2 2x 24 0 2(x2 x 12) 0 x2 x 12 0 (x 4)( x 3) 0 x40 or x30 x4 or x3 Because the length of a side of the triangle cannot be negative, the only viable answer is 3. Therefore, one leg of the triangle is 3 and the other leg is 4. 281

282

Chapter 5


Chapter 5 Review Problem Set For Problems 1– 4, find the greatest common factor of the expressions.

For Problems 29 –52, factor completely. Indicate any polynomials that are not factorable using integers. 29. x2 9x 14

30. 3x2 21x

31. 9x2 4

32. 4x2 8x 5

33. 25x2 60x 36

34. n3 13n2 40n

35. y2 11y 12

36. 3xy2 6x2y

5. 15x4 21x2

37. x4 1

38. 18n2 9n 5

6. 24a3 20a6

39. x2 7x 24

40. 4x2 3x 7

41. 3n2 3n 90

42. x3 xy2

43. 2x2 3xy 2y2

44. 4n2 6n 40

1. 6x2y2, 4x3y, and 14x4y3 2. 9xy2, 21x3y2, and 15x4y3 3 2

2 3

3

3. 24a b , 8a b , and 12ab

4. 36m5n3, 24m3n4, and 60m4n2 For Problems 5 – 8, factor out the greatest common factor.

7. 10x4y 50x3y2 5x2y3 8. 12m2n 20m3n2 24m4n3 For Problems 9 –12, factor completely by grouping. 9. ab 4a 3b 12 10. 2xy 12x 5y 30 11. x3 x2y xy2 y3 12. 8m 2m 12mn 3n 2

45. 5x 5y ax ay 46. 21t2 5t 4

47. 2x3 2x

48. 3x3 108x

49. 16x2 40x 25

50. xy 3x 2y 6 For Problems 13 –16, factor the difference of squares. Be sure to factor completely.

51. 15x2 7xy 2y2

52. 6n4 5n3 n2

13. 25x2 16y2 14. 18x2 50 15. x4 16

For Problems 53 – 72, solve each equation. 53. x2 4x 12 0

54. x2 11x

55. 2x2 3x 20 0

56. 9n2 21n 8 0

57. 6n2 24

58. 16y2 40y 25 0

59. t3 t 0

60. 28x2 71x 18 0

61. x2 3x 28 0

62. 1x 221x 22 21

63. 5n2 27n 18

64. 4n2 10n 14

65. 2x3 8x 0

66. x2 20x 96 0

16. x3 49x For Problems 17–28, factor completely. 17. x2 11x 24 18. x2 13x 12 19. x2 5x 24 20. x2 8x 20 21. 2x 5x 3 2

22. 5x2 17x 6 23. 4x2 5x 6 24. 3x 10x 8 2

25. 9x2 12x 4 26. 4x2 20x 25 27. 16x2 8x 1 28. x2 12xy 36y2

67. 4t2 17t 15 0 68. 31x 22 x1x 22 0 69. 12x 5213x 72 0

Chapter 5 70. 1x 421x 12 50 71. 7n 2n2 15

72. 23x 6x2 20

Set up an equation and solve each of the following problems. 73. The larger of two numbers is one less than twice the smaller number. The difference of their squares is 33. Find the numbers. 74. The length of a rectangle is 2 centimeters less than five times the width of the rectangle. The area of the rectangle is 16 square centimeters. Find the length and width of the rectangle. 75. Suppose that the combined area of two squares is 104 square inches. Each side of the larger square is five times as long as a side of the smaller square. Find the size of each square. 76. The longer leg of a right triangle is one unit shorter than twice the length of the shorter leg. The hypotenuse is one unit longer than twice the length of the shorter leg. Find the lengths of the three sides of the triangle. 77. The product of two numbers is 26, and one of the numbers is one larger than six times the other number. Find the numbers.

Review Problem Set

283

78. Find three consecutive positive odd whole numbers such that the sum of the squares of the two smaller numbers is nine more than the square of the largest number. 79. The number of books per shelf in a bookcase is one less than nine times the number of shelves. If the bookcase contains 140 books, find the number of shelves. 80. The combined area of a square and a rectangle is 225 square yards. The length of the rectangle is eight times the width of the rectangle, and the length of a side of the square is the same as the width of the rectangle. Find the dimensions of the square and the rectangle. 81. Suppose that we want to find two consecutive integers such that the sum of their squares is 613. What are they? 82. If numerically the volume of a cube equals the total surface area of the cube, find the length of an edge of the cube. 83. The combined area of two circles is 53p square meters. The length of a radius of the larger circle is 1 meter more than three times the length of a radius of the smaller circle. Find the length of a radius of each circle. 84. The product of two consecutive odd whole numbers is one less than five times their sum. Find the numbers.

Chapter 5 Test For Problems 1–10, factor each expression completely. 1.

1. x2 3x 10

2.

2. x2 5x 24

3.

3. 2x3 2x

4.

4. x2 21x 108

5.

5. 18n2 21n 6

6.

6. ax ay 2bx 2by

7.

7. 4x2 17x 15

8.

8. 6x2 24

9.

9. 30x3 76x2 48x

10.

10. 28 13x 6x2 For Problems 11–21, solve each equation.

11.

11. 7x2 63

12.

12. x2 5x 6 0

13.

13. 4n2 32n

14.

14. 13x 2212x 52 0

15.

15. 1x 321x 72 9

16.

16. x3 16x2 48x 0

17.

17. 91x 52 x1x 52 0

18.

18. 3t2 35t 12

19.

19. 8 10x 3x2 0

20.

20. 3x3 75x

21.

21. 25n2 70n 49 0 For Problems 22 –25, set up an equation and solve each problem.

22.

22. The length of a rectangle is 2 inches less than twice its width. If the area of the rectangle is 112 square inches, find the length of the rectangle.

23.

23. The length of one leg of a right triangle is 4 centimeters more than the length of the other leg. The length of the hypotenuse is 8 centimeters more than the length of the shorter leg. Find the length of the shorter leg.

24.

24. A room contains 112 chairs. The number of chairs per row is five less than three times the number of rows. Find the number of chairs per row.

25.

25. Suppose the sum of the squares of three consecutive integers is 77. Find the integers.

284

Rational Expressions and Rational Equations

6 6.1 Simplifying Rational Expressions 6.2 Multiplying and Dividing Rational Expressions 6.3 Adding and Subtracting Rational Expressions 6.4 More on Addition and Subtraction of Rational Expressions


6.5 Rational Equations and Problem Solving 6.6 More Rational Equations and Problem Solving

■ If Phil can do a certain job in 15 minutes and Mark can do m m the same job in 10 minutes, then the equation 1 15 10 can be used to determine how long it would take to do the job 1 1 1 if they work together. The equation could also m 15 10 be used.

O

ne day, Jeff rode his bicycle 40 miles out into the country. On the way back, he took a different route that was 2 miles longer, and it took him an hour longer to return. If his rate on the way out into the country was 4 miles per hour faster 42 40 1 than his rate back, find both rates. We can use the fractional equation x x4 to determine that Jeff rode out into the country at 16 miles per hour and returned at 12 miles per hour. In Chapter 1 our study of common fractions led naturally to some work with simple algebraic fractions. Then in Chapters 4 and 5 we discussed the basic operations that pertain to polynomials. Now we can use some ideas about polynomials— specifically the factoring techniques—to expand our study of rational expressions. This, in turn, gives us more techniques for solving equations, which increases our problem-solving capabilities.


285

I N T E R N E T

P R O J E C T

Because fractions in this chapter have denominators that contain variables, certain values for the variable are excluded so the denominator does not equal zero. Zero is a very special number. Do an Internet search on the history of zero to discover the two important but different uses of zero. What Indian mathematician was one of the first to discover some of the properties of zero? Was he able to discover properties regarding division by zero?

6.1

Simplifying Rational Expressions OBJECTIVES 1

Simplify Rational Expressions

2

Simplify Rational Expressions Involving Opposites

1 Simplify Rational Expressions If the numerator and denominator of a fraction are polynomials, then we call the fraction an algebraic fraction or a rational expression. Here are some examples of rational expressions. 4 x2

x2 2x 4 x2 9

y x2 xy 3

x3 2x2 3x 4 x2 2x 6

Because we must avoid division by zero, no values can be assigned to variables that 4 create a denominator of zero. Thus the fraction is meaningful for all real x2 number values of x except for x 2. Rather than making a restriction for each individual fraction, we will simply assume that all denominators represent nonzero real numbers. ak a Recall that the fundamental principle of fractions a b provides the bk b basis for expressing fractions in reduced (or simplified) form, as the next examples demonstrate. 18 3#6 3 # 24 4 6 4

42xy 2#3#7#x#y 6x 77y 7 # 11 # y 11

15x 3#5#x 3 # # 25x 5 5 x 5

28x2y2 63x2y3

4 # 7 # x2 # y2

9 # 7 # x2 # y3

4 9y

y

We can use the factoring techniques from Chapter 5 to factor numerators and/or denominators so that we can apply the fundamental principle of fractions. The following examples should clarify this process.

EXAMPLE 1

Simplify

x2 6x . x2 36

Solution x1x 62 x2 6x x 2 1x 62 1x 62 x 6 x 36 286

6.1

Simplifying Rational Expressions

287


EXAMPLE 2

Simplify

2a2 8a . a2 16

Simplify

a2 . a2 4a 4

■

Solution 11a 22 1 a2 1a 22 1a 22 a 2 a 4a 4 2


EXAMPLE 3

Simplify

x3 . x2 5x 6

Simplify

x2 4x 21 . 2x2 15x 7

■

Solution

1x 32 1x 72 x2 4x 21 x3 2 12x 12 1x 72 2x 1 2x 15x 7


EXAMPLE 4

Simplify

a2 a 20 . 2a2 11a 5

Simplify

a2b ab2 . ab b2

■

Solution

ab1a b 2 a2b ab2 a b1a b 2 ab b2


EXAMPLE 5

Simplify

xy2 x2y x2 xy

■

.

4x3y 36xy 2x2 4x 30

.

Solution 4x3y 36xy 2x2 4x 30

4xy1x2 92 21x2 2x 152 2

4xy1x 321x 32 21x 52 1x 32

2xy1x 32 x5

288

Chapter 6



5a2 8a 4 . 3a3b 12ab

■

Notice in Example 5 that we left the numerator of the final fraction in factored form. We do this when polynomials other than monomials are involved. Either 2xy1x 32 2x2y 6xy or is an acceptable answer. x5 x5

EXAMPLE 6

Simplify

3x 9 . x2 3x

Solution 31x 32 3x 9 x1x 32 x2 3x

In the numerator, factor out a common factor of 3

31x 32 x1x 32 3 3 x x


5x 10 . 6x 12

■

2 Simplify Rational Expressions Involving Opposites Remember that the quotient of any nonzero real number and its opposite is 1. For 9 7 example, 1 and 1. Likewise, the indicated quotient of any polynomial 7 9 and its opposite is equal to 1. Consider these examples. x 1 because x and x are opposites. x xy 1 yx

because x y and y x are opposites.

a2 9 1 9 a2

because a2 9 and 9 a2 are opposites.

Use this property to simplify rational expressions in the final examples of this section.

EXAMPLE 7

Simplify

14 7n . n2

Solution 712 n2 14 7n n2 n2 7112 7

2n 1 n2

6.1

Simplifying Rational Expressions

289


EXAMPLE 8

Simplify

20 5y . y4

Simplify

x2 4x 21 . 15 2x x2

■

Solution

1x 72 1x 32 x2 4x 21 2 15 x 2 13 x 2 15 2x x a

x7 b 112 x5

x7 x5

or

x3 1 3x

x 7 x5


CONCEPT QUIZ

12 a a2 . a2 2a 8

■

For Problems 1–5, answer true or false. 1. A fraction in which both numerator and denominator are polynomials is called a polynomial fraction. 2. For rational expressions, no values can be assigned to variables that create a denominator of zero. 3. The fundamental principle of fractions is applied to reduce rational expressions. x2 4. The rational expression simplifies to 1. x2 5a 5. The rational expression simplifies to 1. a5

Problem Set 6.1 1 Simplify Rational Expressions

13.

For Problems 1– 46, simplify each rational expression. 1.

6x 14y

9xy 3. 24x 5.

7.

9.

11.

15x2y 25x 36x4y3 48x6y2 12a2b5 54a2b3 32xy2z3 72yz4

2.

8y 18x

12y 4. 20xy 6.

8.

10.

12.

16x3y2 28x2y 18x3y 36xy3 24a3b3 39a5b2 27x2y3z4 45x3y3z

xy x 2x 2

14.

x2 5x xy

15.

8x 12y 12

16.

8 12x 16y

17.

x2 2x x2 7x

18.

x2 6x 2x2 6x

19.

x2 1 3x2 3x

20.

5x2 25x x2 25

22.

x3 x x2y

21.

23.

25.

x2 xy 2

x

6x3 15x2y 6x 24xy 2

n2 2n n2 3n 2

24.

26.

3

6x2 42xy 16x3 8x2y n2 9n 18 n2 6n

290

Chapter 6


27.

2n2 5n 3 n2 9

28.

3n2 10n 8 n2 16

29.

2x2 17x 35 3x2 19x 20

30.

5x2 32x 12 4x2 27x 18

For Problems 47– 60, simplify each rational expression.

181x 22 3

47.

7x x7

48.

x9 9x

8x 51x 18 8x2 29x 12

49.

15 3n n5

50.

2n2 8n 4n

6a2 11a 10 36. 8a2 22a 5

51.

4x3 4x 1 x2

52.

9 x2 3x3 27x

x2 3xy 2y2

53.

1 x2 x x2

54.

2x x2 4 x2

31.

91x 12 2

32.

121x 12 3 7x 61x 18 7x2 19x 6 2

33.

10a2 a 3 35. 15a2 4a 3 37.

161x 22 2 2

34.

x2 2xy 3y2

38.

2x2 xy y2

2 Simplify Rational Expressions Involving Opposites

x2 4y2

x2 9 39. x2 3x

x2 2x 40. x2 4

55.

6 x 2x2 12 7x 10x2

56.

15 x 2x2 21 10x x2

n2 14n 49 41. 8n 56

6n 60 42. 2 n 20n 100

57.

x2 7x 18 12 4x x2

58.

3x 21 28 4x

4n2 12n 9 43. 2n2 n 3

9n2 30n 25 44. 3n2 n 10

59.

5x 40 80 10x

60.

x2 x 12 8 2x x2

45.

y2 6y 72

46.

y2 8y 84

y2 20y 96 y2 23y 120

THOUGHTS INTO WORDS 61. Explain the role that factoring plays in simplifying rational expressions.

62. Which of the following simplification processes are correct? Explain your answer. 2x 2 x

x2 2 x

x1x 22 x

x2

FURTHER INVESTIGATIONS For Problems 63 – 66, simplify each fraction. You will need to use the factoring by grouping method. 63.

xy 3x 2y 6 xy 5x 2y 10

64.

xy 4x y 4 xy 4x 4y 16

xy 7x 5y 35 66. xy 9x 5y 45

xy 6x y 6 65. xy 6x 5y 30

link between positive and negative exponents 1 an n aa n b , along with the property m anm, can also be a a

used when reducing fractions. Consider this example: 1 x3 x37 x4 4 x7 x For Problems 67–72, use this approach to express each fraction in reduced form. Give all answers with positive exponents only. 67.

The

70.

x3 x9 x5y2 x6y3

68.

71.

x4 x8 28a2b3 7a5b3


2a a4

2.

1 x2

3.

a4 2a 1

4. y

5.

5a 2 3ab1a 22


2. True


6.

5 6

7. 5

8.

a3 a2

69.

72.

x4y3 x7y5 44a3b4 4a3b6

6.2

6.2

Multiplying and Dividing Rational Expressions

291

Multiplying and Dividing Rational Expressions OBJECTIVES 1

Multiply Rational Expressions

2

Divide Rational Expressions

1 Multiply Rational Expressions In Chapter 1 we defined the product of two rational numbers as

a # c ac . This b d bd

definition extends to rational expressions in general.

Definition 6.1 If

C A and are rational expressions, with B 0 and D 0, then B D AC A # C B D BD

In other words, to multiply rational expressions we multiply the numerators, multiply the denominators, and express the product in simplified form. The following examples illustrate this concept. 1.

2#5#x#y 5 2x # 5y # # # 3y 4x 3 4 x y 6 2

Notice that we used the commutative property of multiplication to rearrange factors in a more convenient form for recognizing common factors of the numerator and denominator

2.

4a # 8b 4#8#a#b 4 # 2 2 # # 6b 12a 9a 6 12 a b

3.

9 # 5 # x2 # y2 9x2 # 5y 3 15xy 7x2y3 15 # 7 # x3 # y24 7xy2

4

3

3

a 3

2

5

x

y

When multiplying rational expressions, we sometimes need to factor the numerators and/or denominators so that we can recognize common factors. Consider the next examples.

EXAMPLE 1

Multiply and simplify

x # x 3. y x2 9

Solution x1x 32 x # x3 y 1x 32 1x 321y2 x 9 2

x y1x 32

x is also an acceptable answer. xy 3y

292

Chapter 6


▼ PRACTICE YOUR SKILL Multiply and simplify

a # a 4. b a 5a 4 2

■

Remember, when working with rational expressions, we are assuming that all denominators represent nonzero real numbers. Therefore, in Example 1, we are x x # x3 claiming that 2 for all real numbers except 3 and 3 for x, y y1x 32 x 9 and 0 for y.

EXAMPLE 2


2 x # x 10x 16 . 5 x 2x 2

Solution 2 x # x 10x 16 x1x 221x 82 x 8 5 x1x 22152 5 x 2x 2


EXAMPLE 3


y2 4y 21 # 2 y . 2 y 3y


a2 3a # a2 3a 10 . a5 a2 5a 6

■

Solution

a1a 32 1a 521a 22 a2 3a # a2 3a 10 a 2 a5 1a 521a 22 1a 32 a 5a 6


EXAMPLE 4


2x 6 # x2 5x 4 . 3x 12 x2 2x 3


6n2 7n 3 # n2 1 . n1 2n2 3n

■

Solution

12n 3213n 12 1n 12 1n 12 6n2 7n 3 # n2 1 n1 1n 121n2 12n 32 2n2 3n

13n 121n 12 n

▼ PRACTICE YOUR SKILL Multiply and simplify

y2 25

2y2 y . 2y2 7y 15 y 5

#

■

6.2


293

2 Divide Rational Expressions a form, we invert the divisor and mulb a c a # d tiply. Symbolically we express this as . Furthermore, we call the b d b c d c numbers and reciprocals of each other because their product is 1. Thus we can also c d describe division as to divide by a fraction, multiply by its reciprocal. We define division of rational expressions in the same way using the same vocabulary. Recall that to divide two rational numbers in

Definition 6.2 If

A C and are rational expressions, with B 0, D 0, and C 0, then B D A C A B D B

#

D AD C BC

Consider some examples. 4 # 14 # x # y2 4y 4x 6x2 4x # 14y2 2 2 2 # # # 7y 7y 6x 3x 14y 7 6 x y 2

1.

y

2

3

x

8ab # 15a2b 18a3 8 # 15 # a3 # b2 20b 8ab # 2 3 9b 9b 27 15a b 18a 9 18 # a3 # b 4

2.

3

9

x2y3 9a2b x2y3 5xy2 9 # x # y3 # a2 # b 9axy # 2 2 2 # # # # # 4ab 4ab 5xy 20 9a b 4 5 a b x y x 2

3.

b

5

y

a

The key idea when dividing fractions is first to convert to an equivalent multiplication problem and second to factor numerator and denominator completely and look for common factors.

EXAMPLE 5

Divide and simplify

x2 4x x2 16 3 . xy y y2

Solution x2 4x x2 16 x2 4x # y3 y2 3 xy xy y y2 x2 16 x1x 42 1y2 2 1y 12 y

xy1x 42 1x 42

y1y 12 x4

▼ PRACTICE YOUR SKILL Divide and simplify

5a 15 a2 9 . 20b ab 4b

■

294

Chapter 6


EXAMPLE 6

Divide and simplify

1 a2 3a 18 2 . a2 4 3a 12

Solution 1 a2 3a 18 # 3a2 12 a2 3a 18 2 2 1 a 4 3a 12 a2 4

1a 621a 32 132 1a2 42 a2 4

31a 621a 32


EXAMPLE 7

Divide and simplify

1 x2 4x 12 . 2x 3 4x 6

Divide and simplify

2n2 7n 4 1n 42 . 6n2 7n 2

■

Solution 2n2 7n 4 # 1 2n2 7n 4 1n 42 2 2 6n 7n 2 6n 7n 2 n 4 12n 121n 42

12n 1213n 221n 42

1 3n 2

▼ PRACTICE YOUR SKILL Divide and simplify

3x2 5x 2 1x 12 . 3x2 x 2

■

In a problem such as Example 7, it may be helpful to write the divisor with a den4 nominator of 1. Thus we can write n 4 as ; its reciprocal then is obviously 1 1 . n4

CONCEPT QUIZ

For Problems 1–5, answer true or false. 1. To multiply rational expressions we multiply the numerators, multiply the denominators, and express the result in simplified form. 2. To divide by a fraction, multiply by its reciprocal. a b 3. The reciprocal of is . a b 4. The product of two numbers that are reciprocals is 1. 1 5. The reciprocal of is x. x

6.2


295

Problem Set 6.2 1 Multiply Rational Expressions For Problems 1– 40, perform the indicated multiplications and divisions and express your answers in simplest form. 1.

5 # 3 9 10

2.

3 6 3. a b a b 4 7 5.

9.

4 5 4. a b a b 6 15

8xy 6x # 12y 14y

7. a

7 # 12 8 14

6.

5n2 27n ba b 18n 25

9x # 20xy 15y 18x

8. a

2 2 y # x y xy xy

10.

4ab 30a b a b 10 22b

x2 9 # 8 6 x3

2x2 xy y # 11. xy 10x 5y

25.

27. 24x3

31.

18rs 9r 34

32.

8rs 6s 3

33.

7a 7b 6ab 2 2 4ab 4b a b2

34.

ab b 4ab 3a 3b 2a 2ab

35.

38.

16.

2

39.

40.

2a2 11a 21 # 3a2 11a 4 3a2 a 2a2 5a 12

2

2x2 3xy y2 4x2y 2x2 xy y2 2

xy

2x2 xy 3y2 1x y2

2

x2 4xy 4y2 2

x

13t 12 2 45t 15

x2 y2 6x2y2

5x2 4xy y2 y

4x2 12xy 9y2 10x 15y x2 4y2 x2 2xy

12t2 5t 3 20t 5

5t2 3t 2 5t2 32t 12 2 1t 12 4t2 3t 1

For Problems 41– 46, perform the indicated operations and express the answers in simplest form. Remember that multiplications and divisions are done in the order that they appear from left to right.

2x2 2xy

2 # x 16 2 17. 2 x 4x 32 5xy 5y

18.

x3 3x2 # x2 5x 14 x 4x 4 x2 3x

41.

30x # 5xy 6 9y 4 12y2

19.

n3 n # 4n 24 n 7n 6 n2 n

42.

5xy2 18x2 3 # 12y 15y 2xy

20.

2x2 6x 36 # x2 5x 24 x2 2x 48 2x2 18

43.

8x2 # x2 1 2 xy 2 xy xy xy 8x 8y

44.

5x 20 # x 3 15 2 x3 x 9 x4

45.

x2 8x x2 9x 18 # x2 5x x2 3x x2 25 x2 3x 40

46.

4x # 25xy 2 10 3x 6y x 4 x 4x 4

2

2

2 Divide Rational Expressions 21. a 23.

17 19 b a b 9 9

3a2 6a 7 28

22. a 24.

7y 9

2 1 7a2b3 9ab4

x2 11x 30 # 5x2 20 13. 2 2 x 4 x 14x 45

a a2 # 1a 15. 2 15a 11a 2 1 a2

28. 14xy2 30.

36.

x 15x 54 # 3x 6 x2 2 x2 10x 9

16x y

24ab2 12ab 25b 15a2

1 1 12a 15ab3

37.

14.

26.

29.

x2 y2 x2 xy # 12. xy 3

2

18a2b2 9a 27a 5b

13 15 b a b 7 14

4x 12x 11y 33

296

Chapter 6


THOUGHTS INTO WORDS 47. Give a step-by-step description of how to do the following multiplication problem:

x2 x is undefined x1 x1 for x 1, x 1, and x 0 but is defined for x 2.

49. Explain why the quotient

x2 x # x2 x 6 x2 1 x2 4x 12 x1 x x1 1 x 1 a b b? x1 x x x1 x x Justify your answer.

48. Is a


a b1a 12

2.

y7 2

3.

2 3

4.

y12y 12 2y 3

5.

a4 41a 32

6. 2(x 6)(x 2)

7.

1 x1


6.3

2. True 3. False

4. True 5. True

Adding and Subtracting Rational Expressions OBJECTIVES 1

Add or Subtract Rational Expressions with Like Denominators

2

Add or Subtract Rational Expressions with Unlike Denominators

1 Add or Subtract Rational Expressions with Like Denominators c a In Chapter 1 we defined addition and subtraction of rational numbers as b b ac a c ac and , respectively. These definitions extend to rational b b b b expressions in general.

Definition 6.3 If

A C and are rational expressions, with B 0, then B B A C AC B B B

and

A C AC B B B

Thus if the denominators of two rational expressions are the same, then we can add or subtract the fractions by adding or subtracting the numerators and placing the result over the common denominator. Here are some examples: 7 57 12 5 x x x x 8 3 83 5 xy xy xy xy

6.3

Adding and Subtracting Rational Expressions

297

14 15 14 15 29 2x 1 2x 1 2x 1 2x 1 3 4 34 1 a1 a1 a1 a1

or

1 a1

In the next example, notice how we put to use our previous work with simplifying polynomials.

EXAMPLE 1

Add or subtract as indicated. (a)

x3 2x 3 4 4

(d)

213n 12 31n 12 n n

Solution (a) (b)

(b)

x5 x2 7 7

(c)

2x 3 3x 1 xy xy

1x 32 12x 32 2x 3 3x x3 4 4 4 4

1x 52 1x 22 x2 x5x2 3 x5 7 7 7 7 7

(c)

13x 12 12x 32 2x 3 5x 4 3x 1 xy xy xy xy

(d)

31n 12 213n 12 31n 12 213n 12 n n n

3n 5 6n 2 3n 3 n n

▼ PRACTICE YOUR SKILL Add or subtract as indicated. (a)

2x 4 x1 3 3

(b)

51y 62 3y

214y 12

■

3y

It may be necessary to simplify the fraction that results from adding or subtracting two fractions.

EXAMPLE 2

Add or subtract as indicated. Express your answer in simplest form. (a)

Solution (a) (b)

4x 3 2x 3 8 8

(b)

3n 1 n5 12 12

(c)

2x 3 3x 1 2 x2 4 x 4

14x 32 12x 32 4x 3 2x 3 6x 3x 8 8 8 8 4

13n 12 1n 52 n5 3n 1 n 5 3n 1 12 12 12 12 21n 22 2n 4 n2 12 12 6

298

Chapter 6


12x 32 13x 12 3x 1 x2 2x 3 2 2 x2 4 x 4 x2 4 x 4

(c)

x2 1x 22 1x 22

1 x2

▼ PRACTICE YOUR SKILL Add or subtract as indicated. 8x 4 5x 4 9 9

(a)

(b)

4x 6 3x 8 2 x 5x 6 x 5x 6 2

■

2 Add or Subtract Rational Expressions with Unlike Denominators Recall that to add or subtract rational numbers with different denominators, we first change them to equivalent fractions that have a common denominator. In fact, we found that by using the least common denominator (LCD), our work was easier. Let’s carefully review the process, because it will also work with algebraic fractions in general.

EXAMPLE 3

Add

3 1 . 5 4

Solution By inspection, we see that the LCD is 20. Thus we can change both fractions to equivalent fractions that have a denominator of 20. 3 1 3 4 1 5 12 5 17 a b a b 5 4 5 4 4 5 20 20 20 Form of 1

Form of 1


EXAMPLE 4

1 4 . 7 3

Subtract

■

7 5 . 18 24

Solution If we cannot find the LCD by inspection, then we can use the prime factorization forms. 18 2 # 3 # 3

  24 2 # 2 # 2 # 3 

LCD 2 # 2 # 2 # 3 # 3 72

7 5 4 7 3 20 21 1 5 a b a b 18 24 18 4 24 3 72 72 72

6.3


299


5 7 . 12 30

■

Now let’s consider adding and subtracting rational expressions with different denominators.

EXAMPLE 5

Add

3x 1 x2 . 4 3

Solution By inspection, we see that the LCD is 12. x2 3x 1 x2 3 3x 1 4 a ba b a ba b 4 3 4 3 3 4

31x 22 12

413x 12 12

31x 22 413x 12 12

3x 6 12x 4 12

15x 2 12


EXAMPLE 6

2x 7 x4 . 5 3

Subtract

■

n6 n2 . 2 6

Solution By inspection, we see that the LCD is 6. n2 n6 n2 3 n6 a ba b 2 6 2 3 6

31n 22 6

1n 62 6

31n 22 1n 62 6

3n 6 n 6 6

2n 6

n 3

Don’t forget to simplify!


x1 5x 3 . 6 2

■

300

Chapter 6


It does not create any serious difficulties when the denominators contain variables; our approach remains basically the same.

EXAMPLE 7

Add

3 7 . 4x 3x

Solution By inspection, we see that the LCD is 12x. 3 7 3 3 7 4 9 28 9 28 37 a b a b 4x 3x 4x 3 3x 4 12x 12x 12x 12x


EXAMPLE 8

1 5 . 2x 3x

Subtract

■

5 11 . 12x 14x

Solution 12x 2 # 2 # 3 # x 14x 2 # 7 # x

  

LCD 2 # 2 # 3 # 7 # x 84x

11 5 11 7 5 6 a b a b 12x 14x 12x 7 14x 6

30 77 30 47 77 84x 84x 84x 84x


EXAMPLE 9

Add

3 7 . 8a 10a

■

4 2 . y y2

Solution By inspection, we see that the LCD is y1y 22 . y 4 2 2 y2 4 a b a b y y y2 y2 y2 y Form of 1

21y 22 y1y 22

Form of 1

4y y1y 22

21y 22 4y y1y 22 2y 4 4y 6y 4 y1y 22 y1y 22

6.3


301


14 4 . n n1

■

Notice the final result in Example 9. The numerator, 6y 4, can be factored into 2(3y 2). However, because this produces no common factors with the denominator, the fraction cannot be simplified. Thus the final answer can be left as 213y 22 6y 4 ; it is also acceptable to express it as . y1y 22 y1y 22

EXAMPLE 10

Subtract

4 7 . x2 x3

Solution By inspection, we see that the LCD is 1x 22 1x 32 . 4 7 4 x3 7 x2 a ba b a ba b x2 x3 x2 x3 x3 x2

41x 32

1x 22 1x 32

71x 22

1x 32 1x 22

41x 32 71x 22 1x 22 1x 32

4x 12 7x 14 1x 22 1x 32

3x 2 1x 22 1x 32


CONCEPT QUIZ

2 3 . a5 a4

■

For Problems 1–5, answer true or false. 1. To add two rational expressions, add the numerators and add the denominators. 2. To add or subtract two rational expressions, the expressions must have a common denominator. 3. After adding rational expressions, the result may need to be simplified. 4. To add or subtract rational expressions with different denominators, the expressions are changed to equivalent expressions that have common denominators. 5 7 5. When adding and , the least common denominator is 6n2. 3n 2n

Problem Set 6.3 1 Add or Subtract Rational Expressions with Like Denominators For Problems 1–34, add or subtract as indicated. Be sure to express your answers in simplest form. 1.

5 12 x x

2.

17 13 x x

3.

7 5 3x 3x

4.

4 3 5x 5x

5.

7 1 2n 2n

6.

5 4 3n 3n

7.

9 13 2 4x2 4x

8.

12 22 2 5x2 5x

302

Chapter 6


x1 3 x x

10.

4 x2 x x

47.

x3 x4 5 2

48.

x2 x1 5 6

11.

3 6 x1 x1

12.

10 8 x4 x4

49.

x6 x2 9 3

50.

x2 x4 4 8

13.

x1 1 x x

14.

3 2x 3 x x

51.

3n 1 2n 5 3 4

15.

3t 1 2t 3 4 4

16.

8t 5 4t 1 7 7

52.

2n 3 4n 1 4 7

17.

7a 2 4a 6 3 3

18.

4a 2 9a 1 6 6

53.

4n 3 3n 5 6 18

19.

4n 3 6n 5 8 8

20.

6n 1 2n 5 10 10

54.

5n 2 4n 7 12 6

21.

3n 7 9n 1 6 6

22.

7n 1 2n 6 5 5

55.

3x x 5x 4 6 8

56.

5x 3x 7x 2 4 6

23.

5x 2 8x 3 7x 7x

24.

4x 1 2x 5 3x 3x

57.

x 3 7x 5 10 12

58.

4x 5 11x 3 9 6

59.

5 1 8x 6x

60.

7 5 8x 12x

61.

5 7 6y 9y

62.

11 8 9y 15y

63.

11 5 12x 16x2

64.

4 7 2 9x 6x

65.

3 2 5 2x 3x 4x

66.

3 5 10 4x 6x 9x

9.

25.

27. 29.

31x 22 4x 61n 12 3n

213x 42 2

7x

61x 12 4x 31n 22

3n 7x 8 7x2

26.

28. 30.

41x 32 5x 21n 42 3n

314x 32 2

8x

21x 62 5x 41n 22

3n 11x 9 8x2

31.

a2 4 a2 a2

33.

18 3x 1x 62 2 1x 62 2

67.

3 7 x5 x

68.

4 9 x8 x

34.

x2 5x 4 1x 12 2 1x 12 2

69.

2 3 n1 n

70.

5 7 n3 n

71.

4 6 n n4

72.

8 3 n n9

73.

6 12 x 2x 1

74.

2 6 x 3x 2

32.

n2 16 n4 n4

2 Add or Subtract Rational Expressions with Unlike Denominators For Problems 35 – 80, add or subtract as indicated and express your answers in simplest form. 35.

3x 5x 8 4

36.

2x 5x 3 9

75.

4 6 x4 x3

37.

7n 4n 12 3

38.

7n n 6 12

76.

7 8 x2 x1

39.

y 3y 6 4

40.

7y 3y 4 5

77.

3 9 x2 x1

41.

8x 3x 3 7

42.

3y 5y 6 8

78.

5 4 x1 x6

43.

2x 3x 6 5

44.

7x 6x 9 12

79.

3 4 2x 1 3x 1

45.

7n 3n 8 9

46.

8n 7n 10 15

80.

4 6 3x 4 2x 3

6.4

More on Addition and Subtraction of Rational Expressions

303

THOUGHTS INTO WORDS 81. Give a step-by-step description of how to do this addition problem:

83. Suppose that your friend does an addition problem as follows: 51122 8172 5 7 60 56 116 29 8 12 81122 96 96 24

2x 3 3x 1 6 9 3 3 and opposites? What should be the x2 2x 3 3 result of adding and ? x2 2x

Is this answer correct? What advice would you offer your friend?

82. Why are

FURTHER INVESTIGATIONS For Problems 84 – 89, use this approach to help with the additions and subtractions.

9 4 . Notice x2 2x that the denominators x 2 and 2 x are opposites; that is, 1(2 x) (x 2). In such cases, add the fractions as follows:

Consider the addition problem

9 4 9 4 1 a b x2 2x x2 2 x 1

4 9 x2 x2 9 142 x2

Form of 1

5 x2

84.

7 2 x1 1x

85.

5 1 x3 3x

86.

x 4 x4 4x

87.

4 2 a1 1a

88.

2 3 1 x3 3x x2 9

89.

n 3 2n 1 1 2n

Answers to the Example Practice Skills 13y 28 x3 x 1 (b) 2. (a) (b) 3 3y 3 x3 a 7 18n 14 9. 10. n1n 12 1a 521a 42 1. (a)

3.

19 21

4.

13 20

5.

11x 1 15

6.

x 3

7.

17 6x

8.

13 40a


6.4

2. True 3. True

4. True 5. False

More on Addition and Subtraction of Rational Expressions OBJECTIVES 1

Add or Subtract Rational Expressions

2

Simplify Complex Fractions

3

Write Rational Expressions to Express Rates

1 Add or Subtract Rational Expressions In this section, we expand our work with adding and subtracting rational expressions, and we discuss the process of simplifying complex fractions. Before we begin, however, this seems like an appropriate time to offer a bit of advice regarding your study of algebra. Success in algebra depends on having a good understanding of the

304

Chapter 6


concepts as well as being able to perform the various computations. As for the computational work, you should adopt a carefully organized format that shows as many steps as you need in order to minimize the chances of making careless errors. Don’t be eager to find shortcuts for certain computations before you have a thorough understanding of the steps involved in the process. This advice is especially appropriate at the beginning of this section. Study Examples 1– 4 very carefully. Note that the same basic procedure is followed when solving each problem:

Step 1 Factor the denominators. Step 2 Find the LCD. Step 3 Change each fraction to an equivalent fraction that has the LCD as its denominator.

Step 4 Combine the numerators and place over the LCD. Step 5 Simplify by performing the addition or subtraction. Step 6 Look for ways to reduce the resulting fraction.

EXAMPLE 1

Add

3 5 . x x2 2x

Solution 1st denominator:

x2 2x x1x 22

2nd denominator: x

  

LCD is x(x 2)

5 3 3 5 x2 a b x x x2 x1x 22 x2 2x This fraction has the LCD as its denominator

Form of 1

51x 22 3 51x 22 3 x1x 22 x1x 22 x1x 22

3 5x 10 5x 13 x1x 22 x1x 22


EXAMPLE 2

2 4 2 . y y 3y

Subtract

■

1 4 . x2 x2 4

Solution x2 4 1x 221x 22 x2x2

  

LCD is (x 2)(x 2)

1 4 4 1 x2 a ba b x2 1x 221x 22 x2 x2 x 4 2

11x 22 4 1x 221x 22 1x 221x 22

6.4


4 11x 22

1x 221x 22

x 2 1x 221x 22

1x 221x 22

11x 22

305

4x2 1x 221x 22

Note the changing of x 2 to 1(x 2)

1 x2


EXAMPLE 3

Add

7 8 2 . a2 a 5a 6

■

3 2 2 . a2 9 a 5a 6

Solution a2 9 1a 32 1a 32

a 5a 6 1a 32 1a 22 2

  

LCD is (a 3)(a 3)(a 2).

3 2 2 a 9 a 5a 6 2

a

a2 3 a3 2 ba b a ba b 1a 32 1a 32 a2 1a 32 1a 22 a3 Form of 1

21a 22

1a 321a 32 1a 22

1a 321a 32 1a 22

5a 5 1a 321a 32 1a 22

21a 22 31a 32

Form of 1

31a 32

1a 321a 321a 22

2a 4 3a 9 1a 321a 321a 22 or

51a 12

1a 321a 32 1a 22


EXAMPLE 4

5 3 2 . x2 4 x 6x 8

■

Perform the indicated operations. 2x 3 2 2 xy xy x y 2

Solution x2 y2 1x y21x y2 xyxy xyxy

  

LCD is (x y)(x y).

306

Chapter 6


2x 3 2 2 xy xy x y 2

xy xy 2 2x 3 b a ba a ba b xy xy 1x y21x y2 xy xy

31x y2 21x y2 2x 1x y21x y2 1x y2 1x y2 1x y21x y2

2x 31x y2 21x y2 1x y2 1x y2

2x 3x 3y 2x 2y 1x y2 1x y2

3x 5y 1x y21x y2

▼ PRACTICE YOUR SKILL Perform the indicated operations. 4y y 2y 3 2

6 5 . y3 y1

■

2 Simplify Complex Fractions Fractional forms that contain fractions in the numerators and/or denominators are called complex fractions. Here are some examples of complex fractions: 2 3 4 5

1 x 3 y

1 1 2 3 5 1 6 4

2 2 x y 1 5 2 x y

It is often necessary to simplify a complex fraction—that is, to express it as a simple fraction. We will illustrate this process with the next four examples.

EXAMPLE 5

2 3 Simplify . 4 5

Solution This type of problem creates no difficulty, because it is merely a division problem. Thus, 2 1 3 2 4 2 # 5 5 4 3 5 3 4 6 2 5

▼ PRACTICE YOUR SKILL 3 8 Simplify . 7 16

■

6.4

EXAMPLE 6


307

1 x Simplify . 3 y

Solution 1 y x 1 3 1 y # x y x 3 3 3x y

▼ PRACTICE YOUR SKILL 2 a Simplify . 5 b

EXAMPLE 7

■

1 1 2 3 Simplify . 1 5 6 4 Let’s look at two possible strategies for such a problem.

Solution A 1 1 2 3 5 2 2 3 6 6 6 5 # 12 10 5 1 10 3 7 6 7 7 1 6 4 12 12 12 Invert divisor and multiply

Solution B The least common multiple of all four denominators (2, 3, 6, and 4) is 12. We 12 multiply the entire complex fraction by a form of 1, specifically . 12 1 1 2 3 12 a b± 5 1 12 6 4

1 1 2 3 ≤ 5 1 6 4

12 a

1 1 1 1 12 a b 12 a b b 2 3 2 3 5 1 5 1 12 a b 12 a b 12 a b 6 4 6 4

64 10 10 3 7

308

Chapter 6


▼ PRACTICE YOUR SKILL 4 1 5 2 Simplify . 3 1 10 5

EXAMPLE 8

■

3 2 x y Simplify . 1 5 2 x y

Solution A 2y 3 3 x 3x 2 2 y a b a b x y x y y x xy xy 2 1 5 5y 5 y2 1 x x 2 a b 2a b 2 x y x y2 y x xy2 xy

2y 3x xy 5y2 x xy2 y

2 2y 3x # xy xy 5y2 x

Invert divisor and multiply

y12y 3x2 5y2 x

Solution B The least common multiple of all four denominators (x, y, x, and y 2 ) is xy2. We xy2 multiply the entire complex fraction by a form of 1, specifically, 2 . xy 3 3 2 2 2 x y x y xy a 2b ± ≤ 1 1 5 5 xy 2 2 x x y y

xy2 a

3 2 b x y

xy2 a

1 5 2b x y

2y2 3xy 5y2 x

or

2 3 xy2 a b xy2 a b x y 5 1 xy2 a b xy2 a 2 b x y y12y 3x2 5y2 x

▼ PRACTICE YOUR SKILL 3 2 2 b a Simplify . 5 1 a b

■

6.4


309

Certainly, either approach (Solution A or Solution B) will work for problems such as Examples 7 and 8. You should carefully examine Solution B of each example. This approach works very effectively with complex fractions when the least common multiple of all the denominators of the simple fractions is easy to find. We can summarize the two methods for simplifying a complex fraction as follows:

Two Methods for Simplifying Complex Fractions 1. Simplify the numerator and denominator of the fraction separately. Then divide the simplified numerator by the simplified denominator. 2. Multiply the numerator and denominator of the complex fraction by the least common multiple of all of the denominators that appear in the complex fraction.

EXAMPLE 9

2 3 x Simplify . 5 4 y

Solution 2 2 3 3 xy x x a b± ≤ xy 5 5 4 4 y y 2 1xy2 a b 1xy2132 x 5 1xy2142 1xy2 a b y

2y 3xy 4xy 5x

or

y12 3x2 x14y 52

▼ PRACTICE YOUR SKILL 6 Simplify

2 m

3 7 n

■

.

3 Write Rational Expressions to Express Rates We can use our skill at translating expressions to express rates, which are essentially a comparison between two quantities. We see the concept of rates being used everyday, and they are used in many fields. The following list shows some examples of rates that you may have encountered. Rate that an automobile is traveling

Miles per hour

Rate of Internet speed

Megabits per second

Exchange rate of money

Euros per dollar

Price of bananas

Cents per pound

Rate of gasoline consumption

Miles per gallon

Rate of doing homework problems

Problems per hour

Rate of pay

Dollars per hour

310

Chapter 6


For our work in algebra we need to be able to write expressions for rates using variables to represent the quantities. Example 10 uses rational expressions to express rates and quantities related to rates.

EXAMPLE 10

Write a rational expression to answer each question. (a) If Jorge drove x miles in y hours, what was his rate in miles per hour? (b) If Megan can clean the entire house in k hours, what fraction of the house has she cleaned in 3 hours? (c) If m pounds of chocolate cost n dollars, what is the price per pound?

Solution (a) To express Jorge’s rate in miles per hour, we divide the number of miles by x the numbers of hours. Therefore Jorge’s rate is expressed as . y 1 (b) If it takes Megan k hours to clean the house, then in 1 hour she gets of the k 3 job done. So if she works for 3 hours, she will get of the job done. k (c) If m pounds of chocolate cost n dollars, then the price per pound can be expressed by dividing the number of pounds by the price. Therefore the price n per pound is expressed as . m

▼ PRACTICE YOUR SKILL Write a rational expression to answer each question. (a) If Selena jogs x miles in y hours, what is her jogging rate? (b) If Danielle purchased m pounds of apples for n dollars, what is the price (in dollars) per pound of the apples? (c) If it takes Taylor a hours to wash b automobiles, what is his rate of washing automobiles? ■

CONCEPT QUIZ

For Problems 1–5, answer true or false. 1. When adding two rational expressions, the result cannot be 0. 2. A complex fraction is a fraction that contains fractions in the numerator or denominator. 1 1 x 2 3. The complex fraction can be simplified by multiplying the numerator by 1 1 y 3 2x and the denominator by 3y. 1 1 3m 2n 4. The complex fraction can be simplified by multiplying the fraction by 1 1 2m 3n 6mn a form of 1, specifically . 6mn 5. If Kevin buys 4 custom wheels for y dollars, then the price per wheels is repre4 sented by . y

6.4


311

Problem Set 6.4 1 Add or Subtract Rational Expressions

27.

x 5 2 x2 6x 8 x 3x 10

28.

x 7 2 x2 x 30 x 7x 6

29.

a b 2 ab b2 a ab

30.

2y 2x 2 2 xy y x xy

31.

3 5 4 2 x5 x5 x 25

32.

2 5 3 2 x1 x1 x 1

33.

10 8 3 2 2 x2 2x x 2x x 4

34.

1 2 5 2 2 x2 7x x 7x x 49

35.

3x 2 3 x2 x5 x2 7x 10

7 4 15. 2 2 a 2a a 2a

36.

4 3 x 2 x3 x6 x 9x 18

5 3 16. 2 2 a 4a a 4a

37.

5x 1 2 3x 5 x 4 3x 7x 20

1 1 17. 2 2 x 6x x 6x

38.

4x 2 4 2x 1 3x 2 6x2 7x 2

4 3 2 x2 5x x 5x

39.

2 1 2x 1 2 x4 x3 x x 12

n 2 19. 2 3n 12 n 16

40.

4 3x 27 3 2 x5 x7 x 2x 35

For Problems 1– 40, perform the indicated operations and express answers in simplest form. 1. 3. 5. 7.

3 4 x x2 4x

2.

5 7 x x2 2x

4.

8 2 2 n n 6n

6.

4 4 n n2 n

7 x 2 9. 2x x x

8.

3 7 x x2 2x 9 1 x x2 5x 4 6 2 n n 6n 8 4 n n2 2n

4 3x 10. 2 5x x 2x

5 3 11. 2 x4 x 16 9 6 12. 2 x3 x 9 4 8x 13. 2 x1 x 1

18.

20.

2 n 3n 15 n2 25

21.

2x 5x 6x 4 9x 6

22.

7x 3x 3x 12 4x 16

23.

x1 x4 5x 5 3x 3

24.

2x 1 3x 4 6x 12 8x 16

25.

2 3 2 x2 7x 12 x 9

26.

x 3 2 x 1 x 5x 4 2

3 6x 14. 2 x2 x 4

2

2 Simplify Complex Fractions For Problems 41– 60, simplify each of the complex fractions. 1 3 2 4 41. 1 1 6 3

3 1 8 4 42. 1 3 2 16

2 1 9 3 43. 5 2 6 3

1 7 8 3 44. 1 3 6 4

2 3 45. 1 2 4 3

4 46.

3 5

1 2 3

312

Chapter 6


3 x 47. 9 y

6 a 48. 8 b

3 2 x y 49. 5 1 x y

2 3 2 x x 50. 4 7 2 x x

64. If Roy traveled m miles in h hours, what was his rate in miles per hour?

4 1 2 y x 51. 7 3 x y

4 2 ab b 52. 8 1 a b

66. If p pounds of candy cost c cents, what is the price per pound?

6 2 x 53. 3 4 x

6 1 y 54. 2 3 y

3 4 2 x 2x 55. 5 7 2 3x x

4 5 2 3x x 56. 7 9 4x x

x2 4 57. 1 3 x 2

3 2 x1 58. 2 4 x1

1 2 x1 59. 3 4 x1

3 x2 x 60. 4 x2 x

62. If Kent can mow the entire lawn in m minutes, what fractional part of the lawn has he mowed at the end of 20 minutes? 63. If Sandy drove k kilometers at a rate of r kilometers per hour, how long did it take her to make the trip?

65. If l liters of gasoline cost d dollars, what is the price per liter?

67. Suppose that the product of two numbers is 34, and one of the numbers is n. What is the other number? 68. If a cold water faucet when opened can fill a tank in 3 hours, how much of the tank is filled at the end of h hours? (See Figure 6.1.)

2 2 5 2

3 Write Rational Expressions to Express Rates

Figure 6.1

69. If the area of a rectangle is 47 square inches, and the length is l inches, what is the width of the rectangle?

For Problems 61–71, answer each question with a rational expression.

70. If the area of a rectangle is 56 square centimeters, and the width is w centimeters, what is the length of the rectangle?

61. If by jogging at a constant rate Joan can complete a race in 40 minutes, how much of the course has she completed at the end of m minutes?

71. If the area of a triangle is 48 square feet and the length of one side is b feet, what is the length of the altitude to that side?

THOUGHTS INTO WORDS 72. Which of the two techniques presented in the text would 1 1 4 3 you use to simplify ? 3 1 4 6

Which technique would you use to simplify 3 5 8 7 ? Explain your choice for each problem. 7 6 9 25

FURTHER INVESTIGATIONS For Problems 73 –76, simplify each complex fraction. 73. 1

n 1 1 n

74. 2

3n 4 1 n

75.

3x 2 4 x

1

76.

5x 3

1 x

2

6.5

Rational Equations and Problem Solving

313


4y 14 y1y 32

8.

3b 2a2 3b 2a2 or a1b 5a2 ab 5a2

2.

5y 21 8x 2 6 4. 5. 1x 221x 22 1x 42 1y 12 1y 32 7 2n13m 12 6mn 2n x n b or 9. 10. (a) (b) (c) y m a 3m 7mn m13 7n2

7a 13 1a 221a 32

3.

6.

2b 5a

7.

3 5


6.5


Rational Equations and Problem Solving OBJECTIVES 1

Solve Rational Equations with Constants in the Denominator

2

Solve Rational Equations with Variables in the Denominator

3

Solve Word Problems

1 Solve Rational Equations with Constants in the Denominator We will consider two basic types of rational equations in this text. One type has only constants as denominators, and the other type has some variables in the denominator. In Chapter 2 we considered rational equations that had only constants in the denominator. Let’s review our approach to these equations, because we will be using that same basic technique to solve any rational equation.

EXAMPLE 1

Solve

x1 1 x2 . 3 4 6

Solution x2 x1 1 3 4 6 12 a 12 a

x2 x1 1 b 12 a b 3 4 6

x2 x1 1 b 12 a b 12 a b 3 4 6 41x 22 31x 12 2 4x 8 3x 3 2 7x 5 2 7x 7 x1

The solution set is {1}. (Check it!)

Multiply both sides by 12, the LCD of all three denominators

314

Chapter 6



2x 7 5 x3 . 2 3 6

■

2 Solve Rational Equations with Variables in the Denominator If an equation contains a variable in one or more denominators, then we proceed in essentially the same way except that we must avoid any value of the variable that makes a denominator zero. Consider the next example.

EXAMPLE 2

Solve

1 5 3 . x x 2

Solution First, we need to realize that x cannot equal zero. Then we can proceed in the usual way. 3 1 5 x x 2 2x a

1 3 5 b 2x a b x x 2

Multiply both sides by 2x, the LCD of all denominators

6 x 10 x4

✔ Check 3 1 5 x x 2

becomes

1 5 3 ⱨ 4 2 4

when x 4

3 2 5 ⱨ 4 4 4 5 5 4 4 The solution set is {4}.


EXAMPLE 3

Solve

1 7 5 . x x 4

Solve

2 5 . x2 x1

■

Solution Because neither denominator can be zero, we know that x 2 and x 1. 5 2 x2 x1 1x 22 1x 12 a

5 2 b 1x 221x 12 a b x2 x1

51x 12 21x 22

Multiply both sides by (x 2)(x 1), the LCD

6.5


315

5x 5 2x 4 3x 9 x3 Because the only restrictions are x 2 and x 1, the solution set is {3}. (Check it!)


EXAMPLE 4

Solve

10 4 . x2 x4

Solve

2 x . 2 x2 x2

■

Solution No denominator can be zero, so x 2. x 2 2 x2 x2 1x 22 a

2 x 2b 1x 22 a b x2 x2

Multiply by x 2, the LCD

2 21x 22 x 2 2x 4 x 2x 2 x x2 Two cannot be a solution because it will produce a denominator of zero. There is no solution to the given equation; the solution set is .


x 5 3 . x5 x5

■

Example 4 illustrates the importance of recognizing the restrictions that must be placed on possible values of a variable. We will indicate such restrictions at the beginning of our solution.

EXAMPLE 5

Solve

10 125 n 4 . n n

Solution 125 n 10 4 , n n na

125 n 10 b na4 b n n

n0

Note the necessary restriction

Multiply both sides by n

125 n 4n 10 115 5n 23 n The only restriction is n 0, and the solution set is {23}.

316

Chapter 6



70 x 84 2 . x x

■

3 Solve Word Problems We are now ready to solve more problems, specifically those that translate into fractional equations.

EXAMPLE 6

Apply Your Skill One number is 10 larger than another number. The indicated quotient of the smaller 3 number divided by the larger number reduces to . Find the numbers. 5

Solution We let n represent the smaller number. Then n 10 represents the larger number. The second sentence in the statement of the problem translates into the following equation: n 3 , n 10 5

n 10

5n 31n 102


5n 3n 30 2n 30 n 15 If n is 15, then n 10 is 25. Thus the numbers are 15 and 25. To check, consider the quotient of the smaller number divided by the larger number. 3#5 3 15 # 25 5 5 5

▼ PRACTICE YOUR SKILL One number is 4 more than another number. The indicated quotient of the smaller 2 number divided by the larger number reduces to . Find the numbers. ■ 3

EXAMPLE 7

Apply Your Skill One angle of a triangle has a measure of 40°, and the measures of the other two angles are in the ratio of 5 to 2. Find the measures of the other two angles.

Solution The sum of the measures of the other two angles is 180° 40° 140°. Let y represent the measure of one angle. Then 140 y represents the measure of the other angle. y 5 , 140 y 2

y 140

2y 51140 y2 2y 700 5y


6.5


317

7y 700 y 100 If y 100, then 140 y 40. Therefore the measures of the other two angles of the triangle are 100° and 40°.

▼ PRACTICE YOUR SKILL One angle of a triangle has a measure of 30°, and the measures of the other angles are in the ratio of 2 to 3. Find the measures of the other two angles. ■ In Chapter 3, we solved some uniform motion problems in which the formula d rt played an important role. Let’s consider another one of those problems; keep d d in mind that we can also write the formula d rt as t or r. r t

Thinkstock Images/Jupiter Images

EXAMPLE 8

Apply Your Skill Wendy rides her bicycle 30 miles in the same time that it takes Kim to ride her bicycle 20 miles. If Wendy rides 5 miles per hour faster than Kim, find the rate of each.

Solution Let r represent Kim’s rate. Then r 5 represents Wendy’s rate. Let’s record the information of this problem in a table.

Distance

Rate

Kim

20

r

Wendy

30

r5

Time

Distance Rate

20 r 30 r5

We can use the fact that their times are equal as a guideline. Kim’s Time

Wendy’s Time

Distance Wendy rides Distance Kim rides Rate Kim rides Rate Wendy rides 20 30 , r r5

r 0 and r 5

201r 52 30r 20r 100 30r 100 10r 10 r Therefore, Kim rides at 10 miles per hour, and Wendy rides at 10 5 15 miles per hour.

▼ PRACTICE YOUR SKILL Javier rides his bicycle 24 miles in the same time that it takes Aldrin to ride his bicycle 18 miles. If Javier rides 2 miles per hour faster than Aldrin, find the rate of each. ■

318

Chapter 6


CONCEPT QUIZ

For Problems 1–5, answer true or false. 2 5 is a proportion that could be solved by using x3 x4 cross-multiplication. x x The equation 3x is a proportion that could be solved by using 8 5 cross-multiplication. 4 x The solution to the equation cannot be 6. 3 x6 x6 Any value of the variable that makes the numerator of an equation equal to zero cannot be a solution of the equation. Zero cannot be a solution to a rational equation.

1. The equation

2.

3. 4. 5.

Problem Set 6.5 1 Solve Rational Equations with Constants in the Denominator For Problems 1– 40, solve each of the equations. 1. 3. 5. 7. 9.

x x 10 2 3

2.

x 4x 1 6 3 9

4.

n n1 5 2 6 2

6.

t3 t1 1 4 9

8.

x x 1 8 6 x 3 3x 4 5 10 n2 n 12 7 3 7 t3 t2 1 4 7

2x 3 3x 4 17 3 4 4

2x 3 3x 1 10. 2 4 5 x4 x5 11. 3 8 4 12.

x6 x2 7 9 5 15

2x 1 2 3x 2 13. 5 6 15

19.

1 4 3 2x 3x

20.

5 2 1 3x 4x

21.

4 3 1 5t 2t

22.

7 1 2 6t 8t

23.

5 7 1 4h 6h 4

24.

5 3 1 h 2h

25.

90 n 2 10 n n

26.

3 51 n 7 n n

27.

1 n 3 49 n 49 n

28.

2 n 10 57 n 57 n

29.

x 3 2 x3 x3

30.

5 4 x2 x6

31.

7 5 x3 x9

32.

4 x 2 x4 x4

33.

1 x 3 x2 x2

34.

3 x 4 x5 x5

40.

h h h 1 4 5 6

35. 1

2x 5 1 4x 1 14. 3 8 6

36.

2 Solve Rational Equations with Variables in the Denominator

3 5 x2 x2

3 x1 2 x 2

37. 1

3 n1 2n 4

15.

1 2 7 x 3 6

16.

2 1 13 x 4 20

38.

2 3 4 n1 n1

17.

5 1 1 3n 9 n

18.

9 1 7 n 4 n

39.

h h h 1 2 4 3

6.5

3 Solve Word Problems For Problems 41–52, set up an equation and solve each problem. 41. The numerator of a fraction is 8 less than the denomi5 nator. The fraction in its simplest form is . Find the 6 fraction.


319

46. The measure of angle A of a triangle is 20° more than the measure of angle B. The measures of the angles are in a ratio of 3 to 4. Find the measure of each angle. 47. The ratio of the measures of the complement of an angle to its supplement is 1 to 4. Find the measure of the angle. 48. One angle of a triangle has a measure of 45°, and the measures of the other two angles are in the ratio of 2 to 1. Find the measures of the other two angles.

42. One number is 12 larger than another number. The indicated quotient of the smaller number divided by 2 the larger reduces to . Find the numbers. 3

49. It took Heidi 3 hours and 20 minutes longer to ride her bicycle 125 miles than it took Abby to ride 75 miles. If they both rode at the same rate, find the rate.

43. What number must be added to the numerator and 2 4 denominator of to produce a fraction equivalent to ? 5 5

50. Two trains left a depot traveling in opposite directions at the same rate. One train traveled 338 miles in 2 hours more time than it took the other train to travel 234 miles. Find the rate of the trains.

44. What number must be subtracted from the numerator 29 and denominator of to produce a fraction equivalent 31 11 to ? 12 45. One angle of a triangle has a measure of 60°, and the measures of the other two angles are in the ratio of 2 to 3. Find the measures of the other two angles.

51. Kent drives his Mazda 270 miles in the same time that Dave drives his Toyota 250 miles. If Kent averages 4 miles per hour faster than Dave, find their respective rates. 52. An airplane travels 2050 miles in the same time that a car travels 260 miles. If the rate of the plane is 358 miles per hour faster than the rate of the car, find the rate of each.

THOUGHTS INTO WORDS 53. (a) Explain how to do the addition problem 5 3 . x2 x1

4 x 54. How can you tell by inspection that has no x4 x4 solution? 55. How would you help someone solve the equation 1 2 3 ? x x x

(b) Explain how to solve the equation 5 3 0. x2 x1

FURTHER INVESTIGATIONS For Problems 56 –59, solve each of the equations. 56. 57.

3 1 5 2n n 3n 1 4 9 2n n 2n

58.

n1 n 1 2 3 2

59.

1 2 3n 7 n2 n3 1n 221n 32

Answers to the Example Practice Skills 1. {4} 2. {8} is 6 mph

3. {6}

4. 5. {14}

6. 8 and 12


2. False

3. True

4. False 5. False

7. 60 and 90 8. Javier’s rate is 8 mph and Aldrin’s rate

320

Chapter 6

6.6


More Rational Equations and Problem Solving OBJECTIVES 1

Solve Rational Equations

2

Solve Word Problems Including Rate-Time Problems

1 Solve Rational Equations Let’s begin this section by considering a few more rational equations. We will continue to solve them using the same basic techniques as in the preceding section.

EXAMPLE 1

Solve

10 6 1 . 8x 2 4x 1 9

Solution 10 6 1 , 8x 2 4x 1 9

1 x 4

10 6 1 214x 12 4x 1 9 1814x 12 a

Do you see why 1 x cannot equal ? 4 Factor the first denominator

10 6 1 b 1814x 12 a b 214x 12 4x 1 9

Multiply both sides by 18(4x 1), the LCD

91102 18162 214x 12 90 108 8x 2 18 8x 2 16 8x Be sure that the solution 2 checks!

2 x The solution set is {2}.


3 1 7 . 6x 2 3x 1 20

■

Remark: In the second step of the solution for Example 1, you may choose to 10 5 to . Then the left 214x 12 4x 1 1 1 . This forms the proportion 4x 1 4x 1 cross-multiplication method. reduce

EXAMPLE 2

Solve

side,

1 , which can be solved easily using the 9

5n 2n 2. 2 n3 n 9

Solution 2n 5n 2, 2 n3 n 9

6 5 , simplifies to 4x 1 4x 1

n 3 and n 3

6.6

More Rational Equations and Problem Solving

321

2n 5n 2 n3 1n 32 1n 32 1n 321n 32 a

2n 5n b 1n 321n 32 122 n3 1n 321n 32 2n1n 32 5n 21n2 92 2n2 6n 5n 2n2 18 6n 5n 18

Add 2n2 to both sides

n 18 n 18 The solution set is {18}.


EXAMPLE 3

x x 1. 2 x2 x 4

Solve n

■

10 1 . n 3

Solution n 3n a n

1 10 , n 3

n0

1 10 b 3n a b n 3

3n2 3 10n 3n2 10n 3 0

13n 121n 32 0

Remember when we used the factoring techniques to help solve equations of this type in Chapter 5?

3n 1 0

or

n30

3n 1

or

n3

n

1 3

1 The solution set is e , 3 f . 3

▼ PRACTICE YOUR SKILL Solve y

1 5 . y 2

■

2 Solve Word Problems Including Rate-Time Problems 2 3 and are called multiplicative inverses, or reciprocals, of each other 3 2 because their product is 1. In general, the reciprocal of any nonzero real number n 1 is the number . Let’s use this idea to solve a problem. n

Recall that

Chapter 6


EXAMPLE 4

Apply Your Skill The sum of a number and its reciprocal is

26 . Find the number. 5

Solution 1 represents its reciprocal. n

We let n represent the number. Then Number

n

Its reciprocal

26 5

1 n

26 , 5

5n a n

n0

1 26 b 5n a b n 5

Multiply both sides by 5n, the LCD

5n2 5 26n 5n2 26n 5 0

15n 121n 52 0 5n 1 0

or

n50

5n 1

or

n5

n

1 5

1 1 1 If the number is , its reciprocal is 5. If the number is 5, its reciprocal is . 5 1 5 5

▼ PRACTICE YOUR SKILL The sum of a whole number and its reciprocal is

50 . Find the number. 7

■

Now let’s consider another uniform motion problem, which is a slight variation of those we studied in the previous section. Again, keep in mind that we always use the distance-rate-time relationships in these problems.

Apply Your Skill To travel 60 miles, it takes Sue, riding a moped, 2 hours less than it takes LeAnn, riding a bicycle, to travel 50 miles (see Figure 6.2). Sue travels 10 miles per hour faster than LeAnn. Find the times and rates of both women. Sue

PE

D

LeAnn

O

EXAMPLE 5

M

322

50 miles 60 miles Figure 6.2

6.6


323

Solution We let t represent LeAnn’s time. Then t 2 represents Sue’s time. We can record the information from the problem in the table.

Distance

Time

LeAnn

50

t

Sue

60

t2

Rate ar

d b t

50 t 60 t2

We use the fact that Sue travels 10 miles per hour faster than LeAnn as a guideline to set up an equation. Sue’s Rate

60 t2

LeAnn’s Rate 10

50 10, t

t 2 and t 0

Solving this equation yields 60 50 10 t2 t t1t 22 a

60 50 b t1t 22 a 10b t2 t 60t 501t 22 10t1t 22 60t 50t 100 10t2 20t 0 10t2 30t 100 0 t2 3t 10

0 1t 52 1t 22 t50

or

t5

or

t20 t 2

We must disregard the negative solution, so LeAnn’s time is 5 hours, and Sue’s time 50 60 is 5 2 3 hours. LeAnn’s rate is 10 miles per hour, and Sue’s rate is 5 3 20 miles per hour. (Be sure that all of these results check back into the original problem!)

▼ PRACTICE YOUR SKILL To jog 16 miles, it takes Kim 1 hour less time than it takes Ryan to bicycle 36 miles. Ryan travels 4 miles per hour faster than Kim. Find the rates of both Kim and Ryan. ■ There is another class of problems that we commonly refer to as work problems, or sometimes as rate-time problems. For example, if a certain machine produces 120 120 items in 10 minutes, then we say that it is producing at a rate of 12 items 10

324

Chapter 6


per minute. Likewise, if a person can do a certain job in 5 hours, then that person 1 is working at a rate of of the job per hour. In general, if Q is the quantity of 5 Q something done in t units of time, then the rate r is given by r . The rate is t stated in terms of so much quantity per unit of time. The uniform-motion problems we discussed earlier are a special kind of rate-time problem in which the quantity is distance. The use of tables to organize information, as we illustrated with the uniformmotion problems, is a convenient aid for some rate-time problems. Let’s consider some examples.

EXAMPLE 6

Apply Your Skill

Hardy James/Jupiter Images

Printing press A can produce 35 flyers per minute, and press B can produce 50 flyers per minute. Printing press A is set up and starts a job, and then 15 minutes later printing press B is started, and both presses continue printing until 2225 flyers are produced. How long would printing press B be used?

Solution We let m represent the number of minutes that printing press B is used. Then m 15 represents the number of minutes that press A is used. The information in the problem can be organized in a table.

Press A Press B

Rate

Time

Quantity Rate Time

35 50

m 15 m

35(m 15) 50 m

Since the total quantity (total number of flyers) is 2225 flyers, we can set up and solve the following equation: 351m 152 50m 2225 35m 525 50m 2225 85m 1700 m 20 Therefore, printing press B must be used for 20 minutes.

▼ PRACTICE YOUR SKILL Derek, a sandwich maker, can make 3 sandwiches per minute. Marlene, another sandwich maker, can make 5 sandwiches per minute. Derek starts making sandwiches, and 10 minutes later Marlene begins to make sandwiches. They both continue making sandwiches until the catering order for 110 sandwiches is completed. How long did Marlene work making the sandwiches? ■

EXAMPLE 7

Apply Your Skill Bill can mow a lawn in 45 minutes, and Jennifer can mow the same lawn in 30 minutes. How long would it take the two of them working together to mow the lawn? (See Figure 6.3.)

6.6


325

Figure 6.3

Remark: Before you look at the solution of this problem, estimate the answer. Remember that Jennifer can mow the lawn by herself in 30 minutes.

Solution 1 1 of the lawn per minute, and Jennifer’s rate is of the lawn 45 30 per minute. If we let m represent the number of minutes that they work together, 1 then represents the rate when working together. Therefore, since the sum of the m individual rates must equal the rate working together, we can set up and solve the following equation: Bill’s rate is

1 1 1 , m 30 45 90m a

m0

1 1 1 b 90m a b m 30 45

Multiply both sides by 90m, the LCD

3m 2m 90 5m 90 m 18 It should take them 18 minutes to mow the lawn when working together. (How close was your estimate?)

▼ PRACTICE YOUR SKILL Dean can wash and detail a car in 60 minutes, and Tasha can wash and detail the same car in 90 minutes. How long would it take the two of them working together to wash and detail the car? ■

EXAMPLE 8

Apply Your Skill

Photodisc/Getty Images

It takes Amy twice as long to deliver papers as it does Nancy. How long would it take each girl by herself if they can deliver the papers together in 40 minutes?

Solution We let m represent the number of minutes that it takes Nancy by herself. Then 1 2m represents Amy’s time by herself. Therefore, Nancy’s rate is , and Amy’s rate m

326

Chapter 6


1 1 . Since the combined rate is , we can set up and solve the following 2m 40 equation: is

Nancy’s rate

Amy’s Combined rate rate

1 m

40m a

1 1 1 b 40m a b m 2m 40

1 2m

1 , 40

m0

40 20 m 60 m Therefore, Nancy can deliver the papers by herself in 60 minutes, and Amy can deliver them by herself in 2(60) 120 minutes.

▼ PRACTICE YOUR SKILL It takes Martin three times as long to clean an office as it does Julio. How long would it take each man to clean the office by himself if they can clean the office together in 30 minutes? ■ One final example of this section outlines another approach that some people find meaningful for work problems. This approach represents the fractional parts of a job. For example, if a person can do a certain job in 7 hours, then at the end 3 of 3 hours, that person has finished of the job. (Again, we assume a constant rate of 7 5 work.) At the end of 5 hours, of the job has been done—in general, at the end of 7 h h hours, of the job has been completed. Let’s use this idea to solve a work problem. 7

Jose Luis Pelaez Inc. /Getty Images

EXAMPLE 9

Apply Your Skill It takes Pat 12 hours to install a wood floor. After he had been working for 3 hours, he was joined by his brother Mike, and together they finished the floor in 5 hours. How long would it take Mike to install the floor by himself?

Solution Let h represent the number of hours that it would take Mike to install the floor by himself. The fractional part of the job that Pat does equals his working rate times his 1 time. Because it takes Pat 12 hours to do the entire floor, his working rate is . 12 He works for 8 hours (3 hours before Mike and then 5 hours with Mike). There8 1 fore, Pat’s part of the job is (8) . The fractional part of the job that Mike 12 12 does equals his working rate times his time. Because h represents Mike’s time to 1 install the floor, his working rate is . He works for 5 hours. Therefore, Mike’s part h 5 1 of the job is (5) . Adding the two fractional parts together results in one entire h h job being done. Let’s also show this information in chart form to set up our guideline. Then we can write the equation and solve.

6.6


Time to do entire job Pat

12

Mike

h

Working rate

Time working

1 12 1 h

Fractional part of the job that Pat does

327

Fractional part of the job done 8 12 5 h

8 5

Fractional part of the job that Mike does

8 5 1 12 h 12h a 12h a

8 5 b 12h112 12 h

8 5 b 12h a b 12h 12 h 8h 60 12h 60 4h 15 h

It would take Mike 15 hours to install the floor by himself.

▼ PRACTICE YOUR SKILL It takes David 16 hours to build a deck. After he had been working for 7 hours, he was joined by his friend Carlos, and together they finished the deck in 5 hours. How long would it take Carlos to build the deck by himself? ■ We emphasize a point made earlier. Don’t become discouraged if solving word problems is still giving you trouble. The development of problem-solving skills is a long-term objective. If you continue to work hard and give it your best shot, you will gradually become more and more confident in your approach to solving problems. Don’t be afraid to try some different approaches on your own. Our problem-solving suggestions simply provide a framework for you to build on.

CONCEPT QUIZ

For Problems 1–5, answer true or false. 1. Multiplicative inverses are called reciprocals. 2. The product of a number and its reciprocal is 0. 1 3. The reciprocal of any nonzero number n is the number . n x 4. If it takes a person x hours to do y tasks, then the person is working at a rate of y tasks per hour. 3 5. If a person can do a certain job in 7 hours, then that person gets of the job done 7 in 3 hours.

Problem Set 6.6 1 Solve Rational Equations

3.

3 4 11 2x 2 x1 12

5.

3 5 1 2n 10 n5

For Problems 1–32, solve each equation. 1.

4 7 1 2 x 6 x 3x

2.

9 25 2 3x x 9

4.

5 1 7 2x 6 x3 2

328

Chapter 6


6.

2 7 2 3x 6 x2

7.

3 5 7 1 2t t 5t

9.

4 x 1 x2 x2

For Problems 33 –50, set up an equation and solve the problem. 9 33. The sum of a number and twice its reciprocal is . Find 2 the number.

10.

3 2x 2 x1 x1

34. The sum of a number and three times its reciprocal is 4. Find the number.

11.

2x x 1 x4 x4

35. A number is

12.

2x x 3 x2 x2

2 Solve Word Problems Including Rate-Time Problems 8.

13.

3 5 2 1 3t 4t 2t

n 3n 2 n3 n3

2n 4n 14. 2 n5 n5 15.

5 2 3 t2 t2 t 4

16.

1 16 t 2 2t 8 2 t 16

3 5 y2 y2 2y

20. 2

4 4 2 y1 y y

21. n

1 17 n 4

23.

15 15 1 4n 41n 42

24.

10 10 1 7x 71x 32

25. x

5x 10 x2 x2

27.

1 5 t 2 4t 4 4 t 1

28.

1 4 x 2 3x 6 3 x 4

36. Suppose that the reciprocal of a number subtracted from 5 the number yields . Find the number. 6

38. To travel 300 miles, it takes a freight train 2 hours longer than it takes an express train to travel 280 miles. The rate of the express train is 20 miles per hour faster than the rate of the freight train. Find the rates of both trains.

4 5 3x 1 2 x3 x3 x 9

19. 8

number.

37. Suppose that Celia rides her bicycle 60 miles in 2 hours less time than it takes Tom to ride his bicycle 85 miles. If Celia rides 3 miles per hour faster than Tom, find their respective rates.

2

6 2x 3 4 17. 2 x1 x1 x 1 18.

39. One day, Jeff rides his bicycle out into the country 40 miles (see Figure 6.4). On the way back, he takes a different route that is 2 miles longer, and it takes him an hour longer to return. If his rate on the way out into the country is 4 miles per hour faster than his rate back, find both rates. 22. n

26.

3 4 n

3 12 x1 2 x3 x x 3x

40

42 N E W S

4 2n 11 3 29. 2 n5 n7 n 2n 35 30.

3 2n 1 2 2 n3 n4 n n 12

31.

3 14 a 2 a2 a4 a 6a 8

32. 3

6 6 2 t3 t 3t

21 larger than its reciprocal. Find the 10

Figure 6.4 40. Rita jogs for 8 miles and then walks an additional 12 miles. She jogs at a rate twice her walking rate, and she

6.6


329

covers the entire distance of 20 miles in 4 hours. Find the rate she jogs and the rate she walks.

longer than Pam to do the job. How long does it take each of them working alone?

41. A water tank can be filled by an inlet pipe in 5 minutes. A drain pipe will empty the tank in 6 minutes. If by mistake the drain is left open as the tank is being filled, how long will it take before the tank overflows?

47. A copy center has two copiers. Copier A can produce copies at a rate of 40 pages per minute, and copier B does 30 pages per minute. How long will copier B need to run if copier A has been copying by itself for 6 minutes, and both copier A and B are used until 520 copies are made?

42. Betty can do a job in 10 minutes. Doug can do the same job in 15 minutes. If they work together, how long will it take them to complete the job? 43. It takes Barry twice as long to deliver papers as it does Mike. How long would it take each of them if they deliver the papers together in 60 minutes?

48. It takes two pipes 3 hours to fill a water tank. Pipe B can fill the tank alone in 8 hours more than it takes pipe A to fill the tank alone. How long would it take each pipe to fill the tank by itself?

44. Working together, Cindy and Sharon can address envelopes in 12 minutes. Cindy could do the addressing by herself in 20 minutes. How long would it take Sharon to address the envelopes by herself?

49. In a survivor competition, the Pachena tribe can shuck 300 oysters in 10 minutes less time than it takes the Tchaika tribe. If the Pachena tribe shucks oysters at a rate of 5 oysters per minute faster than the Tchaika tribe, find the rate of each tribe.

45. Mark can overhaul an engine in 20 hours, and Phil can do the same job by himself in 30 hours. If they both work together for a while, and then Mark finishes the job by himself in 5 hours, how long did they work together?

50. Machine A can wrap 600 pieces of candy in 5 minutes less time than it takes machine B to wrap 600 pieces of candy. If the rate of machine A is 20 candies per minute faster than machine B, find the rate of each machine.

46. Working together, Pam and Laura can complete a job in 1 1 hours. When working alone, it takes Laura 4 hours 2

THOUGHTS INTO WORDS 51. Write a paragraph or two summarizing the new ideas about problem solving that you have acquired thus far in this course.

FURTHER INVESTIGATIONS For Problems 52 –54, solve each equation. 52.

4 3x 1 7 x3 x3 x2 9

53.

3 x2 5 x1 x1 x2 1

54.

5 7x 12 2 x4 x4 x2 16

Answers to the Example Practice Skills 1 3. e , 2 f 4. 7 5. Kim’s rate is 8 mph, and Ryan’s rate is 12 mph 6. 10 minutes 2 7. 36 minutes 8. Julio’s time is 40 minutes, and Martin’s time is 120 minutes 9. 20 hours

1. {3}

2. {4}


2. False

3. True

4. False 5. True


SUMMARY

Simplify rational expressions. (Sec. 6.1, Obj. 1, p. 286)

The fundamental principle of fraca ak tions a b provides the basis bk b for simplifying rational expressions. For many problems, the numerator and denominator will have to be factored before you can apply the fundamental principle of fractions.

Simplify rational expressions involving opposites. (Sec. 6.1, Obj. 2, p. 288)

The quotient of any nonzero real number and its opposite is 1. Likewise, the indicated quotient of any polynomial and its opposite is 1.

EXAMPLE Simplify

3x2 15x . x 3x 10

Solution

3x1x 52 3x2 15x 1x 221x 52 x 3x 10 2

Simplify

12 x2 12 x2 4 x2 2x 4 21x 22 12 x2 12 x2 21x 22

Multiply

2x 2

3y 3 . 8 y 4y 3 y

2

#

Problems 9 –12

Solution

3y 3 8 y 4y 3 y

Division by a fraction: to divide by a fraction, multiply by its reciprocal. Division of rational expressions is done in the same manner.

Problems 5 – 8

Solution

#

2

Divide rational expressions. (Sec. 6.2, Obj. 2, p. 293)

3x x2

4 x2 . 2x 4

To multiply rational expressions, multiply the numerators, multiply the denominators, and then express the product in simplified form.

Problems 1– 4

2

Multiply rational expressions. (Sec. 6.2, Obj. 1, p. 291)


y 1y 12 1y 32

3y 81y 32

#

31y 12 8

Divide 6a 6 a2 36 . 2 2 a 8a 12 a 3a 2

Problems 13 –16

Solution

a2 36 a2 3a 2 # 6a 6 a2 8a 12 1a 621a 62

1a 621a 22 1a 221a 12

#

330

61a 12

a6 6

(continued)

Chapter 6

OBJECTIVE

SUMMARY

Add or subtract rational expressions with like denominators. (Sec. 6.3, Obj. 1, p. 296)

If the denominators of two rational expressions are the same, then the expressions can be added or subtracted by adding or subtracting the numerators and placing the result over the denominator. The final answer should always be in simplified form.

Add or subtract rational expressions with unlike denominators. (Sec. 6.3, Obj. 2, p. 298; Sec. 6.4, Obj. 1, p. 303)

To add or subtract rational expressions with unlike denominators, first change the expressions to equivalent fractions that have a common denominator. Review the steps given on page 304.

Simplify complex fractions. (Sec. 6.4, Obj. 2, p. 306)

Fractional forms that contain fractions in the numerator and/or the denominator are complex fractions. To simplify a complex fraction means to express it as a single fraction. One method for simplifying is to multiply the entire complex fraction by a form of 1. Another method is to simplify the numerator, simplify the denominator, and then proceed as with a division-of-fractions problem.

EXAMPLE Subtract

8n 1 5n 8 . 6 6

Summary

331

CHAPTER REVIEW PROBLEMS Problems 17–18

Solution

5n 8 8n 1 6 6 8n 1 15n 82 6 8n 1 5n 8 6 3n 9 6 31n 32 n3 6 2 Subtract

1 x . x4 x 16

Problems 19 –26

2

Solution

1 x x4 x2 16 1 x 1x 421x 42 x4 x 1x 421x 42 11x 42 1x 421x 42 x 11x 42 1x 421x 42 xx4 1x 421x 42 4 1x 421x 42 2 3 x y Simplify . 1 1 x2 y2

Problems 27–30

Solution

2 2 3 3 x y y x2y2 x 2 2± ≤ 1 1 1 1 xy x2 y2 x2 y2 2 3 x2y2 a b x y 1 1 x2y2 a 2 2 b x y 3xy2 2x2y y2 x2

(continued)

332

Chapter 6



OBJECTIVE

SUMMARY

EXAMPLE

Write rational expressions to express rates. (Sec. 6.4, Obj. 3, p. 309)

To solve certain types of word problems, you will have to write rational expressions to represent rates.

Write a rational expression to answer this question: If it costs x dollars to buy y euros, what is the exchange rate of euros per dollar?

Problems 31–34

Solution

To express the exchange rate, divide the amount of euros by the amount of dollars. Therefore, y the exchange rate is euros per x dollar. Solve rational equations with constants in the denominator. (Sec. 6.5, Obj. 1, p. 313)

To solve rational equations with constants in the denominator, multiply both sides of the equation by the lowest common denominator of all the denominators in the equation.

Solve

x4 21 x3 . 2 5 10

Problems 35 –36

Solution

x3 x4 21 2 5 10 10 a

x3 x4 21 b 10 a b 2 5 10

51x 32 21x 42 21 5x 15 2x 8 21 7x 7 21 7x 14 x2 The solution set is {2}. Solve rational equations with variables in the denominator. (Sec. 6.5, Obj. 2, p. 314; Sec. 6.6, Obj. 1, p. 320)

If an equation contains a variable in one or more of the denominators, avoid any value of the denominator that makes the denominator zero.

Solve

3 5 4 . n 2n 3

Problems 37– 48

Solution

First we need to realize that n cannot equal zero. 6n a

3 5 4 b 6n a b n 2n 3

3 5 4 6n a b 6n a b 6n a b n 2n 3 18 15 8n 33 8n 33 n 8 33 The solution set is e f . 8

(continued)

Chapter 6

Review Problem Set

333


OBJECTIVE

SUMMARY

EXAMPLE

Solve word problems. (Sec. 6.5, Obj. 3, p. 316; Sec. 6.6, Obj. 2, p. 321)

The word problems in this chapter translated into fractional equations. Many of the problems involved work or rate-time problems.

A water tank has two inlet pipes. Working alone, the first inlet pipe can fill the tank in 30 minutes, and the second inlet pipe can fill the tank in 45 minutes. The first pipe starts filling the tank and then 10 minutes later the second pipe is turned on to help fill the tank. How long will it take to fill the water tank?

Problems 49 –54

Solution

Let’s use a table to organize the facts. Let x represent the time to fill the tank.

Pipe A Pipe B

Work done

Rate

Time

1 30

x

x 30

1 45

x 10

x 10 45

So the guideline for the equation is work done by pipe A plus work done by pipe B equals 1 job done. x x 10 1 30 45 Solving this equation gives a result of x 22. Therefore it takes 22 minutes to fill the tank.

Chapter 6 Review Problem Set For Problems 1– 8, simplify each algebraic fraction. 3

1.

56x y 3

72xy

3n n 10 n2 4

2.

x2 9x x 6x 27

4.

16a 24a 9 20a2 7a 6

2

3.

x2 16 5. 4x 7.

24 6a a4

2

2

25 y2 6. y5 8.

For Problems 9 –12, multiply and express your answer in simplest form.

x2 3x 10 2 x x2

9.

7x2y2 18y # 12y3 28x

10.

11.

x2 8x 15 # 3x2 12 x2 4 x2 6x 5

12.

x2 3xy 2y2 x2 xy # xy 3x 6y

x2y

x2 x 6 y x 2x 2

#

334

Chapter 6


For Problems 13 –16, divide and express your answer in simplest form.

For Problems 35 – 48, solve each equation.

n3 6n2 n2 2n 24 2 13. 2 n 11n 28 n 49

35.

2x 1 3x 2 5 3 4 6

4a2 4a 1 6a2 5a 4 14. 2 2 1a 62 3a 14a 24

36.

6x 4 2x 5 1 3 7

37.

1 3 5 2t 4 t

38.

5 7 1 2 3x 2x 5x

39.

67 x 4 6 x x

40.

5 6 2n 3 3n 2

41.

x 5 1 x3 x3

28a2b2 4a 15. 21a 7b

3x2y 6x2 16. 24x3y2 y

For Problems 17–26, add or subtract as indicated, and express your answer in simplest form. 17.

12n 8 2n 7 5 5

18.

31x 42 7x

21x 62 7x

2x 7 3x 4 19. 5 4 7 5 2 20. 2 3x 4x 8x

3 7 21. n n1

3 2 22. a4 a2 3 2x 23. 2 4x x 3x 3 2 24. 2 2 x 7x 10 x 25 3 4 5x 25. 2 x 7 x 3 x 4x 21 26.

3 5 4x 2 x7 x4 x 3x 28

For Problems 27–30, simplify each complex fraction. 4 3 2 x y 27. 4 5 y x

2 1 x 28. 5 3 y

6 3 mn m 29. 5 2 m n

x6 4 30. 3 1 x 4

For Problems 31–34, answer each question with a rational expression. 31. How long would it take to drive x miles when driving at a rate of y miles per hour? 32. If g gallons of heating oil cost d dollars, what is the price per gallon? 33. If Erick can assemble a computer in m minutes, what (fractional) part of the assembly has he done after 40 minutes? 34. Suppose the product of two numbers is p, and one of the factors is k. What is the other factor?

42. n

1 2 n

43.

n n1 4 n9 n2 8n 9

44.

6 1 5 7x 6 6x

46.

n 10 5 n5

47.

1 5 2x 4 2 2x 5 6x 15 4x 25

48. 1

45. n

1 5 n 2

1 1 2 n1 n n

For Problems 49 –54, set up an equation and solve each problem. 49. It takes Nancy three times as long to complete a task as it does Becky. How long would it take each of them to complete the task if working together they can do it in 2 hours? 50. The sum of a number and twice its reciprocal is 3. Find the number. 51. The denominator of a fraction is twice the numerator. If 4 is added to the numerator and 18 to the denominator, 4 a fraction that is equivalent to is produced. Find the 9 original fraction. 52. Lanette can ride her moped 44 miles in the same time that Todd rides his bicycle 30 miles. If Lanette rides 7 miles per hour faster than Todd, find their rates. 53. Jim rode his bicycle 36 miles in 4 hours. For the first 20 miles, he rode at a constant rate, and then for the last 16 miles, he reduced his rate by 2 miles per hour. Find his rate for the last 16 miles. 54. An inlet pipe can fill a tank in 10 minutes. A drain can empty the tank in 12 minutes. If the tank is empty and both the inlet pipe and drain are open, how long will it be before the tank overflows?

Chapter 6 Test For Problems 1– 4, simplify each algebraic fraction. 1.

72x4y5 81x2y4

1.

2.

x2 6x x2 36

2.

3.

2n2 7n 4 3n2 8n 16

3.

4.

2x3 7x2 15x x3 25x

4.

For Problems 5 –14, perform the indicated operations and express answers in simplest form. 5. a

8x2y 21xy3 ba b 7x 12y2

5.

6.

x2 4x 21 x2 49 2 x 7x x2 2x

6.

7.

x2 5x 36 # x2 2x 24 x2 15x 54 x2 7x

7.

8.

2x 3 3x 1 6 8

8.

9.

n2 n1 n6 3 5 6

9.

10.

5 7 3 2x 6 9x

10.

11.

6 4 n n1

11.

12.

3 2x 4x x 6x

12.

13.

9 5 x8 x 4x 32

13.

14.

5 3 2 6x2 7x 20 3x 14x 24

14.

2

2

For Problems 15 –22, solve each of the equations. 15.

x2 23 x3 5 6 30

15.

16.

5 3 2 x 8x

16.

13 4 n 3

17.

17. n 18.

x 6 8 x2

18.

335

336

Chapter 6


19.

19.

x 2 8 x1 x1 3

20.

20.

3 5 2x 1 3x 6

21.

21.

3 4 1 n 1 n n

22.

22.

2n 5 4n 6 3n 1 3 4 9

2

For Problems 23 –25, set up an equation and solve the problem. 23.

2 23. The sum of a number and twice its reciprocal is 3 . Find the number. 3

24.

24. Wendy can ride her bicycle 42 miles in the same time that it takes Betty to ride her bicycle 36 miles. Wendy rides 2 miles per hour faster than Betty. Find Wendy’s rate.

25.

25. Garth can mow a lawn in 20 minutes, and Alex can mow the same lawn in 30 minutes. How long would it take the two of them working together to mow the lawn?

Chapters 1– 6 Cumulative Review Problem Set For Problems 1– 8, evaluate each algebraic expression for the given values of the variables. First you may want to simplify the expression or change its form by factoring. 1. 3x 2xy 7x 5xy

for x

2. 7(a b) 3(a b) (a b) b 5 3.

xy yz y

1 and y 3 2 for a 3 and

2 5 3 for x , y , and z 3 6 4

4. ab b

for a 0.4 and b 0.6

5. x2 y2

for x 6 and y 4

2

for x 9

6. x 5x 36 2

x2 2x 7. 2 x 5x 6 8.

for x 6

x2 3x 10 x2 9x 14

for x 4

11. a

1 1 0 b 2 3

13. 42 15.

1 2 2 a b 5

7 2 3 5x x 2x

31.

24x2y3 48x4y5 8xy2

32. 128x2 19x 202 14x 52 For Problems 33 – 42, factor each polynomial completely. 33. 3x3 15x2 27x

34. x2 100

35. 5x2 22x 8

36. 8x2 22x 63

37. n2 25n 144

38. nx ny 2x 2y

39. 3x3 3x

40. 2x3 6x2 108x

41. 36x2 60x 25

42. 3x2 5xy 2y2

43. 31x 22 21x 62 21x 12 2 10. a b 3 12. a

1

1 1 1 b 3 4

44. x2 11x 45. 0.2x 31x 0.42 1 46.

3n 1 5n 2 4 7

2 2 14. a b 3

48. x2 5x 6 0

16. 132 3

50.

18.

4x 12x2 5y 10y2

52.

5 3 1 2 2x 3x

53. 6t2 19t 7 0 54. 12x 121x 82 0

3 2 2 x 4x x

56.

x 7 1 x2 x1

21.

x2 8x # x2 49 2 3xy x x 56

57.

2 3 1 n n1 n

22.

3 5 x 4 x x 12

2

2

24. 19ab3 2 2

25. 13n2 2 15n2 6n 22 28. 1x 2212x2 3x 12

4 4 n

2x 1 3x 4 1 2 3

20.

26. 15x 1213x 42

49. n

55. 1x 121x 62 24

4 3 19. x6 x4

23. 15x2y2 17x3y4 2

47. 5n2 5 0

51. 21x 12 x1x 12 0

For Problems 17–32, perform the indicated operations and express your answers in simplest form. 17.

30. 12x 1213x 72

For Problems 43 –57, solve each of the equations.

For Problems 9 –16, evaluate each of the expressions. 9. 33

29. 1x2 x 121x2 2x 32

27. 12x 52 2

For Problems 58 – 67, set up an equation or an inequality to help solve each problem. 58. One leg of a right triangle is 2 inches longer than the other leg. The hypotenuse is 4 inches longer than the shorter leg. Find the lengths of the three sides of the right triangle. 59. Twenty percent of what number is 15? 337

338

Chapter 6


60. How many milliliters of a 65% solution of hydrochloric acid must be added to 40 milliliters of a 30% solution of hydrochloric acid to obtain a 55% hydrochloric acid solution? 61. The material for a landscaping border 28 feet long was bent into the shape of a rectangle. The length of the rectangle was 2 feet more than the width. Find the dimensions of the rectangle. 62. Two motorcyclists leave Daytona Beach at the same time and travel in opposite directions. If one travels at 55 miles per hour and the other travels at 65 miles per hour, how long will it take for them to be 300 miles apart? 63. Find the length of an altitude of a trapezoid with bases of 10 centimeters and 22 centimeters and an area of 120 square centimeters.

64. If a car uses 16 gallons of gasoline for a 352-mile trip, at the same rate of consumption, how many gallons will it use on a 594-mile trip? 65. Swati had scores of 89, 92, 87, and 90 on her first four history exams. What score must she get on the fifth exam to have an average of 90 or higher for the five exams? 66. Two less than three times a certain number is less than 10. Find all positive integers that satisfy this relationship. 67. One more than four times a certain number is greater than 15. Find all real numbers that satisfy this relationship.

Coordinate Geometry

7 7.1 Cartesian Coordinate System 7.2 Graphing Linear Equations and Applications 7.3 Graphing Linear Inequalities 7.4 Slope of a Line

Douglas Allen /istockphoto.com

7.5 Writing Equations of Lines

■ Using the concept of slope, we can set up and solve the y 30 proportion to determine how much vertical 100 5280 change a highway with a 30% grade has in a horizontal distance of 1 mile.

R

ené Descartes, a French mathematician of the 17th century, transformed geometric problems into an algebraic setting so that he could use the tools of algebra to solve those problems. This interface of algebraic and geometric ideas is the foundation of a branch of mathematics called analytic geometry, today more commonly called coordinate geometry. We started to make this connection between algebra and geometry in Chapter 2 when graphing the solution sets of inequalities in one variable. For example, the solution set for 3x 1 4 is {x|x 1}, and a geometric picture of this solution set is shown in Figure 7.1. −4 − 3 − 2 − 1

0

1

2

3

4

Figure 7.1

In this chapter we will associate pairs of real numbers with points in a geometric plane. This will provide the basis for obtaining pictures of algebraic equations and inequalities in two variables.


339

I N T E R N E T

P R O J E C T

René Descartes devised the system of associating points on a plane with ordered pairs of real numbers. Do an Internet search to read about the legend associated with Descartes’s inspiration for the rectangular coordinate system. The most commonly used coordinate system today is the latitude and longitude system. What are the reference planes used to define latitude and longitude?

7.1

Cartesian Coordinate System OBJECTIVES 1

Plot Points on a Rectangular Coordinate System

2

Draw Graphs of Equations by Plotting Points

3

Solve Equations for the Specified Variable

1 Plot Points on a Rectangular Coordinate System In Section 1.2 we introduced the real number line (Figure 7.2) as the result of setting up a one-to-one correspondence between the set of real numbers and the points on a line. Recall that the number associated with a point on the line is called the coordinate of that point. − 2

−π

1 2

−1 2

−5 − 4 −3 −2 −1

0

π

2 1

2

3

4

5

Figure 7.2

Now let’s consider two lines (one vertical and one horizontal) that are perpendicular to each other at the point we associate with zero on both lines (Figure 7.3). We refer to these number lines as the horizontal and vertical axes or together as the coordinate axes; they partition the plane into four parts called quadrants. The quadrants are numbered counterclockwise from I to IV as indicated in Figure 7.3. The point of intersection of the two axes is called the origin.

II

5 4 3 2 1

− 5 − 4 − 3 − 2 −1 0 −1 −2 III −3 −4 −5

I

1 2 3 4 5 IV

Figure 7.3

It is now possible to set up a one-to-one correspondence between ordered pairs of real numbers and the points in a plane. To each ordered pair of real numbers there 340

7.1

Cartesian Coordinate System

341

corresponds a unique point in the plane, and to each point there corresponds a unique ordered pair of real numbers. We have indicated a part of this correspondence in Figure 7.4. The ordered pair (3, 1) corresponds to point A and denotes that the point A is located 3 units to the right of and 1 unit up from the origin. (The ordered pair (0, 0) corresponds to the origin.) The ordered pair (2, 4) corresponds to point B and denotes that point B is located 2 units to the left of and 4 units up from the origin. Make sure that you agree with all of the other points plotted in Figure 7.4. A(3, 1) B(−2, 4) C (−4, −3) D (5, −2) O (0, 0) E(1, 3)

B E A O D C

Figure 7.4

Remark: The notation (2, 4) was used earlier in this text to indicate an interval of the real number line. Now we are using the same notation to indicate an ordered pair of real numbers. This double meaning should not be confusing because the context of the material will always indicate which meaning of the notation is being used. Throughout this chapter, we will be using the ordered-pair interpretation. In general, we refer to the real numbers a and b in an ordered pair (a, b) associated with a point as the coordinates of the point. The first number, a, called the abscissa, is the directed distance of the point from the vertical axis measured parallel to the horizontal axis. The second number, b, called the ordinate, is the directed distance of the point from the horizontal axis measured parallel to the vertical axis (see Figure 7.5(a)). Thus in the first quadrant, all points have a positive abscissa and a positive ordinate. In the second quadrant, all points have a negative abscissa and a positive ordinate. We have indicated the signs in all four quadrants in Figure 7.5(b). This system of associating points with ordered pairs of real numbers is called the Cartesian coordinate system or the rectangular coordinate system.

(−, +)

(+, +)

(−, −)

(+, −)

b a

(a) Figure 7.5

(a, b)

(b)

342

Chapter 7

Coordinate Geometry

Plotting points on a rectangular coordinate system can be helpful when analyzing data to determine a trend or relationship. The following example shows the plot of some data.

EXAMPLE 1

The chart below shows the Friday and Saturday scores of golfers in terms of par. Plot the charted information on a rectangular coordinate system. For each golfer, let Friday’s score be the first number in the ordered pair, and let Saturday’s score be the second number in the ordered pair.

Friday’s score Saturday’s score

Mark

Ty

Vinay

Bill

Herb

Rod

1 3

2 2

1 0

4 7

3 4

0 1

Solution Bill

The ordered pairs are as follows: Mark (1, 3)

Ty (2, 2)

Vinay (1, 0)

Bill (4, 7)

Herb (3, 4)

Rod (0, 1)

Mark

The points are plotted on the rectangular coordinate system in Figure 7.6. In the study of statistics, this graph of the charted data would be called a scatterplot. For this plot, the points appear to approximate a straight-line path, which suggests that there is a linear correlation between Friday’s score and Saturday’s score.

Rod Vinay Ty Herb Figure 7.6

▼ PRACTICE YOUR SKILL The chart below shows the increase or decrease (inches) in rainfall for Florida cities in May and June, compared to the average rainfalls. Plot the charted information on a rectangular coordinate system. For each city, let May’s value be the first number in the ordered pair, and let June’s value be the second number in the ordered pair.

May June

Tampa

Orlando

Miami

Naples

Tallahassee

2 3

1 3

0 2

1 4

3 1 ■

2 Draw Graphs of Equations by Plotting Points In Section 2.6, the real number line was used to display the solution set of an inequality. Recall that the solution set of an inequality such as x 3 has an infinite number of solutions. Hence the real number line is an effective way to display the solution set of an inequality. Now we want to find the solution sets for equations with two variables. Let’s begin by considering solutions for the equation y x 3. A solution of an equation in two variables is a pair of numbers that makes the equation a true statement. For the equation y x 3, the pair of numbers in which x 4 and y 7 makes the equation a true statement. This pair of numbers can be written as an ordered pair (4, 7). When using the variables x and y, we agree that the first number of an ordered pair is the

7.1


343

value for x, and the second number is the value for y. Likewise (2, 5) is a solution for y x 3 because 5 2 3. We can find an infinite number of ordered pairs that satisfy the equation y x 3 by choosing a value for x and determining the corresponding value for y that satisfies the equation. The table below lists some of the solutions for the equation y x 3.

Choose x

Determine y from y x 3

Solution for y x 3

0 1 3 5 1 3 5

3 4 6 8 2 0 2

(0, 3) (1, 4) (3, 6) (5, 8) (1, 2) (3, 0) (5, 2)

Because the number of solutions for the equation y x 3 is infinite, we do not have a convenient way to list the solution set. This is similar to inequalities in which the solution set is infinite. For inequalities we used a real number line to display the solution set. Now we will use the rectangular coordinate system to display the solution set. On a rectangular coordinate system where we label the horizontal axis as the x axis and the vertical axis as the y axis, we can locate the point associated with each ordered pair of numbers in the table. These points are shown in Figure 7.7(a). These points are only some of the infinite solutions of the equation y x 3. The straight line in Figure 7.7(b) that connects the points represents all the solutions of the equation and is called the graph of the equation y x 3. y

y

y=x+3

x

x

(a)

(b)

Figure 7.7

The next examples further illustrate the process of graphing equations.

EXAMPLE 2

Graph y 2x 1.

Solution First we set up a table of some of the solutions for the equation y 2x 1. You can choose any value for the variable x and find the corresponding y value. The values we chose for x are in the table and include positive integers, zero, and negative integers.

344

Chapter 7

Coordinate Geometry

x

y

0 1 2 1 2 3

1 3 5 1 3 5

Solutions (x, y) (0, 1) (1, 3) (2, 5) (1, 1) (2, 3) (3, 5)

From the table we plot the points associated with the ordered pairs as shown in Figure 7.8(a). Connecting these points with a straight line produces the graph of the equation y 2x 1 as shown in Figure 7.8(b). y

y

y 2x 1

x

(a)

x

(b)

Figure 7.8

▼ PRACTICE YOUR SKILL Graph y 3x 2.

EXAMPLE 3

■

1 Graph y x 3. 2

Solution 1 First we set up a table of some of the solutions for the equation y x 3. You 2 can choose any value for the variable x and find the corresponding y value; however, the values we chose for x are numbers that are divisible by 2. (This is not necessary but does produce integer values for y.)

x

y

Solutions (x, y)

0 2 4 2 4

3 2 1 4 5

(0, 3) (2, 2) (4, 1) (2, 4) (4, 5)

7.1


345

From the table we plot the points associated with the ordered pairs as shown in Figure 7.9(a). Connecting these points with a straight line produces the graph of the 1 equation y x 3 as shown in Figure 7.9(b). 2 y

y 1 y x3 2

x

(a)

x

(b)

Figure 7.9

▼ PRACTICE YOUR SKILL 1 Graph y x 1 . 3

■

3 Solve Equations for the Specified Variable Sometimes changing the form of an equation makes it easier to find solutions of the equation. In the next two examples, the equation will be solved for x or y.

EXAMPLE 4


Solution 4x 9y 12 9y 4x 12 12 4 y x 9 9 4 4 y x 9 3

Subtracted 4x from each side Divided both sides by 9

▼ PRACTICE YOUR SKILL Solve 2x 3y 9 for y.

EXAMPLE 5

■

Solve 2x 6y 3 for x.

Solution 2x 6y 3 2x 6y 3 3 6 x y 2 2 3 x 3y 2

Added 6y to each side Divided both sides by 2

346

Chapter 7

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▼ PRACTICE YOUR SKILL Solve 3x 5y 7 for x.

■

To graph an equation in two variables, x and y, keep these steps in mind: 1. 2. 3. 4.

Solve the equation for y in terms of x or for x in terms of y, if it is not already in such a form. Set up a table of ordered pairs that satisfies the equation. Plot the points associated with the ordered pairs. Connect the points.

We conclude this section with Example 6, which illustrates the steps.

EXAMPLE 6

Graph 6x 3y 9.

Solution First, let’s change the form of the equation to make it easier to find solutions of the equation 6x 3y 9. We can solve either for x in terms of y or for y in terms of x. Typically the equation is solved for y in terms of x, so that’s what we show here. 6x 3y 9 3y 6x 9 y

9 6x 3 3

y 2x 3 Now we can set up a table of values. Plotting these points and connecting them produces Figure 7.10.

x

y

0 1 2 1

3 1 1 5

y y 2x 3

x

Figure 7.10

▼ PRACTICE YOUR SKILL Solve the equation for y and graph the equation. 4x 2y 8

CONCEPT QUIZ

■

For Problems 1–5, answer true or false. 1. In a rectangular coordinate system, the coordinate axes partition the plane into four parts called quadrants. 2. Quadrants are named with Roman numerals and numbered clockwise.

7.1


347

3. The real numbers in an ordered pair are referred to as the coordinates of the point. 4. The equation y x 3 has an infinite number of ordered pairs that satisfy the equation. 5. The point of intersection of the coordinate axes is called the origin. For Problems 6 –10, match the points plotted in Figure 7.11 with their coordinates. y E

6. (3, 1) 7. (4, 0) 8. (3, 1)

D C x

9. (0, 4) 10. (1, 3)

A

B

Figure 7.11

Problem Set 7.1 1 Plot Points on a Rectangular Coordinate System For Problems 1–10, plot the points on a rectangular coordinate system, and determine if the points fall in a straight line. 1. (1, 4), (2, 1), (4, 1), (3, 0) 2. (0, 2), (3, 1), (5, 3), (3, 5) 3. (0, 3), (2, 7), (3, 6) 4. (1, 4), (1, 6), (6, 3) 5. (3, 1), (3, 3), (0, 2)

Mon. Change in amount of light Change in amount of water

Tue.

Wed.

Thu. Fri. Sat.

1

2

1

4

3

0

3

4

1

0

5

1

12. Chase is studying the monthly percent changes in the stock price for two different companies. Using the data in the table below, plot the points for each month. Let the percent change for XM Inc. be the first number of the ordered pair, and let the percent change for Icom be the second number in the ordered pair.

6. (0, 1), (2, 0), (2, 2) 7. (4, 0), (2, 1), (6, 5) 8. (1, 3), (2, 6), (0, 0)

XM Inc. Icom

Jan.

Feb.

Mar.

Apr.

May

Jun.

1 3

2 4

1 2

4 5

3 1

0 1

9. (2, 4), (2, 4), (1, 0) 10. (2, 2), (3, 6), (1, 1) 11. Maria, a biology student, designed an experiment to test the effects of changing the amount of light and the amount of water given to selected plants. In the experiment, the amounts of water and light given to the plant were randomly changed. The chart shows the amount of light and water above or below the normal amount given to the plant for six days. Plot the charted information on a rectangular coordinate system. Let the change in light be the first number in the ordered pair, and let the change in water be the second number in the ordered pair.

2 Draw Graphs of Equations by Plotting Points For Problems 13 –20, (a) determine which of the ordered pairs satisfy the given equation, and then (b) graph the equation by plotting the points that satisfied the equation. 13. y x 4; 11, 32, 10, 42, 12, 12, 12, 62, 11, 52 14. y 2x 3; 11, 12, 11.5, 02, 13, 12, 10, 32, 11, 52 15. y 2x 3; 11, 22, 13, 02, 10, 32, 12, 12, 11, 52 16. y x 4; 11, 32, 12, 12, 12, 22, 11, 52, 10, 42

348

Chapter 7

Coordinate Geometry

2 17. y x; 11, 22, 13, 22, 10, 02, 13, 62, 16, 22 3

For Problems 31–50, graph each of the equations. 31. y x 1

32. y x 4

33. y x 2

34. y x 1

35. y 3x 4

36. y 2x 1

37. y 4x 4

38. y 2x 6

1 39. y x 3 2

1 40. y x 2 2

41. x 2y 4

42. x 3y 6

43. 2x 5y 10

44. 5x 2y 10

1 45. y x 2 3

2 46. y x 3 3

For Problems 21–30, solve the given equation for the variable indicated.

47. y x

48. y x

21. 3x 7y 13 for y

22. 5x 9y 17 for y

49. y 3x 2

50. 3x y 4

23. x 3y 9 for x

24. 2x 7y 5 for x

25. x 5y 14 for y

26. 2x y 9 for y

27. 3x y 7 for x

28. x y 9 for x

29. 2x 3y 5 for y

30. 3x 4y 7 for y

1 1 1 18. y x; 11, 22, 12, 12, a 0, b, 12, 12, a 1, b 2 2 2 1 1 4 3 2 19. y x ; a 1, b, a 3, b, a 1, b, 5 5 5 5 5 2 a 2, b, 12, 12 5 2 1 1 1 20. y x ; a 1, b, 10, 12, 15, 32, a , 0b, 11, 02 3 3 3 2

3 Solve Equations for the Specified Variable

THOUGHTS INTO WORDS 51. How would you convince someone that there are infinitely many ordered pairs of real numbers that satisfy the equation x y 9?

52. Explain why no points of the graph of the equation y x will be in the second quadrant.

FURTHER INVESTIGATIONS Not all graphs of equations are straight lines. Some equations produce graphs that are smooth curves. For example, the 2 graph of y x will produce a smooth curve that is called a parabola. To graph y x2, let’s proceed as in the previous problems and plot points associated with solutions to the equation y x2, and then connect the points with a smooth curve.

x

y

Solutions (x, y)

0 1 2 3 1 2

0 1 4 9 1 4

(0, 0) (1, 1) (2, 4) (3, 9) (1, 1) (2, 4)

3

9

(3, 9)

We plot the points associated with the solutions as shown in Figure 7.12(a). Finally, we connect the points with the smooth curve shown in Figure 7.12(b).

7.1


y

349

y

y = x2

x

x

(a)

(b)

Figure 7.12 For Problems 53 – 62, graph each of the equations. Be sure to find a sufficient number of solutions so that the graph of the equation can be determined.

57. y x3

58. y x3 4

59. y 1x 32 3

60. y x3

53. y x2 2

54. y x2 3

61. y x4

62. y x4

55. y x2

56. y 1x 22 2


2.

y

Tampa

y

Orlando y = 3x − 2 x Tallahassee Miami

x

Naples

3.

5 2 7 4. y x 3 5. x y 3 3 3

y

y

1 x 3

6. y 2x 4 y

1 y = 2x + 4 x x


2. False

3. True

4. True 5. True 6. D

7. C

8. A

9. E

10. B

350

Chapter 7

7.2

Coordinate Geometry

Graphing Linear Equations and Applications OBJECTIVES 1

Find x and y Intercepts for Linear Equations

2

Graph Linear Equations

3

Apply Linear Equations

1 Find x and y Intercepts for Linear Equations In the preceding section we graphed equations whose graphs were straight lines. In general any equation of the form Ax By C, where A, B, and C are constants (A and B not both zero) and x and y are variables, is a linear equation in two variables, and its graph is a straight line. We should clarify two points about this description of a linear equation in two variables. First, the choice of x and y for variables is arbitrary. We could use any two letters to represent the variables. An equation such as 3m 2n 7 can be considered a linear equation in two variables. So that we are not constantly changing the labeling of the coordinate axes when graphing equations, however, it is much easier to use the same two variables in all equations. Thus we will go along with convention and use x and y as our variables. Second, the statement “any equation of the form Ax By C” technically means “any equation of the form Ax By C or equivalent to the form.” For example, the equation y x 3, which has a straight line graph, is equivalent to x y 3. All of the following are examples of linear equations in two variables. yx3

y 3x 2

2 y x1 5

y 2x

3x 2y 6

x 4y 5

5x y 10

y

2x 4 3

The knowledge that any equation of the form Ax By C produces a straight line graph, along with the fact that two points determine a straight line, makes graphing linear equations in two variables a simple process. We merely find two solutions, plot the corresponding points, and connect the points with a straight line. It is probably wise to find a third point as a check point. Let’s consider an example.

EXAMPLE 1

Graph 2x 3y 6.

Solution We recognize that equation 2x 3y 6 is a linear equation in two variables, and therefore its graph will be a straight line. All that is necessary is to find two solutions and connect the points with a straight line. We will, however, also find a third solution to serve as a check point. Let x 0; then 2102 3y 6 3y 6 y 2

Thus (0, 2) is a solution

Let y 0; then 2x 3102 6 2x 6 x3


7.2

Graphing Linear Equations and Applications

Let x 3; then 2132 3y 6 6 3y 6 3y 12 y 4

351


We can plot the points associated with these three solutions and connect them with a straight line to produce the graph of 2x 3y 6 in Figure 7.13. y 2 x − 3y = 6

x

Figure 7.13

▼ PRACTICE YOUR SKILL Graph 2x 3y 12.

■

Let us review the approach to solving Example 1. • We did not begin by solving the equation for x or y. We know the graph is a straight line, and there is no need for an extensive table of values. • The first two solutions, (0, 2) and (3, 0), indicate where the line intersects the coordinate axes. • The y coordinate of the point (0, 2) is called the y intercept. The y intercept is 2. • The x coordinate of the point (3, 0) is called the x intercept. The x intercept is 3. • To find the y intercept, let x 0 and solve for y. • To find the x intercept, let y 0 and solve for x. • The third solution, (3, 4), serves as a check. If (3, 4) had not been on the line we determined by the two intercepts, then we would have known that we made an error.

2 Graph Linear Equations EXAMPLE 2

Graph x 2y 4 by plotting the intercepts and a check point.

Solution Let’s find the x and y intercepts, and then determine a third point as a check point. To find the y intercept, let x 0.

To find the x intercept, let y 0.

x 2y 4 0 2y 4 2y 4 y2

x 2y 4 x 2102 4 x04 x4

Therefore the points (0, 2) and (4, 0) are on the graph.

352

Chapter 7

Coordinate Geometry

To find a check point, let x 2. x 2y 4 2 2y 4 when x 2 2y 2 y1 Therefore, the point (2, 1) is on the graph. We plot the points (0, 2), (4, 0), and (2, 1) and connect them with a straight line to produce the graph in Figure 7.14. y y intercept checkpoint

x x intercept x + 2y = 4

Figure 7.14

▼ PRACTICE YOUR SKILL Graph 2x y 6 by plotting the x and y intercepts and a check point.

EXAMPLE 3

■

Graph 2x 3y 7.

Solution The intercepts and a check point are given in the table. Finding intercepts may involve fractions, but the computation is usually easy. We plot the points from the table and show the graph of 2x 3y 7 in Figure 7.15.

x

y

0

7 3

7 2 2

y

y intercept

Intercepts

checkpoint

0 1

Check point

2 x + 3y = 7

x intercept

x

Figure 7.15

▼ PRACTICE YOUR SKILL Graph x 2y 5 by plotting the x and y intercepts and a check point.

■

7.2

EXAMPLE 4


353

Graph y 2x.

Solution Notice that (0, 0) is a solution; thus, this line intersects both axes at the origin. Since both the x intercept and the y intercept are determined by the origin, (0, 0), we need another point to graph the line. Then a third point should be found to serve as a check point. These results are summarized in the table; the graph of y 2x is shown in Figure 7.16. y

x

y

0 2 1

0 4 2

y = 2x

Intercept Additional point Check point

x

Figure 7.16

▼ PRACTICE YOUR SKILL Graph y 3x.

EXAMPLE 5

■

Graph x 3.

Solution Since we are considering linear equations in two variables, the equation x 3 is equivalent to x 0(y) 3. Now we can see that any value of y can be used, but the x value must always be 3. Therefore, some of the solutions are (3, 0), (3, 1), (3, 2), (3, 1), and (3, 2). The graph of all of the solutions is the vertical line indicated in Figure 7.17. y

x=3

x

Figure 7.17

▼ PRACTICE YOUR SKILL Graph x 2.

■

354

Chapter 7

Coordinate Geometry

EXAMPLE 6

Graph y 3.

Solution Since we are considering equations in two variables, the equation y 4 is equivalent to 0(x) y 3. Now we can determine that for any value of x, the value of y will be 3. Therefore some of the solutions are (1, 3), (0, 3), (2, 3), and (4, 3). The graph of all the solutions is a horizontal line indicated in Figure 7.18. y

y = −3 x

Figure 7.18

▼ PRACTICE YOUR SKILL Graph y 4.

■

3 Apply Linear Equations Linear equations in two variables can be used to model many different types of realworld problems. For example, suppose that a retailer wants to sell some items at a profit of 30% of the cost of each item. If we let s represent the selling price and c the cost of each item, then we can use the equation s c 0.3c 1.3c to determine the selling price of each item on the basis of the cost of the item. For example, if the cost of an item is $4.50, then the retailer should sell it for s (1.3)(4.5) $5.85. By finding values that satisfy the equation s 1.3c, we can create this table:

c

1

5

10

15

20

s

1.3

6.5

13

19.5

26

From the table we see that if the cost of an item is $15, then the retailer should sell it for $19.50 in order to make a profit of 30% of the cost. Now let’s get a picture (graph) of this linear relationship. We can label the horizontal axis c and the vertical axis s, and we can use the origin along with one ordered pair from the table to produce the straight line graph in Figure 7.19. (Because of the type of application, only nonnegative values for c and s are appropriate.)

s 40 30 20 s = 1.3c 10 0

10

Figure 7.19

20

30

40

c

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355

From the graph we can approximate s values on the basis of given c values. For example, if c 30, then by reading up from 30 on the c axis to the line and then across to the s axis, we see that s is a little less than 40. (We get an exact s value of 39 by using the equation s 1.3c.)

EXAMPLE 7

Apply Your Skill

VStock LLC /Index Open

Kristin has a cell phone plan that charges her 2 and one-half cents per minute of cell phone usage. The equation C 0.025m, where m is the number of minutes used and C is the charge in dollars, describes this cell phone plan. (a) Complete a table that shows the charges for 100 minutes, 300 minutes, 500 minutes, 800 minutes, and 1000 minutes. (b) Graph the equation C 0.025m. (c) Use the graph from part (b) to approximate the charge when m 600. (d) Check the accuracy of your reading from the graph in part (c) by using the equation C 0.025m.

Solution (a) By finding values that satisfy the equation C 0.025m, we can create this table:

m

100

300

500

800

1000

C

2.50

7.50

12.50

20.00

25.00

(b) Label the horizontal axis m and the vertical axis C and plot the points given in the table to create Figure 7.20. Connect the points with a straight line. 30 25 20 15 10 5

C (1000, 25.00) (500, 12.50) C 0.025m m 100

300

500

700

900

Figure 7.20

(c) On the horizontal axis locate where m 600. From that point read up until you intersect the graph and then go across to the vertical axis. We can see that when m 600, C is approximately 15.00. (d) To check the accuracy, use the equation C 0.025m and evaluate when m 600. C 0.025m C 0.02516002 15 when m 600.

▼ PRACTICE YOUR SKILL Alex pays $5 a month and $1 per hour to use a video game program. The equation C m 5, where m is the number of hours played and C is the charge in dollars, describes the charges for this video game.

356

Chapter 7

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(a) Complete a table that shows the charges for 1 hour, 3 hours, 5 hours, and 10 hours. (b) Graph the equation C m 5. (c) Use the graph from part (b) to approximate the charge when m 2.5. (d) Check the accuracy of your reading from the graph in part (c) by using the equation C m 5. ■

Many formulas that are used in various applications are linear equations in 5 two variables. For example, the formula C 1F 322 , which converts tempera9 tures from the Fahrenheit scale to the Celsius scale, is a linear relationship. The next 5 example shows the graph of the formula C 1F 322 . 9

National Geographic/Getty Images

EXAMPLE 8

Apply Your Skill 5 1F 322 , where F is the degrees Fahrenheit and C is the degrees 9 Celsius, describes the relationship between the two temperature scales. The equation C

(a) Complete a table that shows the degrees Celsius for 22°F, 13°F, 5°F, 32°F, 50°F, 68°F, and 86°F. 5 (b) Graph the equation C 1F 322 . 9 (c) Use the graph from part (b) to approximate the temperature in degrees Celsius when the temperature is 77°F. (d) Check the accuracy of your reading from the graph in part (c) by using the 5 equation C 1F 322 . 9

Solution 5 (a) By finding values that satisfy the equation C 1F 322 , we can create 9 this table: 22 30

F C

13 25

5 15

32 0

50 10

68 20

86 30

5 1F 322 , we can label the horizontal axis F 9 and the vertical axis C and plot two points that are given in the table. Figure 7.21 shows the graph of the equation.

(b) To graph the equation C

C 40 20 −20

20 −20 −40

Figure 7.21

40

60

C = 5 (F − 32) 9

80 F

7.2


357

(c) From the graph we can approximate C values on the basis of given F values. For example, if F 77°, then by reading up from 77 on the F axis to the line and then across to the C axis, we see that C is approximately 24°. 5 (d) To check the accuracy, use the equation C 1F 322 and evaluate for 9 F 77°. C

5 1F 322 9

When F 77, C

5 5 177 322 1452 25 9 9

▼ PRACTICE YOUR SKILL Aidan is spending a semester studying in Ireland. While he is there he uses the formula D 1.5e to convert euros, e, to dollars, D. (a) Complete a table that shows the amount of dollars for 2 euros, 3 euros, 5 euros, and 10 euros. (b) Graph the equation D 1.5e. (c) Use the graph from part (b) to approximate the dollars when e 6. (d) Check the accuracy of your reading from the graph in part (c) by using the equation D 1.5e. ■

CONCEPT QUIZ

For Problems 1–7, answer true or false. 1. 2. 3. 4. 5. 6. 7.

The equations 2x y 4 and y 2x 4 are equivalent. The y intercept of the graph of 3x 4y 12 is 4. The x intercept of the graph of 2x 4y 8 is 4. Determining just two points is sufficient to graph a straight line. The graph of y 4 is a vertical line. The graph of x 3 is a vertical line. The graph of every linear equation has a y intercept.

Problem Set 7.2 1 Find x and y Intercepts for Linear Equations For Problems 1–10, find the x and y intercepts and the coordinates of another point. Then graph the equation.

2 Graph Linear Equations For Problems 11– 46, graph each linear equation either by finding the x and y intercepts or by making a table of values. 11. x y 2

12. x y 4

13. x y 3

14. x y 1

15. x y 4

16. x y 5

6. x y 3

17. x 2y 2

18. x 3y 5

7. 3x 2y 6

8. 4x 3y 12

19. 3x y 6

20. 2x y 4

9. 6x 2y 12

10. 5x 2y 10

21. 3x 2y 6

22. 2x 3y 4

1. x 2y 4

2. x 3y 6

3. 2x y 2

4. 3x y 6

5. x y 5

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Chapter 7

Coordinate Geometry

23. x y 0

24. x y 0

25. y 3x

26. y 2x

27. x 2

28. y 3

Hours, t

29. y 0

30. x 0

Dollars and cents, A

31. y 2x 1

32. y 3x 4

1 33. y x 1 2

2 34. y x 2 3

1 35. y x 2 3

3 36. y x 1 4

50. An online grocery store charges for delivery based on the equation C 0.30p, where C represents the cost in dollars, and p represents the weight of the groceries in pounds. Label the horizontal axis p and the vertical axis C, and graph the equation C 0.30p for nonnegative values of p.

37. 4x 5y 10

38. 3x 5y 9

51. a. The equation F

39. 2x y 4

40. 3x y 5

41. 3x 4y 7

42. 4x 3y 10

43. y 4x 0

44. y 5x 0

45. x 2y

46. x 3y

49. At $0.06 per kilowatt-hour, the equation A 0.06t determines the amount, A, of an electric bill for t hours. Complete this table of values:

C

6.5

7.0

8.0

9.0

10.0

Blood glucose, G 48. Suppose that a used-car dealer determines the selling price of his cars by using a markup of 60% of the cost. If s represents the selling price and c the cost, this equation applies: s c 0.6c 1.6c Complete this table using the equation s 1.6c: Dollars, c

250

325

575

0

5

10

15

20

5

b. Graph the equation F

8.5

895

740

775

782

10

15

20

25

F

47. A doctor’s office needs a chart that correlates the Hemoglobin A1c test to average blood glucose levels. The equation G 30h 60 describes the relationship, in which h is the Hemoglobin A1c reading, and G is the average blood glucose reading. Complete this chart of values:

6.0

720

9 C 32 converts temperatures 5 from degrees Celsius to degrees Fahrenheit. Complete this table:

3 Apply Linear Equations

Hemoglobin A1c, h

696

1095

9 C 32. 5

c. Use your graph from part (b) to approximate values for F when C 25°, 30°, 30°, and 40°. d. Check the accuracy of your readings from the graph 9 in part (c) by using the equation F C 32. 5 52. Suppose that the daily profit from an ice cream stand is given by the equation p 2n 4, where n represents the number of gallons of ice cream mix used in a day, and p represents the number of dollars of profit. Label the horizontal axis n and the vertical axis p, and graph the equation p 2n 4 for nonnegative values of n. 53. The cost (c) of playing an online computer game for a time (t) in hours is given by the equation c 3t 5. Label the horizontal axis t and the vertical axis c, and graph the equation for nonnegative values of t. 54. The area of a sidewalk whose width is fixed at 3 feet can be given by the equation A 3l, where A represents the area in square feet, and l represents the length in feet. Label the horizontal axis l and the vertical axis A, and graph the equation A 3l for nonnegative values of l.

Dollars, s

THOUGHTS INTO WORDS 55. Your friend is having trouble understanding why the graph of the equation y 3 is a horizontal line containing the point (0, 3). What might you do to help him? 56. How do we know that the graph of y 4x is a straight line that contains the origin?

57. Do all graphs of linear equations have x intercepts? Explain your answer. 58. How do we know that the graphs of x y 4 and x y 4 are the same line?

7.2


359

FURTHER INVESTIGATIONS From our previous work with absolute value, we know that 0 x y 0 2 is equivalent to x y 2 or x y 2. Therefore, the graph of 0 x y0 2 consists of the two lines x y 2 and x y 2. Graph each of the following equations.

59. 0 x y 0 1

60. 0 x y 0 2

61. 0 2x y0 4

62. 0 3x y 0 6


2.

3. y

y

y 2x + 3y = 12

x − 2y = 5

(0, 4) (3, 2)

(5, 0)

(3, 0)

x

x

(6, 0)

(3, −1)

(2, −2)

(0, − 5) 2

2x − y = 6 (0, −6)

4.

5.

6.

y (−2, 6)

y

y

y = −3x

(2, 4) x = −2

(−2, 4)

(−2, 2)

(0, 0)

(−2, 0)

x

x

(−2, −2) (2, −6)

7. (a) m

1

3

5

10

8. (a) e

2

3

5

10

C

6

8

10

15

D

3

4.5

7.5

15.00

(b)

(b)

18 16 14 12 10 8 6 4 2 0

C

(10, 15) (5, 10) (3, 8) (1, 6)

C

m

5

2 4 6 8 10 12 14 16 18 m

(c) 7 hours (d) C 2.5 5 7.5

x

18 16 14 12 10 8 6 4 2 0

D (10, 15.00)

(5, 7.5) (3, 4.5) (2, 3)

D

1.5e

2 4 6 8 10 12 14 16 18 e

(c) 9 dollars (d) 9 dollars

(0, 4)

y=4 x

360

Chapter 7

Coordinate Geometry


7.3

2. False

3. True 4. True 5. False 6. True

7. False

Graphing Linear Inequalities OBJECTIVE 1

Graph Linear Inequalities

1 Graph Linear Inequalities Linear inequalities in two variables are of the form Ax By C or Ax By C, where A, B, and C are real numbers. (Combined linear equality and inequality statements are of the form Ax By C or Ax By C.) Graphing linear inequalities is almost as easy as graphing linear equations. Our discussion leads to a simple step-bystep process. First we consider the following equation and related inequalities: xy2 xy 2 xy 2 The graph of x y 2 is shown in Figure 7.22. The line divides the plane into two half-planes, one above the line and one below the line. In Figure 7.23(a) we have indicated the coordinates for several points above the line. Note that for each point, the ordered pair of real numbers satisfies the inequality x y 2. This is true for all points in the half-plane above the line. Therefore, the graph of x y 2 is the half-plane above the line, indicated by the shaded region in Figure 7.23(b). We use a dashed line to indicate that points on the line do not satisfy x y 2. y

y

x

Figure 7.22 y

(3, 4) (−1, 3) (1, 2) (− 4, 1) x

(−1, −2)

x

(− 4, −2)

(a) Figure 7.23

(b)

7.3

Graphing Linear Inequalities

361

In Figure 7.24(a), we have indicated the coordinates of several points below the line x y 2. Note that for each point, the ordered pair of real numbers satisfies the inequality x y 2. This is true for all points in the half-plane below the line. Therefore, the graph of x y 2 is the half-plane below the line, indicated by the shaded region in Figure 7.24(b). y

y

(5, 2) x

x

(3, −1) (2, −3) (−1, −4)

(− 2, −5) (a)

(b)

Figure 7.24

Based on this discussion, we suggest the following steps for graphing linear inequalities:

Step 1 Graph the corresponding equality. Use a solid line if the equality is included in the given statement and a dashed line if the equality is not included.

Step 2 Choose a test point not on the line and substitute its coordinates into the inequality statement. (The origin is a convenient point to use if it is not on the line.) Step 3 Determine if the test point makes the inequality a true statement or a false statement. In other words, does the test point satisfy the inequality or not? Step 4 The graph of the inequality is (a) the half-plane containing the test point if the test point makes the inequality a true statement, or (b) the half-plane not containing the test point if the test point makes the inequality a false statement. We can apply these steps to some examples.

EXAMPLE 1

Graph 2x y 4.

Solution Step 1 Graph 2x y 4 as a dashed line, because equality is not included in the given statement 2x y 4; see Figure 7.25(a).

Step 2 Choose the origin as a test point, and substitute its coordinates into the inequality. 2x y 4

becomes 2(0) 0 4

Step 3 The inequality 2(0) 0 4 simplifies to 0 4, which is a false statement. Step 4 Since the test point does not satisfy the given inequality, the graph is the halfplane that does not contain the test point. Thus the graph of 2x y 4 is the half-plane above the line, indicated in Figure 7.25(b).

362

Chapter 7

Coordinate Geometry y

y

x

(a)

x

(b)

Figure 7.25

▼ PRACTICE YOUR SKILL Graph x 2y 6.

EXAMPLE 2

■

Graph y 2x.

Solution Step 1 Graph y 2x as a solid line, because equality is included in the given statement [see Figure 7.26(a)].

Step 2 Since the origin is on the line, we need to choose another point as a test point. Let’s use (3, 2). y 2x

becomes 2 2(3)

Step 3 The inequality 2 2(3) simplifies to 2 6, which is a true statement. Step 4 Since the test point satisfies the given inequality, the graph is the half-plane that contains the test point. Thus the graph of y 2x is the line along with the half-plane below the line indicated in Figure 7.26(b). y

y

x

(a)

x

(b)

Figure 7.26

▼ PRACTICE YOUR SKILL Graph y 3x.

■

7.3

EXAMPLE 3

Graphing Linear Inequalities

363

Graph y 4.

Solution Step 1 Graph y 4 as a dashed line, because equality is not included in the given

statement y 4. Remember the equation y 4 is equivalent to the equation (0)x y 4; hence for any value of x, the y value is always 4. Therefore the graph of y 4 is a horizontal line as shown in Figure 7.27(a). Step 2 Choose the origin as a test point, and substitute the coordinates into the inequality. y 4

becomes 0 4

Step 3 The inequality 0 4 is a true statement. Step 4 Because the test point satisfies the given inequality, the graph is the halfplane that contains the test point. Thus the graph of y 4 is the half-plane below the line as indicated in Figure 7.27(b). y

y

x

(a)

x

(b)

Figure 7.27

▼ PRACTICE YOUR SKILL Graph y 2.

EXAMPLE 4

■

Graph x 2.

Solution Step 1 Graph x 2 as a solid line, because equality is included in the given state-

ment x 2. Remember that the equation x 2 is equivalent to the equation x (0)y 2; hence for any value of y, the x value is always 2. Therefore the graph of x 2 is a vertical line as shown in Figure 7.28(a). Step 2 Choose the origin as a test point, and substitute the coordinates into the inequality. x 2

becomes 0 2

Step 3 The inequality 0 2 is a false statement. Step 4 Because the test point does not satisfy the given inequality, the graph is the half-plane that does not contain the test point. Thus the graph of x 2 is the line along with the half-plane to the left of the line as indicated in Figure 7.28(b).

364

Chapter 7

Coordinate Geometry

y

y

x

(a)

x

(b)

Figure 7.28

▼ PRACTICE YOUR SKILL Graph x 3.

CONCEPT QUIZ

■

For Problems 1–5, answer true or false. 1. A dashed line on the graph of an inequality indicates that the points on the line do not satisfy the inequality. 2. The coordinates of any point can be used as a test point to determine the halfplane that is the solution of the inequality. 3. The ordered pair (2, 3) satisfies the inequality 2x y 1. 4. The solution set of 2x y 4 would include the coordinates of the points on the line 2x y 4. 5. The solution set of x 1 would be the coordinates of the points on the line x 1, because the graph of x 1 is a vertical line.

Problem Set 7.3 1 Graph Linear Inequalities

1 15. y x 2 2

1 16. y x 1 2

17. y x 4

18. y x 3

19. 3x 4y 12

20. 4x 3y 12

21. y 2x

22. y 3x

23. y x 2

24. y x 3

25. y 1

26. y 3

27. x 5

28. x 1

29. y 0

30. x 0

For Problems 1–30, graph each linear inequality. 1. x y 1

2. 2x y 4

3. 3x 2y 6

4. x 3y 3

5. 2x y 4

6. x 2y 2

7. 4x 3y 12

8. 3x 4y 12

9. y x

10. y x

11. 2x y 0

12. 3x y 0

13. x 2y 2

14. 2x y 2

THOUGHTS INTO WORDS 31. Explain how you would graph the inequality x 2y 4.

32. Why is the point (3, 2) not a good test point to use when graphing the inequality 3x 2y 13?

7.4

Slope of a Line

365


2.

y

y

x

x

3.

4.

y

y

x

x


7.4

2. False


Slope of a Line OBJECTIVES 1

Find the Slope of a Line between Two Points

2

Graph a Line Given a Point and the Slope

3

Apply the Concept of Slope to Solve Problems

1

Find the Slope of a Line between Two Points

In Figure 7.29, note that the line associated with 4x y 4 is steeper than the line associated with 2x 3y 6. Mathematically, we use the concept of slope to discuss the steepness of lines. The slope of a line is the ratio of the vertical change to the horizontal change as we move from one point on a line to another point. We indicate this in Figure 7.30 with the points P1 and P2.

366

Chapter 7

Coordinate Geometry y

y

4x − y = 4

2x

P2

6 y= −3

Vertical change P1

x

Q x Horizontal change Slope =

Figure 7.29

Vertical change Horizontal change

Figure 7.30

We can give a precise definition for slope by considering the coordinates of the points P1, P2, and Q in Figure 7.31. Since P1 and P2 represent any two points on the line, we assign the coordinates 1x1, y1 2 to P1 and 1x2, y2 2 to P2. The point Q is the same distance from the y axis as P2 and the same distance from the x axis as P1. Thus we assign the coordinates 1x2, y1 2 to Q (see Figure 7.31). It should now be apparent that the vertical change is y2 y1, and the horizontal change is x2 x1. Thus we have the following definition for slope.

y

P x 1,

) , y2 2 x ( 2

y 1)

( P1

x2 − x1

y2 − y1 Vertical change

Q (x2, y1)

Horizontal change

x

Figure 7.31

Definition 7.1 If points P1 and P2 with coordinates 1x1, y1 2 and 1x2, y2 2 , respectively, are any two different points on a line, then the slope of the line (denoted by m) is m

y2 y1 , x2 x1

x1 x2

Using Definition 7.1, we can easily determine the slope of a line if we know the coordinates of two points on the line.

EXAMPLE 1

Find the slope of the line determined by each of the following pairs of points. (a) (2, 1) and (4, 6) (b) (3, 2) and (4, 5) (c) (4, 3) and (1, 3)

Solution (a) Let (2, 1) be P1 and (4, 6) be P2 as in Figure 7.32; then we have m

y2 y1 61 5 x2 x1 42 2

7.4

Slope of a Line

y

P2 (4, 6)

P1 (2, 1) x

Figure 7.32

(b) Let (3, 2) be P1 and (4, 5) be P2 as in Figure 7.33. m

y2 y1 3 3 52 x2 x1 4 3 7 7

P2 (− 4, 5)

y

P1 (3, 2)

x

Figure 7.33

(c) Let (4, 3) be P1 and (1, 3) be P2 as in Figure 7.34. m

3 132 y2 y1 0 0 x2 x1 1 142 3 y

x P2 (−1, −3) P1 (− 4, −3)

Figure 7.34

367

368

Chapter 7

Coordinate Geometry

▼ PRACTICE YOUR SKILL Find the slope of the line determined by each of the following pairs of points: (a) (1, 2) and (3, 8)

(b) (5, 3 ) and (2, 4) ■

(c) (4, 2) and (1, 2)

The designation of P1 and P2 in such problems is arbitrary and does not affect the value of the slope. For example, in part (a) of Example 1, suppose that we let (4, 6) be P1 and (2, 1) be P2. Then we obtain m

y2 y1 16 5 5 x2 x1 24 2 2

The parts of Example 1 illustrate the three basic possibilities for slope: that is, the slope of a line can be positive, negative, or zero. • Positive slope

A line that has a positive slope rises as we move from left to right, as in Example 1(a). • Negative slope A line that has a negative slope falls as we move from left to right, as in Example 1(b). • Zero slope A line that has a zero slope is a horizontal line, as in Example 1(c). Finally, we need to realize that the concept of slope is undefined for vertical lines. This is because, for any vertical line, the change in x as we move from one point to another y2 y1 is zero. Thus the ratio will have a denominator of zero and be undefined. x2 x1 So in Definition 7.1, the restriction x1 x2 is made. Example 1 showed how to find the slope of a line when you know the coordinates of two points on the line. To find the slope when we are presented with the equation of the line, we need to first determine two points on the line. Example 2 provides the details on finding the slope when given the equation of a line.

EXAMPLE 2

Find the slope of the line determined by the equation 3x 4y 12.

Solution Since we can use any two points on the line to determine the slope of the line, let’s find the intercepts. If x 0, then 3102 4y 12 4y 12 y3

Thus (0, 3) is on the line

If y 0, then 3x 4102 12 3x 12 x4

Thus (4, 0) is on the line

Using (0, 3) as P1 and (4, 0) as P2, we have m

y2 y1 03 3 3 x2 x1 40 4 4

▼ PRACTICE YOUR SKILL Find the slope of the line determined by the equation 2x y 4.

■

7.4

Slope of a Line

369

2 Graph a Line Given a Point and the Slope We need to emphasize one final idea pertaining to the concept of slope. The slope 3 of a line is a ratio of vertical change to horizontal change. A slope of means that 4 for every 3 units of vertical change, there is a corresponding 4 units of horizontal change. So starting at some point on the line, we could move to other points on the line as follows: 6 3 4 8

by moving 6 units up and 8 units to the right

3 15 4 20

by moving 15 units up and 20 units to the right

3 3 2 4 2 3 3 4 4

1 by moving 1 units up and 2 units to the right 2 by moving 3 units down and 4 units to the left

5 Likewise, a slope of indicates that starting at some point on the line, we could 6 move to other points on the line as follows:

EXAMPLE 3

5 5 6 6

by moving 5 units down and 6 units to the right

5 5 6 6

by moving 5 units up and 6 units to the left

10 5 6 12

by moving 10 units down and 12 units to the right

15 5 6 18

by moving 15 units up and 18 units to the left

1 Graph the line that passes through the point (0, 2) and has a slope of . 3

Solution vertical change 1 , horizontal change 3 we can locate another point on the line by starting from the point (0, 2) and moving 1 unit up and 3 units to the right to obtain the point (3, 1). Because two points determine a line, we can draw the line (Figure 7.35). To begin, plot the point (0, 2). And, because the slope

y

x (0, −2)

Figure 7.35

(3, −1)

370

Chapter 7

Coordinate Geometry

1 1 , we can locate another point by moving 1 unit down 3 3 and 3 units to the left from the point (0, 2).

Remark: Because m



EXAMPLE 4

■

Graph the line that passes through the point (1, 3) and has a slope of 2.

Solution To graph the line, plot the point (1, 3). We know that m 2

2 . Furthermore, 1

vertical change 2 , we can locate another point on the line horizontal change 1 by starting from the point (1, 3) and moving 2 units down and 1 unit to the right to obtain the point (2, 1). Because two points determine a line, we can draw the line (Figure 7.36). because the slope

y (1, 3) (2, 1) x

Figure 7.36

2 2 , we can locate another point by moving 1 1 2 units up and 1 unit to the left from the point (1, 3).

Remark: Because m 2



■

3 Apply the Concept of Slope to Solve Problems The concept of slope has many real world applications even though the word “slope” is often not used. For example, the highway in Figure 7.37 is said to have a grade of 17%. This means that for every horizontal distance of 100 feet, the highway rises or drops 17 feet. In other words, the absolute value of the slope of the high17 way is . 100

7.4

Slope of a Line

371

17 feet 100 feet Figure 7.37

EXAMPLE 5

Apply Your Skill


A certain highway has a 3% grade. How many feet does it rise in a horizontal distance of 1 mile?

Solution 3 . Therefore, if we let y represent the unknown 100 vertical distance and use the fact that 1 mile 5280 feet, we can set up and solve the following proportion: A 3% grade means a slope of

y 3 100 5280 100y 3152802 15,840 y 158.4 The highway rises 158.4 feet in a horizontal distance of 1 mile.

▼ PRACTICE YOUR SKILL An exit ramp for an interstate highway has a grade of 2%. How many feet does it rise in a horizontal distance of 1250 feet? ■ A roofer, when making an estimate to replace a roof, is concerned about not only the total area to be covered but also the pitch of the roof. (Contractors do not define pitch the same way that mathematicians define slope, but both terms refer to steepness.) The two roofs in Figure 7.38 might require the same number of shingles, but the roof on the left will take longer to complete because the pitch is so great that scaffolding will be required.

Figure 7.38

The concept of slope is also used in the construction of flights of stairs. The terms “rise” and “run” are commonly used, and the steepness (slope) of the stairs can be expressed as the ratio of rise to run. In Figure 7.39, the stairs on the left with the 7 10 the ratio of are steeper than the stairs on the right, which have a ratio of . 11 11

372

Chapter 7

Coordinate Geometry

rise of 10 inches rise of 7 inches run of 11 inches

run of 11 inches

Figure 7.39

Technically, the concept of slope is involved in most situations involving an incline. Hospital beds are constructed so that both the head-end and the foot-end can be raised or lowered; that is, the slope of either end of the bed can be changed. Likewise, treadmills are designed so that the incline (slope) of the platform can be raised or lowered as desired. Perhaps you can think of other applications of the concept of slope.

CONCEPT QUIZ

For Problems 1– 8, answer true or false. 1. The concept of slope of a line pertains to the steepness of the line. 2. The slope of a line is the ratio of the horizontal change to the vertical change moving from one point to another point on the line. 3. A line that has a negative slope falls as we move from left to right. 4. The slope of a vertical line is 0. 5. The slope of a horizontal line is 0. 6. A line cannot have a slope of 0. 5 5 7. A slope of is the same as a slope of . 2 2 8. A slope of 5 means that for every unit of horizontal change there is a corresponding 5 units of vertical change.

Problem Set 7.4 1 Find the Slope of a Line between Two Points For Problems 1–20, find the slope of the line determined by each pair of points. 1. (7, 5), (3, 2)

2. (9, 10), (6, 2)

3. (1, 3), (6, 4)

4. (2, 5), (7, 1)

5. (2, 8), (7, 2)

6. (3, 9), (8, 4)

7. (2, 5), (1, 5)

8. (3, 4), (2, 6)

9. (4, 1), (4, 7)

10. (5, 3), (5, 9)

11. (3, 4), (2, 4)

12. (3, 6), (5, 6)

13. (6, 1), (2, 7)

14. (8, 3), (2, 11)

15. (2, 4), (2, 6)

16. (4, 5), (4, 9)

17. (1, 10), (9, 2)

18. (2, 12), (10, 2)

19. (a, b), (c, d)

20. (a, 0), (0, b)

21. Find y if the line through the points (7, 8) and (2, y) 4 has a slope of . 5 22. Find y if the line through the points (12, 14) and (3, y) 4 has a slope of . 3 23. Find x if the line through the points (2, 4) and 3 (x, 2) has a slope of . 2 24. Find x if the line through the points (6, 4) and (x, 6) 5 has a slope of . 4

7.4 For Problems 25 –32, you are given one point on a line and the slope of the line. Find the coordinates of three other points on the line. 25. 13, 22, m

26. 14, 12, m

2 3

2 28. 16, 22, m 5

29. 13, 42, m

30. 12, 62, m

31. 14, 52, m 2

373

2 Graph a Line Given a Point and the Slope For Problems 61– 68, graph the line that passes through the given point and has the given slope.

5 6

1 27. 12, 42, m 2 3 4

Slope of a Line

3 7

32. 16, 22, m 4

For Problems 33 – 40, sketch the line determined by each pair of points and decide whether the slope of the line is positive, negative, or zero. 33. (2, 8), (7, 1)

34. (1, 2), (7, 8)

35. (1, 3), (6, 2)

36. (7, 3), (4, 6)

37. (2, 4), (6, 4)

38. (3, 4), (5, 4)

39. (3, 5), (2, 7)

40. (1, 1), (1, 9)

61. 13, 12, m

62. 11, 02, m

2 3

3 4

63. 12, 32, m 1

64. 11, 42, m 3

65. 10, 52, m

1 4

66. 13, 42, m

67. 12, 22, m

3 2

68. 13, 42, m

3 2

5 2

3 Apply the Concept of Slope to Solve Problems 69. Suppose that a highway rises a distance of 135 feet in a horizontal distance of 2640 feet. Express the grade of the highway to the nearest tenth of a percent.

For Problems 41– 60, find the coordinates of two points on the given line, and then use those coordinates to find the slope of the line.

70. The grade of a highway up a hill is 27%. How much change in horizontal distance is there if the vertical height of the hill is 550 feet? Express the answer to the nearest foot.

41. 3x 2y 6

42. 4x 3y 12

71. If the ratio of rise to run is to be

43. 5x 4y 20

44. 7x 3y 21

45. x 5y 6

46. 2x y 4

47. 2x y 7

48. x 4y 6

49. y 3

50. x 6

51. 2x 5y 9

52. 3x 7y 10

53. 6x 5y 30

54. 7x 6y 42

55. y 3x 1

56. y 2x 5

57. y 4x

58. y 6x

2 1 59. y x 3 2

3 1 60. y x 4 5

3 for some stairs, 5 and the measure of the rise is 19 centimeters, find the measure of the run to the nearest centimeter. 2 for some stairs, 3 and the measure of the run is 28 centimeters, find the measure of the rise to the nearest centimeter.


1 73. A county ordinance requires a 2 % fall for a sewage pipe 4 from the house to the main pipe at the street. How much vertical drop must there be for a horizontal distance of 45 feet? Express the answer to the nearest tenth of a foot.

THOUGHTS INTO WORDS 74. How would you explain the concept of slope to someone who was absent from class the day it was discussed? 2 75. If one line has a slope of , and another line has a slope 3 of 2, which line is steeper? Explain your answer.

76. Why do we say that the slope of a vertical line is undefined? 3 and contains the 4 point (5, 2). Are the points (3, 4) and (14, 9) also on the line? Explain your answer.

77. Suppose that a line has a slope of

374

Chapter 7

Coordinate Geometry

Answers to the Example Problem Skills 1. (a) m 3 (b) m

1 (c) m 0 3

2. m 2

y

3.

4.

y (− 3, 4)

(3, 1) (−1, 1)

(−2, −1)

x

x

5. 25 feet


7.5

2. False


6. False

7. False 8. True

Writing Equations of Lines OBJECTIVES 1

Find the Equation of a Line Given a Point and a Slope

2

Find the Equation of a Line Given Two Points

3

Find the Equation of a Line Given the Slope and y Intercept

4

Use the Point-Slope Form to Write Equations of Lines

5

Apply the Slope-Intercept Form of an Equation

1 Find the Equation of a Line Given a Point and a Slope There are two basic types of problems in analytic or coordinate geometry: 1.

Given an algebraic equation, find its geometric graph.

2.

Given a set of conditions pertaining to a geometric figure, determine its algebraic equation.

We discussed problems of the first type in the first two sections of this chapter. Now we want to consider a few problems of the second type that deal with straight lines. In other words, given certain facts about a line, we need to be able to write its algebraic equation.

7.5

EXAMPLE 1

Find the equation of the line that has a slope of

Writing Equations of Lines

375

3 and contains the point (1, 2). 4

Solution First, let’s graph the line as indicated in Figure 7.40. To draw the line, plot the point 3 (1, 2) and then because we know that the slope is , we can find a second point by mov4 ing 3 units up and 4 units to the right. Now that we have a graph of the line, we can choose an arbitrary point (x, y) on the line that represents any point on the line other 3 than the point (1, 2). The slope between (1, 2) and (x, y) is ; therefore we can write 4 the following equation: y2 3 x1 4

y

Now simplify this equation to write the equation of the line in standard form.

(x, y) (1, 2) given point

y2 3 x1 4 31x 12 41y 22

(5, 5)

(5, 5) is 3 units up and 4 units to the right of (1, 2)

x Cross-multiply

3x 3 4y 8 3x 4y 5 Figure 7.40

▼ PRACTICE YOUR SKILL Find the equation of the line that has a slope of


■

2 Find the Equation of a Line Given Two Points EXAMPLE 2

Find the equation of the line that contains (3, 4) and (2, 5).

Solution First, we draw the line determined by the two given points in Figure 7.41. Since we know two points, we can find the slope. m

y

y2 y1 1 1 54 x2 x1 2 3 5 5

Now we can use the same approach as in Example 1. We form an equation using a variable point (x, y), one of the two given points 1 (we choose P1), and the slope of . 5 y4 1 x3 5

1 1 5 5

x 3 5y 20

Cross-multiply

x 5y 23

P2 (−2, 5)

P(x, y) P1 (3, 4)

x

Figure 7.41

376

Chapter 7

Coordinate Geometry


Find the equation of the line that contains (2, 3) and (1, 5).

3 Find the Equation of a Line Given the Slope and y Intercept EXAMPLE 3


1 and a y intercept of 2. 4

Solution A y intercept of 2 means that the point (0, 2) 1 is on the line. Since the slope is , we 4 can find another point by moving 1 unit up and 4 units to the right of (0, 2). The line is drawn in Figure 7.42. We choose a variable point (x, y) and proceed as in the preceding examples.

y (x, y)

(0, 2) x

y2 1 x0 4 11x 02 41y 22 x 4y 8

Figure 7.42

x 4y 8

▼ PRACTICE YOUR SKILL Find the equation of the line that has a slope of


■

4 Use the Point-Slope Form to Write Equations of Lines It may be helpful for you to pause for a moment and look back over Examples 1, 2, and 3. Notice that we used the same basic approach in all three examples; that is, we chose a variable point (x, y) and used it, along with another known point, to determine the equation of the line. You should also recognize that the approach we take in these examples can be generalized to produce some special forms of equations for straight lines. y To develop a general form, let us find (x, y) the equation of the line that has a slope of m and contains the point (x1, y1). Then we can (x1, y1) choose (x, y) to represent any other point on the line in Figure 7.43. The slope of the line is given by x

m

y y1 x x1

Simplifying by cross-multiplication enables us to obtain y y1 m1x x1 2

Figure 7.43

7.5


377

We refer to the equation y y1 m1x x1 2 as the point-slope form of the equation of a straight line. Instead of the approach we used in Example 1, we could use the point-slope form to write the equation of a line with a given slope that contains a given point, as Example 4 illustrates.

EXAMPLE 4

Write the equation of the line that has a slope of


Solution Substituting

3 for m and (2, 4) for (x1, y1) in the point-slope form, we obtain 5

y y1 m1x x1 2 y 142

3 1x 22 5

y4

3 1x 22 5

51y 42 31x 22


5y 20 3x 6 26 3x 5y

▼ PRACTICE YOUR SKILL Write the equation of a line that has a slope of


■

5 Apply the Slope-Intercept Form of an Equation Now consider the equation of a line that has a slope of m and a y intercept of b (see Figure 7.44). A y intercept of b means that the line contains the point (0, b); therefore, we can use the point-slope form. y y1 m1x x1 2 y b m1x 02

y1 b and x1 0

y b mx y mx b y

(0, b)

x

Figure 7.44

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Chapter 7

Coordinate Geometry

The equation y mx b is called the slope-intercept form of the equation of a straight line. We use it for three primary purposes, as the next three examples illustrate.

EXAMPLE 5



Solution This is a restatement of Example 3, but this time we will use the slope-intercept form of a line (y mx b) to write its equation. From the statement of the problem 1 we know that m and b 2. Thus substituting these values for m and b into 4 y mx b, we obtain y mx b 1 y x2 4 4y x 8 x 4y 8

Same result as in Example 3


2 Find the equation of a line that has a slope of and a y intercept of 5. 3

■

Remark: It is acceptable to leave answers in slope-intercept form. We did not do that in Example 5 because we wanted to show that it was the same result as in Example 3.

EXAMPLE 6

Find the slope of the line with the equation 2x 3y 4.

Solution We can solve the equation for y in terms of x, and then compare it to the slopeintercept form to determine its slope. 2x 3y 4 3y 2x 4 4 2 y x 3 3 2 Compare this result to y mx b, and you see that the slope of the line is . 3 4 Furthermore, the y intercept is . 3

▼ PRACTICE YOUR SKILL Find the slope of the line whose equation is 4x 2y 3.

■

7.5

EXAMPLE 7


379

2 Graph the line determined by the equation y x 1. 3

Solution Comparing the given equation to the general slope-intercept form, we see that the 2 slope of the line is , and the y intercept is 1. Because the y intercept is 1, we can 3 2 plot the point (0, 1). Then because the slope is , let’s move 3 units to the right and 3 2 units up from (0, 1) to locate the point (3, 1). The two points (0, 1) and (3, 1) determine the line in Figure 7.45. y y=

2 x 3

–1

(3, 1) x (0, –1)

Figure 7.45


3 Graph the line determined by the equation y x 2. 5

■

In general, if the equation of a nonvertical line is written in slope-intercept form, the coefficient of x is the slope of the line, and the constant term is the y intercept. (Remember that the concept of slope is not defined for a vertical line.) Let’s consider a few more examples.

EXAMPLE 8

Find the slope and y intercept of each of the following lines and graph the lines. (a) 5x 4y 12

(b) y 3x 4

(c) y 2

Solution (a) We change 5x 4y 12 to slope-intercept form to get 5x 4y 12 4y 5x 12 4y 5x 12 5 y x3 4 5 (the coefficient of x), and the y intercept is 3 (the 4 constant term). To graph the line, we plot the y intercept, 3. Then because 5 the slope is , we can determine a second point, (4, 2), by moving 5 units 4 The slope of the line is

380

Chapter 7

Coordinate Geometry

up and 4 units to the right from the y intercept. The graph is shown in Figure 7.46. y

y=

5 x 4

−3 (4, 2)

x

(0, −3)

Figure 7.46

(b) We multiply both sides of the given equation by 1 to change it to slopeintercept form. y 3x 4 y 3x 4 The slope of the line is 3, and the y intercept is 4. To graph the line, we 3 plot the y intercept, 4. Then because the slope is 3 , we can find a 1 second point, (1, 1), by moving 3 units down and 1 unit to the right from the y intercept. The graph is shown in Figure 7.47. y (0, 4) y = −3x + 4 (1, 1) x

Figure 7.47

(c) We can write the equation y 2 as y 0(x) 2 The slope of the line is 0, and the y intercept is 2. To graph the line, we plot the y intercept, 2. Then because a line with a slope of 0 is horizontal, we draw a horizontal line through the y intercept. The graph is shown in Figure 7.48.

7.5


381

y

y=2 (0, 2)

x

Figure 7.48

▼ PRACTICE YOUR SKILL Find the slope and y intercept of each of the following lines and graph the lines. (a) 3x 2y 6

CONCEPT QUIZ

(b) y x 3

(c) y 4

■

For Problems 1–5, answer true or false. y y1 1. The equations m and y y1 m(x x1) are equivalent. x x1 2. In the slope-intercept form of an equation of a line y mx b, m is the slope. 3. The slope of a line determined by the equation 5x 2y 1 is 5. 3 4. The slope of a line determined by the equation 3x 2y 4 is . 2 5. The concept of slope is not defined for the line x 3.

Problem Set 7.5 2 Find the Equation of a Line Given Two Points

1 Find the Equation of a Line Given a Point and a Slope For Problems 1–12, find the equation of the line that contains the given point and has the given slope. Express equations in the form Ax By C, where A, B, and C are integers. 1. 12, 32, m

2. 15, 22, m

2 3

3 7

3. 13, 52, m

1 2

4. 15, 62, m

5. 14, 82, m

1 3

6. 12, 42, m

7. 13, 72, m 0 9. 10, 02, m

4 9

11. 16, 22, m 3

3 5 5 6

8. 13, 92, m 0 10. 10, 02, m

5 11

12. 12, 102, m 2

For Problems 13 –22, find the equation of the line that contains the two given points. Express equations in the form Ax By C, where A, B, and C are integers. 13. (2, 3) and (7, 10)

14. (1, 4) and (9, 10)

15. (3, 2) and (1, 4)

16. (2, 8) and (4, 2)

17. (1, 2) and (6, 7)

18. (8, 7) and (3, 1)

19. (0, 0) and (3, 5)

20. (5, 8) and (0, 0)

21. (0, 4) and (7, 0)

22. (2, 0) and (0, 9)

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3 Find the Equation of a Line Given the Slope and y intercept For Problems 23 –32, find the equation of the line with the given slope and y intercept. Leave your answers in slope-intercept form. 3 23. m and b 2 5 25. m 2 and b 1

5 24. m and b 4 9 26. m 4 and b 3

1 27. m and b 4 6 5 28. m and b 1 7 5 29. m 1 and b 2 7 30. m 2 and b 3

35. m 3 and passes through (1, 4) 36. m 4 and passes through (2, 6) 37. Passes through (0, 6) and (3, 0) 38. Passes through (3, 0) and (0, 2) 39. Passes through (1, 3) and (2, 3) 40. Passes through (1, 6) and (4, 0)

5 Apply the Slope-Intercept Form of an Equation For Problems 41–52, determine the slope and y intercept of the line represented by the given equation, and graph the line.

1 5 31. m and b 9 2 2 7 32. m and b 12 3

4 Use the Point-Slope Form to Write Equations of Lines For Problems 33 – 40, use the point-slope form to determine the equation of the line for the given conditions. Leave your answers in slope-intercept form.

41. y 2x 5

2 42. y x 4 3

43. 3x 5y 15

44. 7x 5y 35

45. 4x 9y 18

46. 6x 7y 14

3 47. y x 4 4

48. 5x 2y 0

49. 2x 11y 11

2 11 50. y x 3 2

51. 9x 7y 0

52. 5x 13y 26

2 33. m and passes through (0, 5) 5 3 34. m and passes through (0, 2) 4

THOUGHTS INTO WORDS 53. Explain the importance of the slope-intercept form (y mx b) of the equation of a line.

55. What does it mean to say that two points “determine” a line?

54. How would you describe coordinate geometry to a group of elementary algebra students?

56. How can you tell by inspection that y 2x 4 and y 3x 1 are not parallel lines?

FURTHER INVESTIGATIONS The following two properties pertain to parallel and perpendicular lines.

59. a

4x 3y 12 b 3x 4y 12

60. a

9x 2y 18 b 2x 9y 13

(a) Two nonvertical lines are parallel if and only if their slopes are equal.

61. a

x 3y 7 b 5x 15y 9

62. a

7x 2y 14 b 7x 2y 5

(b) Two nonvertical lines are perpendicular if and only if the product of their slopes is 1. For Problems 57– 62, determine whether each pair of lines is (a) parallel, (b) perpendicular, or (c) intersecting lines that are not perpendicular. 57. a

5x 2y 6 b 2x 5y 9

58. a

2x y 9 b 6x 3y 5

63. Write the equation of the line that contains the point (4, 3) and is parallel to the line 2x 3y 6. 64. Write the equation of the line that contains the point (1, 3) and is perpendicular to the line 3x y 4.

7.5

Answers to the Example Practice Skills 1. 2x 3y 2

2. 2x 3y 13

3. x 3y 12

2 5. y x 5 3

4. 5x 2y 11

3 8. (a) m , y intercept 3 2

y

7.


y

3 x 2 5

y

(5, 1)

(0, 3) x (2, 0)

(0, 2)

x

(b) m 1, y intercept 3

(c) m 0, y intercept 4 y

y

x

x (0, − 3)

(1, −2)


2. True 3. False

4. True 5. True

(0, − 4)

(3, − 4)

6. m 2

383

Chapter 7 Summary CHAPTER REVIEW PROBLEMS

OBJECTIVE

SUMMARY

EXAMPLE

Find the x and y intercepts for linear equations. (Sec. 7.2, Obj. 1, p. 350)

The x intercept is the x coordinate of the point where the graph intersects the x axis. The y intercept is the y coordinate of the point where the graph intersects the y axis. To find the x intercept, substitute 0 for y in the equation and then solve for x. To find the y intercept, substitute 0 for x in the equation and then solve for y.

Find the x and y intercepts for the line with the equation 2x y 4.

Problems 1– 6

Solution

Let y 0. 2x 0 4 2x 4 x 2 So the x intercept is 2. Let x 0. 2102 y 4 y 4 y4 So the y intercept is 4.

Graph linear equations. (Sec. 7.2, Obj. 2, p. 351)

To graph a linear equation, you must determine two points. It is advisable to find a third point as a checkpoint. The two points that are determined could be the x and y intercepts or any other two points found by substituting values into the equation. Another method for graphing is to plot a point on the line and then locate another point on the line by using the slope of the line.

Graph y 2x 5.

Problems 7–20

Solution

From the equation y 2x 5 we know the slope is 2 and the y intercept is 5. Plot the point (0, 5); from the point (0, 5) move up two units and to the right 1 unit because the slope is 2. Draw a straight line connecting the points. y

y = 2x − 5

x (1, − 3) (0, − 5)

384

(continued)

Chapter 7

Summary

385


OBJECTIVE

SUMMARY

EXAMPLE

Apply linear equations. (Sec. 7.2, Obj. 3, p. 354)

Linear equations in two variables can be used to model many types of real world problems.

A trucking firm has calculated the equation C 0.15d to determine the fuel cost, C, depending on the distance, d, of the trip. Draw a graph of the equation and approximate the cost for a trip of 150 miles.

Problems 21–24

Solution 90 80 70 60 50 40 30 20 10

C

d 100

200

300

400

Reading the graph, a trip of 150 miles would have approximately $21 in fuel cost. Graph linear inequalities. (Sec. 7.3, Obj. 1, p. 360)

To graph a linear inequality, first graph the line for the corresponding equality. Use a solid line if the equality is included in the given statement or a dashed line if the equality is not included. Then a test point is used to determine which halfplane is included in the solution set. See page 361 for the detailed steps.

Graph 2x y 4.

Problems 25 – 40

Solution

First graph 2x y 4. Choose (0, 0) as a test point. Substituting (0, 0) into the inequality yields 0 4. Because the test point (0, 0) makes the inequality a false statement, the half-plane not containing the point (0, 0) is the solution. y

x

(continued)

386

Chapter 7

Coordinate Geometry and Linear Systems


OBJECTIVE

SUMMARY

EXAMPLE

Find the slope of a line between two points. (Sec. 7.4, Obj. 1, p. 365)

The slope of a line is a ratio of vertical change to horizontal change. The slope can be negative, positive, or zero. The concept of slope is not defined for vertical lines. The y2 y1 formula for slope is m , x2 x1 where the points (x1, y1) and (x2, y2) are any two points on the line and x1 x2.

Find the slope of a line that passes though the points (1, 4) and (3, 5).

The concept of slope in most situations involves an incline. The grade of a highway and the incline of a treadmill are a couple of examples involving slope.

The backyard of a house built 3 on a hillside has a slope of . 2 How much does the vertical distance change for a 40-foot change in horizontal distance?

Apply the concept of slope. (Sec. 7.4, Obj. 3, p. 370)

Problems 41– 48

Solution

m

5 142 31

9 2

Problems 49 –52

Solution

Let y represent the change in vertical distance. Then solve the following proportion: y 3 2 40 2y 120 y 60 The vertical change would be 60 feet. Find the equation of a line given a point and a slope. (Sec. 7.5, Obj. 1, p. 374)

The point-slope form of the equation of a line, y y1 m(x x1), can be used to find the equation. Substitute the slope and the coordinates of a point into the form. Your answer can be left in standard form or slope-intercept form depending upon the instructions.

Find the equation of a line that has a slope of 3 and passes through the point (2, 5). Write the result in slope-intercept form.

Problems 53 –56

Solution

y y1 m1x x1 2

y 5 3 3x 122 4

y 5 31x 22 y 5 3x 6 y 3x 11

(continued)

Chapter 7

Review Problem Set

387


OBJECTIVE

SUMMARY

EXAMPLE

Find the equation of a line given two points. (Sec. 7.5, Obj. 2, p. 375)

To use the point-slope form we must first determine the slope. Then substitute the slope and the coordinates of a point into the form.

Find the equation of a line that passes through (2, 5) and (1, 3). Write the result in standard form.

Problems 57– 60

Solution

m

35 2 1 122 3

Now substitute the slope and either point into the point-slope form. y y1 m1x x1 2 y1

2 1x 32 3

31y 12 21x 32 3y 3 2x 6 2x 3y 9 Apply the slopeintercept form of an equation. (Sec. 7.5, Obj. 5, p. 377)

The equation y mx b is called the slope-intercept form of the equation of a straight line. If the equation of a nonvertical line is written in slope-intercept form, then the coefficient of x is the slope of the line, and the constant term is the y intercept.

Determine the slope and y intercept of the line with equation 5x 3y 12.

Problems 61– 66

Solution

Solve for y to have the equation in slope-intercept form. 5x 3y 12 3y 5x 12 5 y x4 3 Therefore, the slope of the line 5 is and the y intercept is 4. 3

Chapter 7 Review Problem Set For Problems 1– 6, find the x and y intercepts for the line with the given equation.

For Problems 7–20, graph each of the equations.

1. x 3y 6

7. 2x 5y 10

2. x 2y 4

9. y 2x

3. 4x 5y 20 4. 2x y 1 5. y 3x 6 6. y 2x 8

1 8. y x 1 3 10. 3x 4y 12

11. 2x 3y 0

12. 2x y 2

13. x y 4

14. x 2y 2

2 15. y x 1 3

16. y 3x

388

Chapter 7

Coordinate Geometry

17. y 3

18. y 2

19. x 4

20. x 1

21. The equation C 1.5e is used to convert euros, e, to Canadian dollars, C. Graph the equation and from the graph approximate the value of 9 euros in Canadian dollars. 22. A store that is going out of business is offering everything in the store at a sale price of 30% off the regular price. The equation s 0.7p gives the sale price, s, for any given price, p. Graph the equation and from the graph approximate the sale price of an item that was priced at $15. 23. A laundry service charges $4 for pickup and 50 cents per pound to wash clothes. The equation C 4 0.5p, where C is the charge in dollars and p is the pounds of clothes, can be used to determine the charge. Graph the equation and approximate the charge for washing 18 pounds of clothes. 24. A u-pick strawberry patch charges $1.75 a pound for the 7 strawberries you pick. The equation C p, where C is 4 the cost in dollars and p is the pounds of strawberries, can be used to determine the cost. Graph the equation and approximate the cost for 7 pounds of strawberries. For Problems 25 – 40, graph each of the inequalities. 25. 2x y 2

26. 3x y 3

27. x 3y 3

28. x 4y 4

29. y 2x 1

30. y 2x 5 3 32. y x 1 2 34. y x

31. y x 3 33. y 3x 1 35. y x 2

36. y x

37. x 4

38. x 2

39. y 2

40. y 0

For Problems 41– 48, find the slope of the line determined by the two points. 41. (3, 2) and (1, 4) 42. (1, 2) and (0, 5) 43. (4, 1) and (7, 0) 44. (2, 4) and (4, 7)

49. The county building code stipulates that the rise to run for a wheelchair ramp has to have a ratio of 1 to 10. Find the length required for a ramp that has a rise of 5 feet. 7 for some stairs and the 10 measure of the rise is 30 centimeters, find the measure of the run to the nearest centimeter.


3 to be consid20 ered a beginner’s slope. One hill in consideration for the beginner’s slope has a rise of 320 feet and a run of 1500 feet. Find the slope of the hill and decide if it can be classified as a beginner’s slope.

51. A ski slope must have a slope of less than

52. The grade of a highway up a mountain is 8%. How much change in horizontal distance is there if the vertical height of the hill is 600 feet? Express the answer to the nearest foot. 3 and 4 contains the point (1, 2). Express the answer in standard form.

53. Write the equation of a line that has a slope of

54. Write the equation of a line that has a slope of 2 and contains the point (5, 3). Express the answer in standard form. 55. Write the equation of a line that has a slope of 4 and contains the point (1, 2). Express the answer in slopeintercept form. 3 56. Write the equation of a line that has a slope of and 5 contains the point (5, 3). Express the answer in slopeintercept form. 57. Write the equation of a line that contains the points (2, 1) and (0, 6). Express the answer in standard form. 58. Write the equation of a line that contains the points (1, 6) and (1, 3). Express the answer in standard form. 59. Write the equation of a line that contains the points (0, 0) and (4, 1). Express the answer in slope-intercept form. 60. Write the equation of a line that contains the points (1, 5) and (2, 5). Express the answer in slope-intercept form. For Problems 61– 66, determine the slope and y intercept and graph the line. 61. 2x 5y 10

1 62. y x 1 3

47. (3, 6) and (3, 1)

63. x 2y 2

64. 3x y 2

48. (2, 12) and (4, 7)

65. 2x y 4

66. 3x 4y 12

45. (1, 5) and (3, 5) 46. (2, 5) and (3, 6)

Chapter 7 Test For Problems 1– 4, determine the slope and y intercept for each equation. 1. 5x 3y 15

1.

2. 2x y 4

2.

1 3. y x 2 2

3.

4. 3x y 0

4.

5. Find the slope of the line determined by the points (5, 1) and (3, 4).

5.


6.

3 7. Find x if the line through the points (4, 7) and (x, 13) has a slope of . 2

7.

3 8. Find y if the line through the points (1, y) and (6, 2) has a slope of . 5

8.

1 and passes through the point (3, 5), find the coordinates 4 of two other points on the line.

9. If a line has a slope of

9.

10. If a line has a slope of 3 and passes through the point (2, 1), find the coordinates of two other points on the line.

10.

11. For the graphs given in Figure 7.49, match the lines with the following descriptions of their slopes.

11.

(a) Positive slope (c) Slope of 0

(b) Negative slope (d) Slope is undefined y

B

x

C

A D Figure 7.49

12. Suppose that a highway rises a distance of 85 feet over a horizontal distance of 1850 feet. Express the grade of the highway to the nearest tenth of a percent.

12.

13. Find the x intercept of the graph of y 4x 8.

13.

14. Find the y intercept of the graph of 2x 3y 12.

14.

389

390

Chapter 7

Coordinate Geometry

For Problems 15 –21, graph each equation or inequality. 15.

15. 5x 3y 15

16.

2 16. y x 3

17.

17. y 2x 3

18.

18. x 2

19.

19. y 4

20.

20. y 2x 4

21.

21. x 3y 3

22.

22. An online tutoring service charges $4 for a question plus 25 cents per minute for 1 the time spent. The equation C t 4, where C is the cost in dollars and t is 4 the time in minutes, describes the cost depending on the time used. Graph the equation and approximate the cost for 10 minutes of tutoring.

23.

For Problems 23 –25, express each equation in Ax By C form, where A, B, and C are integers. 3 23. Determine the equation of the line that has a slope of and a y intercept of 4. 5

24.

24. Determine the equation of the line containing the point (4, 2) and having a 4 slope of . 9

25.

25. Determine the equation of the line that contains the points (4, 6) and (2, 3).

Systems of Equations

8 8.1 Solving Linear Systems by Graphing 8.2 Elimination-by-Addition Method 8.3 Substitution Method

Andrea Gingerich /istockphoto.com

8.4 3 3 Systems of Equations

■ A system of equations can be solved to determine how much of each solution should be mixed to obtain the desired strength and amount of the final solution.

I

n Chapter 7 we studied linear equations in two variables. Now in Chapter 8 we will continue that work by considering two or more linear equations at the same time. The equations considered together form a system of equations. In this chapter we will discuss various techniques for solving systems of equations and also consider systems of linear inequalities. Systems of equations can readily be applied to solve many of the word problem types we’ve encountered in previous chapters. Many students find that using two variables and a system of equations to solve a word problem is easier than using only one variable. Although some of the word problems presented in this chapter could be solved without a system of equations, we encourage you to expand your problem-solving abilities by using a system of equations to solve them.


391

I N T E R N E T

P R O J E C T

In this chapter three methods for solving systems of equations are presented. Another method for solving systems of equations is Gaussian elimination and uses matrices. Do an Internet search on matrices in mathematics to find a definition of a matrix. What ancient civilizations were among the first to use matrices?

8.1

Solving Linear Systems by Graphing OBJECTIVES 1

Determine If an Ordered Pair Is a Solution of a System of Equations

2

Solve Linear Systems of Equations by Graphing

3

Solve Linear Systems of Inequalities by Graphing

1 Determine If an Ordered Pair Is a Solution of a System of Equations Suppose we graph x 2y 4 and x 2y 8 on the same set of axes, as shown in Figure 8.1. The ordered pair (6, 1), which is associated with the point of intersection of the two lines, satisfies both equations. That is to say, (6, 1) is the solution for x 2y 4 and x 2y 8. To check this, we can substitute 6 for x and 1 for y in both equations. x 2y 4

becomes 6 2(1) 4

x 2y 8

becomes 6 2(1) 8

Thus we say that 516, 12 6 is the solution set of the system a

EXAMPLE 1

y x + 2y = 8 (6, 1) x x − 2y = 4

Figure 8.1

x 2y 4 b x 2y 8

Determine if the ordered pair (2, 1) is a solution of the system of equations 3x 4y 5 a b. 3x 4y 6

Solution Substitute 2 for x and 1 for y in both equations. 3x 4y 5 becomes 3(2) 1 5

a true statement

3x 4y 6 becomes 2 4(1) 6

a false statement

Because the ordered pair does not satisfy both equations, it is not a solution of the system of equations. 392

8.1

Solving Linear Systems by Graphing

393

▼ PRACTICE YOUR SKILL Determine if the ordered pair (3, 1) is a solution of the system of equations 2x y 5 ■ a b. 5x 2y 17

2 Solve Linear Systems of Equations by Graphing Two or more linear equations in two variables considered together are called a system of linear equations. Here are three systems of linear equations: a

x 2y 4 b x 2y 8

a

5x 3y 19 b 3x 7y 12

4x 2y 15 ° 2x 2y 19 ¢ 7x 2y 13

To solve a system of linear equations means to find all of the ordered pairs that are solutions of all of the equations in the system. There are several techniques for solving systems of linear equations. We will use three of them in this chapter—a graphing method in this section and two other methods in the following sections. To solve a system of linear equations by graphing, we proceed as in the opening discussion of this section. We graph the equations on the same set of axes, and then the ordered pairs associated with any points of intersection are the solutions to the system. Let’s consider another example.

EXAMPLE 2

Solve the system a

x 2y 5 b. x 2y 4

Solution We can find the intercepts and a check point for each of the lines.

x 2y 4

xy5 x

y

0 5 2

5 0 3

Intercepts Check point

x

y

0 4 2

5 0 1

Intercepts Check point

Figure 8.2 shows the graphs of the two equations. It appears that (2, 3) is the solution of the system. To check it we can substitute 2 for x and 3 for y in both equations. y x+y=5

(2, 3) x − 2y = − 4 x

Figure 8.2

394

Chapter 8


xy

becomes 2 3 5

x 2y 4

A true statement

becomes 2 2(3) 4

A true statement

Therefore, 512, 32 6 is the solution set.

▼ PRACTICE YOUR SKILL Solve the system a

x 2y 2 b. x 2y 6

■

It should be evident that solving systems of equations by graphing requires accurate graphs. In fact, unless the solutions are integers it is really quite difficult to obtain exact solutions from a graph. For this reason the systems in this section have integer solutions. Furthermore, checking a solution takes on additional significance when the graphing approach is used. By checking you can be absolutely sure that you are reading the correct solution from the graph. Figure 8.3 shows the three possible cases for the graph of a system of two linear equations in two variables.

Case I

The graphs of the two equations are two lines intersecting at one point. There is one solution, and we call the system a consistent system.

Case II

The graphs of the two equations are parallel lines. There is no solution, and we call the system an inconsistent system.

Case III The graphs of the two equations are the same line. There are infinitely many solutions to the system. Any pair of real numbers that satisfies one of the equations also satisfies the other equation, and we say the equations are dependent.

y

y

x

Case I

y

x

Case II

x

Case III

Figure 8.3

Thus as we solve a system of two linear equations in two variables, we know what to expect. The system will have no solutions, one ordered pair as a solution, or infinitely many ordered pairs as solutions. (Most of the systems that we will be working with in this text have one solution.) An example of case I was given in Example 2 (Figure 8.2). The next two examples illustrate the other cases.

8.1

EXAMPLE 3

Solve the system a


395

2x 3y 16 b. 2x 3y 12 y

Solution 2x 3y 6

2x 3y 12

x

y

x

y

0 3 3

2 0 4

0 6 3

4 0 2

2 x + 3y = 12

x 2 x + 3y = 6

Figure 8.4 shows the graph of the system. Since the lines are parallel, there is no solution to the system. The solution set is . Figure 8.4


EXAMPLE 4

Solve the system a

3x y 6 b. 3x y 0

Solve the system a

2x 2y 3 b. 2x 2y 6

Solution xy3

■

y

2x 2y 6

x

y

x

y

0 3 1

3 0 2

0 3 1

3 0 2

x+y=3 x 2x + 2y = 6

Figure 8.5 shows the graph of this system. Since the graphs of both equations are the same line, there are infinitely many solutions to the system. Any ordered pair of real numbers that satisfies one equation also satisfies the other equation.

Figure 8.5


2x 4y 8 b. 2x 2y 4

■

3 Solve Linear Systems of Inequalities by Graphing You now have the skills to use a graphing approach to solve a system of linear inequalities. For example, the solution set of a system of linear inequalities, such as a

xy 1 b xy 1

396

Chapter 8


is the intersection of the solution sets of the individual inequalities. In Figure 8.6(a) we indicated the solution set for x y 1, and in Figure 8.6(b) we indicated the solution set for x y 1. Then in Figure 8.6(c) we shaded the region that represents the intersection of the two shaded regions in parts (a) and (b), which is the solution of the given system. The shaded region in Figure 8.6(c) includes all points that are below the line x y 1 and also that are below the line x y 1. y

y

x

y

x

(a)

x

(b)

(c)

Figure 8.6

Let’s solve another system of linear inequalities.

EXAMPLE 5

Solve the system a

2x 3y 6 b. 2x 4y 4

Solution Let’s first graph the individual inequalities. The solution set for 2x 3y 6 is shown in Figure 8.7(a), and the solution set for x 4y 4 is shown in Figure 8.7(b). (Note the solid line in part (a) and the dashed line in part (b).) Then in Figure 8.7(c) we shaded the intersection of the graphs in parts (a) and (b). Thus we represent the solution set for the given system by the shaded region in Figure 8.7(c). This region includes all points that are on or below the line 2x 3y 6 and also that are above the line x 4y 4. y

y

y

x

(a)

x

(b)

x

(c)

Figure 8.7


2x 3y 5 b. 2x 3y 6

■

Remark: Remember that the shaded region in Figure 8.7(c) represents the solution set of the given system. Parts (a) and (b) were drawn only to help determine the final shaded region. With some practice, you will be able to go directly to part (c) without actually sketching the graphs of the individual inequalities.

8.1

CONCEPT QUIZ


397

For Problems 1–7, answer true or false. 1. To solve a system of equations means to find an ordered pair that satisfies any of the equations in the system of equations. 2. A consistent system of equations will have more than one solution. 3. If the graph of a system of two linear equations results in two distinct parallel lines, then the system has no solutions. 4. Every system of equations has a solution. 5. If the graphs of the two equations in a system are the same line, then the equations in the system are dependent. 6. To solve a system of two equations in variables x and y, it is sufficient to just find a value for x. 2x y 2 7. The ordered pair (1, 4) satisfies the system of linear inequalities a b. 2x y 3

Problem Set 8.1 1 Determine If an Ordered Pair Is a Solution of a System of Equations

13. a

2x 2y 4 b 2x 2y 3

14. a

2x y 8 b 2x y 2

For Problems 1–10, decide whether the given ordered pair is a solution of the given system of equations.

15. a

x 3y 6 b x 3y 3

16. a

y 2x b y 3x 0

1. a

5x 2y 9 b, 11, 42 3x 2y 4

17. a

xy0 b xy0

18. a

3x y 3 b 3x y 3

2. a

3x 2y 8 b, 12, 12 2x 2y 3

19. a

3x 2y 5 b 2x 5y 3

20. a

2x 3y 1 b 4x 3y 7

3. a

x 3y 17 b, 12, 52 2x 5y 21

21. a

22. a

y 2x 5 b x 3y 6

4. a

2x 7y 5 b, 11, 12 4x 7y 6

y 2x 3 b 6x 3y 9

23. a

24. a

5. a

y 2x b, 11, 22 3x 4y 5

y 5x 2 b 4x 3y 13

yx2 b 2x 2y 4

25. a

26. a

6. a

x 2y 8 b, 12, 32 x 2y 8

y 4 2x b y 7 3x

y 3x 4 b y 5x 8

y 2x b 3x 2y 2

28. a

7. a

6x 5y 5 b, 10, 12 3x 4y 4

27. a

y 3x b 4x 3y 5

8. a

y 3x 1 b, 13, 82 5x 2y 1

29. a

7x 2y 8 b x 2

30. a

3x 8y 1 b y 2

3x 3y 24 9. a b, 14, 52 2x 3y 23 10. a

5x 4y 15 b, 13, 02 2x 7y 16

3 Solve Linear Systems of Inequalities by Graphing

2 Solve Linear Systems of Equations by Graphing For Problems 11–30, use the graphing method to solve each system. 11. a

xy1 b xy3

12. a

x y 2 b x y 4

For Problems 31– 40, indicate the solution set for each system of linear inequalities by shading the appropriate region. 31. a

3x 2y 4 b 3x y 3

32. a

3x 3y 3 b 3x y 3

33. a

4x 3y 12 b 4x y 4

34. a

y 2x b y 2x

398

Chapter 8

35. a

y x b y 3x

Systems of Equations 36. a

y x2 b 37. a y x3

y x 1 b y x 1

1 y x2 2 39. ± ≤ 1 y x1 2

1 y x2 2 ≤ 38. ± 1 y x1 2

40. a

2x 3y 6 b x y 2

THOUGHTS INTO WORDS 41. Discuss the strengths and weaknesses of solving a system of linear equations by graphing.

43. Is it possible for a system of two linear equations to have exactly two solutions? Defend your answer.

42. Determine a system of two linear equations for which the solution is (5, 7). Do any other systems have the same solution set? If so, find at least one more system.

Answers to the Example Practice Skills 1. Yes

2. {(2, 4)} 3.

4. Infinitely many 5.

y

x


8.2

2. False 3. True

4. False

5. True

6. False 7. False

Elimination-by-Addition Method OBJECTIVES 1

Use Elimination-by-Addition to Solve a System of Equations

2

Solve Word Problems Using a System of Equations

1 Use Elimination-by-Addition to Solve a System of Equations We have used the addition property of equality (if a b, then a c b c) to help solve equations that contain one variable. An extension of the addition property forms the basis for another method of solving systems of linear equations. Property 8.1 states that when we add two equations, the resulting equation will be equivalent to the original ones. We can use this property to help solve a system of equations.

8.2

Elimination-by-Addition Method

399

Property 8.1 For all real numbers a, b, c, and d, if a b and c d, then acbd

EXAMPLE 1

Solve a

x y 12 b. x y 12

Solution 2x y 12 2x y 12 2x y 14

This is the result of adding the two equations

Solving this new equation in one variable, we obtain 2x 14 x7 Now we can substitute the value of 7 for x in one of the original equations. Thus x y 12 7 y 12 y5 To check, we see that x y 12

and

xy2

7 5 12

and

752

Thus the solution set of the system is 517, 526.

▼ PRACTICE YOUR SKILL Solve a

2x y 6 b. xy6

■

Note in Example 1 that by adding the two original equations, we obtained a simple equation that contains only one variable. Adding equations to eliminate a variable is the key idea behind the elimination-by-addition method for solving systems of linear equations. The next example further illustrates this point.

EXAMPLE 2

Solve a

2x 3y 26 b. 4x 3y 22

Solution 2x 3y 26 4x 3y 22 6x 3y 24 x 4

Added the two equations

400

Chapter 8


We substitute 4 for x in one of the two original equations. 2x 3y 26 2142 3y 26 8 3y 26 3y 18 y 6

The solution set is 514, 62 6.


x 2y 17 b. 3x 2y 13

■

It may be necessary to change the form of one, or perhaps both, of the original equations before adding them. The next example demonstrates this idea.

EXAMPLE 3

Solve a

y x 21 b. x y 3

Solution y x 21

Subtract x from both sides

x y 3 Leave alone

x y 21 x y 13 2y 24 y 12

We substitute 12 for y in one of the original equations. x y 3

x 1122 3 x9



yx7 b. x 2y 5

■

Frequently, the multiplication property of equality needs to be applied first so that adding the equations will eliminate a variable, as the next example illustrates.

EXAMPLE 4

Solve a

2x 5y 29 b. 3x 5y 21

Solution Notice that adding the equations as they are would not eliminate a variable. However, we observe that multiplying the bottom equation by 5 and then adding this newly formed, but equivalent, equation to the top equation will eliminate the y variable terms.

8.2


2x 5y 29

Leave alone

3x 5y 21


401

2x 5y 29 15x 5y 25 17x 5y 34 x2

We substitute 2 for x in one of the original equations. 3x y 1 3122 y 1 6y1 y 5 y5



3x 2y 3 b. 4x 2y 18

■

Notice in these problems that after finding the value of one of the variables, we substitute that number into one of the original equations to find the value of the other variable. It doesn’t matter which of the two original equations you use, so pick the easiest one to solve. Sometimes we need to apply the multiplication property to both equations. Let’s look at an example of this type.

EXAMPLE 5

Solve a

2x 3y 14 b. 9x 2y 13

Solution A If we want to eliminate the x variable terms, then we want their coefficients to be equal in number (absolute value) and opposite in sign. To eliminate the x variable terms we can multiply the top equation by 9 and multiply the bottom equation by 2. 2x 3y 14 Multiply both sides by 9

18x 27y 36

9x 2y 13 Multiply both sides by 2

18x 24y 26 31y 62 y2

We substitute 2 for y in one of the original equations. 2x 3y 4 2x 3122 4 2x 6 4 2x 2 x 1

The solution set is 511, 22 6 .

402

Chapter 8


Solution B If we want to eliminate the y variable terms, then we want their coefficients to be equal in number (absolute value) and opposite in sign. To eliminate the y variable terms we can multiply the top equation by 2 and multiply the bottom equation by 3. 2x 3y 14 Multiply both sides by 2

24x 6y 38


27x 6y 39 31x

31 x 1

We substitute 1 for x in one of the original equations. 2x 3y 4 2112 3y 4 2 3y 4 3y 6 y2



3x 2y 14 b. 4x 5y 13

■

Look carefully at Solutions A and B for Example 5. Especially notice the first steps, where we applied the multiplication property of equality to the two equations. In Solution A we multiplied by numbers so that adding the resulting equations eliminated the x variable. In Solution B we multiplied so that the y variable was eliminated when we added the resulting equations. Either approach will work; pick the one that involves the easiest computation.

EXAMPLE 6

Solve a

3x 4y 7 b. 5x 3y 9

Solution Let’s eliminate the y variable terms by multiplying the top equation by 3 and the bottom equation by 4. 3x 4y 7 Multiply both sides by 3

29x 12y 21


20x 12y 36 29x

57 x

57 29

57 for x in one of the original equations will produce some messy 29 calculations, let’s solve for y by eliminating the x variable terms.

Since substituting

8.2



15x 20y 35


15x 29y 27

403

29y 8 y

The solution set is e a

8 29

57 8 , b f. 29 29


5x 2y 13 b. 2x 3y 10

■

2 Solve Word Problems Using a System of Equations Many word problems that we solved earlier in this text using one equation in one variable can also be solved using a system of two linear equations in two variables. In fact, many times you may find that it seems natural to use two variables. It may also seem more meaningful at times to use variables other than x and y. Let’s consider two examples.

EXAMPLE 7

Apply Your Skill The difference of two numbers is 9. If four times the smaller number is subtracted from three times the larger number, the result is 21. Find the numbers.

Solution We let x represent the larger number and y represent the smaller number. The problem translates into the following equations. x 2y 29 3x 4y 21

The difference of two numbers is nine Four times the smaller is subtracted from 3 times the larger

Solving this system by the elimination-by-addition method, we obtain 3x 2y 29 Multiply both sides by 3 3x 4y 21 Leave alone

3x 3y 27 3x 4y 21 y 26 y 16

Substitute 6 for y in one of the original equations. xy9 x69 x 15 Therefore, the larger number is 15 and the smaller number is 6.

▼ PRACTICE YOUR SKILL The sum of two numbers is 31. If twice the smaller number is subtracted from the larger number, the result is 1. Find the numbers. ■

404

Chapter 8


EXAMPLE 8

Apply Your Skill The cost of 3 tennis balls and 2 golf balls is $7. Furthermore, the cost of 6 tennis balls and 3 golf balls is $12. Find the cost of 1 tennis ball and the cost of 1 golf ball.

Koji Aoki /Photolibrary

Solution We can use t to represent the cost of one tennis ball and g the cost of one golf ball. The problem translates into the following system of equations. 3t 2g 7

The cost of 3 tennis balls and 2 golf balls is $7

6t 3g 12

The cost of 6 tennis balls and 3 golf balls is $12

Solving this system by the elimination-by-addition method, we obtain 3t 2g 7 Multiply both sides by 2

6t 4g 14

6t 3g 12 Leave alone

6t 3g 12 g 2 g2

We substitute 2 for g in one of the original equations. 3t 2g 7 3t 2122 7 3t 4 7 3t 3 t1 The cost of a tennis ball is $1 and the cost of a golf ball is $2.

▼ PRACTICE YOUR SKILL The cost of 2 pizzas and 3 baskets of wings is $29. Also, the cost of 5 pizzas and 2 baskets of wings is $45. Find the cost of one pizza and the cost of a basket of wings. ■ Before you tackle the word problems in this next problem set, it might be helpful to review the problem-solving suggestions we offered in Section 2.4. Those suggestions continue to apply here except that now you have the flexibility of using two equations and two unknowns. Don’t forget that to check a word problem you need to see whether your answers satisfy the conditions stated in the original problem.

CONCEPT QUIZ

For Problems 1– 6, answer true or false. 1. When two equations are added, the result is an equation that is equivalent to the original equations. 2. The objective of the elimination-by-addition method is to produce an equivalent equation in only one variable. 3. When solving a system of equations using the elimination-by-addition method, eliminate the x variable so that the x coefficients are equal in number (absolute value) and the same in sign. x 2y 5 4. For the system of equations a b , the y variable can be eliminated by x 2y 3 just adding the equations.

8.2


405

xy5 b , after determining that x 4 xy3 you can find the value of y by substituting x 4 into either equation. a 2b 5 6. When solving the system of equations a b , you must solve for the a 2b 3 variable a first because it is the first variable in alphabetic order.

5. When solving the system of equations a

Problem Set 8.2 1 Use Elimination-by-Addition to Solve a System of Equations For Problems 1–24, solve each system using the eliminationby-addition method.

2 Solve Word Problems Using a System of Equations For Problems 25 –38, solve each problem by setting up and solving a system of two linear equations in two variables.

1. a

x y 14 b x y 2

2. a

x y 14 b x y 16

25. The sum of two numbers is 30, and their difference is 12. Find the numbers.

3. a

3x 4y 21 b 3x 4y 21

4. a

3x 2y 21 b 3x 7y 36

26. Suppose that the sum of two numbers is 46, and the difference of the numbers is 22. Find the numbers.

5. a

y6x b x y 18

6. a

x y 10 b xy6

27. The sum of two numbers is 21. If three times the smaller number is subtracted from the larger number the result is 5. Find the numbers.

7. a

5x 2y 23 b 3x 2y 19

8. a

4x 5y 36 b 4x 2y 30

9. a

3x 2y 5 b 3x 2y 6

10. a

2x 5y 3 b 2x 5y 7

y x 11. a b 2x y 2

3x 2y 12. a b 8x 20y 19

28. The sum of two numbers is 20. If twice the smaller is subtracted from the larger, the result is 2. Find the numbers. 29. The difference of two numbers is 7. If three times the smaller is subtracted from twice the larger, the result is 6. Find the numbers. 30. The difference of two numbers is 17. If the larger is increased by three times the smaller, the result is 37. Find the numbers.

13. a

4x 5y 59 b 5x 6y 50

14. a

2x 7y 46 b 3x 4y 18

15. a

9x 7y 29 b 5x 3y 17

16. a

7x 5y 6 b 4x 3y 4

17. a

6x 5y 6 b 8x 3y 21

18. a

3x 8y 23 b 7x 4y 48

19. a

2x 7y 1 b 9x 4y 2

20. a

21. a

0.07x 0.08y 750 b 0.07x 0.08y 757.5

22. a

0.06x 0.09y 700 b 0.06x 0.09y 754

23. a

0.09x 0.11y 31 b y x 100

34. Chip used two sites and downloaded 19 songs for his MP3 player for a total cost of $20.00. The songs from Muchmore Music cost $1.00 each, and the songs from iTunes cost $1.20 each. How many songs did Chip download from each service?

24. a

0.08x 0.1y 56 b y 2x

35. Suppose that a library buys a total of 35 books, which cost $546. The hardcover books cost $22 each, and the

6x 2y 3 b 5x 9y 1

31. One number is twice another number. The sum of three times the smaller and five times the larger is 78. Find the numbers. 32. One number is three times another number. The sum of four times the smaller and seven times the larger is 175. Find the numbers. 33. Three lemons and 2 apples cost $2.35. Two lemons and 3 apples cost $2.40. Find the cost of 1 lemon and the cost of 1 apple.

406

Chapter 8


softcover books cost $6 per book. How many books of each type did the library buy?

were rented one day for $1600, how many of each kind were rented?

36. The income from a student production was $101,000. The price of a student ticket was $13, and nonstudent tickets were $25 each. Five thousand people attended the production. How many tickets of each kind were sold?

38. The total receipts from ticket sales at a concert amounted to $15,600. Student tickets were sold at $24 each and nonstudent tickets at $36 each. The number of student tickets sold was five times the number of nonstudent tickets sold. How many student tickets and how many nonstudent tickets were sold?

37. Suppose that a youth hostel rents double rooms at $35 per day and single rooms at $25 per day. If a total of 50 rooms

THOUGHTS INTO WORDS 39. Explain how you would solve the system a

2x 3y 5 b 4x 7y 9

using the elimination-by-addition method.

40. Give a general description of how to apply the elimination-by-addition method.

FURTHER INVESTIGATIONS For Problems 41– 44, solve each system using the eliminationby-addition method. 1 x 2 41. ± 3 x 2

1 y 2 3 ≤ 2 y 34 3

42. a

43. a

2x 4y 6 b 2x 8y 3

44. a

y 1 2x b 6x 3y 3

3x 2y 0 b 3x 5y 0


59 24 1. {(4, 2)} 2. {(5, 1)} 3. {(3, 4)} 4. {(3, 6)} 5. {(2, 1)} 6. b a , b r 19 19 8. Pizza $7 and wings $5

7. 21 and 10


8.3

2. True 3. False

4. False 5. True

6. False

Substitution Method OBJECTIVES 1

Use the Substitution Method to Solve a System of Equations

2

Determine Which Method to Use to Solve a System of Equations

3

Solve Mixture, Interest, and Geometry Word Problems Using a System of Equations

1 Use the Substitution Method to Solve a System of Equations A third method of solving systems of equations is called the substitution method. Like the addition method, it produces exact solutions and can be used on any system of linear equations; however, some systems lend themselves more to the substitution method than others. We will consider a few examples to demonstrate the use of the substitution method.

8.3

EXAMPLE 1

Solve a

Substitution Method

407

y x 10 b. x y 14

Solution Because the first equation states that y equals x 10, we can substitute x 10 for y in the second equation. x y 14 Substitute x 10 for y x (x 10) 14 Now we have an equation with one variable that can be solved in the usual way. x 1x 102 14 2x 10 14 2x 4 x2 Substituting 2 for x in one of the original equations, we can find the value of y. y x 10 y 2 10 y 12

The solution set is 512, 1226.


EXAMPLE 2

Solve a

x y 11 b. x 2y 10

Solve a

3x 5y 7 b. x 2y 5

■

Solution Because the second equation states that x equals 2y 5, we can substitute 2y 5 for x in the first equation. 3x 5y 7 Substitute 2y 5 for x 3(2y 5) 5y 7 Solving this equation, we have 312y 52 5y 7 6y 15 5y 7 11y 15 7 11y 22 y 2 Substituting 2 for y in one of the two original equations produces x 2y 5 x 2122 5 x 4 5 x1


408

Chapter 8



4x 3y 15 b. y 2x 5

■

Note that the key idea behind the substitution method is the elimination of a variable, but the elimination is done by a substitution rather than by addition of the equations. The substitution method is especially convenient to use when at least one of the equations is of the form y equals or x equals. In Example 1 the first equation is of the form y equals, and in Example 2 the second equation is of the form x equals. Let’s consider another example using the substitution method.

EXAMPLE 3

Solve

°

2x 3y 30 . ¢ 2 y x6 3

Solution 2 The second equation allows us to substitute x 6 for y in the first equation. 3 2 3

2x 3y 30 Substitute x 6 for y

2 2x 3 a x 6b 30 3

Solving this equation produces 2 2x 3 a x 6b 30 3 2x 2x 18 30 4x 18 30 4x 12 x 3 Now we can substitute 3 for x in one of the original equations. 2 y x6 3 y

2 132 6 3

y 2 6 y 8



°

x 4y 22 . ¢ 1 y x4 2

■

It may be necessary to change the form of one of the equations before we make a substitution. The following examples clarify this point.

8.3

EXAMPLE 4

Solve a

Substitution Method

409

4x 5y 55 b. 4x 5y 2

Solution We can easily change the form of the second equation to make it ready for the substitution method. x y 2 y 2 x

Added x to both sides

Now, we can substitute 2 x for y in the first equation. 4x 5y 55 Substitute 2 x for y

4x 5(2 x) 55

Solving this equation, we obtain 4x 512 x2 55 4x 10 5x 55 9x 10 55 9x 45 x5 Substituting 5 for x in one of the original equations produces x y 2 5 y 2 y 7



2x 3y 4 b. 2x 3y 1

■

In Example 4, we could have started by changing the form of the first equation to make it ready for substitution. However, you should be able to look ahead and see that this would produce a fractional form to substitute. We were able to avoid any messy calculations with fractions by changing the form of the second equation instead of the first. Sometimes when using the substitution method, you cannot avoid fractional forms. The next example is a case in point.

EXAMPLE 5

Solve a

3x 2y 38 b. 2x 3y 38

Solution Looking ahead, we see that changing the form of either equation will produce a fractional form. Therefore, we will merely pick the first equation and solve for y. 3x 2y 8 2y 8 3x y

8 3x 2

Added 3x to both sides Multiplied both sides by

1 2

410

Chapter 8


Now we can substitute

8 3x for y in the second equation and determine the 2

value of x. 2x 3y 38 2x 3 a

8 3x b 38 2

2x

24 9x 38 2

4x 24 9x 76


13x 24 76 13x 52 x 4 Substituting 4 for x in one of the original equations, we have 3x 2y 8 3142 2y 8 12 2y 8 2y 20 y 10



2x 3y 18 b. 3x 2y 14

■

2 Determine Which Method to Use to Solve a System of Equations We have now studied three methods of solving systems of linear equations—the graphing method, the elimination-by-addition method, and the substitution method. As we indicated earlier, the graphing method is quite restrictive and works well only when the solutions are integers or when we need only approximate answers. Both the elimination-by-addition method and the substitution method can be used to obtain exact solutions for any system of linear equations in two variables. The method you choose may depend upon the original form of the equations. Next we consider two examples to illustrate this point.

EXAMPLE 6

Solve a

7x 5y 52 b. y 3x 4

Solution Because the second equation indicates that we can substitute 3x 4 for y, this system lends itself to the substitution method. 7x 5y 52 Substitute 3x 4 for y 7x 5(3x 4) 52

8.3

Substitution Method

411

Solving this equation, we obtain 7x 513x 42 52 7x 15x 20 52 8x 20 52 8x 72 x9 Substituting 9 for x in one of the original equations produces y 3x 4 y 3192 4 y 27 4 y 23

The solution set is 519, 2326.


EXAMPLE 7

Solve a

x 2y 6 b. 4x 3y 16

Solve a

10x 7y 19 b. 12x 6y 11

■

Solution Because changing the form of either of the two equations in preparation for the substitution method would produce a fractional form, we are probably better off using the elimination-by-addition method. Furthermore, notice that the coefficients of x lend themselves to this method. 10x 7y 19 Leave alone

10x 37y 19

2x 6y 11 Multiply by 5 10x 30y 55 37y 74 y2 Substituting 2 for y in the first equation of the given system produces 10x 7y 19 10x 7122 19 10x 14 19 10x 5 x

5 1 10 2

1 The solution set is e a , 2b f . 2


9x 5y 15 b. 3x 2y 27

■

412

Chapter 8


In Section 8.1, we explained that you can tell by graphing the equations whether the system has no solutions, one solution, or infinitely many solutions. That is, the two lines may be parallel (no solutions), they may intersect in one point (one solution), or they may coincide (infinitely many solutions). From a practical viewpoint, the systems that have one solution deserve most of our attention. However, we do need to be able to deal with the other situations as they arise. The next two examples demonstrate what occurs when we hit a “no solution” or “infinitely many solutions” situation when we are using either the elimination-by-addition or substitution method.

EXAMPLE 8

Solve the system a

y 2x 1 b. 6x 3y 7

Solution Because the first equation indicates that we can substitute 2x 1 for y, this system lends itself to the substitution method. 6x 3y 7 Substitute 2x 1 for y 6x 3(2x 1) 7 Now we solve this equation. 6x 6x 3 7 037 37 The false numerical statement, 3 7, implies that the system has no solutions. Thus the solution set is . (You may want to graph the two lines to verify this conclusion!)


EXAMPLE 9

y 3x 2 b. 3x y 4

Solve the system a

■

12x 13y 24 b. 10x 15y 20

Solution We use the elimination-by-addition method. 12x 13y 24 Multiply both sides by 5

10x 15y 20

10x 15y 20 Leave alone

10x 15y 20 0 0 20

The true numerical statement, 0 0 0, implies that the system has infinitely many solutions. Any ordered pair that satisfies one of the equations also satisfies the other equation.


3x 2y 34 b. 3x 6y 12

■

8.3

Substitution Method

413

3 Solve Mixture, Interest, and Geometry Word Problems Using a System of Equations We will conclude this section with three word problems.

EXAMPLE 10

Apply Your Skill Niki invested $5000, part of it at 4% interest and the rest at 6%. Her total interest earned for a year was $265. How much did she invest at each rate?

Solution We let x represent the amount invested at 4%, and we let y represent the amount invested at 6%. The problem translates into the following system: x y 5000 0.04x 0.06y 265

Niki invested $5000 Her total interest earned for a year was $265

Multiplying the second equation by 100 produces 4x 6y 26,500. Then we have the following equivalent system to solve: a

4x 6y 2,5000 b 4x 6y 26,500

Using the elimination-by-addition method, we can proceed as follows: x y

5000

4x 6y 26,500

Multiply both sides by 4 4x 4y 20,000 Leave alone

4x 6y 26,500 2y

6500

y

3250

Substituting 3250 for y in x y 5000 yields x y 5000 x 3250 5000 x 1750 Therefore, we know that Niki invested $1750 at 4% interest and $3250 at 6%.

▼ PRACTICE YOUR SKILL Simoni invested $7000 in two accounts. One account paid 5% interest and the other account paid 7% interest. Her total interest earned for the year was $440. How much did she invest at each rate? ■

EXAMPLE 11

Apply Your Skill A 25% chlorine solution is to be mixed with a 40% chlorine solution to produce 12 gallons of a 35% chlorine solution. How many gallons of each solution should be mixed?

Solution Let x represent the gallons of 25% chlorine solution, and let y represent the gallons of 40% chlorine solution. Then one equation of the system will be x y 12. For the other equation we need to multiply the number of gallons of each solution by its percentage of chlorine, which gives us the equation 0.25x 0.40y 0.35(12). So we need to solve the following system: a

0.25x 0.40y 12 b 0.25x 0.40y 00.351122

414

Chapter 8


Let’s simplify the second equation by finding the product of 0.35(12) and multiplying both sides of the equation by 100. Then the system becomes a

25x 40y 412 b 25x 40y 420

Using the elimination-by-addition method, we can proceed as follows: 25x 40y 412


25x 40y 420

Leave alone

25x 25y 300 25x 40y 420 15y 120 y 128

Substituting 8 for y in x y 12 yields x y 12 x 8 12 x4 Therefore, we need 4 gallons of the 25% chlorine solution and 8 gallons of the 40% chlorine solution to make 12 gallons of 35% chlorine solution.

▼ PRACTICE YOUR SKILL A 6% saline solution is to be mixed with a 12% saline solution to produce 30 liters of a 10% saline solution. How many liters of each solution should be mixed? ■

EXAMPLE 12

Apply Your Skill The length of a rectangle is 1 centimeter less than three times the width. The perimeter of the rectangle is 94 centimeters. Find the length and width of the rectangle.

Solution We let w represent the width of the rectangle, and we let l represent the length of the rectangle (see Figure 8.8). The problem translates into the following system of equations: l 3w 1

The length of the rectangle is 1 centimeter less than three times the width

2l 2w 94

The perimeter of the rectangle is 94 centimeters

l

w

w

l Figure 8.8

Multiplying both sides of the second equation by one-half produces the equivalent equation l w 47, so we have the following system to solve: a

l 3w 1 b l w 47

8.3

Substitution Method

415

The first equation indicates that we can substitute 3w 1 for l in the second equation. l w 47 Substitute 3w 1 for l 3w 1 w 47 Solving this equation yields 3w 1 w 47 4w 1 47 4w 48 w 12 Substituting 12 for w in one of the original equations produces l 3w 1 l 31122 1 l 36 1 l 35 The rectangle is 12 centimeters wide and 35 centimeters long.

▼ PRACTICE YOUR SKILL The length of a rectangle is 6 inches longer than twice the width. The perimeter of the rectangle is 168 inches. Find the length and width of the rectangle. ■

CONCEPT QUIZ

For Problems 1–7, answer true or false. 1. Every system of equations can be solved by the substitution method. 2. The substitution method eliminates a variable by substitution rather than by adding the equations. 3. To use the substitution method, one of the equations has to be solved for x. 4. The graphing method for solving a system of equations is practical for finding an exact solution. 5. Both the elimination-by-addition method and the substitution method will find exact solutions. 6. When solving a system of equations, obtaining a false numerical equation such as 1 5 implies that the system has infinitely many solutions. 7. When solving a system of equations, obtaining a true numerical equation such as 3 3 implies that any ordered pair that satisfies one equation also satisfies the other equation.

Problem Set 8.3 1 Use the Substitution Method to Solve a System of Equations For Problems 1–26, solve each system using the substitution method.

7. a

3x 6y 1 b 3x 6y 7

°

2x y 12 ¢ 3 x y 4

°

3 y x ¢ 2 6x 5y 15

9.

y 2x 1 b x y 14

2. a

3. a

x y 14 b y 3x 2

4. a

x y 23 b y 2x 5

11.

5. a

4x 3y 6 b y 2x 7

6. a

8x y 8 b y 4x 5

13. a

1. a

y 3x 4 b x y 52

4y 1y x b 2x 8y 3

8. a 10.

12.

2x 4y 9 b 2x 4y 3

°

4x 5y 6 ¢ 2 y x 3

°

3 x y 5 ¢ 4x 3y 12

14. a

y 5x 2 b y 2x 7

416

Chapter 8


15. a

7x 2y 2 b 6x 5y 18

16. a

2x 3y 31 b 3x 5y 20

17. a

8x 3y 19 b 8x 5y 71

18. a

9x 2y 18 b 4x 2y 17

19. a

4x 6y 1 b 2x 3y 4

20. a

2x 3y 1 b 4x 6y 2

21. a

5x 7y 3 b 3x 2y 0

22. a

7x 3y 4 b 2x 5y 0

23. a

0.05x 0.07y 33 b y x 300

24. a

0.06x 0.08y 15 b y x 100

25. a

0.05x 0.1y 1.13 b 0.05x 0.1y 1.15

26. a

0.1x 0.25y .17 b 0.1x 0.25y 3.2

For Problems 47– 64, solve each problem by setting up and solving a system of two linear equations in two variables. Use either the elimination-by-addition or the substitution method to solve the systems. 47. The sum of two numbers is 34. The larger number is 4 more than twice the smaller number. Find the numbers. 48. The sum of two numbers is 52. The larger number is 2 more than four times the smaller number. Find the numbers. 49. In a class of 50 students, the number of women is 2 more than five times the number of men. How many women are there in the class? 50. In a recent survey, 1000 registered voters were asked about their political preferences. The number of men in the survey was 5 less than one-half the number of women. Find the number of men in the survey.

2 Determine Which Method to Use to Solve a System of Equations For Problems 27– 46, solve each system using either the elimination-by-addition or the substitution method, whichever seems more appropriate to you. 27. a

5x 4y 14 b 7x 3y 32

28. a

2x 3y 13 b 3x 5y 28

29. a

2x 9y 6 b y x

30. a

y 3x 2 b 4x 3y 21

0.6x 0.5y 22 b 31. a 0.6x 0.5y 12 32. a

0.4x 0.5y 13 b 0.4x 0.5y 16

33. a

4x 2y 0 b 7x 2y 9

34. a

5x 2y 10 b 5x 8y 12

35. a

2x 7y 51 b 6x 7y 57

36. a

7x 9y 11 b x 3y 13

37. a

16x 2y 1 b 10x 2y 13

38. a

6x 4y 5 b 6x 5y 1

39. a

4x 8y 20 b 4x 2y 25

40. a

2x 15y 5 b 2x 10y 2

41. a

3x 8y 5 b x 2y

42. a

5x 6y 3 b 5x 6y 22

43. a

5y 2x 4 b 10y 3x 4

44. a

3x 7 2y 15 b 3x 6y 18

x y 1 b 45. a 6x 5y 4

3 Solve Mixture, Interest, and Geometry Word Problems Using a System of Equations

4x 7y 1 b 46. a y 2x 3

51. Sidney invested $13,000, part of it at 5% and the rest at 6%. If his total yearly interest was $730, how much did he invest at each rate? 52. Heather invested $1100, part of it at 8% and the rest at 9%. Her total interest earned for a year was $95. How much did she invest at each rate? 53. Tina invested some money at 8% interest and some money at 9%. She invested $250 more at 9% than she invested at 8%. Her total yearly interest from the two investments was $48. How much did Tina invest at each rate? 54. George invested some money at 4% and some money at 6%. He invested $500 more at 6% than he invested at 4%. His total yearly interest from the two investments was $130. How much did George invest at each rate? 55. One solution contains 40% alcohol, and a second solution contains 70% alcohol. How many liters of each solution should be mixed to make 30 liters that contain 50% alcohol? 56. One solution contains 30% alcohol, and a second solution contains 70% alcohol. How many liters of each solution should be mixed to make 10 liters containing 40% alcohol? 57. Some punch that contains 10% grapefruit juice is mixed with some punch that contains 5% grapefruit juice to produce 10 quarts of punch that is 8% grapefruit juice. How many quarts of 10% and 5% grapefruit juice must be used? 58. A 10% salt solution is to be mixed with a 15% salt solution to produce 10 gallons of a 13% salt solution. How many gallons of the 10% solution and how many gallons of the 15% solution will be needed? 59. The length of a rectangle is 3 meters less than four times the width. The perimeter of the rectangle is 74 meters. Find the length and width of the rectangle.

8.4 60. The perimeter of a rectangle is 94 inches. The length of the rectangle is 7 inches more than the width. Find the dimensions of the rectangle. 61. A deposit slip listed $700 in cash to be deposited. There were 100 bills, some of them five-dollar bills and the remainder ten-dollar bills. How many bills of each denomination were deposited? 62. Suppose that Larry has a number of dimes and quarters totaling $12.05. The number of quarters is 5 more than

3 3 Systems of Equations

417

twice the number of dimes. How many coins of each kind does he have? 63. Suppose that Sue has three times as many nickels as pennies in her collection. Together her pennies and nickels have a value of $4.80. How many pennies and how many nickels does she have? 64. Two angles are supplementary, and the measure of one of them is 20° less than three times the measure of the other one. Find the measure of each angle.

THOUGHTS INTO WORDS 65. Explain how you would solve the system a

5x 4y 10 b 3x 4y 16

using the substitution method.

67. What do you see as the strengths and weaknesses of the elimination-by-addition method and the substitution method?

66. How do you decide whether to solve a system of linear equations by using the elimination-by-addition method or the substitution method?

FURTHER INVESTIGATIONS For Problems 68 –71, use the substitution method to solve each system. 2 1 y x 3 2 68. ± ≤ 3 1 xy 2 3

69. a

70. a

y 2x 1 b 6x 3y 3

71. a

x 4y 5 b 2x 8y 1

5x 3y 0 b 4x 7y 0

Answers to the Example Practice Skills 1. {(4, 7)} 2. {(3, 1)} 3. {(2, 5)} 4. {(7, 6)} 5. {(2, 4)} 6. {(10, 8)} 7. {(5, 6)} 8. 9. Infinitely many 10. $2500 at 5% and $4500 at 7% 11. 10 liters of 6% solution and 20 liters of 12% solution 12. Length is 58 inches and width is 26 inches


8.4

2. True 3. False

4. False 5. True

6. False

7. True

3 3 Systems of Equations OBJECTIVES 1

Solve 3 3 Systems of Equations

2

Solve Word Problems Using 3 3 Systems of Equations

1 Solve 3 3 Systems of Equations When we find the solution set of an equation in two variables, such as 2x + y 9, we are finding the ordered pairs that make the equation a true statement. Plotted in two dimensions, the graph of the solution set is a line. Now consider an equation with three variables, such as 2x y 4z 8. A solution set of this equation is an ordered triple, (x, y, z), that makes the equation a true statement. For example, the ordered triple (3, 2, 1) is a solution of 2x y 4z 8,

418

Chapter 8


because 2(3) 2 4(1) 8. The graph of the solution set of an equation in three variables is a plane, not a line. In fact, graphing equations in three variables requires the use of a three-dimensional coordinate system. A 3 3 (read “3 by 3”) system of equations is a system of three linear equations in three variables. To solve a 3 3 system such as 2x 2y 4z 15 ° 3x 2y 5z 14 ¢ 4x 3y 5z 11 means to find all the ordered triples that satisfy all three equations. In other words, the solution set of the system is the intersection of the solution sets of all three equations in the system. Using a graphing approach to solve systems of three linear equations in three variables is not at all practical. However, a graphic analysis will provide insight into the types of possible solutions. In general, each linear equation in three variables produces a plane. A system of three such equations produces three planes. There are various ways that the planes can intersect. In this course, however, you need to realize that a system of three linear equations in three variables produces one of the following possible solution sets. 1.

There is one ordered triple that satisfies all three equations. The three planes have a common point of intersection as indicated in Figure 8.9.

Figure 8.9

2.

There are infinitely many ordered triples in the solution set, all of which are coordinates of points on a line common to the planes. This can happen when three planes have a common line of intersection, as in Figure 8.10(a), or when two of the planes coincide and the third plane intersects them, as in Figure 8.10(b).

(a)

(b)

Figure 8.10

3.

There are infinitely many ordered triples in the solution set, all of which are coordinates of points on a plane. This happens when the three planes coincide, as illustrated in Figure 8.11.

Figure 8.11

8.4

4.


419

The solution set is empty; it is . This can happen in various ways, as you can see in Figure 8.12. Notice that in each situation there are no points common to all three planes.

(a) Three parallel planes

(b) Two planes coincide and the third one is parallel to the coinciding planes.

(c) Two planes are parallel and the third intersects them in parallel lines.

(d) No two planes are parallel, but two of them intersect in a line that is parallel to the third plane.

Figure 8.12

Now that you know what possibilities exist, we can consider finding the solution sets for some systems. Our approach will be the elimination-by-addition method, whereby systems are replaced with equivalent systems until we get a system that allows us to easily determine the solution set. We will start with an example that allows us to determine the solution set without changing to another, equivalent system.

EXAMPLE 1

Solve the system 4x 2y 2z 5 ° 4x 3y 2z 1 ¢ 4x 2y 2z 10

Solution From equation (3), we can find the value of z. 2z 10 z5 Now we can substitute 5 for z in equation (2). 3y z 1 3y 5 1 3y 6 y 2

(1) (2) (3)

420

Chapter 8


Finally, we can substitute 2 for y and 5 for z in equation (1). 4x 2y z 5 4x 2122 5 5 4x 9 5 4x 4 x1

The solution set is 511, 2, 52 6.

▼ PRACTICE YOUR SKILL Solve the system °

3x y 2z 15 y 2z 66 ¢ . 3z 12

■

Notice the format of the equations in the system in Example 1. The first equation contains all three variables, the second equation has only two variables, and the third equation has only one variable. This allowed us to solve the third equation and then use “back substitution” to find the values of the other variables. Let’s consider another example where we replace one equation to make an equivalent system.

EXAMPLE 2


(1) (2) (3)

Solution In order to achieve the same format as in Example 1, we will need to eliminate the term with the y variable in equation (3). Using the concept of elimination, we can replace equation (3) with an equivalent equation we form by multiplying equation (2) by 5 and then adding that result to equation (3). The equivalent system is 2x 4y 15z 38 ° 2x 4y 14z 37 ¢ 2x 4y 17z 34 From equation (6) we can find the value of z. 17z 34 z2 Now we can substitute 2 for z in equation (5). y 4z 7 y 4122 7 y 1 Finally, we can substitute 1 for y and 2 for z in equation (4). 2x 4y 5z 8 2x 4112 5122 8

(4) (5) (6)

8.4


421

2x 4 10 8 2x 14 8 2x 6 x3


▼ PRACTICE YOUR SKILL Solve the system °

3x 4y 4z 14 y 4z 17 ¢ . 2y 3z 19

■

Now let’s consider some examples where we replace more than one equation to make an equivalent system.

EXAMPLE 3


(1) (2) (3)

Solution We start by picking a pair of equations to form a new equation by eliminating a variable. We will use equations (1) and (2) to form a new equation while eliminating the x variable. We can replace equation (2) with an equation formed by multiplying equation (1) by 3 and adding the result to equation (2). The equivalent system is 2x 2y 13z 10 ° 2x 7y 11z 10 ¢ 2x 3y 15z 40

(4) (5) (6)

Now we take equation (4) and equation (6) and eliminate the same variable, x. We can replace equation (6) with a new equation formed by multiplying equation (4) by 2 and adding the result to equation (6). The equivalent system is x 2y 13z 1 ° x 7y 11z 10 ¢ x 7y 11z 2

(7) (8) (9)

Now we take equations (8) and (9) and form a new equation by eliminating a variable. Either y or z can be eliminated. For this example we will eliminate y. We can replace equation (8) with a new equation formed by multiplying equation (9) by 7 and adding the result to equation (8). The equivalent system is x 2y 3z 1 ° x 2y 4z 4 ¢ y 4z 2 From equation (11), we can find the value of z. 4z 4 z1

(10) (11) (12)

422

Chapter 8


Now we substitute 1 for z in equation (12) and determine the value of y. y z 2 y 1 2 y 3 y3 Finally, we can substitute 3 for y and 1 for z in equation (10). x 2y 3z 1 x 2132 3112 1 x 6 3 1 x 3 1 x 4


▼ PRACTICE YOUR SKILL x 3y 2z 1 Solve the system ° 2x 3y 3z 5 ¢ . 3x 2y 3z 8

EXAMPLE 4

■


(1) (2) (3)

Solution Studying the coefficients in the system indicates that eliminating the z terms from equations (2) and (3) would be easy to do. We can replace equation (2) with an equation formed by multiplying equation (1) by 3 and adding the result to equation (2). The equivalent system is 2x 3y 2z 8 ° x 7y 2z 3 ¢ 3x 4y 2z 5

(4) (5) (6)

Now we replace equation (6) with an equation formed by multiplying equation (4) by 2 and adding the result to equation (6). The equivalent system is 2x 3y z 8 ° x 7y z 3 ¢ 7x 2y z 21

(7) (8) (9)

Now we can eliminate the x term from equation (9). We replace equation (9) with an equation formed by multiplying equation (8) by 7 and adding the result to equation (9). The equivalent system is 2x 3y z 8 ° x 7y z 3 ¢ 47y z 0

(10) (11) (12)

8.4


423

From equation (12), we can determine the value of y. 47y 0 y0 Now we can substitute 0 for y in equation (11) and find the value of x. x 7y 3 x 7102 3 x 3 x3 Finally, we can substitute 3 for x and 0 for y in equation (10). 2x 3y z 8 2132 3102 z 8 6z8 z 2 z 2


▼ PRACTICE YOUR SKILL 2x 3y 2z 19 Solve the system ° 3x 2y 2z 16 ¢ . 5x 4y 3z 19

EXAMPLE 5

■

Solve the system x 3y 2z 3 ° 3x 4y 2z 4 ¢ 2x 6y 4z 9

(1) (2) (3)

Solution Studying the coefficients indicates that it would be easy to eliminate the x terms from equations (2) and (3). We can replace equation (2) with an equation formed by multiplying equation (1) by 3 and adding the result to equation (2). Likewise, we can replace equation (3) with an equation formed by multiplying equation (1) by 2 and adding the result to equation (3). The equivalent system is x 13y 2z 3 ° x 13y 5z 5 ¢ 0 130 0z 3

(4) (5) (6)

The false statement 0 3 in equation (6) indicates that the system is inconsistent, and therefore the solution set is . [If you were to graph this system, equations (1) and (3) would produce parallel planes, which is the situation depicted in Figure 8.12(c).]

▼ PRACTICE YOUR SKILL x 2y 2z 14 Solve the system ° 3x 4y 2z 15 ¢ . 2x 4y 2z 10

■

424

Chapter 8


EXAMPLE 6

Solve the system 3x y 3z 6 ° 3x y 3z 2 ¢ 5x y 3z 2

(1) (2) (3)

Solution Studying the coefficients indicates that it would be easy to eliminate the y terms from equations (2) and (3). We can replace equation (2) with an equation formed by multiplying equation (1) by 1 and adding the result to equation (2). Likewise, we can replace equation (3) with an equation formed by multiplying equation (1) by 1 and adding the result to equation (3). The equivalent system is 2x y 2z 6 ° 2xy 2z 4 ¢ 4x y 4z 8

(4) (5) (6)

Now we replace equation (6) with an equation formed by multiplying equation (5) by 2 and adding the result to equation (6). The equivalent system is 2x y 2z 6 ° 2xy 2z 4 ¢ 2x 0 0z 0

(7) (8) (9)

The true numerical statement 0 0 0 in equation (9) indicates that the system has infinitely many solutions.

▼ PRACTICE YOUR SKILL 2x y 5z 8 Solve the system ° x y 5z 5 ¢ . x y 5z 1

■

2 Solve Word Problems Using 3 3 Systems of Equations Now we will use the techniques we have presented to solve a geometric problem.

EXAMPLE 7

Apply Your Skill In a certain triangle, the measure of ⬔A is 5 more than twice the measure of ⬔B. The sum of the measures of ⬔B and ⬔C is 10 more than the measure of ⬔A. Find the measures of all three angles.

Solution We can solve this problem by setting up a system of three linear equations in three variables. We let x measure of ⬔A y measure of ⬔B z measure of ⬔C Knowing that the sum of the measures of the angles in a triangle is 180 gives us the equation x y z 180. The information “the measure of ⬔A is 5 more than twice the measure of ⬔B” gives us the equation x 2y 5 or an equivalent form x 2y 5.

8.4


425

The information “the sum of the measures of ⬔B and ⬔C is 10 more than the measure of ⬔A” gives us the equation y z x 10 or an equivalent form x y z 10. Putting the three equations together, we get the system of equations x 2y z 180 ° x 2y z 185 ¢ x 2y z 10

(1) (2) (3)

To solve the system, we first replace equation (3) with an equation formed by adding equation (1) and equation (3). The equivalent system is 2x 2y z 180 ° 2x 2y z 185 ¢ 2x 2y z 170

(4) (5) (6)

From equation (6), we can determine that x 85. Now we can substitute 85 for x in equation (5) and find the value of y. x 2y 5 85 2y 5 2y 80 y 40 Finally, we can substitute 40 for y and 85 for x in equation (4) and find the value of z. x y z 180 85 40 z 180 125 z 180 z 55 The measures of the angles are ⬔A 85 , ⬔B 40 , and ⬔C 55 .

▼ PRACTICE YOUR SKILL In a certain triangle, the measure of ⬔A is 15° less than ⬔B. The sum of the measures of ⬔A and ⬔B is 10° more than the measure of ⬔C. Find the measure of all three angles. ■

CONCEPT QUIZ

For Problems 1– 8, answer true or false. 1. The graph of a linear equation in three variables is a line. 2. A system of three linear equations in three variables produces three planes when graphed. 3. Three planes can be related by intersecting in exactly two points. 4. One way three planes can be related is if two of the planes are parallel, and the third plane intersects them in parallel lines. 5. A system of three linear equations in three variables always has an infinite number of solutions. 6. A system of three linear equations in three variables can have one ordered triple as a solution. 2x y 3z 14 7. The solution set of the system ° y 3z 12 ¢ is {(3, 15, 5)}. 2z 16 x 3y 2z 4 8. The solution set of the system ° x 3y 2z 6 ¢ is {(3, 1, 2)}. 3y 2z 9

426

Chapter 8


Problem Set 8.4 1 Solve 3 3 Systems of Equations For Problems 1–16, solve each system of equations. 3x 6y 2z 6 1. ° 3x 6y 5z 4 ¢ 3x 6y 4z 8 2x 3y 3z 20 2. ° 2x 2y 5z 98 ¢ 2x 3y 3z 12 x 2y 2z 11 2y 2z 11 ¢ 3. ° 2y 2z 12 3x y 2z 0 4. ° 3x y 3z 5 ¢ 2x y 5z 1 4x 3y 2z 29 5. ° 2x 3y2z 27 ¢ 3x 2y 2z 21 2x 5y 3z 43 6. ° 3x 5y 3z 25 ¢ 3x 3y 3z 5 2x 2y 3z 11 7. ° 2x 3y 2z 13 ¢ 4x 3y 2z 16 2x 2y 2z 10 8. ° 3x 2y 2z 15 ¢ 2x 3y 3z 19 4x 3y 3z 14 9. ° 2x 3y 3z 16 ¢ 3x 4y 2z 19

3x 3y 2z 3 15. ° 2x 3y 4z 2 ¢ 4x 3y 4z 6 2x 2y 3z 11 16. ° 2x 2y 4z 10 ¢ 3x 2y 2z 11

2 Solve Word Problems Using 3 3 Systems of Equations For Problems 17–26, solve each problem by setting up and solving a system of three linear equations in three variables. 17. Brooks has 20 coins consisting of quarters, dimes, and nickels worth $3.40. The sum of the number of dimes and nickels is equal to the number of quarters. How many coins of each kind are there? 18. One binder, 2 reams of paper, and 5 spiral notebooks cost $14.82. Three binders, 1 ream of paper, and 4 spiral notebooks cost $14.32. Two binders, 3 reams of paper, and 3 spiral notebooks cost $19.82. Find the cost for each item. 19. In a certain triangle, the measure of ⬔A is five times the measure of ⬔B. The sum of the measures of ⬔B and ⬔C is 60 less than the measure of ⬔A. Find the measure of each angle. 20. Shannon purchased a skirt, blouse, and sweater for $72. The cost of the skirt and sweater was $2 more than six times the cost of the blouse. The skirt cost twice the sum of the costs of the blouse and sweater. Find the cost of each item.

5x 2y 2z 13 10. ° 2x 3y 2z 15 ¢ 2x 2y 3z 10

21. The wages for a crew consisting of a plumber, an apprentice, and a laborer are $80 an hour. The plumber earns $20 an hour more than the sum of the wages of the apprentice and the laborer. The plumber earns five times as much as the laborer. Find the hourly wage of each.

2x 2y 4z 15 11. ° 5x 2y 2z 10 ¢ 3x 3y 2z 14

22. Martha has 24 bills consisting of $1, $5, and $20 bills worth $150. The number of $20 bills is twice the number of $5 bills. How many bills of each kind are there?

4x 2y 2z 11 12. ° 3x 2y 2z 11 ¢ 2x 3y 4z 5 3x 3y 4z 11 13. ° 3x 3y 2z 15 ¢ 2x 5y 2z 18 4x 2y 3z 13 14. ° 4x 2y 2z 8 ¢ 2x 3y 2z 3

23. Two pounds of peaches, 1 pound of cherries, and 3 pounds of pears cost $5.64. One pound of peaches, 2 pounds of cherries, and 2 pounds of pears cost $4.65. Two pounds of peaches, 4 pounds of cherries, and 1 pound of pears cost $7.23. Find the price per pound for each item. 24. In a certain triangle, the measure of ⬔C is 40 more than the sum of the measures of ⬔A and ⬔B. The measure of ⬔A is 20 less than the measure of ⬔B. Find the measure of each angle.

8.4 25. Mike bought a motorcycle helmet, jacket, and gloves for $650. The jacket costs $100 more than the helmet. The cost of the helmet and gloves together was $50 less than the cost of the jacket. How much did each item cost?


427

costs $5 per hour more than the server. The chef costs the same as the salad maker and the server together. Find the cost per hour of each.

26. A catering group that has a chef, a salad maker, and a server costs the customer $70 per hour. The salad maker

THOUGHTS INTO WORDS 27. Give a step-by-step description of how to solve this system of equations. 2x 3y 6z 13 ° 2x 4y 3z 12 ¢ 2x 3y 6z 12

28. Describe how you would solve this system of equations. 2x 3y z 7 ° 2x 2y z 4 ¢ 3x 4y z 2

FURTHER INVESTIGATIONS For Problems 29 –32, solve each system of equations, indicating whether the solution is or contains infinitely many solutions. 2x 2y 3z 1 29. ° 2x 2y 2z 3 ¢ 3x 2y 2z 4

3x 3y 2z 1 31. ° 2x 3y 2z 8 ¢ 8x 2y 5z 4 2x 2y 2z 16 32. ° 3x 2y 4z 10 ¢ 5x 3y 2z 12

2x 3y 2z 7 30. ° 2x 4y 2z 3 ¢ 3x 3y 3z 5

Answers to the Example Practice Skills 1. {(3, 2, 4)} 2. {(1, 5, 3)} 3. {(4, 3, 2)} 4. {(2, 4, 1)} 5. 7. 40°, 55°, and 85°

6. Infinitely many solutions



6. True

7. False 8. False


OBJECTIVE

SUMMARY

EXAMPLE

Determine if an ordered pair is a solution of a system of equations. (Sec. 8.1, Obj. 1, p. 392)

An ordered pair is a solution to a system of equations if it satisfies every equation in the system.

Determine if (3, 5) is a solution of 2x 3y 1 the system a b. 4x 3y 3

Problems 1– 4

Solution

Substitute 3 for x and 5 for y in each equation. 2(3) (5) 1

True

4(3) 3(5) 3

True

Because (3, 5) satisfies both equations, it is a solution of the system. Solve a linear system of two equations by graphing. (Sec. 8.1, Obj. 2, p. 393)

2x y 1 b by 2x y 2 graphing.

Solving a system of two linear equations by graphing produces one of the three following possibilities:

Solve a

1. The graphs of the two equations are intersecting lines, which indicates one solution for the system. 2. The graphs of the two equations are parallel lines, which indicates no solution for the system. 3. The graphs of the two equations are the same line, which indicates infinitely many solutions for the system.

Solution

Problems 5 –10

Graph both lines on the same coordinate plane. y

x

The lines intersect at the point (3, 5), which is the solution for the system. (continued)

428

Chapter 8

OBJECTIVE

SUMMARY

EXAMPLE

Solve linear systems of inequalities by graphing. (Sec. 8.1, Obj. 3, p. 395)

The solution set of a system of linear inequalities is the intersection of the solution sets of the individual inequalities.

Solve a

Summary

429


x 2y 4 b. 2x 2y 4

Problems 11–16

Solution

Graph both inequalities on the same coordinate plane. y

x

The shaded region represents the solution set for the system of inequalities. Use the eliminationby-addition method to solve a system of equations. (Sec. 8.2, Obj. 1, p. 398)

The elimination-by-addition method works by eliminating one variable when the equations in the system are added together. It may be necessary to change the form of one or both of the original equations before adding them.

Solve a

2x 5y 11 b. 3x 5y 18

Problems 17–22

Solution

Multiply the second equation by 5 to eliminate the y variable. 2x 5y 11 15x 5y 40 17x 5y 51 x3 To find y, substitute 3 for x in either of the original equations. 2132 5y 11 6 5y 11 5y 5 y1 The solution set is {(3, 1)}.

(continued)

430

Chapter 8


OBJECTIVE

SUMMARY

EXAMPLE

Use the substitution method to solve a system of equations. (Sec. 8.3, Obj. 1, p. 406)

To solve a system by the substitution method, solve one of the equations for x or y. Then substitute for that variable into the other equation.

Solve a


3x 2y 9 b. 3x 2y 5

Problems 23 –28

Solution

Solve the first equation for x. x 2y 9 Substitute 2y 9 for x in the second equation. 312y 92 2y 5 6y 27 2y 5 8y 32 y4 To find x, substitute 4 for y in the equation already solved for x. x 2142 9 x 1 The solution set is {(1, 4)}.

Determine which method to use to solve a system of equations. (Sec. 8.3, Obj. 2, p. 410)

Solve 3 3 systems of equations. (Sec. 8.4, Obj. 1, p. 417)

We have studied three methods for solving systems of equations: 1. The graphing method works well if you want a visual representation of the problem. However, it may be impossible to get an exact solution from the graph. 2. The elimination-by-addition method gives exact solutions. 3. The substitution method works well if one of the equations is already solved for a variable. It gives exact solutions. The solution set of a 3 3 system of equations is an ordered triple, such as (x, y, z). To solve a 3 3 system, the eliminationby-addition method is used until we can easily determine the solution set.

Decide which method, eliminationby-addition or substitution, would be the most fitting to solve the given system of equations. a

Problems 29 –32

2x y 8 b y 3x 1

Solution

The substitution method would work well for this problem, because the second equation is already solved for y. Using the substitution method we can determine that the 7 26 solution set is e a , b f . 5 5 x 2y z 9 y z 3¢. Solve ° yz9

Problems 33 –36

Solution

Adding the second and third equations and solving for y gives us y 6. Substituting 6 for y in the third equation allows us to determine that z 3. Now substituting 6 for y and 3 for z in the first equation, we can determine that x 0. Therefore, the solution set is {(0, 6, 3)}.

(continued)

Chapter 8

Review Problem Set

431


OBJECTIVE

SUMMARY

EXAMPLE

Solve word problems using a system of equations. (Sec. 8.2, Obj. 2, p. 403; Sec. 8.3, Obj. 3, p. 413; Sec. 8.4, Obj. 2, p. 424)

Many word problems that were worked out earlier in the text could be solved using a system of equations. These problems may seem easier to solve using a system of equations. Either the elimination-by-addition or the substitution method can be used to solve the resulting system of equations.

In a recent survey, 500 students were asked about their college majors. The number of women in the survey was 40 less than twice the number of men in the survey. Find the number of women in the survey.

Problems 37– 44

Solution

Let x represent the number of women in the survey and y represent the number of men in the survey. Then with the information provided we can write the following system of equations. a

x y 500 b x 2y 40

Solving this system we can determine that the number of women in the survey is 320.

Chapter 8 Review Problem Set For Problems 1– 4, decide whether the given ordered pair is a solution of the given system of equations.

For Problems 11–16, indicate the solution set for each system of inequalities by shading the appropriate region on a graph.

1. a

3x 2y 10 b , (1, 3) 6x 2y 12

11. a

4x y 4 b 2x y 4

12. a

3x y 3 b 3x y 5

2. a

2x 3y 10 b , (4, 2) 2x 2y 10

13. a

3x 2y 3 b 3x 2y 4

14. a

4x 2y 12 b 3x 2y 12

3. a

2x 2y 13 b , (6, 1) 2x 3y 10

2 y x 2 5 ¢ y 3x 3

16. a

y x 1 b y 2x 3

15.

x 5y 2 b , (8, 2) 4. a 4x 3y 26 For Problems 5 –10, use the graphing method to solve each system. 5. a 7.

°

9. a

°

For Problems 17–22, solve each system of equations using the elimination-by-addition method.

2x y 2 b 2x y 5

6. a

2x 2y 8 b 2x 2y 6

17. a

18. a

2x 3y 2 ¢ 2 y x4 3

9x 2y 140 b x 5y 135

8. a

y 2x 2 b 2x y 4

2x 2y 1 b 3x 2y 5

19. a

3x 2y 14 b 2x 2y 10

20. a

4x 3y 8 b 2x 2y 3

3x y 6 b y 3x 6

10. a

2x 2y 4 b 4x 2y 8

21. a

3x 2y 7 b 4x 5y 17

22. a

4x 3y 16 b 3x 5y 10

432

Chapter 8


For Problems 23 –28, solve each system of equations using the substitution method. 23. a

2x 5y 7 b x 3y 1

2 y x 3 ≤ 25. ± 1 x y 9 3 27. a

3x 2y 29 b 3x 5y 16

5x 7y 9 b y 3x 2

37. The sum of two numbers is 113. The larger number is 1 less than twice the smaller number. Find the numbers.

26. a

y 5x 2 b 10x 2y 1

38. Last year Mark invested a certain amount of money at 6% annual interest and $50 more than that amount at 8%. He received $39.00 in interest. How much did he invest at each rate?

28. a

3x 5y 49 b 2x 3y 30

39. Cindy has 43 coins consisting of nickels and dimes. The total value of the coins is $3.40. How many coins of each kind does she have?

24. a

For Problems 29 –32, decide which method, elimination-byaddition or substitution, would be the most fitting to solve the given system of equations. 29. a

2x 5y 7 b 3x 5y 3

1 y x 3 ¢ ° 31. 2x 3y 9

30. a

32. a

2x 4y 8 b 2x 4y 1 3x 2y 6 b xy1

For Problems 33 –36, solve the system of equations. 2x 3y 2z 21 33. ° 2x 2y 2z 23 ¢ 3x 2y 2z 12 x 2y 2z 24 35. °3x 2y 5z 21 ¢ 5x 3y 7z 30 2x 2y 2z 7 36. °5x 2y 3z 17 ¢ 3x 2y 7z 5

For Problems 37– 44, solve each problem by setting up and solving a system of two linear equations in two variables.

2x 3y 2z 24 34. ° 2x 2y 2z 27 ¢ 3x 2y 2z 13

40. The length of a rectangle is 1 inch longer than three times the width. If the perimeter of the rectangle is 50 inches, find the length and width. 41. Two angles are complementary, and one of them is 6° less than twice the other one. Find the measure of each angle. 42. Two angles are supplementary, and the larger angle is 20° less than three times the smaller angle. Find the measure of each angle. 43. Four cheeseburgers and five milkshakes cost a total of $25.50. Two milkshakes cost $1.75 more than one cheeseburger. Find the cost of a cheeseburger, and also find the cost of a milkshake. 44. Three bottles of orange juice and two bottles of water cost $6.75. On the other hand, two bottles of juice and three bottles of water cost $6.15. Find the cost per bottle of each.

Chapter 8 Test For Problems 1–3, determine if the given ordered pair or ordered triple is a solution of the system of equations. 1. a

3x 2y 2 b , (1, 1) 3x 2y 9

1.

2. a

5x 2y 17 b , (3, 1) 3x 2y 17

2.

2x 3y 2z 8 3. ° 4x 2y 3z 4 ¢ , (2, 0, 4) 7x 5y 7z 6

3.

For Problems 4 –7, solve the system of equations by graphing. 4. a

x 2y 5 b x 2y 4

4.

5. a

3x 2y 6 b 3x 2y 3

5.

6. a

y 2x 4 b 2x y 3

6.

1 y x1 2 7. ° ¢ x 2y 2

7.

For Problems 8 –10, indicate the solution set for each system by shading the appropriate region. 8. a

3x 2y 6 b x 2y 2

8.

9. a

x 4y 4 b 2x 3y 6

9.

1 y x3 2 10. ° ¢ y 2x 2

10.

For Problems 11–13, solve the system of equations by the elimination-by-addition method. 11. a

2x 3y 26 b x 3y 14

11.

12. a

4x 3y 9 b 4x 7y 40

12.

13. a

4x 3y 11 b 5x 2y 18

13.

For Problems 14 –17, solve the system of equations by the substitution method. 14. a

y 5x 16 b 6x 5y 18

14.

15. a

2x y 2 b 4x y 5

15.

433

434

Chapter 8


16.

1 y x2 3 16. ° ¢ x 3y 6

17.

17. a

x 3y 4 b x 3y 2

For Problems 18 –20, solve the system of equations. 18.

18. °

3x 3y 2z 11 2y 2z 18 ¢ 5z 10

19.

19. °

2y 3z 9 y 3z 8 ¢ 3x 4y 2z 5

20.

x 2y 2z 16 20. ° 2x 2y 3z 16 ¢ 3x 5y 2z 10 For Problems 21–25, solve each problem by setting up and solving a system of equations.

21.

21. The sum of two numbers is 96. The larger number is four less than three times the smaller number. Find the numbers.

22.

22. Three reams of paper and 4 notebooks cost $19.63. Four reams of paper and 1 notebook cost $16.25. Find the cost of each item.

23.

23. The length of a rectangle is 1 inch less than twice the width of the rectangle. If the perimeter of the rectangle is 40 inches, find the length of the rectangle.

24.

24. One solution contains 30% alcohol and another solution contains 80% alcohol. Some of each of the two solutions are mixed to produce 5 liters of a 60% alcohol solution. How many liters of the 80% alcohol solution are used?

25.

25. Last year Julio invested a certain amount of money at 5% (annual interest rate) and $500 more than that amount at 6%. He received $305 in interest. How much did he invest at each rate?

Roots and Radicals

9 9.1 Roots and Radicals 9.2 Simplifying Radicals 9.3 Simplifying Radicals of Quotients 9.4 Products and Quotients Involving Radicals

Ian Hamilton /istockphoto.com

9.5 Solving Radical Equations

■ How far a car skids on wet pavement can be predicted by a formula involving a radical expression.

S

uppose that a car is traveling at 65 miles per hour on a highway during a rainstorm. Suddenly, something darts across the highway, and the driver hits the brake pedal. How far will the car skid on the wet pavement? We can use the formula S 230Df, where S represents the speed of the car, D the length of skid marks, and f a coefficient of friction, to determine that the car will skid approximately 400 feet. In Section 1.6 we used 22 and 23 as examples of irrational numbers. Irrational numbers in decimal form are nonrepeating decimals. For example, 22 1.414213562373 . . . , where the three dots indicate that the decimal expansion continues indefinitely. In Chapter 1, we stated that we would return to the irrationals in Chapter 9. The time has come for us to expand our set of skills relative to the set of irrational numbers.


435

I N T E R N E T

P R O J E C T

Search the Internet for sites on square roots or cube roots. Find three sites with activities such as flash cards, matching, or timed drills to learn the roots of numbers. Write the urls of these sites and a brief description of their activities to share with other students and the instructor.

9.1

Roots and Radicals OBJECTIVES 1

Find the Square Root of a Perfect Square

2

Find the Cube Root of a Perfect Cube

3

Find an Approximation of a Square or Cube Root

4

Add and Subtract Radical Expressions

5

Solve Applications Using Formulas with Radicals

1 Find the Square Root of a Perfect Square To square a number means to raise it to the second power—that is, to use the number as a factor twice. 42 4 # 4 16

Read “four squared equals sixteen”

102 10 # 10 100 1 1 2 1 1 a b # 2 2 2 4 (3)2 (3)(3) 9 A square root of a number is one of its two equal factors. Thus 4 is a square root of 16 because 4 # 4 16. Likewise, 4 is also a square root of 16 because (4)(4) 16. In general, a is a square root of b if a2 b. The following generalizations are a direct consequence of the previous statement. 1.

Every positive real number has two square roots; one is positive and the other is negative. They are opposites of each other.

2.

Negative real numbers have no real number square roots because any nonzero real number is positive when squared.

3.

The square root of 0 is 0.

The symbol 2 , called a radical sign, is used to designate the nonnegative square root. The number under the radical sign is called the radicand. The entire expression (such as 216 ) is called a radical. 216 4

216 indicates the nonnegative or principal square root of 16

216 4

216 indicates the negative square root of 16

20 0

Zero has only one square root. Technically, we could write 20 0 0

24 is not a real number. 24 is not a real number. 436

9.1

Roots and Radicals

437

In general, the following definition is useful.

Definition 9.1 If a 0 and b 0, then 2b a if and only if a2 b; a is called the principal square root of b. If a is a number that is the square of an integer, then 2a and 2a are rational numbers. For example, 21, 24, and 225 are the rational numbers 1, 2, and 5, respectively. The numbers 1, 4, and 25 are called perfect squares because each represents the square of some integer. The following chart contains the squares of the whole numbers from 1 through 20, inclusive. You should know these values so that you can immediately recognize such square roots as 281 9, 2144 12, 2289 17, and so on from the list. Furthermore, perfect squares of multiples of 10 are easy to recognize. For example, because 302 900, we know that 2900 30. 12 1

82 64

152 225

22 4

92 81

162 256

32 9

102 100

172 289

42 16

112 121

182 324

5 25

12 144

192 361

62 36

132 169

202 400

72 49

142 196

2

2

Knowing this listing of perfect squares can also help you with square roots of some fractions. Consider the next example.

EXAMPLE 1

Evaluate the following square roots: (a) 2196

(b)

16 B 25

(c)

64 B 49

(d) 20.09

Solution (a) 2196 14 because 142 196 (c)

64 8 B 49 7

8 2 64 because a b 7 49

(b)

4 16 5 B 25

4 2 16 because a b 5 25

(d) 20.09 0.3 because 10.32 2 0.09

▼ PRACTICE YOUR SKILL Evaluate the following square roots: (a) 2289

(b)

25 B 81

(c) 20.04

■

2 Find the Cube Root of a Perfect Cube To cube a number means to raise it to the third power—that is, to use the number as a factor three times. 23 2 # 2 # 2 8 4

3

4 # 4 # 4

Read “two cubed equals eight”

64

3

2 2 2 2 8 a b # # 3 3 3 3 27 (2)3 (2)(2)(2) 8

438

Chapter 9

Roots and Radicals

A cube root of a number is one of its three equal factors. Thus 2 is a cube root of 8 because 2 # 2 # 2 8. (In fact, 2 is the only real number that is a cube root of 8.) Furthermore, 2 is a cube root of 8 because (2)(2)(2) 8. (In fact, 2 is the only real number that is a cube root of 8.) In general, a is a cube root of b if a3 b. The following generalizations are a direct consequence of the previous statement. 1.

Every positive real number has one positive real number cube root.

2.

Every negative real number has one negative real number cube root.

3.

The cube root of 0 is 0.

Remark: Technically, every nonzero real number has three cube roots, but only one of them is a real number. The other two roots are classified as complex numbers. We are restricting our work at this time to the set of real numbers. 3

The symbol 2 designates the cube root of a number. Thus we can write 3 2 82

3 2 8 2

1 3 1 3 B 27 3

B

1 1 27 3

In general, the following definition is useful.

Definition 9.2 3 2 b a if and only if a3 b.

In Definition 9.2, if b 0 then a 0, whereas if b 0 then a 0. The number a is called the principal cube root of b or simply the cube root of b. In the radical 3 2b, the 3 is called the index of the radical. When working with square roots, we com2 monly omit writing the index; so we write 2b instead of 2 b. The concept of root can be extended to fourth roots, fifth roots, sixth roots, and, in general, nth roots. However, in this text we will restrict our work to square roots and cube roots. You should become familiar with the following perfect cubes so that you can recognize their roots without a calculator or a table. 23 8

53 125

83 512

33 27

63 216

93 729

43 64

73 343

103 1000

3 For example, you should recognize that 2 343 7. Consider the following example on finding cube roots.

EXAMPLE 2

Evaluate the following cube roots: 3 (a) 2 512

(b)

8 B 27 3

3 (c) 2 0.125

Solution 3 (a) 2 512 8 because 83 512

(b)

8 2 B 27 3

3 (c) 20.125 0.5 because 10.52 3 0.125

3

2 3 8 because a b 3 27

9.1

Roots and Radicals

439

▼ PRACTICE YOUR SKILL Evaluate the following cube roots: 3 (a) 2 64

(b)

3 27 B 1000

3 (c) 2 0.008

■

3 Find an Approximation of a Square or Cube Root So far we have found square and cube roots of perfect squares and perfect cubes. The square roots of a positive integer, which is not the square of an integer, are irrational numbers. Likewise the cube root of an integer, which is not the cube of an integer, is 3 3 an irrational number. For example, 22, 22, 223, 210, 250, and 275 are irrational numbers. Remember that irrational numbers have nonrepeating, nonterminating decimal representations. For example, 22 1.414213562373 . . . , where the decimal never terminates and never repeats a block of digits. For practical purposes, we often need to use a rational approximation of an irrational number. The calculator is a very useful tool for finding such approximations. Be sure you can use your calculator to approximate roots. Most calculators have a n square root function key. To find a cube root, many calculators have a 2x key, where you would enter 3 for n to find a cube root. Once you have found the root with your calculator, you will have to round off the result. In the following example, the roots have been rounded off to the nearest thousandth.

EXAMPLE 3

Use a calculator to find a rational approximation for each root. Express your answers to the nearest thousandth. (a) 219

(b) 238

3 (c) 2 72

3 (d) 2 81

Solution (a) 219 4.359

(b) 238 6.164

3 (c) 2 72 4.160

3 (d) 281 4.327

▼ PRACTICE YOUR SKILL Use a calculator to find a rational approximation for each root. Express your answers to the nearest thousandth. (a) 2425

(b) 2138

3 (c) 2 450

3 (d) 2 212

■

4 Add and Subtract Radical Expressions Recall our use of the distributive property as the basis for combining similar terms. Here are three examples: 3x 2x 13 22x 5x 7y 4y 17 42y 3y 9a2 5a2 19 52a2 14a2

440

Chapter 9

Roots and Radicals

In a like manner, we can often simplify expressions that contain radicals by using the distributive property. 522 7 22 15 72 22 12 22 825 2 25 18 22 25 6 25 427 6 27 3 211 211 14 62 27 13 12 211 10 27 2 211 627 4 27 2 27 16 4 22 27 8 27 3

3

3

3

3

Note that if we want to simplify when adding or subtracting radical expressions, the radicals must have the same radicand and index. Also note the form we use to indicate multiplication when a radical is involved. For example, 5 # 22 is written as 522.

EXAMPLE 4

Simplify the following radical expressions: (a) 625 9 25 7 25

(b) 427 6 27 2 23 7 23

Solution (a) 625 9 25 7 25 16 9 72 25 4 25

(b) 427 6 27 2 23 7 23 14 62 27 12 72 23 1027 9 23


Simplify 625 12 23 2 25 4 23 .

Now suppose that we need to evaluate 522 22 422 222 to the nearest tenth. We can either evaluate the expression as it stands or first simplify it by combining radicals and then evaluate that result. Let’s use the latter approach. (It would probably be a good idea for you to do it both ways for checking purposes.) 522 22 4 22 2 22 15 1 4 22 22 622 8.5 to the nearest tenth

EXAMPLE 5

Find a rational approximation, to the nearest tenth, for 723 9 25 2 23 3 25 13 23

Solution First, we simplify the given expression and then evaluate that result. 723 9 25 2 23 3 25 13 23 17 2 132 23 19 32 25 2223 6 25 51.5 to the nearest tenth

▼ PRACTICE YOUR SKILL Find a rational solution, to the nearest tenth, for 622 4 27 522 ■ 227 3 22 .

9.1

441

5 Solve Applications Using Formulas with Radicals

XII

IX

Roots and Radicals

III

VI

Many real world applications involve radical expressions. For example, the period of a pendulum is the time it takes to swing from one side to the other and back (see Figure 9.1). A formula for the period is L B 32

T 2p

where T is the period of the pendulum expressed in seconds, and L is the length of the pendulum in feet. Figure 9.1

EXAMPLE 6

Apply Your Skill

Leonard Lessin /Peter Arnold

Find the period, to the nearest tenth of a second, of a pendulum 2.5 feet long.

Solution We use 3.14 as an approximation for p and substitute 2.5 for L in the formula. T 2p

L B 32

213.142 1.8

2.5 B 32

to the nearest tenth

The period is approximately 1.8 seconds.

▼ PRACTICE YOUR SKILL Find the period, to the nearest tenth, of a pendulum 3.2 feet long.

■

Police use the formula S 230Df to estimate a car’s speed based on the length of skid marks (see Figure 9.2). In this formula, S represents the car’s speed in miles per hour, D the length of skid marks measured in feet, and f a coefficient of friction. For a particular situation, the coefficient of friction is a constant that depends on the type and condition of the road surface.

Figure 9.2

442

Chapter 9

Roots and Radicals

EXAMPLE 7

Apply Your Skill Using 0.40 as a coefficient of friction, find how fast a car was moving if it skidded 225 feet. Express the answer to the nearest mile per hour.

Solution We substitute 0.40 for f and 225 for D in the formula S 230Df. S 2301225210.402 52 to the nearest whole number The car was traveling at approximately 52 miles per hour.

▼ PRACTICE YOUR SKILL Using 0.35 as a coefficient of friction, find how fast a car was moving if it skidded 200 feet. Round the answer to the nearest mile per hour. ■

CONCEPT QUIZ

For Problems 1–10, answer true or false. 1. 2. 3. 4. 5. 6. 7. 8.

Every positive real number has two square roots. Negative real numbers have no real number square roots. Numbers called “perfect squares” represent the square of some integer. The square root of a real number is always a positive number. The symbol 2 is called a radical sign. The 25 is a rational number. The cube root of a number is one of its three equal factors. Every negative real number has one negative cube root and two positive cube roots. 9. The symbol 2 is used to designate all the square roots of a number. 10. The cube root of 0 is 0.

Problem Set 9.1 1 Find the Square Root of a Perfect Square For Problems 1–18, evaluate each square root without using a calculator or a table. 1. 249

2. 2100

3. 264

4. 236

5. 2121

6. 2144

7. 23600

8. 22500

9. 21600

10. 2900

11. 26400

12. 2400

13. 2324

14. 2361

15.

25 B9

17. 20.16

16.

1 B 225

18. 20.0121

2 Find the Cube Root of a Perfect Cube For Problems 19 –26, evaluate each cube root without using a calculator. 1 3 3 19. 227 20. 3 21. 2 8 B8 125 3 3 22. 3 23. 2 729 24. 2 125 B 8 3 25. 2216

3 26. 264

3 Find an Approximation of a Square or Cube Root For Problems 27–38, use a calculator to find a rational approximation of each root. Express your answers to the nearest hundredth. 27. 219

28. 234

29. 250

30. 266

31. 275

32. 290

9.1 33. 295

34. 298

3 35. 2 100

3 36. 2 252

3 37. 2 81

3 38. 2 428

For Problems 39 – 48, use a calculator to find a whole number approximation for each expression.

443

For Problems 61–74, find a rational approximation, to the nearest tenth, for each radical expression. 61. 923 23

62. 622 1422

63. 925 325

64. 1826 1226

65. 1422 1522

66. 723 1223

39. 24325

40. 27500

41. 21175

67. 827 427 627

42. 21700

43. 29501

44. 28050

68. 925 225 25

3 45. 2 7814

3 46. 2 1456

47. 21000

48. 23400

Roots and Radicals

69. 423 222

70. 322 23 25

71. 926 325 226 725 26 72. 827 2210 427 3210 727 4210

4 Add and Subtract Radical Expressions For Problems 49 – 60, simplify each expression by using the distributive property.

73. 4211 5211 7211 2211 3211 74. 14213 17213 3213 4213 5213

5 Solve Applications Using Formulas with Radicals

49. 7 22 14 22 50. 9 23 4 23

75. Use the formula from Example 6 to find the periods of pendulums that have lengths of 2 feet, 3.5 feet, and 4 feet respectively. Express the answers to the nearest tenth of a second.

51. 17 27 9 27 52. 19 25 8 25 3 3 53. 4 2 2 72 2

76. Use the formula from Example 7 with a coefficient of friction of 0.35 to find the speeds of cars that leave skid marks of lengths 150 feet, 200 feet, and 275 feet respectively. Express the answers to the nearest mile per hour.

3 3 54. 8 2 6 102 6 3 3 3 55. 9 2 7 22 5 62 7

77. The time T, measured in seconds, that it takes for an object to fall d feet (neglecting air resistance) is given

56. 8 27 13 27 9 27

d . Find the times that it takes B 16 objects to fall 75 feet, 125 feet, and 5280 feet respectively. Express the answers to the nearest tenth of a second. by the formula T

57. 8 22 4 23 9 22 6 23 58. 7 25 9 26 14 25 2 26 59. 6 27 5 210 8 210 4 27 11 27 210 60. 23 25 425 3 23 9 25 16 23

THOUGHTS INTO WORDS 79. How could you find a whole number approximation for 21450 if you did not have a calculator available?

78. Why is 24 not a real number?


5 3 (c) 0.2 2. (a) 4 (b) (c) 0.2 3. (a) 20.616 (b) 11.747 (c) 7.663 (d) 5.963 9 10 4. 425 8 23 5. 16.6 6. 2.0 seconds 7. 46 mph

1. (a) 17 (b)


2. True 3. True

4. False


8. False

9. False 10. True

444

Chapter 9

9.2

Roots and Radicals

Simplifying Radicals OBJECTIVES 1

Change Radicals to Simplest Radical Form

2

Change Radicals That Contain Variables to Simplest Radical Form

3

Combine Radicals after Simplifying

1 Change Radicals to Simplest Radical Form Note the following facts that pertain to square roots: 24 # 9 236 6 24 29 2 # 3 6 Thus we observe that 24 # 9 2429 . This illustrates a general property.

Property 9.1 For any nonnegative real numbers a and b, 2ab 2a2b

In other words, we say that the square root of a product is equal to the product of the square roots. Property 9.1 and the definition of square root provide the basis for expressing radical expressions in simplest radical form. For now, simplest radical form means that the radicand contains no factors other than 1 that are perfect squares. We present some examples to illustrate this meaning of simplest radical form.

EXAMPLE 1

Change each radical to simplest radical form. (a) 28

(b) 245

(c) 248

Solution (a) 28 24 # 2 2422 2 22

4 is a perfect square

24 2

(b) 245 29 # 5 2925 3 25 9 is a perfect square

29 3

(c) 248 216 # 3 21623 4 23 16 is a perfect square

216 4

9.2

Simplifying Radicals

445

▼ PRACTICE YOUR SKILL Change each radical to simplest radical form. (a) 224

(b) 263

(c) 232

■

The first step in each example is to express the radicand of the given radical as the product of two factors, at least one of which is a perfect square other than 1. Observe the radicands of the final radicals. In each case, the radicand cannot be expressed as the product of two factors, at least one of which is a perfect square other than 1. We say that the final radicals, 222, 325, and 423, are in simplest radical form. You may vary the steps somewhat in changing to simplest radical form, but the final result should be the same. Consider another sequence of steps to change 248 to simplest form. 248 24 # 12 24212 2 212 2 24 # 3 2 2423 2 # 2 23 4 23 4 is a perfect square

This is not in simplest form


Same result as in example 3

Another variation of the technique for changing radicals to simplest form is to prime-factor the radicand and then to look for perfect squares in exponential form. We will redo the previous examples. 1. 28 22 # 2 # 2 222 # 2 222 22 2 22 Prime factors of 8


2. 245 23 # 3 # 5 232 # 5 232 25 3 25

Prime factors of 45


3. 248 22 # 2 # 2 # 2 # 3 224 # 3 224 23 22 23 4 23


Prime factors of 48

224 22 because 22 # 22 24

The next example further illustrates the process of changing to simplest radical form. Only the major steps are shown, so be sure that you can fill in the details.

EXAMPLE 2

Change each radical to simplest radical form. (a) 256

(b) 275

(c) 2108

(d) 5212

Solution (a) 256 24214 2 214 (b) 275 22523 5 23

(c) 2108 22 # 2 # 3 # 3 # 3 222 # 32 23 6 23 (d) 5 212 5 2423 5 # 2 # 23 10 23

446

Chapter 9

Roots and Radicals

▼ PRACTICE YOUR SKILL Change each radical to simplest radical form. (a) 220

(b) 272

(c) 6 250

■

We can extend Property 9.1 to apply to cube roots.

Property 9.2 For any real numbers a and b, 3

3

3

2ab 2a2b

Now, using Property 9.2, we can simplify radicals involving cube roots. Here it is helpful to recognize the perfect cubes that we listed in the previous section.

EXAMPLE 3

Change each radical to simplest radical form. 3 (a) 2 24

3 (b) 2 108

3 (c) 2 375

Solution 3 3 3 3 (a) 2 24 2 82 3 22 3

Perfect cube 3 3 3 3 (b) 2108 22724 3 24

Perfect cube 3 3 3 3 (c) 2 375 2 1252 3 52 3

Perfect cube

▼ PRACTICE YOUR SKILL Change each radical to simplest radical form. 3 (a) 216

3 (b) 240

3 (c) 281

■

2 Change Radicals That Contain Variables to Simplest Radical Form Before we discuss the process of simplifying radicals that contain variables, there is one technicality that we should call to your attention. Let’s look at some examples to clarify the point. Consider the radical 2x2. Let x 3; then 2x2 232 29 3 Let x 3; then 2x2 2132 2 29 3

9.2


447

Thus if x 0, then 2x2 x, but if x 0, then 2x2 x. Using the concept of absolute value, we can state that for all real numbers, 2x2 0x 0 . Now consider the radical 2x3. Because x 3 is negative when x is negative, we need to restrict x to the nonnegative reals when working with 2x3. Therefore, we can write: If x 0, then 2x3 2x2 2x x2x, and no absolute value sign is necessary. Finally, 3 3 let’s consider the radical 2 x. 3 3 3 3 3 Let x 2; then 2 x 2 2 2 82 3 3 3 3 Let x 2; then 2 x 2 122 3 2 8 2 3 3 Therefore, it is correct to write: 2 x x for all real numbers, and again no absolute value sign is necessary. The previous discussion indicates that technically every radical expression involving variables in the radicand needs to be analyzed individually as to the necessary restrictions imposed on the variables. However, to avoid considering such restrictions on a problem-to-problem basis, we shall merely assume that all variables represent positive real numbers.

EXAMPLE 4

Change each radical to simplest radical form. (a) 2x2y

(b) 24x3

(c) 28xy3

(d) 227x5y3

3 (e) 2 40x4y5

Solution (a) 2x2y 2x2 2y x 2y (b) 24x3 24x2 2x 2x 2x 4x2 is a perfect square because (2x)(2x) 4x2

(c) 28xy3 24y2 22xy 2y 22xy (d) 227x5y3 29x4y2 23xy 3x2y 23xy

29x 4y 2 3x2y 3 3 3 3 (e) 2 40x4y5 2 8x3y3 2 5xy2 2xy 2 5xy2

Perfect cube

▼ PRACTICE YOUR SKILL Change each radical to simplest radical form. (a) 212x2y

(b) 275a3b2

3 (c) 216x4y3

■

3 Combine Radicals after Simplifying When simplifying expressions that contain radicals, we must often first change the radicals to simplest form, and then apply the distributive property.

448

Chapter 9

Roots and Radicals

EXAMPLE 5

Simplify 528 3 22.

Solution 528 3 22 5 2422 3 22 5 # 2 # 22 3 22 1022 3 22 110 32 22 1322

▼ PRACTICE YOUR SKILL Simplify 2245 4 25 .

EXAMPLE 6

■

Simplify 2227 5 248 4 23.

Solution 2227 5 248 4 23 2 2923 5 21623 4 23 2132 23 5142 23 4 23 623 20 23 4 23 16 20 42 23 1023

▼ PRACTICE YOUR SKILL Simplify 218 3 250 272 .

EXAMPLE 7

■

1 1 Simplify 245 220. 4 3

Solution 1 1 1 1 245 220 29 25 24 25 4 3 4 3 1 1 132 25 122 25 4 3 3 2 25 25 4 3 3 2 9 8 a b 25 a b 25 4 3 12 12 17 25 12

▼ PRACTICE YOUR SKILL 1 3 Simplify 227 23 . 5 4

■

9.2

EXAMPLE 8


449

3 3 Simplify 52 24 7 2 375.

Solution 3 3 3 3 3 3 52 24 7 2 375 5 2 82 3 72 1252 3 3 3 5122 2 3 7152 2 3 3 3 10 2 3 35 2 3 3 45 2 3

▼ PRACTICE YOUR SKILL 3 3 Simplify 92 16 2 2 54 .

CONCEPT QUIZ

■

For Problems 1–5, answer true or false. 1. The property 2ab 2a2b can be used to express the product of two radicals as one radical. 2. One condition for simplest radical form is that the radicand contains no factors other than 1 that are perfect squares. 3. For all real numbers, 2x2 x. 3 3 4. For all real numbers, 2 x x. 5. 4 22 6 23 10 25

Problem Set 9.2 1 Change Radicals to Simplest Radical Form

2 Change Radicals That Contain Variables to Simplest Radical Form

For Problems 1–30, change each radical to simplest radical form.

For Problems 31–52, express each radical in simplest radical form. All variables represent nonnegative real numbers.

1. 224

2. 254

3. 218

31. 2x2y3

32. 2xy4

33. 22x2y

4. 250

5. 227

6. 212

34. 23x2y2

35. 28x2

36. 224x3

7. 240

8. 290

3 9. 254

37. 227a3b

38. 245a2b4

3 39. 2 64x4y2

3 10. 232

11. 280

13. 2117 16. 8 298

14. 2126 3

17. 3 240

12. 2125 15. 4 272 18. 4 2375

20. 6245

22. 4 254

23.

3 28 2

24.

26.

4 227 5

2 27. 245 3

3 212 4

3 28. 2125 5

1 3 29. 232 4

41. 263x4y2

42. 228x3y

43. 3248x2

44. 5212x2y2

45. 6272x7

46. 8280y9

47.

2 254xy 9 1 3 2250x4 8

3

19. 5 220

25.

3 40. 2 56x3y5

21. 8296

30.

5 232 2

2 3 281 3

48.

4 220xy 3

49.

50.

1 3 281x3 3

2 51. 2169a8 3

2 52. 2196a10 7

450

Chapter 9

Roots and Radicals

3 Combine Radicals after Simplifying

62. 4218 6250 3272

For Problems 53 – 66, simplify each expression. 53. 7 232 522

54. 6 248 5 23

55. 4 245 925

56. 7 224 12 26

3 3 57. 2 2 54 62 16

3 3 58. 2 2 24 4 2 81

63.

2 1 1 220 245 280 2 3 4

64.

1 3 3 212 248 2108 3 2 4

65. 328 5220 7218 92125 59. 4 263 7228

60. 2 240 7 290 66. 5227 3224 8254 7275

61. 5 212 3227 2 275

THOUGHTS INTO WORDS 67. Explain how you would help someone express 5272 in simplest radical form.

68. Explain your thought process when expressing 2153 in simplest radical form.

FURTHER INVESTIGATIONS 69. Express each of the following in simplest radical form. The divisibility rules given in Section 1.4 should be of some help. (a) 2162

(b) 2279

(c) 2275

(d) 2212

following, first make a whole number estimate, and then use a calculator to see how well you estimated. (a) 3210 4224 6265 (b) 9227 5237 3280 (c) 1225 13218 9247

70. Use a calculator to evaluate each expression in Problems 53 – 66. Then evaluate the simplified expression that you obtained when doing those problems. Your two results for each problem should be the same. 71. Sometimes we can reach a fairly good estimate of a radical expression by using whole number approximations. For example, 5 235 7 250 is approximately 5(6) 7(7) 79. Using a calculator, we find that 5235 7 250 79.1 to the nearest tenth. In this case our whole number estimate is very good. For each of the

(d) 3298 4283 72120 (e) 42170 22198 52227 (f) 32256 62287 112321 72. Evaluate 2x2 for x 5, x 4, x 3, x 9, x 8, and x 11. For which values of x does 2x2 x? For which values of x does 2x2 x? For which values of x does 2x2 兩 x兩?

Answers to the Example Practice Skills 3 3 3 3. (a) 22 2 (b) 22 5 (c) 32 3 27 3 3 4. (a) 2x23y (b) 5ab23a (c) 2xy 2 2x 5. 1025 6. 1222 7. 23 8. 122 2 20

1. (a) 2 26 (b) 327 (c) 422

2. (a) 225 (b) 622 (c) 3022


2. True 3. False

4. True 5. False

9.3

9.3

Simplifying Radicals of Quotients

451

Simplifying Radicals of Quotients OBJECTIVES 1

Simplify Roots of Quotients

2

Rationalize Denominators of Square Roots

3

Combine Radicals after Rationalizing the Denominators

1 Simplify Roots of Quotients We now continue our work of simplifying radicals. We will now look at some examples that simplify the root of a quotient. 36 24 2 B9 Thus we see that

236 29

6 2 3

4 2 2

36 236 . B9 29

64 3 2 82 B8 3

Thus we see that

and

and

3 2 64 3

28

3 64 64 2 3 . B8 28 3

We can state the following general property.

Property 9.3 For any nonnegative real numbers a and b, b 0, For any real numbers a and b, b 0,

a 2a . Bb 2b

3 a 2a 3 . Bb 2b 3

25 , in which the numerator and denominator B4 of the fractional radicand are perfect squares, you may use Property 9.3 or merely rely on the definition of square root. To evaluate a radical such as

25 225 5 B4 2 24 or 25 5 B4 2

because

5 # 5 25 2 2 4

Sometimes it is easier to do the indicated division first and then find the square root, as you can see in the next example. 324 236 6 B 9

452

Chapter 9

Roots and Radicals

Now we can extend our concept of simplest radical form. An algebraic expression that contains a radical is said to be in simplest radical form if the following conditions are satisfied: 1.

No fraction appears within a radical sign. a

2.

No radical appears in the denominator. a

2 violates this condition.b B3

violates this condition.b 28 No radicand when expressed in prime factored form contains a factor raised to a power greater than or equal to the index. ( 28 223 violates this condition.)

3.

5

The next examples show how to simplify expressions that do not meet these three stated conditions.

EXAMPLE 1

Simplify

13 . B4

Solution Apply Property 9.3 to write the expression as the quotient of two radicals. 13 213 213 B4 2 24


EXAMPLE 2

Simplify

7 . B9

Simplify

16 . B 27

■

3

Solution Apply Property 9.3 to write the expression as the quotient of two cube roots. 3 3 3 3 3 2 22 16 2 16 2 16 82 2 2 3 B 27 3 3 3 227 3

Don’t stop here. The radical in the numerator can be simplified


3 81 . B8

■

For Examples 3, 4, and 5, we show two methods of simplifying for each example. Both methods are correct but you may prefer one approach over the other.

9.3

EXAMPLE 3

Simplify

212 216


453

.

Solution A 212 216

212 24 23 223 23 4 4 4 2

Solution B 212 216

12 3 23 23 B 16 B 4 2 24 Reduce the fraction


18 . B 81

■

The two approaches in this example show how important it is to think about the approach to a problem before you push the pencil. Now let’s consider an example where neither the numerator nor the denominator of the radicand is a perfect square. Keep in mind that an expression is not simplified if there is a radical in the denominator.

EXAMPLE 4

Simplify

2 . B3

Solution A 2 22 22 B3 23 23

Form of 1

#

23 23

26 3

Solution B 2 2 # 3 6 26 26 B3 B3 3 B9 3 29


3 . B5

■

2 Rationalize Denominators of Square Roots We refer to the process we used to simplify the radical in Example 4 as rationalizing the denominator. Notice that the denominator becomes a rational number. There is more than one way to rationalize the denominator, as the next example shows.

454

Chapter 9

Roots and Radicals

EXAMPLE 5

25

Simplify

28

.

Solution A 25

28

25 28

#

28

#

22

28

240 24 210 2210 210 8 8 8 4

Solution B 25

28

25 28

22

210 216

210 4

Solution C 25

28

25 24 22

25 222

25

#

222

22 22

210 4


5 . B 12

■

Study the following examples, and check that the answers are in simplest radical form according to the three conditions we listed on page 452.

EXAMPLE 6

Simplify each of these expressions. 3

(a)

2x B 3y

(b)

2x

(c)

325 26

(d)

4x2 B 9y

Solution (a)

(b)

(c)

(d)

3 2x

3 2x

2x

#

2x

3 2x x

2x 22x 22x B 3y 23y 23y 3 25 26

3 25 26

#

26 26

#

23y 23y

26xy 3y

230 3 230 6 2

4x2 24x2 2x 2x 2x B 9y 29y 29 2y 32y 32y

#

2y 2y

2x 2y 3y

▼ PRACTICE YOUR SKILL Simplify each of the expressions. (a)

4 2a

(b)

5m B 2n

(c)

523 210

(d)

9a2 B 16b

■

9.3


455

3 Combine Radicals after Rationalizing the Denominators Let’s return again to the idea of simplifying expressions that contain radicals. Sometimes it may appear as if no simplifying can be done; however, after the individual radicals have been changed to simplest form, the distributive property may apply.

EXAMPLE 7

Simplify 5 22

3 22

.

Solution 522

3 22

5 22

3 22

#

22 22

5 22

3 22 2

3 10 3 a5 b 22 a b 22 2 2 2 13 22 2

or

1322 2

▼ PRACTICE YOUR SKILL Simplify 723

EXAMPLE 8

Simplify

2 23

■

.

3 224. B2

Solution 3 23 23 224 224 B2 22 22

#

22 22

2426

26 226 2

a

1 2b 26 2

a

1 4 b 26 2 2

5 26 2


2 290 . B5

■

456

Chapter 9

Roots and Radicals

CONCEPT QUIZ

For Problems 1–5, answer true or false. 1. A radical is not in simplest radical form if it has a fraction within the radical sign. 1 2. The radical is in simplest radical form. 25 3. The process of rationalizing the denominator means the denominator has been changed from an irrational number to a rational number. 6 4. The expression could have the denominator rationalized by multiplying 212 212 23 the expression by or . 23 212 28 5. The radical is in simplest form because it does not have a radical in the 5 denominator.

Problem Set 9.3 1 Simplify Roots of Quotients

25.

For Problems 1–10, evaluate each radical. 1.

16 B 25

4.

7.

64 B 16

3 125 B 64

2.

4 B 49

3.

81 B9

5.

1 B 64

6.

8.

3 27 B 8

9.

100 B 121 25 B 256

28.

31.

34.

289 10. B 225

37.


40.

11.

19 B 25

24 14. B 25 17.

212 236

263 27 23 232 1 B 24 322 26 327

26.

29.

32.

35.

15.

18.

17 B4 3 2 375 3 2 216

13.

16.

8 B 49 3 2 16 3 2 54

41.

44.

2 Rationalize Denominators of Square Roots For Problems 19 – 40, change each radical to simplest radical form. 19.

3 B2

20.

22.

7 B 12

23.

2 B5 256 28

21.

24.

5 B8 255 211

26 24

30.

227 1 B 12

33.

422

36.

38.

25 218 29 248 223 25 225 728

6212

39.

5224

1 4 B 9

1 3 B 4

For Problems 41– 60, change each radical to simplest radical form. All variables represent positive real numbers.

220 264

27.

323

4212

12.

296

47.

50.

53.

3 2x 7 23y 12 B x2 23y 232x 22x3 28y

42.

2

43.

2xy

45.

3 Bx

46.

48.

27 B 4y2

49.

51.

54.

25x 227y 25x2 245y

3

52.

55.

5 22x 8 Bx 22x 25y 23x2 25y2 9 B x3

9.3

56.

59.

25 B y5

4

57.

3 2x

2x

2 2y

1 B5

2x

67. 326

7 2xy

3 Combine Radicals after Rationalizing the Denominators

69. 4212

For Problems 61–72, simplify each expression. 61. 7 23

1 B3

63. 4 210

71.

62. 322

2 B5

1 B2

925 23

66. 627 4

5 22

68. 428

23 3 23

457

1 B7

65. 225 5

5

5 2x

60.

3

14

58.

7


5227

6260

10 23 25

6 22

70. 328 4218

72. 223 3248

6 22 3 23

1 B5

64. 8 25 3

THOUGHTS INTO WORDS 73. Your friend simplifies 6 26 26 B8 28 28

#

Could you show him a much shorter way to simplify this expression?

6 as follows: B8

28 28

74. Is the expression 3 22 250 in simplest radical form? Why or why not?

248 21623 8 8

4 23 23 8 2

FURTHER INVESTIGATIONS To rationalize the denominator of algebraic expressions that have cube roots, you must multiply by a factor in the form of 1 that makes the radicand in the denominator a perfect cube. Consider the following example and refer to a list of perfect cubes if you do not have them memorized. 5 Simplify 3 . B2 3 3 5 2 5 2 5 3 3 B2 22 22 3

#

3 2 4 3 2 4

3 2 20 3 2 8

For Problems 75 – 80, change each radical to simplest form. 75.

78.

3 2 7 3

23 3 2 B9

76.

79.

3 2 6 3 2 4 3 7 B 32

3 2 20 2

Form of 1

Answers to the Example Practice Skills 3 27 32 3 22 2. 3. 3 2 3 16210 2323 7. 8. 3 5

1.

4.

215 5

5.

215 6


2. False

3. True

4. True 5. False

6. (a)

42a 210mn 230 3a2b (b) (c) (d) a 2n 2 4b

77.

3 8 B 25

80.

3 9 B 10

458

Chapter 9

9.4

Roots and Radicals

Products and Quotients Involving Radicals OBJECTIVES 1

Multiply Radicals

2

Multiply Radicals Using the Distributive Property

3

Use Conjugates to Rationalize Denominators

1 Multiply Radicals 3 3 3 We use Property 9.1 1 2ab 2a2b2 and Property 9.2 1 2 ab 2 a2 b2 to multiply radical expressions, and in some cases, to simplify the resulting radical. The following examples illustrate several types of multiplication problems that involve radicals.

EXAMPLE 1

Multiply and simplify where possible. (a) 23 212

(b) 23 215

(c) 27 28

3 3 (d) 2 42 6

(e) 13 222 14232

3 3 (f) 12 2 92142 62

Solution (a) 23 212 236 6 (b) 23 215 245 2925 3 25 (c) 27 28 256 24214 2 214 3 3 3 3 3 3 (d) 24 26 224 28 23 2 23

(e) 13 222 14232 3 # 4 # 22 # 23 12 26 (f) 12 29214262 2 # 4 # 29 # 26 3

3

3

3

3 82 54 3 3 82 272 2 3 8#3# 2 2 3 24 2 2

▼ PRACTICE YOUR SKILL For each of the following, multiply and simplify where possible. 3 3 3 3 (a) 22 214 (b) 1523212272 (c) 2 (d) 122 102 4 52 132 502

■

2 Multiply Radicals Using the Distributive Property Recall how we use the distributive property when we find the product of a monomial and a polynomial. For example, 2x(3x 4) 2x(3x) 2x(4) 6x2 8x. Likewise, the distributive property and Properties 9.1 and 9.2 provide the basis for finding certain special products involving radicals. The next examples demonstrate this idea.

9.4

EXAMPLE 2

Products and Quotients Involving Radicals

459

Multiply and simplify where possible. (a) 221 23 252

(b) 231 212 262

(c) 281 22 32

(d) 2x1 2x 2y2

3

3

3

(e) 221 24 2102

Solution (a) 221 23 252 2223 2225 26 210 (b) 231 212 262 23212 2326 236 218 6 2922 6 322 (c) 281 22 32 2822 1 282 132 216 3 28 4 32422 4 622 (d) 2x1 2x 2y2 2x2x 2x2y 2x2 2xy x 2xy 3

3

3

3 3 3 (e) 221 24 2102 28 220 2 220

▼ PRACTICE YOUR SKILL For each of the following, multiply and simplify where possible. (a) 231 27 252

(b) 221 218 52

(c) 2a1 2a 2b2

3 3 3 (d) 2 31 2 3 2 182

■

The distributive property plays a central role when we find the product of two binomials. For example, (x 2)(x 3) x(x 3) 2(x 3) x2 3x 2x 6 x2 5x 6. We can find the product of two binomial expressions involving radicals in a similar fashion.

EXAMPLE 3

Multiply and simplify. (a) 1 23 252 1 22 262

(b) 1 27 32 1 27 62

Solution (a) 1 23 252 1 22 262 231 22 262 251 22 262 2322 2326 2522 2526 26 218 210 230 26 3 22 210 230 (b) 1 27 321 27 62 271 27 62 31 27 62 2727 6 27 3 27 18 7 627 3 27 18 11 327

460

Chapter 9

Roots and Radicals

▼ PRACTICE YOUR SKILL For each of the following, multiply and simplify: (a) 1 26 222 1 23 272

(b) 1 25 42 1 25 22

■

If the binomials are of the form (a b)(a b), then we can use the multiplication pattern (a b)(a b) a2 b2.

EXAMPLE 4

Multiply and simplify. (b) 13 252 13 252

(a) 1 26 221 26 22

(c) 1 28 2521 28 252

Solution (a) 1 26 221 26 22 1 262 2 22 6 4 2

(b) 13 25213 252 32 1 252 2 9 5 4

(c) 1 28 2521 28 252 1 282 2 1 252 2 8 5 3

▼ PRACTICE YOUR SKILL For each of the following, multiply and simplify: (b) 14 232 14 232

(a) 1 25 621 25 62

(c) 1 26 2221 26 222

■

Note that in each part of Example 4, the final product contains no radicals. This happens whenever we multiply expressions such as 2a 2b and 2a 2b, where a and b are rational numbers. 1 2a 2b21 2a 2b2 1 2a2 2 1 2b2 2 a b Expressions such as 28 25 and 28 25 are called conjugates of each other. Likewise, 26 2 and 26 2 are conjugates, as are 3 25 and 3 25. Now let’s see how we can use conjugates to rationalize denominators.

3 Use Conjugates to Rationalize Denominators EXAMPLE 5

Simplify

4 25 22

.

Solution To simplify the expression, the denominator needs to be a rational number. Let’s multiply the numerator and denominator by 25 22, which is the conjugate of the denominator. 4 25 22

4 25 22

#

25 22

25 22

25 22

25 22

41 25 222

1 25 2221 25 222

is merely a form of 1

9.4


461

41 25 222 52 41 25 222 3

425 4 22 3

or

Either answer is acceptable

▼ PRACTICE YOUR SKILL 3

Simplify

27 22

■

.

The next four examples illustrate further the process of rationalizing and simplifying expressions that contain binomial denominators.

EXAMPLE 6

Rationalize the denominator and simplify

23 26 2

.

Solution To simplify, we want to multiply the numerator and denominator by the conjugate of the denominator, which is 26 2. 23 26 2

23 26 2

#

26 2 26 2

231 26 22

1 26 221 26 22

218 2 23 64

3 22 2 23 2

218 2922 3 22

▼ PRACTICE YOUR SKILL Simplify:

EXAMPLE 7

22 210 3

■

.


2x 2 2x 3

.

Solution To rationalize the denominator, we need to multiply the numerator and denominator by 2x 3, which is the conjugate of the denominator. 2x 2 2x 3

2x 2 2x 3

#

2x 3 2x 3

1 2x 22 1 2x 32

1 2x 32 1 2x 32

462

Chapter 9

Roots and Radicals

x 22x 3 2x 6 x9

x 5 2x 6 x9


EXAMPLE 8

2a 3 2a 4

■

.


3 22 22 6

.

Solution To change the denominator to a rational number, we can multiply the numerator and denominator by 22 6, which is the conjugate of the denominator. 3 22 22 6

3 22 22 6

#

22 6 22 6

13 2221 22 62

1 22 621 22 62

322 18 2 6 22 2 36

922 20 34

922 20 34

a a b b


CONCEPT QUIZ

5 26 26 2

.

■

For Problems 1–5, answer true or false. 1. The product of two radicals always results in an expression that has a radical, even after simplifying. 2. Just like multiplying binomials, the product of 1 23 421 25 62 can be found by applying the distributive property. 3. The conjugate of 4 23 is 4 23. 4. The product of a binomial radical expression and its conjugate is a rational number. 523 5. To rationalize the denominator for the expression , we would multiply 6 23 23 the expression by . 23

9.4


463

Problem Set 9.4 41. 15 23 22621523 2262

1 Multiply Radicals For Problems 1–20, multiply and simplify where possible. 1. 27 25

2. 23210

3. 26 28

4. 26212

5. 25 210

6. 22212

3

3

3

For Problems 43 –54, find each product and express your answers in simplest radical form. All variables represent nonnegative real numbers.

3

7. 29 26

8. 2929

9. 28 212

10. 212 220

42. 1425 52721425 5272

43. 2xy2x

44. 2x2y 2y

3 3 45. 225x2 25x

3 3 46. 29x23x

47. 14 2a2132ab2

48. 152a2162a2

11. 13 232 15 272

3 3 12. 152102 17 232

3 3 13. 1262 15 242

14. 15 232 16 2102

49. 22x1 23x 26y2

15. 13 262 14 262

16. 12 272 15 272

50. 26x1 22x 24y2

17. 15 222 14 2122

18. 13 222 12 2272

19. 14 232 12 2152

20. 17 272 12 2142

For Problems 21– 42, find the products by applying the distributive property. Express your answers in simplest radical form. 21. 221 23 252

22. 231 25 272

23. 261 22 52

24. 281 23 42

3

3

3

3

3

25. 221 24 2102

26. 231 29 282

27. 2121 26 282

28. 2151 23 252

29. 4 231 22 2252

30. 32213 25 2 272

31. 1 22 62 1 22 92

52. 13 2x217 2x2 53. 1 2x 721 2x 72 54. 1 2x 921 2x 92

2 Multiply Radicals Using the Distributive Property

3

51. 1 2x 521 2x 32

3 Use Conjugates to Rationalize Denominators For Problems 55 –70, rationalize the denominators and simplify. All variables represent positive real numbers. 55.

57.

59.

32. 1 23 42 1 23 72 33. 1 26 52 1 26 32 34. 1 27 62 1 27 12 35. 1 23 262 1 26 282 36. 1 22 262 1 28 252

61.

63.

65.

37. 15 2102 15 2102 38. 1 211 22 1 211 22

67.

39. 13 22 232 1322 232 40. 15 26 222 1526 222

69.

3 22 4 8 26 2 2 25 23 4 26 22 4 2x 2 2x 2x 3 2a 2 2a 5 2 23 3 22

56.

58.

60.

62.

64.

66.

68.

70.

5 23 7 10 3 27 3 26 25 5 22 23 7 2x 4 2y 2y 6 2a 3 2a 1 3 25 4 28

464

Chapter 9

Roots and Radicals

THOUGHTS INTO WORDS 71. Explain how the distributive property has been used in this chapter.

72. How would you help someone rationalize the denom24 inator and simplify the expression ? 212 28

FURTHER INVESTIGATIONS 73. Return to Problems 55 – 62 and use a calculator to evaluate each expression. Then evaluate the result

that you obtained when you rationalized the denominators.

Answers to the Example Practice Skills 3 3 1. (a) 2 27 (b) 10221 (c) 22 5 (d) 302 2 3

3

(d) 29 3 22 5.

2. (a) 221 215 (b) 6 5 22 (c) a 2ab

3. (a) 322 242 26 214 (b) 13 625

31 27 222

6. 2 25 3 22

5

7.

a 72a 12 a 16

8.

4. (a) 31 (b) 13 (c) 4

4 326 2


9.5

2. True


Solving Radical Equations OBJECTIVES 1

Solve Radical Equations of the Form 2ax b c

2

Solve Radical Equations of the Form 2ax b 2cx d

3

Solve Radical Equations of the Form 2ax b cx d

4

Solve and Apply Radical Formulas

1 Solve Radical Equations of the Form 2ax b c Equations that contain radicals with variables in the radicand are called radical equations. Here are some examples of radical equations that involve one variable. 2x 3

22x 1 5

23x 4 4

25s 2 22s 19

22y 4 y 2

2x 6 x

In order to solve such equations we need the following property of equality.

Property 9.4 For real numbers a and b, if a b, then a2 b2

9.5

Solving Radical Equations

465

Property 9.4 states that we can square both sides of an equation. However, squaring both sides of an equation sometimes produces results that do not satisfy the original equation. Let’s consider two examples to illustrate the point.

EXAMPLE 1

Solve 2x 3.

Solution 2x 3 1 2x2 2 32

Square both sides

x9 Since 29 3, the solution set is 596 .

▼ PRACTICE YOUR SKILL Solve 2y 9.

EXAMPLE 2

■

Solve 2x 3.

Solution 2x 3 1 2x2 2 132 2

Square both sides

x9 Because 29 3, 9 is not a solution and the solution set is .

▼ PRACTICE YOUR SKILL Solve 2a 12 .

■

In general, squaring both sides of an equation produces an equation that has all of the solutions of the original equation, but it may also have some extra solutions that do not satisfy the original equation. (Such extra solutions are called extraneous solutions or roots.) Therefore, when using the “squaring” property (Property 9.4), you must check each potential solution in the original equation. We now consider some examples to demonstrate different situations that arise when solving radical equations.

EXAMPLE 3

Solve 22x 1 5.

Solution 22x 1 5 1 22x 12 2 52 2x 1 25 2x 24 x 12

Square both sides

466

Chapter 9

Roots and Radicals

✔ Check 22x 1 5 221122 1 ⱨ 5 224 1 ⱨ 5 225 ⱨ 5 55


▼ PRACTICE YOUR SKILL Solve 23y 1 4 .

EXAMPLE 4

■

Solve 23x 4 4.

Solution 23x 4 4 1 23x 42 2 142 2

Square both sides

3x 4 16 3x 12 x4

✔ Check 23x 4 4 23142 4 ⱨ 4 216 ⱨ 4 4 4 Since 4 does not check (4 is an extraneous root), the equation 23x 4 4 has no real number solutions. The solution set is .

▼ PRACTICE YOUR SKILL Solve 22y 8 10 .

EXAMPLE 5

■

Solve 3 22y 1 5.

Solution 322y 1 5 22y 1

5 3

5 2 1 22y 12 2 a b 3 2y 1

25 9


9.5

2y

25 1 9

2y

25 9 9 9

2y

16 9

1 1 16 12y2 a b 2 2 9 y


Multiplied both sides by

467

1 2

8 9

✔ Check 3 22y 1 5 3

B

8 2a b 1 ⱨ 5 9 16 9 ⱨ5 B9 9

3

25 ⱨ5 B9

3

5 3a b ⱨ 5 3 55 8 The solution set is e f . 9

▼ PRACTICE YOUR SKILL Solve 223x 6 24 .

■

2 Solve Radical Equations of the Form 2ax b 2cx d EXAMPLE 6

Solve 25s 2 22s 19.

Solution 25s 2 22s 19 1 25s 22 2 1 22s 192 2 5s 2 2s 19 3s 21 s7

Square both sides

468

Chapter 9

Roots and Radicals

✔ Check 25s 2 22s 19 25172 2 ⱨ 22172 19 233 233


▼ PRACTICE YOUR SKILL Solve 22y 5 24y 7.

■

3 Solve Radical Equations of the Form 2ax b cx d EXAMPLE 7

Solve 22y 4 y 2.

Solution 22y 4 y 2

1 22y 42 2 1y 22 2

Square both sides

2y 4 y 4y 4 2

0 y2 6y 8

0 1y 42 1y 22 y40

or

y20

y4

or

y2

Factor the right side Remember the property: ab 0 if and only if a 0 or b 0

✔ Check 22y 4 y 2 or 22142 4 ⱨ 4 2

or

22y 4 y 2 22122 4 ⱨ 2 2

24 ⱨ 2

or

20 ⱨ 0

22

or

00


▼ PRACTICE YOUR SKILL Solve 22y 8 y 4 .

EXAMPLE 8

■

Solve 2x 6 x.

Solution 2x 6 x 2x x 6

1 2x2 1x 62 2

2

We added 6 to both sides so that the term with the radical is alone on one side of the equation

x x2 12x 36 0 x2 13x 36

0 1x 421x 92

9.5

x40 x4


469

x 9 0 Apply ab 0 if and only if a 0 or b 0 x9

or or

✔ Check 2x 6 x

or

2x 6 x

24 6 ⱨ 4 26ⱨ4 84

or or or

29 6 ⱨ 9 36ⱨ9 99


▼ PRACTICE YOUR SKILL Solve 2 2y y .

■

Note in Example 8 that we changed the form of the original equation, 1x 6 x, to 1x x 6 before we squared both sides. Squaring both sides of 1x 6 x produces x 121x 36 x2, a more complex equation that still contains a radical. So again, it pays to think ahead a few steps before carrying out the details of the problem.

4 Solve and Apply Radical Formulas In Section 9.1 we used the formula S 230Df to approximate how fast a car was traveling based on the length of skid marks. (Remember that S represents the speed of the car in miles per hour, D the length of skid marks measured in feet, and f a coefficient of friction.) This same formula can be used to estimate the lengths of skid marks that are produced by cars traveling at different rates on various types of road surfaces. To use the formula for this purpose, we change the form of the equation by solving for D. 230Df S 30Df S2 D

Brand X Pictures/Jupiter Images

EXAMPLE 9

The result of squaring both sides of the original equation 2

S 30f

D, S, and f are positive numbers, so this final equation and the original one are equivalent

Apply Your Skill Suppose that for a particular road surface, the coefficient of friction is 0.35. How far will a car traveling at 60 miles per hour skid when the brakes are applied?

Solution We substitute 0.35 for f and 60 for S in the formula D D

S2 . 30f

602 343 to the nearest whole number 3010.352

The car will skid approximately 343 feet.

▼ PRACTICE YOUR SKILL Suppose that for a particular road surface, the coefficient of friction is 0.40. How far will a car traveling at 55 miles per hour skid when the brakes are applied? ■

470

Chapter 9

Roots and Radicals

Remark: Pause for a moment and think about the result in Example 9. The coefficient of friction of 0.35 applies to a wet concrete road surface. Note that a car traveling at 60 miles per hour will skid farther than the length of a football field.

CONCEPT QUIZ

For Problems 1–5, answer true or false. 1. Equations that contain radicals with variables in the radicand are called radical equations. 2. Squaring both sides of a radical equation produces an equivalent equation. 3. Extra solutions that do not satisfy the original equation are called simultaneous solutions. 4. Because squaring both sides of an equation can produce extraneous solutions, all potential solutions must be checked. 5. The equation 2x 3 has no real number solutions.

Problem Set 9.5 1 Solve Radical Equations of the Form 2ax b c For Problems 1–20, solve each equation. Be sure to check all potential solutions in the original equation.

3 Solve Radical Equations of the Form 2ax b cx d For Problems 27– 40, solve each equation. Be sure to check all potential solutions in the original equation. 28. 2x 7 x 7

1. 2x 7

2. 2x 12

27. 2x 3 x 3

3. 22x 6

4. 23x 9

29. 22x 28 x 2

5. 23x 6

6. 22x 8

30. 22x x 4

31. 23x x 6

7. 24x 3

8. 25x 4

32. 22x x 3

33. 42x 5 x

9. 3 2x 2

10. 4 2x 3

34. 2x 6 x

35. 2x2 27 x 3

11. 22n 3 5

12. 23n 1 7

13. 25y 2 1

14. 24n 3 4 0

15. 26x 5 3 0

16. 25x 3 4

17. 5 2x 30

18. 6 2x 42

19. 2 2y 1 5

20. 3 2y 2 4

2 Solve Radical Equations of the Form 2ax b 2cx d For Problems 21–26, solve each equation. Be sure to check all potential solutions in the original equation. 21. 23a 2 22a 4 22. 24a 3 25a 4 23. 27x 3 24x 3 24. 28x 6 24x 11 25. 23n 4 2n 26. 25n 1 22n

36. 2x2 35 x 5 37. 2x2 2x 3 x 2 38. 2x2 x 4 x 3 39. 28x 2 x

40. 22x 4 x 6

4 Solve and Apply Radical Formulas 41. Go to the formula given in the solution to Example 9; use a coefficient of friction of 0.95. How far will a car skid at 40 miles per hour? At 55 miles per hour? At 65 miles per hour? Express the answers to the nearest foot. L for L. (Remember that B 32 in this formula, which was used in Section 9.1, T represents the period of a pendulum expressed in seconds, and L represents the length of the pendulum in feet.)

42. Solve the formula T 2p

43. In Problem 42 you should have obtained the equation 8T2 L 2 . What is the length of a pendulum that has a p period of 2 seconds? Of 2.5 seconds? Of 3 seconds? Use 3.14 as an approximation for p and express the answers to the nearest tenth of a foot.

9.5


471

THOUGHTS INTO WORDS 13 22x2 2 x2

44. Explain in your own words why possible solutions for radical equations must be checked.

9 122x 4x x2

45. Your friend attempts to solve the equation

At this step, she stops and doesn’t know how to proceed. What help can you give her?

3 2 2x x as follows:

FURTHER INVESTIGATIONS For Problems 46 – 49, solve each of the equations.

To solve an equation with cube roots, we raise each side of the equation to the third power. Consider the following example.

46. 2x 1 5 2x 4 47. 2x 2 2x 7 1

3 2 x45

48. 22n 1 2n 5 3

3 12 x 42 3 53

49. 22n 1 2n 3 2

x 4 125

50. Suppose that we solve the equation x a for x as follows: 2

2

For Problems 51–55, solve each equation.

x2 a2 0

1x a21x a2 0 xa0 x a

x 129

or

xa0

or

xa

3 51. 2 x74

3 52. 2 x 12 2

3 53. 2 2x 4 6

3 54. 2 4x 5 3

3 55. 2 4x 8 10

What is the significance of this result?


2.

3. {5}

4.

5. {50}

6. {6}

7. {4, 6}


2. False


8. {4}

9. Approximately 252 feet


OBJECTIVE

SUMMARY

EXAMPLE

Find the square root of a perfect square. (Sec. 9.1, Obj. 1, p. 436)

A square root of a number is one of its two equal factors. For example, a square root of 81 is 9 because 9 # 9 81. Every positive real number has two square roots. Negative real numbers have no real number square roots.

Find the following square roots without the use of a calculator. 36 (a) 2121 (b) B 49

Problems 1– 6

Solution

(a) 2121 11 36 6 (b) B 49 7

Find the cube root of a perfect cube. (Sec. 9.1, Obj. 2, p. 437)

A cube root of a number is one of its three equal factors. Every positive real number has one positive real number cube root. Every negative real number has one negative real number cube root.

Find the following cube roots without the use of a calculator. 3 (a) 2 125

Problems 7–10

3 (b) 2 64

Solution 3 (a) 2 125 5 3

(b) 264 4 Find an approximation of a square or cube root. (Sec. 9.1, Obj. 3, p. 439)

Many square and cube roots are irrational numbers. A calculator can be used to find decimal approximations for these irrational numbers. Be sure you know how to use your calculator to find square and cube roots.

Use a calculator to find a decimal approximation of each root. Express the answer to the nearest hundredth. (a) 2427

Problems 11–16

3 (b) 2 400

Solution

(a) 2427 20.66 3 (b) 2400 7.37

Change radicals to simplest radical form. (Sec. 9.2, Obj. 1, p. 444)

Change radicals that contain variables to simplest radical form. (Sec. 9.2, Obj. 2, p. 446)

The property 2ab 2a2b 3 3 3 ( 2ab 2a2b) provides the basis for changing radicals to simplest radical form. For square (cube) roots one condition for simplest radical form is that the radicand contains no factors other than 1 that are perfect squares (cubes).

Change each radical to simplest radical form.

Here we maintain the assumption that all variables represent positive real numbers.

Change the radical 28x2y3 to simplest radical form.

(a) 2240

Problems 17–22

3 (b) 2160

Solution

(a) 2240 216215 4 215 3 3 3 3 (b) 2160 28220 2 220

Problems 23 –28

Solution

28x2y3 24x2y2 22y 2xy 22y

472

(continued)

Chapter 9

Summary

473


OBJECTIVE

SUMMARY

EXAMPLE

Combine radicals after simplifying. (Sec. 9.2, Obj. 3, p. 447)

First change each radical to simplest radical form. Then use the distributive property to combine like radicals.

Simplify the expression 250 218 232

Problems 29 –34

Solution

250 218 232 22522 2922 21622 5 22 3 22 422 15 3 42 22 6 22

Simplify the roots of quotients. (Sec. 9.3, Obj. 1, p. 451)

The properties

a 2a and Bb 2b

3 a 2 a 3 can be used to simplify Bb 2b some quotients of radicals. 3

Rationalize the denominator of a square root. (Sec. 9.3, Obj. 2, p. 453)

Rationalizing the denominator may be necessary to ensure that no fraction appears within a radical sign and that no radical appears in the denominator. To rationalize the denominator we multiply the radical expression by a form of 1 that will result in a perfect square or a rational number in the denominator.

Simplify the quotient. 7 B 25 Solution

7 27 27 B 25 5 225 Change to simplest radical form. 3 B7

Change each radical to simplest radical form. This may involve rationalizing the denominator. Then use the distributive property to combine like radicals.

Problems 39 –50

Solution

3 23 B7 27

Combine radicals after rationalizing the denominators. (Sec. 9.3, Obj. 3, p. 455)

Problems 35 –38

Simplify 2

23 27

27

#

221 249

27

221 7 Problems 50 –54

25

6 245

Solution

2 25

6 245

2 25

#

25 25

6 2925

2 25 6132 25 5

2 25 1825 5

2 25 9025 9225 5 5 5

(continued)

474

Chapter 9

Roots and Radicals


OBJECTIVE

SUMMARY

EXAMPLE

Multiply radicals. (Sec. 9.4, Obj. 1, p. 458)

Use the properties 2ab 2a2b 3 3 3 and 2ab 2a2b to multiply radicals and to simplify the resulting radical when appropriate.

Find the product and express the answer in simplest radical form. 3 3 2 4xy2 2 4xy

Problems 55 –58

Solution 3

3

3

24xy2 24xy 216x2y3 3 3 2 8y3 2 2x2 3 2y 2 2x2

Multiply radicals using the distributive property. (Sec. 9.4, Obj. 2, p. 458)

The process of multiplying radical expressions that are in binomial form is the same as multiplying binomials.

Find the product and express the answer in simplest radical form. 1 25 2221325 222

Problems 59 – 64

Solution

1 25 2221325 222 1 25 222 1325 222 3225 210 3 210 24 3152 210 3 210 2 15 2210 2 13 2210

Use conjugates to rationalize the denominator. (Sec. 9.4, Obj. 3, p. 460)

To rationalize the denominator when the denominator of a radical expression is in binomial form, multiply by the conjugate of the denominator. The conjugate of 2a 2b is 2a 2b.

Rationalize the denominator. 8

Problems 65 – 68

25 3 Solution

8 25 3

8 25 3

25 3

#

25 3

81 25 32

1 25 32 1 25 32 81 25 32 225 3 25 3 25 9 81 25 32 59

21 25 32

81 25 32 4 (continued)

Chapter 9

Summary

475

OBJECTIVE

SUMMARY

EXAMPLE


Solve radical equations of the form 2ax b c. (Sec. 9.5, Obj. 1, p. 464)

Use the property of equality, if a b, then a2 b2. Squaring both sides of an equation produces an equation that may have some solutions that do not satisfy the original equation.

Solve 23x 2 5.

Problems 69 –74

Therefore, when using the “squaring” property, you must check each potential solution in the original equation.

Solution

23x 2 5 1 23x 22 2 52 3x 2 25 x

23 3

Check: 23x 2 5 B

3a

23 b25 3 225 5 55

The solution set is e Solve radical equations of the form 2ax b 2cx d. (Sec. 9.5, Obj. 2, p. 467)

See the previous remark about solving radical equations.

23 f. 3

Solve 22x 1 23x.

Problems 75 –76

Solution

22x 1 23x

1 22x 12 2 1 23x2 2 2x 1 3x 1x Check: 22112 1 23112 23 23 The solution set is {1}.

(continued)

476

Chapter 9

Roots and Radicals

OBJECTIVE

SUMMARY

EXAMPLE


Solve radical equations of the form 2ax b cx d. (Sec. 9.5, Obj. 3, p. 468)

See the previous remark about solving radical equations.

Solve 22y 4 y 2.

Problems 77– 80

Solution

22y 4 y 2 1 22y 42 2 1y 22 2 2y 4 y2 4y 4 0 y2 6y 0 y1y 62 y 0 or y 6 Check: When y 0 22102 4 0 2 24 2

2 2 1Does not check2

Check: When y 6 22162 4 6 2 216 4

4 4 1Checks2

The solution set is {6}. Solve and apply radical formulas. (Sec. 9.1, Obj. 5, p. 441; Sec. 9.5, Obj. 4, p. 469)

The formula S 230Df is used to approximate the speed of a car by measuring the length of the skid marks. This formula can also be used to determine the distance of the skid knowing the speed of the car and the coefficient of friction of the road surface.

Solve S 230Df for D.

Problems 81– 82

Solution

S 230Df S2 1 230Df2 2 S2 30Df S2 D 30f

Chapter 9 Review Problem Set For Problems 1–10, evaluate each expression without the use of a calculator. 1. 264 4.

81 B 25

3 7. 2 27

8 10. B 125 3

2. 249 4 B9

3. 21600 49 B 36

5.

6.

3 8. 2 125

3 9. 2 64

For Problems 11–16, use a calculator to find a rational approximation of each root. Round each answer to the nearest hundredth. 11. 2158

12. 21250

13. 2860

14. 2250

3 15. 2 612

3 16. 2 789

Chapter 9

Review Problem Set

477


For Problems 55 – 64, find the products and express your answers in simplest radical form.

17. 220

55. 1 2621 2122

56. 12 23213262

57. 1528212222

3 3 58. 12 2 72152 42

18. 232

19. 5 28

3

20. 280

3

21. 224

22. 2250

For Problems 23 –28, change each radical to simplest radical form. All variables represent positive real numbers. 23. 212x

2

25. 3 220a b 3

61. 1 23 2521 23 272

26. 248a

62. 12 23 32221 23 5222

4

3

27. 28x y

63. 1 26 22721326 272

28. 227x

2 3

60. 3251 28 22122

24. 250x y 2

3 2

3 3 3 59. 2 21 2 3 2 42

4

For Problems 29 –34, simplify each radical expression. All variables represent positive real numbers.

64. 13 225214 3 252 For Problems 65 – 68, rationalize the denominator and simplify.

29. 2 250 3 272 5 28 30. 248 2 275 212

65.

31. 28x 3 218x 32. 220y 245y 3 25y

66.

3 3 33. 2 27x4 x2 8x 3 3 34. 2 27x2 32 8x2

67.

For Problems 35 –38, change each radical to simplest radical form. All variables represent positive real numbers. 3 3 2 40 2 40 35. 36. 3 3 25 28 2x3 37. B 9

38.

272a 216b2

For Problems 39 –50, rationalize the denominator and simplify. All variables represent positive real numbers. 39.

236

40.

27 8 42. B 24 45. 48.

3 22

25 4 43. Bx

4 23

46.

212 4 26

49.

3 212

5 22 2 23 3 2x 4 2y3

41. 44. 47. 50.

7 B8 22x 25y

68.

5 27 25 26 23 22 2 322 26 26 327 2210

For Problems 69 – 80, solve each of the equations. 69. 25x 6 6

70. 22x 1 13

71. 28 x 6

72. 22x 8

73. 22x 3 2

74. 23x 1 5

75. 26x 1 23x 13 76. 23x 4 2x 7

322

77. 2y 5 y 5

227

78. 22x 4 x 6

22x2y 5 2xy

79. 23a 10 a 2 80. 3 22x 1 2

For Problems 51–54, simplify each expression. 51.

2 2254 B3

53. 4 220 54.

3 25

3 3260 B5

52. 3 210 245

2 B5

81. The time T, measured in seconds, that it takes an object to fall d feet (neglecting air resistance) is given by the d . Find the times that it takes objects to B 16 fall 100 feet, 350 feet, and 500 feet, respectively. Express the answers to the nearest tenth of a second. formula T

82. Solve the formula 222ab c for b.

Chapter 9 Test 1.

64 1. Evaluate . B 49

2.

2. Evaluate 20.0025. For Problems 3 –5, use a calculater to evaluate each of the following to the nearest tenth.

3.

3. 28

4.

4. 232

5.

5.

3 22

For Problems 6 –14, change each radical expression to simplest radical form. All variables represent positive real numbers. 6.

6. 245

7.

7. 4254

8.

8.

3

223 326 25 B2

9.

9.

10.

10.

11.

11.

12.

3 12. 2250x4y3

13.

13.

14.

14.

224 236 5 B8

23x 25y 3 248x3y2 4

For Problems 15 –18, find the indicated products and express the answers in simplest radical form. 15.

15. 1 2821 2122

16.

3 3 16. 1625214222

17.

17. 2612 212 3 282

18.

18. 12 25 2321 25 3 232

19.

19. Rationalize the denominator and simplify: 26 212 22

20.

478

20. Simplify 2224 4 254 3 296.

Chapter 9

For Problems 21–24, solve each equation. 21. 23x 1 4

21.

22. 22x 5 4

22.

23. 2n 3 3 n

23.

24. 23x 6 x 2

24.

L 25. Use the formula T 2p , in which T is the period of a pendulum expressed B 32 in seconds, and L is the length of the pendulum in feet, to find the period of a pendulum 3.5 feet long. Express your answer to the nearest tenth.

25.

Test

479

Chapters 1–9 Cumulative Review Problem Set For Problems 1– 6, evaluate each of the numerical expressions. 1. 26 3. a

1 2 1 b 3 4

4 5. B9

1 3 2. a b 4

30. Find the total surface area of a sphere that has a radius 5 inches long. Use 3.14 as an approximation for p.

4. 264

31. Write each number in scientific notation.

1

6. 3 3 0

2

3

For Problems 7–10, evaluate each algebraic expression for the given values of the variables. 7. 312x 12 412x 32 1x 62

for x 4

8. 13x2 4x 62 13x2 3x 12

for x 6

9. 21a b2 312a b2 21a 3b2 and b 3 10. x2 2xy y2

for a 2

for x 5 and y 2

For Problems 11–25, perform the indicated operations, and express your answers in simplest form using positive exponents only. 11.

3 5 7 4x 2x x

12.

13.

6x 3x 7y 35y2

14.

x2 x2 6x 9 # x2 x 6 x2 x 12

15.

8 7 x3 x2 3x 18

4 3 x2 x3

(c) 0.00000104

(d) 53,000,000

For Problems 32 –37, factor each expression completely. 32. 12x3 14x2 40x

33. 12x2 27

34. xy 3x 2y 6

35. 30 19x 5x2

36. 4x4 4

37. 21x2 22x 8

For Problems 38 – 43, change each radical expression to simplest radical form. 38. 4228

39. 245

36 B5

41.

42. 272xy5

43.

40.

528 6212 2 2ab2 52b

45. 6 221928 32122

46. 1322 2721322 272

3

12a2b3 4a5b4

18.

19. 13n4 2 1

20. 19x 2213x 42

21. 1x 1215x 72

22. 13x 1212x2 x 42

15x6y8 20x3y5 3 2

5x y

24. 110x 8x2 17x 32 15x 12 3

1 1 x y 25. 1 xy 26. If 2 gallons of paint will cover 1500 square feet of walls, how many gallons are needed for 3500 square feet? 27. 18 is what percent of 72? 1 28. Solve V Bh for B if V 432 and h 12. 3 480

(b) 0.0009

44. 13 28214222

17. 14x5 2 12x3 2

23.

(a) 85,000

For Problems 44 – 46, find each product, and express your answer in simplest radical form.

16. 13xy2 14y 2 15x y2 2

29. How many feet of fencing are needed to enclose a rectangular garden that measures 25 feet by 40 feet?

For Problems 47 and 48, rationalize the denominator and simplify. 4 6 47. 48. 23 22 325 26

For Problems 49 and 50, simplify each of the radical expressions. 49. 3250 7272 4298 50.

3 2 220 245 280 3 4

For Problems 51–55, graph each of the equations. 1 51. 3x 6y 6 52. y x 4 3 2 53. y x 3 5 55. y x

54. y 2x 0

Chapters 1–9

Cumulative Review Problem Set

For Problems 56 –58, graph each linear inequality.

For Problems 81– 86, solve each of the inequalities.

56. y 2x 6

81. 3n 4 11

57. 3x 2y 6

58. 2x 4y 8

83. 21x 22 31x 42 6

For Problems 59 – 64, solve each of the problems.

84.

1 2 n n 1 2 3


86.

x3 x2 9 7 4 14

60. Find the slope of the line determined by the equation 4x 7y 12. 2 61. Write the equation of the line that has a slope of and 3 contains the point (7, 2). 62. Write the equation of the line that contains the points (4, 1) and (1, 3). 1 63. Write the equation of the line that has a slope of 4 and a y intercept of 3.

481

82. 5 3n 4 7n

85.

x1 x2 3 2 6 8

For Problems 87–97, set up an equation, an inequality, or a system of equations to help solve each problem. 87. If two angles are supplementary and the larger angle is 15

less than twice the smaller angle, find the measure of each angle. 88. The sum of two numbers is 50. If the larger number is 2 less than three times the smaller number, find the numbers.

64. Find the slope of a line whose equation is 3x 2y 12.

89. The sum of the squares of two consecutive odd whole numbers is 130. Find the numbers.

For Problems 65 – 68, solve each of the systems by using either the substitution method or the addition method. 4x 3y 20 b 3x 5y 14

90. Suppose that Nick has 47 coins consisting of nickels, dimes, and quarters. The number of dimes is 1 more than twice the number of nickels, and the number of quarters is 4 more than three times the number of nickels. Find the number of coins of each denomination.

2x 7y 22 b 68. a 4x 5y 13

91. A home valued at $140,000 is assessed $2940 in real estate taxes; at the same rate, what would be the taxes on a home assessed at $180,000?

65. a

y 3x 5 b 3x 4y 5

1 x 2 67. ± 1 x 3

2 y 11 3 ≤ 5 y8 6

66. a

For Problems 69 – 80, solve each equation. 69. 21n 12 412n 32 41n 62 70.

1 4 x1 x6

71.

t2 5 t1 3 4 12

n5 n4 3 2 5

74. 0.11x 0.141x 4002 181 75.

4 x 7 60 x 60 x

76. 1

x1 3 2x 4

78. 2x2 8 0 80. 23n 2 7

93. Rosa leaves a town traveling in her car at a rate of 45 miles per hour. One hour later Polly leaves the same town traveling the same route at a rate of 55 miles per hour. How long will it take Polly to overtake Rosa? 94. How many milliliters of pure acid must be added to 100 milliliters of a 10% acid solution to obtain a 20% solution?

72. 7 2n 6n 7n 5n 12 73.

92. A retailer has some skirts that cost her $30 each. She wants to sell them at a profit of 60% of the cost. What price should she charge for the skirts?

77. x2 4x 12 0 79. 23x 6 9

95. Suppose that Andy has scores of 85, 90, and 86 on his first three algebra tests. What score must he get on the fourth algebra test to have an average of 88 or higher for the four tests? 96. The Cubs have won 70 games and lost 72 games. They have 20 more games to play. To win more than 50% of all their games, how many of the 20 games remaining must they win? 97. Seth can do a job in 20 minutes. Butch can do the same job in 30 minutes. If they work together, how long will it take them to complete the job?

10

Quadratic Equations

10.1 Quadratic Equations 10.2 Completing the Square 10.3 Quadratic Formula 10.4 Solving Quadratic Equations—Which Method?

Travel Ink /Getty Images

10.5 Solving Problems Using Quadratic Equations

■ The quadratic equation t(2t 8) 330 can be used to determine the number of trees per row and the number of rows, given that the number of trees per row is 8 less than twice the number of rows in an orchard of 330 trees.

T

he area of a singles tennis court is 2106 square feet. The length of the court is

26 times the width. Find the length and width of the court. We 9

can use the quadratic equation x a

26 xb 2106 to determine that the court is 27 feet 9

wide and 78 feet long. Solving equations has been a central theme of this text. We now pause for a moment and reflect on the different types of equations that we have solved.

Type of equation

Examples

First-degree equations of one variable

4x 3 7x 1; 3(x 6) 9

Second-degree equations of one variable that are factorable

x2 3x 0; x2 5x 6 0; x2 4 0; x2 10x 25 0

Rational equations

2 3 6 5 4; ; x x a2 a3 5 6 2 x2 x2 x2 4 (continued)


483

I N T E R N E T

P R O J E C T

Quadratic equations have two solutions. The derivation of the word quadratic is from Latin. In the English language, the prefix quad- means four. Do an Internet search to discover the origins of the name quadratic for these equations that have two answers.

Type of equation

Examples 1x 4; 1y 2 3;

Radical equations

1a 1 12a 7 2x 3y 4 b; 5x y 7 3a 5b 9 b a 7a 9b 12 a

Systems of equations

As indicated in the chart, we have learned how to solve some second-degree equations, but only those for which the quadratic polynomial is factorable. In this chapter, we will expand our work to include more general types of second-degree equations in one variable and thus broaden our problem-solving capabilities.

10.1 Quadratic Equations OBJECTIVES 1

Solve Quadratic Equations by Factoring

2

Solve Quadratic Equations by Applying the Square-Root Property

3

Solve Word Problems Involving the Pythagorean Theorem

1 Solve Quadratic Equations by Factoring A second-degree equation in one variable contains the variable with an exponent of 2, but no higher power. Such equations are also called quadratic equations. Here are some examples of quadratic equations. x2 25

y2 6y 0

x2 7x 4 0

4y2 2y 1 0

5x2 2x 1 2x2 6x 5

We can also define a quadratic equation in the variable x as any equation that can be written in the form ax 2 bx c 0, where a, b, and c are real numbers and a 0. We refer to the form ax 2 bx c 0 as the standard form of a quadratic equation. In Chapter 5, you solved quadratic equations (we didn’t use the term “quadratic” at that time) by factoring and applying the property, ab 0 if and only if a 0 or b 0. Let’s review a few examples of that type.

EXAMPLE 1

Solve x 2 13x 0.

Solution x2 13x 0 x1x 132 0 484

Factor left side of equation

10.1

x0

or

x0

or

x 13 0

Quadratic Equations

485

Apply ab 0 if and only if a 0 or b 0

x 13

The solution set is 50, 136 . Don’t forget to check these solutions!

▼ PRACTICE YOUR SKILL Solve y2 7y 0.

EXAMPLE 2

■

Solve n2 2n 24 0.

Solution n2 2n 24 0

1n 62 1n 42 0 n60 n 6

Factor left side

or

n40

or

n4



▼ PRACTICE YOUR SKILL Solve x2 4x 21 0.

EXAMPLE 3

■

Solve x2 6x 9 0.

Solution x2 6x 9 0

1x 32 1x 32 0 x30 x 3

or or


Factor left side

x30


x 3

▼ PRACTICE YOUR SKILL Solve a2 10a 25 0.

EXAMPLE 4

■

Solve y2 49.

Solution y2 49 y2 49 0

1y 721y 72 0 y70 y 7

Factor left side

or

y70

or

y7



▼ PRACTICE YOUR SKILL Solve n2 144.

■

486

Chapter 10

Quadratic Equations

Note the type of equation that we solved in Example 4. We can generalize from that example and consider the equation x2 a, where a is any nonnegative real number. We can solve this equation as follows: x2 a

x2 1 1a2 2

1 1a2 2 a

x2 1 1a2 2 0

1x 1a2 1x 1a2 0 x 1a 0 x 1a

or

Factor left side

x 1a 0

or

x 1a


The solutions are 1a and 1a. The process we used to solve this equation is a general property (stated formally below), which can be applied to solve certain types of quadratic equations.

2 Solve Quadratic Equations by Applying the Square-Root Property Property 10.1 For any nonnegative real number a, x 2 a if and only if x 1a or x 1a (The statement “x 1a or x 1a” can be written as x 1a.)

Property 10.1 is sometimes referred to as the square-root property. This property, along with our knowledge of square roots, makes it very easy to solve quadratic equations of the form x2 a.

EXAMPLE 5

Solve x2 81.

Solution x2 81 x 181

Apply Property 10.1

x 9


▼ PRACTICE YOUR SKILL Solve y2 64.

EXAMPLE 6

■

Solve x2 8.

Solution x2 8 x 18 x 2 12

18 1412 212


10.1

Quadratic Equations

487

▼ PRACTICE YOUR SKILL Solve n2 12.

EXAMPLE 7

■

Solve 5n2 12.

Solution 5n2 12 n2

12 5

n

12 A5

n

2115 5


12 112 112 A5 15 15


#

15 2115 160 5 5 15

2 115 2 115 , f. 5 5

▼ PRACTICE YOUR SKILL Solve 3y2 8.

EXAMPLE 8

■

Solve 1x 22 2 16.

Solution 1x 22 2 16 x 2 4 x24

or

x 2 4

x6

or

x 2


▼ PRACTICE YOUR SKILL Solve (a 5)2 9.

EXAMPLE 9

■

Solve 1x 52 2 27.

Solution 1x 52 2 27 x 5 127 x 5 313

127 1913 313

x 5 3 13 x 5 313

or or

x 5 313 x 5 3 13

The solution set is 55 3 13, 5 3136.

▼ PRACTICE YOUR SKILL Solve (y 7)2 20.

■

488

Chapter 10

Quadratic Equations

It may be necessary to change the form before we can apply Property 10.1. The next example illustrates this procedure.

EXAMPLE 10

Solve 312x 32 2 8 44.

Solution 312x 32 2 8 44 312x 32 2 36 12x 32 2 12 2x 3 112 2x 3 112

or

Apply Property 10.1

2x 3 112

2x 3 112

or

x

3 112 2

or

x

3 112 2

x

3 213 2

or

x

3 213 2


2x 3 112

112 1413 2 13

3 2 13 3 213 , f. 2 2

▼ PRACTICE YOUR SKILL Solve 3(5n 1)2 4 58.

■

Note that quadratic equations of the form x2 a, where a is a negative number, have no real number solutions. For example, x2 4 has no real number solutions, because any real number squared is nonnegative. In a like manner, an equation such as 1x 32 2 14 has no real number solutions.

3 Solve Word Problems Involving the Pythagorean Theorem Our work with radicals, Property 10.1, and the Pythagorean theorem merge to form the basis for solving a variety of problems that pertain to right triangles. First, let’s restate the Pythagorean theorem.

Pythagorean Theorem If for a right triangle, a and b are the measures of the legs, and c is the measure of the hypotenuse, then

a

c

a 2 b2 c 2 (The hypotenuse is the side opposite the right angle, and the legs are the other two sides, as shown in Figure 10.1.)

b Figure 10.1

10.1

EXAMPLE 11

Quadratic Equations

489

Apply Your Skill Find c in Figure 10.2.

c

3 centimeters

Solution Applying the Pythagorean theorem, we have

4 centimeters

c2 a2 b2 c2 32 42

Figure 10.2

c2 9 16 c2 25 c5 The length of c is 5 centimeters.

▼ PRACTICE YOUR SKILL Find c in the figure below.

c

5 inches

■

12 inches

Remark: Don’t forget that the equation c2 25 does have two solutions, 5 and 5. However, because we are finding the lengths of line segments, we can disregard the negative solutions.

EXAMPLE 12

Apply Your Skill A 50-foot rope hangs from the top of a flagpole. When pulled taut to its full length, the rope reaches a point on the ground 18 feet from the base of the pole. Find the height of the pole to the nearest tenth of a foot.

Solution We can sketch Figure 10.3 and record the given information. Using the Pythagorean theorem, we solve for p as follows: 50 feet

p

p2 182 502 p2 324 2500 p2 2176 p 12176 46.6

18 feet p represents the height of the flagpole. Figure 10.3


The height of the flagpole is approximately 46.6 feet.

▼ PRACTICE YOUR SKILL A guy wire that is 30 feet in length hangs from the top of a tower. When pulled taut to its full length, the wire reaches a point 12 feet from the base of the tower. Find the height of the tower to the nearest tenth of a foot. ■

490

Chapter 10

Quadratic Equations

An isosceles triangle has two sides of the same length. Thus an isosceles right triangle is a right triangle that has both legs of the same length. The next example presents a problem involving an isosceles right triangle.

EXAMPLE 13

Apply Your Skill Find the length of each leg of an isosceles right triangle if the hypotenuse is 8 meters long.

8 meters

x

Solution We sketch an isosceles right triangle in Figure 10.4 and let x represent the length of each leg. Then we can determine x by applying the Pythagorean theorem.

x

x2 x2 82

Figure 10.4

2x2 64 x2 32 x 132 11612 4 12 Each leg is 412 meters long.

▼ PRACTICE YOUR SKILL Find the length of each leg of an isosceles right triangle if the hypotenuse is 12 feet long. ■

Remark: In Example 12, we made no attempt to express 12176 in simplest radical form because the answer was to be given as a rational approximation to the nearest tenth. However, in Example 13 we left the final answer in radical form and expressed it in simplest radical form. Another special kind of right triangle is one that contains acute angles of 30 and 60 . In such a right triangle, often referred to as a 30 – 60 right triangle, the side opposite the 30 angle is equal in length to one-half the length of the hypotenuse. This relationship, along with the Pythagorean theorem, provides us with another problem-solving technique.

EXAMPLE 14

Apply Your Skill Suppose that a 20-foot ladder is leaning against a building and makes an angle of 60

with the ground. How far up the building does the top of the ladder reach? Express your answer to the nearest tenth of a foot.

h

20

fee t

Ladder

30°

Solution Figure 10.5 illustrates this problem. The side opposite the 30 angle equals one-half 1 of the hypotenuse, so the length of that side is 1202 10 feet. Now we can apply 2 the Pythagorean theorem. h2 102 202

60° 10 feet ( 12 (20) = 10) Figure 10.5

h2 100 400 h2 300 h 1300 17.3


The top of the ladder touches the building at approximately 17.3 feet from the ground.

10.1

Quadratic Equations

491

▼ PRACTICE YOUR SKILL Suppose that a 16-foot ladder is leaning against a building and makes an angle of 60° with the ground. How far up the building does the top of the ladder reach? Express your answer to the nearest tenth of a foot. ■

CONCEPT QUIZ

For Problems 1–10, answer true or false. 1. If an equation has any variable raised to the second power, it is a quadratic equation. 2. Any quadratic equation in the variable x can be written in the form ax2 bx c 0, where a, b, and c are real numbers and a 0. 3. We refer to the form ax2 bx c 0 as the normal form of a quadratic equation. 4. The quadratic equation 2x2 6x 9 0 is in standard form. 5. The statement x 12 means that x 12 or x 12. 6. The Pythagorean theorem applies to any triangle. 7. In a right triangle, the hypotenuse is the side opposite the right angle. 8. If for a right triangle, a and b are the measures of the legs, and c is the measure of the hypotenuse, then a b c. 9. An isosceles right triangle is a right triangle in which two legs are of equal length. 10. In a 30°– 60° right triangle, the side opposite the 30° angle is equal in length to one-third the length of the hypotenuse.

Problem Set 10.1 23. 4x2 64

24. 5x2 500

25. n2 14

26. n2 22

27. n2 16 0

28. n2 24

29. y2 32

30. y2 25 0

31. 3x2 54 0

32. 4x2 108 0

33. 2x2 9

34. 3x2 16

35. 8n2 25

36. 12n2 49

10. x2 5x 36 0

37. 1x 12 2 4

38. 1x 22 2 9

11. n 5n 6 0

12. n 3n 28 0

39. 1x 32 2 25

40. 1x 52 2 36

13. 6y2 7y 5 0

14. 4y2 21y 18 0

41. 13x 22 2 49

42. 14x 32 2 1

15. 30x2 37x 10 0

16. 42x2 67x 21 0

43. 1x 62 2 5

44. 1x 72 2 6

17. 4x2 4x 1 0

18. 9x2 12x 4 0

45. 1n 12 2 8

46. 1n 12 2 12

47. 12n 32 2 20

48. 13n 22 2 28

49. 14x 12 2 2

50. 15x 32 2 32 0

1 Solve Quadratic Equations by Factoring For Problems 1–18, solve each quadratic equation by factoring and applying the property ab 0 if and only if a 0 or b 0. 1. x2 15x 0

2. x2 11x 0

3. n 12n

4. n 21n

5. 3y2 15y

6. 8y2 56y

7. x 9x 8 0

8. x 16x 48 0

2

2

9. x2 5x 14 0 2

2

2

2

2 Solve Quadratic Equations by Applying the Square-Root Property For Problems 19 –56, use Property 10.1 to help solve each quadratic equation. Express irrational solutions in simplest radical form. 19. x2 64

20. x2 169

25 21. x 9

4 22. x 81

2

2

51. 13x 52 2 40 0 52. 12x 92 2 6 0 53. 217x 12 2 5 37 55. 21x 82 2 9 91 56. 21x 72 2 7 101

54. 314x 52 2 50 25

492

Chapter 10

Quadratic Equations

3 Solve Word Problems Involving the Pythagorean Theorem For Problems 57– 62, a and b represent the lengths of the legs of a right triangle, and c represents the length of the hypotenuse. Express your answers in simplest radical form. 57. Find c if a 1 inch and b 7 inches. 58. Find c if a 2 inches and b 6 inches. 59. Find a if c 8 meters and b 6 meters. 60. Find a if c 11 centimeters and b 7 centimeters.

71. If c 9 meters, find a and b. 72. If c 5 meters, find a and b. 73. An 18-foot ladder resting against a house reaches a windowsill 16 feet above the ground. How far is the base of the ladder from the foundation of the house? Express your answer to the nearest tenth of a foot. 74. A 42-foot guy wire makes an angle of 60 with the ground and is attached to a telephone pole, as in Figure 10.8. Find the distance from the base of the pole to the point on the pole where the wire is attached. Express your answer to the nearest tenth of a foot.

61. Find b if c 12 feet and a 10 feet.

42

For Problems 63 – 68, refer to Figure 10.6. Express your answers in simplest radical form.

fee

t

62. Find b if c 10 yards and a 4 yards.

60° B 60°

c

a

30° b

A

C

Figure 10.6 63. If c 8 inches, find a and b. 64. If c 6 inches, find a and b.

Figure 10.8 75. A rectangular plot measures 18 meters by 24 meters. Find the distance, to the nearest meter, from one corner of the plot to the corner diagonally opposite. 76. Consecutive bases of a square-shaped baseball diamond are 90 feet apart (see Figure 10.9). Find the distance, to the nearest tenth of a foot, from first base diagonally across the diamond to third base.

65. If a 6 feet, find b and c. 66. If a 5 feet, find b and c. 67. If b 12 meters, find a and c. 68. If b 5 centimeters, find a and c.

fe 90

fe e

t

t

90

c

et

a

First base

e fe

a=b

Third base

90

B

fe

For Problems 69 –72, refer to the isosceles right triangle of Figure 10.7. Express the answers in simplest radical form.

90

et

Second base

Home plate C

b

A

Figure 10.7 69. If b 10 inches, find a and c. 70. If a 7 inches, find b and c.

Figure 10.9 77. A diagonal of a square parking lot is 50 meters. Find, to the nearest meter, the length of a side of the lot.

10.2

Completing the Square

493

THOUGHTS INTO WORDS 78. Explain why the equation 1x 42 2 14 2 has no real number solutions.

x2 6x 16 0

1x 821x 22 0

79. Suppose that your friend solved the equation 1x 32 2 25 as follows: 1x 32 2 25

x80 x 8

or

x20

or

x2

Is this a correct approach to the problem? Can you suggest an easier approach to the problem?

x2 6x 9 25

FURTHER INVESTIGATIONS 80. Sometimes we simply need to determine whether a particular radical expression is positive or negative. For example, is 6 139 a positive or negative number? We can determine this by approximating a value for 139. Since 62 36, we know that 139 is a little larger than 6. Therefore, 6 139 has to be positive. Determine whether each of the following is positive or negative. (a) 8 156

(b) 7 147

(c) 9 177

(d) 12 1130

(e) 6 5 12

(f) 10 6 13

(g) 13 1150

(h) 14 1200

81. Find the length of an altitude of an equilateral triangle if each side of the triangle is 6 centimeters long. Express your answer to the nearest tenth of a centimeter. 82. Suppose that we are given a cube in which each edge is 12 centimeters long. Find the length of a diagonal from a

lower corner to the diagonally opposite upper corner. Express your answer to the nearest tenth of a centimeter. 83. Suppose that we are given a rectangular box with a length of 8 centimeters, a width of 6 centimeters, and a height of 4 centimeters. Find the length of a diagonal from a lower corner to the diagonally opposite upper corner. Express your answer to the nearest tenth of a centimeter. 84. The converse of the Pythagorean theorem is also true. It states that “if the measures a, b, and c of the sides of a triangle are such that a2 b2 c2, then the triangle is a right triangle with a and b the measures of the legs and c the measure of the hypotenuse.” Use the converse of the Pythagorean theorem to determine which of the following are right triangles, given the measurements of three sides. (a) 9, 40, 41

(b) 20, 48, 52

(c) 19, 21, 26

(d) 32, 37, 49

(e) 65, 156, 169

(f) 21, 72, 75

Answers to the Example Practice Skills 1. 57, 06 2. 53, 76 9. 57 2 15, 7 2 156

3. 556 10. e

4. 512, 126

5. 58, 86

1 3 12 1 3 12 , f 5 5

6. 5213 , 2136

7. e

216 2 16 , f 3 3

8. 58, 26

11. 13 inches 12. 27.5 feet 13. 6 22 feet

14. 13.9 feet


2. True 3. False 4. True 5. True

6. False

7. True

8. False 9. True

10. False

10.2 Completing the Square OBJECTIVE 1

Solve Quadratic Equations by Completing the Square

1 Solve Quadratic Equations by Completing the Square Thus far we have solved quadratic equations by factoring or by applying Property 10.1 (if x2 a, then x 1a or x 1a). In this section, we will consider another method

494

Chapter 10

Quadratic Equations

called completing the square, which will give us the power to solve any quadratic equation. We studied a factoring technique in Chapter 5 that was based on recognizing perfect square trinomials. In each of the following equations, the trinomial on the right side, which is the result of squaring a binomial on the left side, is a perfect square trinomial. 1x 52 2 x2 10x 25 1x 72 2 x2 14x 49 1x 32 2 x2 6x 9

1x 62 2 x2 12x 36 We need to pay attention to the following special relationship. In each of these perfect square trinomials, the constant term is equal to the square of one-half of the coefficient of the x term. For example, x2 10x 25

1 1102 5 2

and 5 2 25

x2 12x 36

1 1122 6 2

and 6 2 36

This relationship allows us to form a perfect square trinomial by adding the proper constant term. For example, suppose that we want to form a perfect square trinomial 1 from x2 8x. Because 182 4 and 42 16, we can form the perfect square trinomial 2 x2 8x 16. Now we can use the preceding ideas to help solve some quadratic equations.

EXAMPLE 1

Solve x2 8x 1 0 by the method of completing the square.

Solution x2 8x 1 0 x 2 8x 1

Isolated the x2 and x terms 1 of the coefficient of the x term and then 2 squared the result

1 182 4 and 42 16 2

Took

x2 8x 16 1 16

Added 16 to both sides of the equation

1x 42 17 2

Factored the perfect square trinomial

Now we can proceed as we did with similar equations in the last section. x 4 117 x 4 117 x 4 117

or or

x 4 117 x 4 117

The solution set is 54 117, 4 1176.

▼ PRACTICE YOUR SKILL Solve y2 10y 6 0 by the method of completing the square.

■

Observe that the method of completing the square to solve a quadratic equation is just what the name implies. We form a perfect square trinomial; then we change the

10.2


495

equation to the necessary form for using the property, if x2 a, then x 1a or x 1a. Let’s consider another example.

EXAMPLE 2

Solve x 2 2x 11 0 by the method of completing the square.

Solution x2 2x 11 0 x 2 2x 11 1 122 1 and 12 1 2

Isolated the x2 and x terms 1 Took of the coefficient of the x term and then 2 squared the result

x2 2x 1 11 1

Added 1 to both sides of the equation

1x 12 12 2


x 1 112 x 1 213 x 1 2 13

or

x 1 2 13

or

x 1 213 x 1 213

The solution set is 51 2 13, 1 2 136.

▼ PRACTICE YOUR SKILL Solve y2 6y 2 0 by the method of completing the square.

■

In the next example, the coefficient of the x term is odd, which means that taking one-half of it puts us into the realm of fractions. The use of common fractions rather than decimals makes our previous work with radicals relevant.

EXAMPLE 3

Solve x 2 3x 1 0 by the method of completing the square.

Solution x2 3x 1 0 x2 3x 1 3 2 9 a b 2 4

1 3 132 2 2

and

x2 3x

9 9 1 4 4

Added

3 2 5 ax b 2 4

x

9 to both sides of the equation 4


x

3 5 2 A4

x

3 15 2 2

3 15 2 2

1 of the coefficient of the x term and then 2 squared the result Took

or

x

3 15 2 2

496

Chapter 10

Quadratic Equations

x

3 15 2 2

or

x

3 15 2 2

x

3 15 2

or

x

3 15 2


3 15 3 15 , f. 2 2

▼ PRACTICE YOUR SKILL Solve a2 5a 3 0 by the method of completing the square.

■

The relationship for a perfect square trinomial that states the constant term is equal to the square of one-half of the coefficient of the x term holds only if the coefficient of x2 is 1. Thus we need to make a slight adjustment when solving quadratic equations that have a coefficient of x2 other than 1. The next example shows how to make this adjustment.

EXAMPLE 4

Solve 2x2 12x 3 0 by the method of completing the square.

Solution 2x2 12x 3 0 2x2 12x 3 x2 6x x2 6x 9 1x 32 2

x3


3 9 2

2 1 c 162 d 32 9; add 9 to both sides of the equation 2

21 A2

x3

142 2

142 2

21 121 121 A2 12 12

or

x3

#

12 142 2 12

142 2

142 2

or

x 3

6 142 2

or

x


1 2

21 2

x3

x 3 x

3 2

142 2

6 142 2

6 142 6 142 , f. 2 2

▼ PRACTICE YOUR SKILL Solve 3y2 6y 2 0 by the method of completing the square.

■

As we mentioned earlier, we can use the method of completing the square to solve any quadratic equation. To illustrate this point, we will use this method to solve an equation that we could also solve by factoring.

10.2

EXAMPLE 5


497

Solve x2 2x 8 0 by the method of completing the square and by factoring.

Solution A By completing the square: x2 2x 8 0 x2 2x 8 2 1 c 122 d 12 1; add 1 to both sides of the equation 2

x2 2x 1 8 1 1x 12 2 9

x 1 3 x13

or

x 1 3

x2

or

x 4


Solution B By factoring: x2 2x 8 0

1x 42 1x 22 0 x40 x 4

or

x20

or

x2


▼ PRACTICE YOUR SKILL Solve n2 n 12 0 by the method of completing the square and by factoring.

■

We don’t claim that using the method of completing the square with an equation such as the one in Example 5 is easier than the factoring technique. However, it is important for you to recognize that the method of completing the square will work with any quadratic equation. Our final example of this section demonstrates that the method of completing the square will identify those quadratic equations that have no real number solutions.

EXAMPLE 6

Solve x2 10x 30 0 by the method of completing the square.

Solution x2 10x 30 0 x2 10x 30 x2 10x 25 30 25 1x 52 2 5

We can stop here and reason as follows: Any value of x will yield a nonnegative value for 1x 52 2; thus, it cannot equal 5. The original equation, x2 10x 30 0, has no solutions in the set of real numbers.

▼ PRACTICE YOUR SKILL Solve y 2 6y 22 0 by the method of completing the square.

■

498

Chapter 10

Quadratic Equations

CONCEPT QUIZ

For Problems 1–7, answer true or false. 1. In a perfect square trinomial, the constant term is equal to one-half the coefficient of the x term. 2. The method of completing the square can be used to solve any quadratic equation. 3. Every quadratic equation solved by completing the square will have real number solutions. 4. The completing-the-square method cannot be used if the equation could be solved by factoring. 5. To use the completing-the-square method for solving the equation 3x 2 2x 5, we would first divide both sides of the equation by 3. 6. The equation x 2 2x 0 cannot be solved by using the completing-the-square method because there is no constant term. 7. To solve the equation x 2 5x 1 by completing the square, we would start by 25 adding to both sides of the equation. 4

Problem Set 10.2 1 Solve Quadratic Equations by Completing the Square For Problems 1–32, use the method of completing the square to help solve each quadratic equation. 1. x2 8x 1 0

2. x2 4x 1 0

3. x2 10x 2 0

4. x2 8x 3 0

5. x2 4x 4 0

6. x2 6x 11 0

7. x2 6x 12 0

8. n2 10n 7

21. 5n2 10n 6 0

22. 2n2 5n 1 0

23. n2 9n 4

24. n2 7n 2

25. 2x2 3x 1 0

26. 3x2 x 3 0

27. 3x2 2x 2 0

28. 9x 3x2 1

29. n1n 22 168

30. n1n 42 140

31. n1n 42 165

32. n1n 22 288

10. n2 12n 40 0

For Problems 33 – 42, solve each quadratic equation by using (a) the factoring method and (b) the method of completing the square.

11. x2 x 3 0

12. x2 3x 5 0

33. x2 4x 12 0

34. x2 6x 40 0

13. a2 5a 2

14. a2 7a 4

35. x2 12x 27 0

36. x2 18x 77 0

15. 2x2 8x 3 0

16. 2x2 12x 1 0

37. n2 3n 40 0

38. n2 9n 36 0

17. 3x2 12x 2 0

18. 3x2 6x 2 0

39. 2n2 9n 4 0

40. 6n2 11n 10 0

19. 2t2 4t 1 0

20. 4t2 8t 5 0

41. 4n2 4n 15 0

42. 4n2 12n 7 0

9. n2 2n 17

THOUGHTS INTO WORDS 43. Give a step-by-step description of how to solve the equation 3x2 10x 8 0 by completing the square. 44. An error has been made in the following solution. Find it and explain how to correct it. 4x2 4x 1 0 4x2 4x 1 4x2 4x 4 1 4 12x 22 2 3

2x 2 13 2x 2 13

or

2x 2 13

or

2 13 2

or

x


2x 2 13 2x 2 13

2 13 2 13 , f. 2 2

x

2 13 2

10.3

Quadratic Formula

499

FURTHER INVESTIGATIONS Now use your calculator to evaluate each of these expressions to the nearest tenth. The solution set is 56.2, 0.26 .

45. Use the method of completing the square to solve ax2 bx c 0 for x, where a, b, and c are real numbers and a 0.

Solve each of the following equations and express the solutions to the nearest tenth.

46. Suppose that in Example 4 we wanted to express the solutions to the nearest tenth. Then we would probably 21 proceed from the step x 3 as follows: B2 x3

(a) x2 6x 4 0 (b) x2 8x 4 0 (c) x2 4x 4 0

21 B2

x 3

(d) x2 2x 5 0 21 B2

21 x 3 B2

(e) x2 14x 2 0 (f) x2 12x 1 0

21 x 3 B2

or

Answers to the Example Practice Skills 1. 55 131 , 5 1316 4. e

3 115 3 115 , f 3 3

2. 53 17 , 3 176

3. e

5 137 5 137 , f 2 2

5. 54, 36 6. No real number solutions



5. True

6. False 7. True

10.3 Quadratic Formula OBJECTIVE 1

Solve Quadratic Equations by Using the Quadratic Formula

1 Solve Quadratic Equations by Using the Quadratic Formula We can use the method of completing the square to solve any quadratic equation. The equation ax2 bx c 0, where a, b, and c are real numbers with a 0, can represent any quadratic equation. These two ideas merge to produce the quadratic formula, a formula that we can use to solve any quadratic equation. The merger is formed when using the method of completing the square to solve the equation ax2 bx c 0 as follows: ax2 bx c 0 ax2 bx c c b x2 x a a b2 b2 b c x2 x 2 2 a a 4a 4a ax

b 2 b2 4ac b 2a 4a2


1 a

b2 Complete the square by adding 2 4a to both sides The right side is combined into a single term with LCD 4a2

500

Chapter 10

Quadratic Equations

x

b2 4ac b 2a B 4a2

x

2b2 4ac b 2a 24a2

x

2b2 4ac b 2a 2a x x

The solutions are

24a2 02a 0 but 2a can be used because of

b 2b2 4ac 2a 2a

b 2b2 4ac 2a

b 2b2 4ac b 2b2 4ac and . 2a 2a

We usually state the quadratic formula as follows:

x

b 2b2 4ac 2a

We can use it to solve any quadratic equation by expressing the equation in standard form and by substituting the values for a, b, and c into the formula. Consider the following examples.

EXAMPLE 1

Solve x2 7x 10 0 by using the quadratic formula.

Solution The given equation is in standard form with a 1, b 7, and c 10. Let’s substitute these values into the quadratic formula and simplify. x x

b 2b2 4ac 2a 7 272 4112 1102 2112

x

7 19 2

x

7 3 2

x

7 3 2

x 2

7 3 2

or

x

or

x 5


▼ PRACTICE YOUR SKILL Solve x2 3x 18 0 by using the quadratic formula.

■

10.3

EXAMPLE 2

Quadratic Formula

501

Solve x2 3x 1 by using the quadratic formula.

Solution First we need to change the equation to the standard form of ax 2 bx c 0. x2 3x 1 x2 3x 1 0 We need to think of x 2 3x 1 0 as x 2 (3)x (1) 0 to determine the values a 1, b 3, and c 1. Let’s substitute these values into the quadratic formula and simplify. x

132 2132 2 4112112 2112

x

3 19 4 2

x

3 113 2


3 113 3 113 , f. 2 2

▼ PRACTICE YOUR SKILL Solve x2 5x 3 by using the quadratic formula.

EXAMPLE 3

■

Solve 15n2 n 2 0 by using the quadratic formula.

Solution Remember that although we commonly use the variable x in the statement of the quadratic formula, any variable could be used. Writing the equation as 15n2 (1)n (2) 0 gives us the standard form of an2 bn c 0 with a 15, b 1, and c 2. Now we can solve the equation by substituting into the quadratic formula and simplifying. n

112 2112 2 41152 122 21152

n

1 11 120 30

n

1 1121 30

n

1 11 30

n

1 11 30

or

n

1 11 30

502

Chapter 10

Quadratic Equations

10 30

n

12 30

or

n

n

2 5

or

n

1 3


▼ PRACTICE YOUR SKILL Solve 6y2 5y 4 0 by using the quadratic formula.

EXAMPLE 4

■

Solve t2 2t 4 0 by using the quadratic formula.

Solution Writing the equation as t 2 (2)t (4) 0 gives us the standard form of at 2 bt c 0, with a 1, b 2, and c 4. Now we can solve by substituting into the quadratic formula and simplifying. t

122 2122 2 4112 142 2112

t

2 14 16 2

t

2 1415 2 215 2 120 2 2 2

Now the fraction can be simplified. One way to simplify is to factor the common factor of 2 out of the numerator and then reduce the fraction. t

211 152 2 215 1 15 2 2


▼ PRACTICE YOUR SKILL Solve x2 4x 6 0 by using the quadratic formula.

■

We can easily identify quadratic equations that have no real number solutions when we use the quadratic formula. The final example of this section illustrates this point.

EXAMPLE 5

Solve x2 2x 8 0 by using the quadratic formula.

Solution x

122 2122 2 4112182 2112

x

2 14 32 2

x

2 128 2

Since 128 is not a real number, we conclude that the given equation has no real number solutions.

10.3

Quadratic Formula

503

▼ PRACTICE YOUR SKILL Solve y2 3y 11 0 by using the quadratic formula.

CONCEPT QUIZ

■

For Problems 1–5, answer true or false. 1. The quadratic formula can be used to solve any quadratic equation. 2. When solving quadratic equations using the quadratic formula, there are always two distinct different answers. 3. The quadratic formula cannot be used if the quadratic equation can be solved by factoring. 4. To solve 3x2 5x 6 0 by the quadratic formula, we would substitute 3 for a, 5 for b, and 6 for c in the quadratic formula. 5. To solve 2x2 7x 5 by the quadratic formula, we would substitute 2 for a, 7 for b, and 5 for c in the quadratic formula.

Problem Set 10.3 1 Solve Quadratic Equations by Using the Quadratic Formula For Problems 1– 48, use the quadratic formula to solve the following quadratic equations.

21. t2 4t 4 0

22. t2 6t 5 0

23. 6x2 x 2 0

24. 4x2 x 3 0

25. 5x2 3x 2 0

26. 6x2 x 2 0

27. 12x2 19x 5

28. 2x2 7x 6 0

29. 2x2 5x 6 0

30. 2x2 3x 3 0

31. 3x2 4x 1 0

32. 3x2 2x 4 0

33. 16x2 24x 9 0

34. 9x2 30x 25 0

35. 4n2 8n 1 0

36. 4n2 6n 1 0

37. 6n2 9n 1 0

38. 5n2 8n 1 0

39. 2y2 y 4 0

40. 3t2 6t 5 0

1. x2 5x 6 0

2. x2 3x 4 0

3. x2 5x 36

4. x2 8x 12

5. n 2n 5 0

6. n 4n 1 0

7. a 5a 2 0

8. a2 3a 1 0

9. x 2x 6 0

10. x 8x 16 0

11. y 4y 2 0

12. n 6n 11 0

13. x2 6x 0

14. x2 8x 0

41. 4t2 5t 3 0

42. 5x2 x 1 0

15. 2x2 7x

16. 3x2 10x

43. 7x2 5x 4 0

44. 6x2 2x 3 0

17. n2 34n 288 0

18. n2 27n 182 0

45. 7 3x2 x

46. 2x2 3x 4

19. x2 2x 80 0

20. x2 15x 54 0

47. n2 23n 126

48. n2 2n 195

2

2

2

2

2

2

2

THOUGHTS INTO WORDS 49. Explain how to use the quadratic formula to solve the equation x2 2x 6.

applied. Is he right about this, and if not, how would you convince him that he is wrong?

50. Your friend states that the equation x2 6x 16 0 must be changed to x2 6x 16 0 (by multiplying both sides by 1) before the quadratic formula can be

51. Another of your friends claims that the quadratic formula can be used to solve the equation x2 4 0. How would you react to this claim?

FURTHER INVESTIGATIONS For Problems 52 – 61, use the quadratic formula to solve the following equations. Express the solutions to the nearest hundredth.

56. 2x2 3x 7 0

52. x2 7x 13 0

53. x2 5x 19 0

59. 4x2 9x 19 0

54. x2 9x 15 0

55. x2 6x 17 0

60. 3x2 2x 11 0

58. 5x2 11x 14 0

57. 3x2 7x 13 0

504

Chapter 10

Quadratic Equations

61. 5x2 x 21 0

1 2 15n2 n 2 0 and obtained solutions of and . 3 5 Let’s check these solutions using the sum and product relationships.

62. Let x1 and x2 be the two solutions of ax2 bx c 0 obtained by the quadratic formula. Thus, we have x1

b 2b2 4ac 2a

x2

b 2b2 4ac 2a

1 2 5 6 1 Sum of solutions 3 5 15 15 15 b 1 1 a 15 15

Find the sum x1 x2 and the product 1x1 21x2 2 . Your answers should be x1 x2

b a

and

1x1 2 1x2 2

Product of solutions

1 2 2 a b a b 3 5 15 2 2 c a 15 15

c a

These relationships provide another way of checking potential solutions when solving quadratic equations. For example, back in Example 3, we solved the equation

Use the sum and product relationships to check at least 10 of the problems that you worked in this problem set.

Answers to the Example Practice Skills 1. 56, 36 2. e

5 137 f 2

4 1 3. e , f 3 2

4. 52 1106

5. No real number solutions


2. False


10.4 Solving Quadratic Equations—Which Method? OBJECTIVE 1

Solve Quadratic Equations by Any of the Three Basic Methods

1 Solve Quadratic Equations by Any of the Three Basic Methods We now summarize the three basic methods of solving quadratic equations by solving a specific quadratic equation using each technique. Consider the equation x2 4x 12 0.

Factoring Method x2 4x 12 0

1x 62 1x 22 0 x60 x 6

or

x20

or

x2


Completing the Square Method x2 4x 12 0 x2 4x 12

10.4

Solving Quadratic Equations—Which Method?

505

x2 4x 4 12 4 1x 22 2 16

x 2 116 x24

or

x 2 4

x2

or

x 6


Quadratic Formula Method x2 4x 12 0 x

x

2112

x

4 164 2

x

4 8 2

4 8 2

x2

4 242 41121122

4 8 2

or

x

or

x 6

The solution set is {6, 2}. We have also discussed the use of Property 10.1 (x2 a if and only if x 1a) for certain types of quadratic equations. For example, we can solve x2 4 easily by applying Property 10.1 and obtaining x 14 or x 14; thus, the solutions are 2 and 2. Which method should you use to solve a particular quadratic equation? Let’s consider some examples in which the different techniques are used. Keep in mind that this is a decision you must make as the need arises. So become as familiar as you can with the strengths and weaknesses of each method.

EXAMPLE 1

Solve 2x2 12x 54 0.

Solution First, it is very helpful to recognize a factor of 2 in each of the terms on the left side. 2x2 12x 54 0 x2 6x 27 0


1 2

Now you should recognize that the left side can be factored. Thus we can proceed as follows: 1x 92 1x 32 0 x90 x 9

or

x30

or

x3


▼ PRACTICE YOUR SKILL Solve 3x2 3x 6 0.

■

506

Chapter 10

Quadratic Equations

EXAMPLE 2

Solve 14x 32 2 16.

Solution The form of this equation lends itself to the use of Property 10.1 (x2 a if and only if x 1a). 14x 32 2 16

4x 3 116 4x 3 4

or

4x 3 4

4x 1

or

4x 7

1 4

or

x

x

7 4


▼ PRACTICE YOUR SKILL Solve (5x 1)2 49.

EXAMPLE 3

Solve n

■

1 5. n

Solution First, we need to clear the equation of fractions by multiplying both sides by n. n n an

1 5, n

n0

1 b 51n2 n

n2 1 5n Now we can change the equation to standard form. n2 5n 1 0 Because the left side cannot be factored using integers, we must solve the equation by using either the method of completing the square or the quadratic formula. Using the formula, we obtain n n

152 2152 2 4112 112 2112 5 121 2


5 121 5 121 , f. 2 2

▼ PRACTICE YOUR SKILL Solve y

2 3. y

■

10.4

EXAMPLE 4

Solving Quadratic Equations—Which Method?

507

Solve t2 12t.

Solution A quadratic equation without a constant term can be solved easily by the factoring method. t2 12t t2 12t 0 t1t 122 0 t0

or

t0

or

t 12 0 t 12

The solution set is 50, 126 . (Check each of these solutions in the given equation.)

▼ PRACTICE YOUR SKILL Solve x2 10x.

EXAMPLE 5

■

Solve x2 28x 192 0.

Solution Determining whether or not the left side is factorable presents a bit of a problem because of the size of the constant term. Therefore, let’s not concern ourselves with trying to factor; instead we will use the quadratic formula. x2 28x 192 0 x

x

1282 21282 2 411211922 2112

x

28 1784 768 2

x

28 116 2

28 4 2

x 16

28 4 2

or

x

or

x 12


▼ PRACTICE YOUR SKILL Solve t 2 30t 216 0.

EXAMPLE 6

■

Solve x2 12x 17.

Solution The form of this equation, and the fact that the coefficient of x is even, makes the method of completing the square a reasonable approach. x2 12x 17 x2 12x 36 17 36

508

Chapter 10

Quadratic Equations

1x 62 2 53 x 6 153 x 6 153


▼ PRACTICE YOUR SKILL Solve y 2 10y 15.

CONCEPT QUIZ

■

Match the equation with the method most appropriate for solving the equation. 1. 2. 3. 4.

x2 5x 3 0 y2 6y 14 12x 12 2 25 m2 2m 15 0

A. B. C. D.

Factoring Square root property Completing the square Quadratic formula

Problem Set 10.4 1 Solve Quadratic Equations by Any of the Three Basic Methods For Problems 1– 46, solve the following quadratic equations using the method that seems most appropriate to you.

26. 2x2 10x 28 0 27. 2x2 3x 7 0 28. 3x2 2x 5 0 29. n1n 462 480

1. x2 4x 45

2. x2 4x 60

3. 15n 62 2 49

4. 13n 12 2 25

30. n1n 422 432

5. t2 t 2 0

6. t2 2t 3 0

31. n

3 1 n

32. n

3 2 n 4

7. 8x 3x2

8. 5x2 7x 33. x

1 25 x 12

34. x

1 65 x 8

9. 9x2 6x 1 0

10. 4x2 36x 81 0

11. 5n2 10n

12. 3n 2n2

35. t2 12t 36 49

13. n2 14n 19

14. n2 10n 14

37. x2 28x 187 0

15. 5x2 2x 7 0

16. 3x2 4x 2 0

38. x2 33x 266 0

17. 15x2 28x 5 0

39.

x2 1 x 3 2

40.

2x 1 5 3 x2

41.

2 1 3 x2 x

42.

3 2 3 x1 x 2

43.

2 n2 3n 1 6

44.

x2 1 x 2 4

18. 20y2 7y 6 0 19. x2 8x 7 0 20. x2 5x 5 0 21. y2 5y 84 23. 2n 3

3 n

22. y2 7y 60 24. n

1 7 n

36. t2 10t 25 16

45. 1n 221n 42 7

46. 1n 321n 82 30

25. 3x2 9x 12 0

THOUGHTS INTO WORDS 47. Which method would you use to solve the equation x 2 30x 216? Explain your reasons for making this choice. 48. Explain how you would solve the equation 0 x 2 x 6.

49. How can you tell by inspection that the equation x 2 x 4 0 has no real number solutions?

10.5

Solving Problems Using Quadratic Equations

509

Answers to the Example Practice Skills 1. 51, 26

6 8 2. e , f 5 5

3. e

3 117 f 2

4. 50, 106

5. 518, 126

6. 55 21106

Answers to the Concept Quiz 1. D 2. C

3. B

4. A

10.5 Solving Problems Using Quadratic Equations OBJECTIVE 1

Solve Word Problems Involving Quadratic Equations

1 Solve Word Problems Involving Quadratic Equations The following diagram indicates our approach in this text. Develop skills

Use skills to solve equations

Use equations to solve word problems

Now you should be ready to use your skills relative to solving systems of equations (Chapter 8) and quadratic equations to solve additional types of word problems. Before you consider such problems, let’s review and update the problem-solving suggestions we offered in Chapter 2.

Suggestions for Solving Word Problems 1. Read the problem carefully and make certain that you understand the meanings of all the words. Be especially alert for any technical terms used in the statement of the problem. 2. Read the problem a second time (perhaps even a third time) to get an overview of the situation being described and to determine the known facts as well as what is to be found. 3. Sketch any figure, diagram, or chart that might be helpful in analyzing the problem. *4. Choose meaningful variables to represent the unknown quantities. Use one or two variables, whichever seems easiest. The term “meaningful” refers to the choice of letters to use as variables. Choose letters that have some significance for the problem under consideration. For example, if the problem deals with the length and width of a rectangle, then l and w are natural choices for the variables. *5. Look for guidelines that you can use to help set up equations. A guideline might be a formula such as area of a rectangular region equals length times width, or a statement of a relationship such as the product of the two numbers is 98. *6. (a) Form an equation that contains the variable and translates the conditions of the guideline from English into algebra; or (b) form two equations that contain the two variables and translate the guidelines from English into algebra.

510

Chapter 10

Quadratic Equations

*7. Solve the equation (system of equations) and use the solution (solutions) to determine all facts requested in the problem. 8. Check all answers back in the original statement of the problem.

The asterisks indicate those suggestions that have been revised to include using systems of equations to solve problems. Keep these suggestions in mind as you study the examples and work the problems in this section.

EXAMPLE 1

Apply Your Skill The length of a rectangular region is 2 centimeters more than its width. The area of the region is 35 square centimeters. Find the length and width of the rectangle.

Solution Area is 35

cm2.

w

l Figure 10.10

We let l represent the length, and we let w represent the width (see Figure 10.10). We can use the area formula for a rectangle, A lw, and the statement “the length of a rectangular region is 2 centimeters greater than its width” as guidelines to form a system of equations. a

lw 35 b l w2

The second equation indicates that we can substitute w 2 for l. Making this substitution in the first equation yields 1w 22 1w2 35 Solving this quadratic equation by factoring, we get w2 2w 35 w2 2w 35 0

1w 72 1w 52 0 w70 w 7

or

w50

or

w5

The width of a rectangle cannot be a negative number, so we discard the solution 7. Thus the width of the rectangle is 5 centimeters, and the length (w 2) is 7 centimeters.

▼ PRACTICE YOUR SKILL The length of a rectangular piece of sheet metal is 3 inches more than the width. The area of the sheet metal is 88 square inches. Find the length and width of the rectangle. ■

EXAMPLE 2

Apply Your Skill Find two consecutive whole numbers whose product is 506.

Solution We let n represent the smaller whole number. Then n 1 represents the next larger whole number. The phrase “whose product is 506” translates into the equation n1n 12 506

10.5


511

Changing this quadratic equation into standard form produces n2 n 506 n2 n 506 0 Because of the size of the constant term, let’s not try to factor; instead, we use the quadratic formula. n

1 212 4112 15062 2112

n

1 12025 2

n

1 45 2

n

1 45 2

n 22

12025 45

1 45 2

or

n

or

n 23

Since we are looking for whole numbers, we discard the solution 23. Therefore, the whole numbers are 22 and 23.

▼ PRACTICE YOUR SKILL Find two consecutive whole numbers whose product is 342.

EXAMPLE 3

■

Apply Your Skill The perimeter of a rectangular lot is 100 meters, and its area is 616 square meters. Find the length and width of the lot.

Solution We let l represent the length and w represent the width (see Figure 10.11).

Area is 616 m 2. Perimeter is 100 m.

w

l Figure 10.11

Then a

lw 616 b 2l 2w 100

Area is 616 m2 Perimeter is 100 m

1 produces l w 50, which can be changed to 2 l 50 w. Substituting 50 w for l in the first equation produces the quadratic equation

Multiplying the second equation by

150 w2 1w2 616 50w w2 616 w2 50w 616

512

Chapter 10

Quadratic Equations

Using the method of completing the square, we have w2 50w 625 616 625 1w 252 2 9

w 25 3 w 25 3 w 28

or

w 25 3 w 22

or

If w 28, then l 50 w 22. If w 22, then l 50 w 28. The rectangle is 28 meters by 22 meters or 22 meters by 28 meters.

▼ PRACTICE YOUR SKILL The perimeter of a rectangular garden is 84 feet, and its area is 360 square feet. Find the length and width of the garden. ■

EXAMPLE 4

Apply Your Skill Find two numbers such that their sum is 2 and their product is 1.

Solution We let n represent one of the numbers and m represent the other number. a

nm2 b nm 1

Their sum is 2 Their product is 1

We can change the first equation to m 2 n; then we can substitute 2 n for m in the second equation. n 12 n2 1 2n n2 1 n2 2n 1 0 n2 2n 1 0 n n


122 2122 2 4112 112 2112 2 18 2 212 1 12 2 2

If n 1 12, then m 2 11 122 2 1 12 1 12

If n 1 12, then m 2 11 122 2 1 12 1 12 The numbers are 1 12 and 1 12.

Perhaps you should check these numbers in the original statement of the problem!

▼ PRACTICE YOUR SKILL Find two numbers such that their sum is 10 and their product is 6.

■

10.5


513

Finally, let’s consider a uniform motion problem similar to those we solved in Chapter 7. Now we have the flexibility of using two equations in two variables.

EXAMPLE 5

Apply Your Skill

Photodisc/Photolibrary

Larry drove 156 miles in 1 hour more than it took Mike to drive 108 miles. Mike drove at an average rate of 2 miles per hour faster than Larry. How fast did each one travel?

Solution We can represent the unknown rates and times like this: Let r represent Larry’s rate. Let t represent Larry’s time. Then r 2 represents Mike’s rate, and t 1 represents Mike’s time. Because distance equals rate times time, we can set up the following system: a

rt 156 b 1r 22 1t 12 108

Solving the first equation for r produces r

156 156 . Substituting for r in the second t t

equation and simplifying, we obtain a

156 2b 1t 12 108 t

156

156 2t 2 108 t

2t

156 154 108 t

2t

156 46 0 t

2t2 156 46t 0

Multiply both sides by t, t 0

2t 46t 156 0 2

t2 23t 78 0


We can solve this quadratic equation by factoring. 1t 262 1t 32 0 t 26 0 t 26

or

t30

or

t3

We must disregard the negative solution. So Mike’s time is 3 1 2 hours. Larry’s 156 rate is 52 miles per hour, and Mike’s rate is 52 2 54 miles per hour. 3

▼ PRACTICE YOUR SKILL In an experimental solar car race, John’s car traveled 90 miles in 1 hour less time than Mark’s car, which traveled 80 miles. John’s car traveled at an average rate of 10 miles per hour faster than Mark’s car. How fast did each car travel? ■

514

Chapter 10

Quadratic Equations

CONCEPT QUIZ

For Problems 1– 6, answer true or false. 1. When solving a word problem, a guideline helps to determine the equation for solving the problem. 2. When solving a word problem, you must choose x as the variable to represent the unknown quantities. 3. When reading a word problem, you should take note of any formulas that can be applied in solving the problem. 4. Some word problems could be solved using only one variable or solved using two variables and a system of equations. 5. When solving a word problem, a sketch, diagram, or chart is useless. 6. When solving a word problem, it is only necessary to check the answers in the equation determined to solve the problem.

Problem Set 10.5 1 Solve Word Problems Involving Quadratic Equations For Problems 1–32, solve the problems. 1. Find two consecutive whole numbers whose product is 306. 2. Find two consecutive whole numbers whose product is 702. 3. Suppose that the sum of two positive integers is 44 and their product is 475. Find the integers. 4. Two positive integers differ by 6. Their product is 616. Find the integers.

region is 45 square centimeters. Find the length and width of the rectangle. 15. The perimeter of a rectangle is 80 centimeters and its area is 375 square centimeters. Find the length and width of the rectangle. 16. The perimeter of a rectangle is 132 yards and its area is 1080 square yards. Find the length and width of the rectangle. 17. The area of a tennis court is 2106 square feet (see 26 Figure 10.12). The length of the court is times the 9 width. Find the length and width of a tennis court.

5. Find two numbers such that their sum is 6 and their product is 4. 6. Find two numbers such that their sum is 4 and their product is 1. 41 7. The sum of a number and its reciprocal is . Find the 20 number. 73 8. The sum of a number and its reciprocal is . Find the 24 number. 9. Each of three consecutive even whole numbers is squared. The three results are added and the sum is 596. Find the numbers. 10. Each of three consecutive whole numbers is squared. The three results are added, and the sum is 245. Find the three whole numbers. 11. The sum of the square of a number and the square of one-half of the number is 80. Find the number. 12. The difference between the square of a positive number and the square of one-half the number is 243. Find the number. 13. Find the length and width of a rectangle if its length is 4 meters less than twice the width, and the area of the rectangle is 96 square meters. 14. Suppose that the length of a rectangular region is 4 centimeters greater than its width. The area of the

Figure 10.12

18. The area of a badminton court is 880 square feet. The length of the court is 2.2 times the width. Find the length and width of the court. 19. An auditorium in a local high school contains 300 seats. There are 5 fewer rows than the number of seats per row. Find the number of rows and the number of seats per row. 20. Three hundred seventy-five trees were planted in rows in an orchard. The number of trees per row was 10 more than the number of rows. How many rows of trees are in the orchard?

10.5 21. The area of a rectangular region is 63 square feet. If the length and width are each increased by 3 feet, the area is increased by 57 square feet. Find the length and width of the original rectangle. 22. The area of a circle is numerically equal to twice the circumference of the circle. Find the length of a radius of the circle. 23. The sum of the lengths of the two legs of a right triangle is 14 inches. If the length of the hypotenuse is 10 inches, find the length of each leg. 24. A page for a magazine contains 70 square inches of type. The height of a page is twice the width. If the margin around the type is to be 2 inches uniformly, what are the dimensions of the page? 25. A 5-by-7-inch picture is surrounded by a frame of uniform width (see Figure 10.13). The area of the picture and frame together is 80 square inches. Find the width of the frame.


515

26. A rectangular piece of cardboard is 3 inches longer than it is wide. From each corner, a square piece 2 inches on a side is cut out. The flaps are then turned up to form an open box that has a volume of 140 cubic inches. Find the length and width of the original piece of cardboard. 27. A class trip cost $3000. If there had been 10 more students, it would have cost each student $25 less. How many students took the trip? 28. Simon mowed some lawns and earned $40. It took him 3 hours longer than he anticipated, and thus he earned $3 per hour less than he anticipated. How long did he expect it to take him? 29. A piece of wire 56 inches long is cut into two pieces, and each piece is bent into the shape of a square. If the sum of the areas of the two squares is 100 square inches, find the length of each piece of wire. 30. Suppose that by increasing the speed of a car by 10 miles per hour, it is possible to make a trip of 200 miles in 1 hour less time. What was the original speed for the trip?

7 inches

31. On a 50-mile bicycle ride, Irene averaged 4 miles per hour faster for the first 36 miles than she did for the last 14 miles. The entire trip of 50 miles took 3 hours. Find her rate for the first 36 miles. 32. One side of a triangle is 1 foot more than twice the length of the altitude to that side. If the area of the triangle is 18 square feet, find the length of a side and the length of the altitude to that side.

5 inches Figure 10.13

THOUGHTS INTO WORDS 33. Return to Example 1 of this section and explain how the problem could be solved using one variable and one equation.

34. Write a page or two on the topic “using algebra to solve problems.”

Answers to the Example Practice Skills 1. Length is 11 inches, width is 8 inches 2. 18 and 19 3. Length is 30 feet, width is 12 feet 4. 55 119, 5 1196 5. Mark’s car 20 mph and John’s car 30 mph


2. False

3. True

4. True 5. False

6. False


SUMMARY

EXAMPLE


Solve quadratic equations by factoring. (Sec. 10.1, Obj. 1, p.484)

A quadratic equation in the variable x is any equation that can be written in the form ax2 bx c 0, where a, b, and c are real numbers and a 0. We can solve quadratic equations that are factorable using integers by factoring and applying the property ab 0 if and only if a 0 or b 0.

Solve x2 4x 21 by factoring.

Problems 1– 4

Solve quadratic equations by applying the square-root property. (Sec. 10.1, Obj. 2, p.486)

The square-root property (Property 10.1), x2 a if and only if x 1a, can be applied to solve certain types of quadratic equations.

Solution

First set the equation equal to 0 and then factor. x2 4x 21 0 1x 72 1x 32 0 x70 or x 3 0 x 7 or x 3 The solution set is {7, 3}. Solve (2x 3)2 15.

Problems 5 – 8

Solution

Applying the square root property gives 2x 3 215 2x 3 115 3 115 x 2 The solution set is 3 115 3 115 e , f. 2 2

Solve quadratic equations by completing the square. (Sec. 10.2, Obj. 1, p.493)

Completing the square allows you to write the equation in the format necessary to apply the square-root property. To use the completing-the-square method, we want the coefficient of the squared term to be 1.

Solve x2 10x 6 0 by completing the square. Solution

First, add 6 to both sides of the equation: x2 10x 6 1 Now take of the coefficient 2 of the x term, 10, and square the result: 1 1102 5 and 52 25 2 Add 25 to both sides of the equation: x2 10x 25 6 25 x2 10x 25 31 Now factor and then apply the square-root property: 1x 52 2 31 x 5 231 x 5 231 The solution set is 55 231,5 2316.

516

Problems 9 –12

(continued)

Chapter 10

OBJECTIVE

SUMMARY

EXAMPLE

Solve quadratic equations by using the quadratic formula. (Sec. 10.3, Obj. 1, p.499)

The quadratic formula b 2b2 4ac x 2a can be used to solve any quadratic equation in the form ax2 bx c 0. The final answer should be reduced and have the radical in lowest terms. Be careful when reducing the final answer because this step can often be the source of an incorrect answer.

Solve x2 6x 2 0 by the quadratic formula.

Solve quadratic equations by any of the three basic methods. (Sec. 10.4, Obj. 1, p.504)

The three basic methods for solving quadratic equations are factoring, completing the square, and the quadratic formula. Factoring only works if the expression is factorable over the integers. Completing the square can be very efficient in situations where you can complete the square without working with fractions. The quadratic formula will work for any quadratic equation.

Solve m2 6m 8 0 by using the method that seems most appropriate.

Solve word problems involving the Pythagorean theorem. (Sec. 10.1, Obj. 3, p.488)

For right triangles, the Pythagorean theorem is stated as a2 b2 c2, where a and b are legs and c is the hypotenuse of the triangle. In an isosceles right triangle, the lengths of the two legs are equal. In a 30°– 60° right triangle, the length of the leg opposite the 30° angle is one-half the length of the hypotenuse.

Summary

517

CHAPTER REVIEW PROBLEMS Problems 13 –16

Solution

For this problem, a 1, b 6, and c 2. 162 2162 2 4112122 x 2112 6 136 8 x 2 6 128 x 2 213 172 6 217 x 2 2 x 3 17 The solution set is 53 17, 3 176 . Problems 17–38

Solution

The expression does not factor, so let’s solve by completing the square. m2 6m 8 0 m2 6m 8 2 m 6m 9 8 9 1m 32 2 17 m 3 117 m 3 117 The solution set is 53 117,3 1176 . A rectangular football field measures approximately 50 yards by 100 yards. Find the length of the diagonal of the football field to the nearest tenth of a yard.

Problems 39 – 42

Solution

The diagonal divides the rectangle into two right triangles. Use the Pythagorean theorem to find the hypotenuse of a triangle with legs measuring 50 yards and 100 yards. a2 b2 c2 2 50 1002 c2 12,500 c2 c 112500 ⬇ 111.8 The length of the diagonal to the nearest tenth of a yard is 111.8 yards.

(continued)

518

Chapter 10

Quadratic Equations


OBJECTIVE

SUMMARY

EXAMPLE

Solve word problems involving quadratic equations. (Sec. 10.5, Obj. 1, p.509)

Many word problems require knowledge of how to solve a quadratic equation.

Find two numbers such that their sum is 8 and their product is 6.

Problems 43 – 48

Solution

Let n represent one number; then the other number will be represented by 8 n. Now write an equation showing the product: n18 n2 6 8n n2 6 0 n2 8n 6 Using the quadratic formula, we find that the numbers are 4 110 and 4 110.

Chapter 10 Review Problem Set For Problems 1– 4, solve each equation by factoring.

25. 5x2 6x 7 0

26. 3x2 18x 15 0

1. x2 2x 24 0

2. x2 13x 40 0

27. 31x 22 2 2 4

28. x2 4x 14 0

3. x2 5x 14

4. 2x2 8x 0

29. y2 45

30. x1x 62 27

31. x2 x

32. n2 4n 3 6

33. n2 44n 480 0

34.

x2 x1 4

35.

5x 2 2 3 x1

36.

2x 1 1 3x 1 2

37.

5 4 6 x3 x

38.

1 2 3 x2 x

For Problems 5 – 8, solve each equation by applying the square-root property. 5. 1x 52 2 64

6. 1x 42 2 64

7. 5x2 80

8. 13x 12 2 40

For Problems 9 –12, solve each equation by completing the square. 9. y2 14y 20 11. w 2w 7 0 2

10. x2 8x 24 12. n 10n 4 0 2

For Problems 13 –16, solve by using the quadratic formula. 13. x2 3x 5 0

14. x2 7x 6 0

15. 3y2 2y 4 0

16. 2x2 4x 3 0

For Problems 39 – 48, set up an equation or a system of equations to help solve each problem. 39. Standing at a point 16 yards from the base of a tower, it is obvious that the distance to the top of the tower is 4 yards more than the height of the tower. Find the height of the tower. 40. The difference in the lengths of the two legs of a right triangle is 2 yards. If the length of the hypotenuse is 2113 yards, find the length of each leg.

For Problems 17–38, solve each quadratic equation. 17. 12x 72 2 25

18. x2 8x 3

19. 21x2 13x 2 0 20. x 17x 2

4 21. n 3 n

22. n2 26n 165 0 23. 3a2 7a 1 0

24. 4x2 4x 1 0

41. The length of the hypotenuse of an isosceles right triangle is 12 inches. Find the length of each leg. 42. In a 30 – 60 right triangle, the side opposite the 60 angle is 8 centimeters long. Find the length of the hypotenuse. 43. The perimeter of a rectangle is 42 inches, and its area is 108 square inches. Find the width and length of the rectangle.

Chapter 10 44. Find two consecutive whole numbers whose product is 342. 45. Each of three consecutive odd whole numbers is squared. The three results are added and the sum is 251. Find the numbers. 46. The combined area of two squares is 50 square meters. Each side of the larger square is three times as long as a side of the smaller square. Find the lengths of the sides of each square.

Review Problem Set

519

47. A company has a rectangular parking lot 40 meters wide and 60 meters long. They plan to increase the area of the lot by 1100 square meters by adding a strip of equal width to one side and one end. Find the width of the strip to be added. 48. Jay traveled 225 miles in 2 hours less time than it took Jean to travel 336 miles. If Jay’s rate was 3 miles per hour slower than Jean’s rate, find each rate.

Chapter 10 Test 1.

1. The two legs of a right triangle are 4 inches and 6 inches long. Find the length of the hypotenuse. Express your answer in simplest radical form.

2.

2. A diagonal of a rectangular plot of ground measures 14 meters. If the width of the rectangle is 5 meters, find the length to the nearest meter.

3.

3. A diagonal of a square piece of paper measures 10 inches. Find, to the nearest inch, the length of a side of the square.

4.

4. In a 30 – 60 right triangle, the side opposite the 30 angle is 4 centimeters long. Find the length of the side opposite the 60 angle. Express your answer in simplest radical form. For Problems 5 –20, solve each equation.

5.

5. 13x 22 2 49

6.

6. 4x2 64

7.

7. 8x2 10x 3 0

8.

8. x2 3x 5 0

9.

9. n2 2n 9

10.

10. 12x 12 2 16

11.

11. y2 10y 24

12.

12. 2x2 3x 4 0

13.

13.

x2 4 3 x1

14.

14.

2 5 1 x x1 2

15.

15. n1n 282 195

16.

16. n

17.

17. 12x 1213x 22 2

18.

18. 17x 22 2 4 21

19.

19. 14x 12 2 27

20.

20. n2 5n 7 0

19 3 n 4

For Problems 21–25, set up an equation or a system of equations to help solve each problem. 21.

21. A room contains 120 seats. The number of seats per row is 1 less than twice the number of rows. Find the number of seats per row.

22.

22. Abu rode his bicycle 56 miles in 2 hours less time than it took Stan to ride his bicycle 72 miles. If Abu’s rate was 2 miles per hour faster than Stan’s rate, find Abu’s rate.

23.

23. Find two consecutive odd whole numbers whose product is 255.

24.

24. The combined area of two squares is 97 square feet. Each side of the larger square is 1 foot more than twice the length of a side of the smaller square. Find the length of a side of the larger square.

25.

25. The perimeter of a rectangle is 56 inches and its area is 180 square inches. Find the length and width of the rectangle.

520

Practice Your Skills Solutions

CHAPTER 1 12. 3xy 2x 4xyz 3162122 2162 4162 122 142

Section 1.1

when x 6, y 2, z 4

1. 5 3 12 2 1 2 12 2 1 14 2 1

18122 12 24122142

12 1

36 12 48142

11

36 12 192 24 192

2. 1216 42 12122

216

24 3. 19 12 19 52 182 142

4.

18 52 11 32 6

2

13.

7142 2 7c d 2c d 2142 2

132142 12 12 11 7 6 4 12 3 6 23 5

28 2 82

30 6

5

14. 1x 3y2 14x 2y2 18 3 # 32 14 # 8 2 # 32 when x 8, y 3

5. 16 # 2 6 32 6

18 92 132 62 1172 1262

38 6.

3 5 # 2 10 2 3 10 5 13 5 8

7.

when c 4, d 2

442

15. a 2ab b 172 2 2172142 142 2 2

2

when a 7, b 4

24 6 # 2 4 # 8 2 4 # 2 32 2

49 14142 16

8 16

49 56 16

24

105 16

8. 12 4 3512 82 4 12 4351102 4 12 4 350 4 12 200 212 9. 3 # 52 7 # 23 3 # 15 # 52 7 # 12 # 2 # 22 3 # 25 7 # 8

121 16. A

1152142 bh 2 2

when b 15, h 4

60 2 30

75 56 19 10. 4x 3y 4152 3122

when x 5, y 2

20 6 26 11. 10a 5b 10182 5132 80 15 65

when a 8, b 3

Section 1.2

1. (a) 24 1322 1 0 24 0 0 32 0 2 124 322 56 (b) 40 75 075 0 040 0 75 40 35

521

522


Chapter 1

(c) 86 1312 086 0 0 31 0

7 7 14

86 31 55 (d) 15 1152 015 0 0 15 0

6.

20 4122 3

15 15

0 2. (a) 20 1382 20 38 18

0 2 0 2

22 8. 3a 4b 3112 4152 17 9.

51F 322 9

12 1#5

0 5 0

#

51362 9 180 9

105,000 1120,0002 225,000

Over four years the boat manufacturer lost $225,000.

0 18 0 18 a b 6 06 0

11. 4122 3112 2132 8 132 6 11 6

18 a b 6

5 Kay was 5 under par.

3

Section 1.3

1. 37 337 182 4 3 37 374 182

0 0 (b) 8

0 182

0 50 0 50 (c) 2 0 2 0

8

2. 503 1221372 4 3 1502122 4 1372 110021372

50 2

3700

25

3. 18 24 182 1122 32

0 80 0 80 a b (d) 4 0 4 0 a

when F 4

515,000 410,000 1120,0002

06 0 2

24

4. (a)

9

10. 835,000 320,000 410,000 1120,0002

(c) 102182 0 14 # 62

514 322

20

5 (d) 142162 1 04 0

#

when a 1, b 5

3 20

16 # 22

(b) 112152 01 0

when x 2, y 6

8 30

50 3. (a) 162122 1 06 0

12 3

8 1302

(d) 25 1252 25 25

#

3

7. 4x 5y 4122 5162

40 112

20 182

4

(b) 35 75 35 1752 (c) 81 1312 81 31

24 32 56

80 b 4

20 5. 5132 4122 7112 15 182 7 15 8 7

18 8 12 38

56 38 18

4. 115212 102 1152122 11521102 30 11502 120


(b) 42 2 # 21

5. 19134 342 19102

2#3#7

0

6. 271122 27122 27 312 122 4

2. 42 2 # 3 # 7

60 2 # 2 # 3 # 5 22 # 3 # 5

271102

The GCF 2 # 3 6.

270 7. (a) 6a 3a 7 5a 1 6a 3a 5a 7 1

16 3 52a 17 12

3. 63 3 # 3 # 7 32 # 7

54 2 # 3 # 3 # 3 2 # 33

The GCF 32 9.

2a 8 (b) 2y 3x 5x 6y 3x 5x 2y 6y

13 52x 12 62y

4. 49 7 # 7 72

80 2 # 2 # 2 # 2 # 5 24 # 5 The GCF 1.

2x 4y 8. (a) 21a 22 51a 62 21a2 2122 51a2 5162

5. 36 2 # 2 # 3 # 3 22 # 32

72 2 # 2 # 2 # 3 # 3 23 # 32

2a 4 5a 30

90 2 # 3 # 3 # 5 2 # 32 # 5

2a 5a 4 30

The GCF 2 # 32 18.

3a 34 (b) 714 2b2 415 b2

6. 12 2 # 2 # 3 22 # 3

30 2 # 3 # 5 2 # 3 # 5

7142 712b2 4152 41b2

The LCM 22 # 3 # 5 4 # 3 # 5 60.

28 14b 20 4b 14b 4b 28 20

7. 20 2 # 2 # 5 22 # 5

75 3 # 5 # 5 3 # 52

10b 8

The LCM 22 # 3 # 52 4 # 3 # 25 300.

9. 5a 3b 2a 7b 5a 2a 3b 7b 3a 4b for a 6, b 5 3162 4152

8. 12 2 # 2 # 3 22 # 3

18 2 # 3 # 3 2 # 32

18 1202

30 2 # 3 # 5 2 # 3 # 5

The LCM 22 # 32 # 5 4 # 9 # 5 180.

18 20 38 10. 2xy 3x 2y 4xy 2xy 4xy 3x 2y

9. 15 3 # 5 3 # 5

8 2 # 2 # 2 23

The LCM 23 # 3 # 5 8 # 3 # 5 120.

2xy 3x 2y for x 3, y 1 2132112 3132 2112

6112 192 2

6 192 2

15 2 13

Section 1.5 1.

4#7 4 7 7 7 28 # # 1# 32 4 8 4 8 8 8

2.

5#5 5 25 # 35 5 7 7

11. 61a 32 21a 42 6a 18 2a 8 6a 2a 18 8 4a 26 when a 7 4172 26

3.

28 26 2

Section 1.4

1. (a) 36 2 # 18

2#2#9

2#2#3#3 22 # 32

Chapter 1

4.

60 2#2#3#5 5 # # # # # 144 2 2 2 2 3 3 12

5#3 5#3 3 5 # 3 # # # 7 10 7 10 7 2 5 14

5 # 12 2 5 12 5. # # 6 25 6 25 5 1

2

1

5

6 10 6 # 10 5 6. a b a b # 18 14 18 14 21 1

3

5

7

523

524


7. a

Chapter 1

10 42 2#5#2#3#7 70 b a b 3 18 3#2#3#3 9 3

2 1 15 15 3 1 # # 8. (a) 5 15 5 2 5 2 2 1

3 3 1 3 (b) 4 # 8 8 4 32

13.

11.

16 9 30

7 30

3 7 a b 10 12

10 2 # 5

12 2 # 2 # 3 22 # 3

1 2 1 3 2 7 9. # # 7 3 7 3 3 7

10.

The LCD 22 # 3 # 5 4 # 3 # 5 60. 3 7 3 # 6 7 # 5 a b a b 10 12 10 6 12 5

3 14 21 21

3 14 21

17 21

5 6 15 15

65 15

11 15

14.

5 122 8

LCD 8

5 2 8 5 122 a # b 8 8 1 8

13 5 60 18

60 2 # 2 # 3 # 5 22 # 3 # 5

18 2 # 3 # 3 2 # 32

The LCD 22 # 32 # 5 4 # 9 # 5 180. 5 13 # 3 5 # 10 13 60 18 60 3 18 10

15.

5 16 a b 8 8 5 1162 8 11 11 8 8

3 1 4 1 3 3 1 3 # # 14 4 7 2 7 14 7 14

39 50 180 180

3 3 1 14 14 7

0

39 50 180

89 180 16.

8 3 12. 15 10

60

17 17 60 60

2 1 2 3 1 5 # # 5 3 5 3 3 5

18 35 a b 60 60 18 1352

1 7

1 7

2 1 1 5 3 a ba b 11 11 3 4 6

15 3 # 5

3 # 11 1 1 5 a ba b 11 2 3 4 6

The LCD 2 # 3 # 5 30.

3 1 1 5 a ba b 2 3 4 6

3 8 # 2 3 # 3 8 15 10 15 2 10 3

3 1 5 a b 2 12 6

3 # 6 1 5 2 a b # 2 6 12 6 2

10 2 # 5

9 16 30 30

LCD 12


18 1 10 a b 12 12 12 18 112 10

12 3#3#3 9 27 # # 12 2 2 3 4

17. 20 a

1 1 1 1 b 20 a b 20 a b 2 5 2 5 10 4 14

1 3 1 3 1 3 1 18. a b a b a b 8 3 5 8 3 8 5

1 3 8 40

3 1 # 5 8 5 40 5 3 40 40

53 40

Chapter 1

(b) 12.45 1.8 0.04 1.261 12.45 1.8 0.04 1.261 15.551 2. (a) 13.6 10.35 13.6 10.35 3.25 (b) 8.427 0.32 8.427 0.32 8.107 3. (a) 28.4(1.32) 28.4 one digit 1.32 two digits 568 8520 28400 37.488 three digits

1

8 8 1 # 40 8 5 5 1

19. Divide 120 by

3 to find the number of installations. 4

120 # 4 3 120 4 1 3 40

120 # 4 1 3 40 # 4 # 1 1 160

1

There will be 160 installations. 20. First add the lengths of the pieces needed to be cut. 3 1 11 13 11 52 63 1 6 8 2 8 2 8 8 8 63 To find the remaining piece, subtract from 12. 8 96 63 33 1 63 4 12 8 8 8 8 8 1 The length of the remaining piece is 4 feet. 8

Section 1.6 1. (a) 17.3 21.05 0.4 17.3 21.05 0.4 38.75

(b) 0.017(0.02) 0.017 three digits 0.02 two digits 0.00034 five digits 4. (a) 29.16 1.2 S 291.6 12 24.3 12 291.6 24 51 48 36 36 0 (b) 20.528 0.04 S 2052.8 4 513.2 4 2052.8 20 5 4 12 12 8 8 0 5. 5.3a 3.06b 0.08b 0.01a 5.3a 0.01a 3.06b 0.08b

15.3 0.012a 13.06 0.082b

5.29a 13.14b2 5.29a 3.14b

525

526

6.


2 1 2 3 1 x y a b 122 3 4 3 7 4 2 2 7 4 2 1 7 2 2 2 1 7 # # 7 2 2 7

Chapter 2

7 4 14 14 7.

31 # 6 # 60 7 1

3 when x , y 2 7

10

31 70

8. 2x 4y when x 2.1, y 0.07 212.12 410.072 4.2 0.28 3.92 9. 5.6x 0.8x 1.8x 0.4

15.6 0.8 1.82x 0.4

11 14

4.6x 0.4 when x 0.3

1 3 1 1 3 1 x x x a bx 4 5 3 4 5 3 1 15 3 12 1 20 a # # # bx 4 15 5 12 3 20 15 36 20 a bx 60 60 60 6 31 x when x 60 7 31 6 a b 60 7

4.610.32 0.4 1.38 0.4 1.78 10. Add the number of ounces. 4.6 1.25 8 0.02 5.85 8 0.02 13.85 0.02 13.87 The pharmacist prepared 13.87 ounces.

CHAPTER 2 Section 2.1 1.

5.

x 11 4

2 x 12 3 3 3 2 a xb 1122 2 3 2

x 11 11 4 11 x7 The solution set is {7}. 2.

x

y62


y6626 y 4

6. 3x 49

The solution set is {4}. 3.

49 3x 3 3

1 1 x 5 2

x

1 1 1 1 x 5 5 2 5

7 10

The solution set is e 4.

7. 7 f. 10

0.08 x 0.45 0.08 0.45 x 0.45 0.45 0.37 x The solution set is {0.37}.

49 3


5 2 x 10 10 x

36 18 2

49 f. 3

1 3 y 5 4 5 1 5 3 a yb a b 3 5 3 4 y

5 12


5 f. 12


8.

P 16x 62 inches

x 3 4 a b 4 a b 4 2

11. Let s represent the length of the side of a square in yards. Because 1 yard 3 feet, the side of the square is (3s) feet. To find the area, use A s2.

12 6 2

A (3s)2


A (9s2) square feet

9. 0.3y 18.6 0.3y 18.6 0.3 0.3

Section 2.3

y 62

1.

4x 3 17 4x 3 3 17 3


4x 20 4x 20 4 4

Section 2.2 1. The sum of two numbers is 56. One of the numbers is y. The other number is 56 y.


2. The difference of two numbers is 20. The smaller number is x. The larger number is 20 + x.

2.

3x 15 3x 15 3 3 x5

4. In 10 minutes the fax transmits 12(10) 120 pages. In 20 minutes the fax transmits 12(20) 240 pages. In y minutes the fax transmits 12(y) (12y) pages.


5y 35 5y 35 5 5

6. Suppose an electric car travels at 30 miles per hour for 2 hours. Using Distance Rate # Time,

y 7

d 1302122


d 60 The car travels 60 miles. Then an electric car traveling p miles per hour for 4 hours would travel d p(4) (4p) miles. e euros. 8

8. Because 1 centimeter 10 millimeters, multiply x by 10. The bolts are (10x) millimeters apart. 9. To change from inches to feet, divide p inches by 12. p So the height of the plasma TV is a b feet. 12 10. Let x represent the width of a rectangle. Then 2x 3 represents the length of the rectangle. P 212x 32 21x2

1 5y 36 1 5y 1 36 1

5. Four quarters and six dimes would be 4(25) 6(10) 160 cents. So q quarters and d dimes would be (25q 10d) cents.

Use P 2l 2w.

3x 2 13 3x 2 2 13 2

3. The product of two numbers is 45. One of the numbers is x. 45 The other number is . x

7. The cost of the trip is e. The cost per day is

527

P 4x 6 2x

3 x 4 2

y

Chapter 2

4.

24 4a 3 24 3 4a 3 3 27 4a 4a 27 4 4 27 a 4 The solution set is e

27 f. 4

5. Let n represent the number to be found. n 21 78 n 21 21 78 21 n 57 The number is 57.

528


Chapter 2

5x 5 45

6. Let a represent Dahlia’s age now. Then a 5 represents her age 5 years ago.

5x 5 5 45 5

a 5 19

5x 40

a 5 5 19 5

5x 40 5 5 x8

a 24 Dahlia is 24 years old now.


7. Let x represent the price per minute. a

Number of b minutes

#

a

34 # x 2.72

Price per Total b a b minute price

2a 24 6a 6a 44 6a 4a 24 24 44 24

34x 2.72 34 34

4a 20 4a 20 4 4

x 0.08 Damien pays $0.08 per minute. 8. Let n represent the number to be found. 4n 6 30


n 1n 22 126

4n 36 4n 36 4 4

2n 2 126 2n 2 2 126 2

n9

2n 124

The number is 9. 9. Let x represent the cost of each DVD movie. Cost of b5 DVD player

a 5 4. Let n represent the first even number. Then n 2 represents the next even number.

4n 6 6 30 6

#

a

Cost of b 439.00 DVD movie

349 5x 439 349 5x 349 439 349 5x 90 5x 90 5 5 x 18 Each DVD movie costs $18.

2n 124 2 2 n 62 If n 62, then n 2 64. So the numbers are 62 and 64. 5. Let x represent the number of deliveries. 50 0.75x 77 50 0.75x 50 77 50 0.75x 27 0.75x 27 0.75 0.75 x 36 The driver must make 36 deliveries.

Section 2.4 1. 2x 7 5x 14 3x 7 14 3x 7 7 14 7 3x 21 21 3x 3 3 x 7 The solution set is {7}. 2.

2a 24 6a 44 4a 24 44

34x 2.72

a

3.

x 5 6x 45 x 5 6x 6x 45 6x

6. Let a represent the measure of the larger angle. Then a 56 represents the measure of the smaller angle. The sum of the measures of supplementary angles is 180°. a 1a 562 180 2a 56 180 2a 236 236 2a 2 2 a 118 If a 118, then a 56 62. The two supplementary angles have measures of 118° and 62°.


529

3 12x 12 12

7. Let a represent the measure of the first angle. Then 3a represents the measure of the third angle, and 5 3a represents the measure of the second angle. The sum of the measures of the angles of a triangle is 180°.

x

a 15 3a2 3a 180

1 4


7a 5 180 7a 5 5 180 5

Chapter 2

7y 5 3 12 6 8

4.

7a 175

The LCD of 6, 8, and 12 is 24

7a 175 7 5

24 a

a 25

24 a

If a 25, then the first angle measures 25°, the second angle measures 5 3a 80°, and the third angle measures 3a 75°.

7y 5 3 b 24 a b 12 6 8

7y 5 b 24 a b 3132 12 6 217y2 4152 9 14y 20 9 14y 29 14y 29 14 14

Section 2.5 1. 31x 42 51x 22 3x 12 5x 10

y

8x 12 10


8x 2 2 8x 8 8

29 f. 14

x4 x2 7 3 5 10

5.

1 x 4

29 14



30 a 30 a

2. 21x 42 71x 22 3 2x 8 7x 14 3

x4 x2 7 b 30 a b 3 5 10

x4 x2 b 30 a b 3172 3 5 101x 42 61x 22 21

5x 22 3

10x 40 6x 12 21

5x 25

16x 28 21

25 5x 5 5

16x 49 49 16x 16 16


x 3.

4 2 7 x 5 3 15

The solution set is e The LCD of 3, 5, and 15 is 15

2 7 4 15 a x b 15 a b 5 3 15

The LCD of 3, 4, and 8 is 24 24 a

314x2 5122 7 12x 3

49 f. 16

x1 x4 5 8 3 4

6.

4 2 15 a xb 15 a b 1172 5 3 12x 10 7

49 16

24 a

x4 5 x1 b 24 a b 8 3 4

x1 x4 b 24 a b 6152 8 3

530


Chapter 2

n 48 0

31x 12 81x 42 30

n 48

3x 3 8x 32 30 5x 29 30 5x 1 1 5x 5 5 x

7.

The number is 48. 3 11. Let x represent the hourly rate. Then x represents 2 1 1 times his hourly rate. 2

1 5

a

Wages for Wages for b a b 696 first 40 hours overtime hours


3 40x 12 a xb 696 2

6x 1 213x 22

40x 613x2 696 40x 18x 696

6x 1 6x 4 6x 1 6x 6x 4 6x

58x 696

14

58x 696 58 58

The result is a false statement. Therefore, the equation is a contradiction, and the solution set is . 8. 31x 12 2x 7 5x 4 3x 3 2x 7 5x 4 5x 4 5x 4

x 12 His hourly rate is $12.00. 12. Let n represent the first whole number. Then n 1 represents the second whole number, and n 2 represents the third whole number. 3n 21n 12 1n 22 76

5x 4 5x 5x 4 5x 44

3n 2n 2 n 2 76 6n 4 76

The result is a true statement. Therefore, the equation is an identity, and any real number is a solution. The solution set is {all reals}.

6n 72 72 6n 6 6

9. Let q represent the number of quarters. Then 20 q represents the number of dimes.

n 12

Value of quarters Value of dimes a b a b 395 in cents in cents

If n 12, then n 1 13 and n 2 14. So the numbers are 12, 13, and 14.

25q 10120 q2 395 25q 200 10q 395

Section 2.6

15q 200 395

1.

15q 195

x55 25

15q 195 15 15

x 3


q 13 If q 13, then 20 q 7, so Michaela has 13 quarters and 7 dimes.

−4 −3 −2−1 0 1 2 3 4

2.

x2

The solution set is 5x 0 x 26 .

The LCD of 2 and 4 is 4

3 1 4 a n 12b 4 a nb 4 2 3 4 a nb 41122 21n2 4 3n 48 2n

x 3 1 x 3 3 1 3

10. Let n represent the number to be found. 3 1 n 12 n 4 2

x5 2

−3 −2 −1 0

3.

1

2

3 4

8 5x 85 5x5 3 x means x 3.


The solution set is 5x 0 x 36 . −2 −1

0

1

2

3

5.

531

3x 24 3x 24 3 3

4

4. 7x 21 7x 21

7 7 x 3 The solution set is 5x 0 x 36 .

Chapter 2

x8

The solution set is 5x 0 x 86 . 4 5 7 y y 3 6 6

3.

The LCD of 3 and 6 is 6

4 5 7 6 a y yb 6 a b 3 6 6

4 2 x 3 7

4 5 6 a yb 6 a yb 7 3 6

3 4 3 2 a xb a b 2 3 2 7

214y2 5y 7

6 x 7

8y 5y 7 3y 7 3y 7

3 3

6 The solution set is e x 0 x f . 7 6. 5x 10

y

5x 10 5 5

7 The solution set is e y 0 y f . 3

x 2


x4 x2 5 3 4 12

4.

7. 2x 1 7

7 3

2x 8


2x 8

2 2 x 4 The solution set is 5x 0 x 46.

x2 5 x4 12 a b 12 a b 3 4 12 12 a

x4 x2 b 12 a b5 3 4 41x 42 31x 22 5

8. 4y 5 17 4y 12 4y 12

4 4 y 3

The solution set is 5y 0 y 36.

4x 16 3x 6 5 x 10 5 x 5

The solution set is 5x 0 x 56. 5. x 0 and

x 5

x 0

Section 2.7 1. 7x 8 3x 12 4x 8 12

− 3 −2 − 1 0 1 2 3 4 5 6

x 5

4x 4 4 4x 4 4

− 3 − 2− 1 0 1 2 3 4 5 6

x 0 and x 5

x1

The solution set is 5x 0 x 16. 2. 21x 12 41x 22 31x 62 2x 2 4x 8 3x 18 6x 6 3x 18 3x 6 18

− 3 −2 − 1 0 1 2 3 4 5 6

The solution set is 5x 0 0 x 56. 6. x 3 and

x5

x3 −2 −1 0

1 2

3 4

5

532


Chapter 3

x 1 or x 1

x5

−4 −3 −2 −1 0 1 2 3 4

−3 −2 −1 0 1 2 3 4 5 6

The solution set is 5x 0 x 1 or x 16.

x 3 and x 5 −2 −1 0

1 2

3 4

5

The solution set is 5x 0 x 36. 7. x 2 or

x 4

x 2 −2 −1

0

1

2

3

4

5

x 4

9. Let y represent the score on the fourth exam. 88 95 90 y 92 4 273 y 92 4 273 y 4a b 41922 4 273 y 368 y 95

−3 −2 −1 0 1 2 3 4 5 6

x 2 or x 4 −2 −1

0

1

2

Felix must score 95 or higher on the fourth exam. 3

4

5

The solution set is 5x 0 x 26. 8. x 1 or

10. Let w represent the number of wins for the remainder of the season. Seventy percent of 160 games is the same as 0.70(160).

x1

a

x 1

70% of Number b b a total wins of wins w 32 0.7011602

−3 −2 −1 0 1 2 3 4 5 6

w 32 112

x1

w 80 They need to win at least 80 games.

− 3 − 2− 1 0 1 2 3 4 5 6


a

x 5 18 6


6x 90 x 15 The solution set is {15}. 2.

x2 x4 3 5 51x 22 31x 42 5x 10 3x 12 2x 10 12


3 8 a1 a2 81a 22 31a 12 8a 16 3a 3 5a 16 3 5a 13

13 f. 5

x x 1 4 2

4. 4a

x x 1b 4 a b 4 2

x 4 a b 4112 21x2 4 x 4 2x

2x 2 x 1

13 5

4 x The solution set is {4}. 5. 1 inch represents 12 feet. Let x represent the distance in feet between the two driveways. 12 1 x 1 3 4 1 x112 a 3 b 1122 4


x a

2 r 5

13 12 ba b 4 1

x 39

2 40 5 100

The distance between the two driveways is 39 feet. 6. Let a represent the amount of money in one account. Then 1500 a represents the amount of money in the second account. a 2 1500 a 3

r 40% Thirty-two is 40% of 80. 12. Sixty is what percent of 40? Let r represent the percent to be found. 60 r1402

3a 211500 a2

60 r 40

3a 3000 2a 5a 3000

3 r 2 150 3 2 100

a 600 If a 600, then 1500 a 900. Therefore, one account has $600 and the other has $900. 7.

r 150%

n 13 100 25

Sixty is 150% of 40.

25n 1300 n 52 13 52% 25 8.

Section 3.2 1. 0.4x 36 4x 360

n 7 100 12

x 90

12n 700 175 1 700 58 n 12 3 3


3x 0.2x 128 1013x 0.2x2 1011282

7 1 58 % 12 3

30x 2x 1280 32x 1280

9. What is 12% of 58? n 112% 21582

x 40

n 10.1221582 n 6.96

Twelve percent of 58 is 6.96. 10. Eight percent of what number is 56? Let n represent the number to be found.


0.04a 0.07a 16.5 10010.04a 0.07a2 100116.52 4a 7a 1650 11a 1650

18% 21n2 56 0.08n 56

8n 5600 n 700 Eight percent of 700 is 56. 11. Thirty-two is what percent of 80? Let r represent the percent to be found. 32 r1802 32 r 80

Chapter 3

a 150 The solution set is {150}. 4. 0.08y 0.151y 2002 99 8y 151y 2002 9900 8y 15y 3000 9900 8y 15y 6900 23y 6900 y 300 The solution set is {300}.

533

534


Chapter 3

5. Let x represent the original price of the computer.

715 r 0.13

Original Discount a b 1Discount2 a b selling price sale price

71,500 r 13

1100% 21x2 120% 21x2 640

5500 r

1x 0.2x 640 10x 2x 6400 8x 6400

The principal must be $5500. 11. i Prt

6400 8x 8 8

i 14,00010.062 a

x 800

i 840 a

The original price of the computer was $800.

x 98 The sale price of the sunglasses is $98.

The monthly interest will be $70.

Section 3.3 1.

s 450 180% 214502

s 450 10.80214502 s 450 360 s 810

The selling price of the wheels is $810. 8. Let n represent the selling price of the photo. Selling Profit a b 1Cost2 a b price 1% of selling price2 n 80 10.6021n2

n 80 0.60n 10n 800 6n 4n 800 n 200 The selling price of the photo must be $200. 9.

i Prt 1620 10,000r132 1620 30,000r 1620 r 30,000

2.5 t 2.

1 A 150, h 10 A h1B1 182 2 1 150 1102 1B1 182 2 150 51B1 182 150 5B1 90 60 5B1 12 B1

1 3. Use A bh. 2 1 A bh A 52, b 16 2 52

1 1162 1h2 2

52 81h2 6.5 h The height of the triangle is 6.5 feet. 1 4. A h1B1 B2 2 2 A

h 10, B1 12, B2 16

1 1102 112 162 2

54 r 1000

A 51282

0.054 r

The area of the deck is 140 square feet.

The annual interest rate is 5.4%. 10.

d rt for d 150, r 60 150 60t

7. Let s represent the selling price of the wheels. Selling Profit b 1Cost2 a a b price 1% of cost2

1 b 12

i 70

6. Let x represent the discount sale price. Because the sunglasses are 30% off, we must pay 70% of the original price. x 10.70211402

1 b 12

i Prt

A 140 5. Use S 2pr2 2prh. S 2pr2 2prh

h 24, r 15

715 P10.0652122

S 2p1152 2p1152 1242

715 P10.132

S 2p12252 2p13602

2


S 500p 720p 1170p

3x 8y 5 8y 5 3x

Area of Area of deck Area of b a b a b deck and pool pool

Area of deck lw and pool

130 4 42120 4 42 13821282 1064

Area of lw pool

Subtract 3x

y

5 3x 8

y

5 3x # 1 5 3x 1 Multiply by 8 1 8 1

y

3x 5 8

Divide by 8

12. Solve y ax b for a. y ax b

13021202 600

Area of Area of deck Area of deck and pool pool 1064 600 464

y b ax yb ax x x

Subtract b Divide by x

yb a x

The area of the deck is 464 square feet. 7. Solve i Prt for P.

Section 3.4

i Prt i Prt rt rt

1. (a) Divide by rt

7 b 4 39 5

1 8. Solve A bh for b. 2

7 b 35 5

1 A bh 2

5 7 5 a bb 1352 7 5 7 Multiply by 2

b 25

2A bh 2A

bh h

2 b b 4 39 5 2 5 b b 4 39 5 5

i P rt

1 2A 2 a bhb 2

The solution set is {25}. Divide by h

(b) n 1n 12 1n 22 153 n n 1 n 2 153

2A b h

3n 3 153 3n 150

9. Solve S 2pr2 2prh for h.

n 50

S 2pr2 2prh S 2pr2 2prh

Subtract 2pr 2

S 2pr2 2prh 2pr 2pr

Divide by 2pr

S 2pr2 h 2pr

The solution set is {50}. (c)

1 2 c 1c 42 15 3 3 1 2 3 c c 1c 42 d 31152 3 3

10. Solve 2x y 3 for x. 2x y 3 2x 3 y 3 y x 2

535

11. Solve 3x 8y 5 for y.

The total surface area is 1170p square feet. 6. a

Chapter 3

Subtract y Divide by 2

2 1 3 a cb 3 c 1c 42 d 45 3 3 c 21c 42 45 c 2c 8 45 3c 8 45

536


Chapter 3

3c 37 c

Rate # Time Distance

37 3

37 The solution set is e f . 3 2. Let t represent the time needed to earn $525. Use i Prt. 525 300010.102t

a

Rate

Time

Distance

Student

60

t

60t

Mom

70

t

1 70 a t b 4

Student’s Mom’s b a b distance distance 1 60t 70 a t b 4 70 60t 70t 4

525 300t 525 t 300 7 t 4

4160t2 4 a 70t

1.75 t It will take 1.75 years to earn $525. 3. Let w represent the width and l the length of the computer screen. The length is twice the width, so l 2w. Use P 2l 2w. 48 212w2 2w

70 b 4

240t 280t 70 40t 70 70 7 t 40 4 7 6 3 1 If t , then t . 4 4 4 2 His mom drove for 1.5 hours.

48 4w 2w

Section 3.5

48 6w

1. Let x represent the amount of pure orange juice added.

8w If w 8, then l 2w 16. The width is 8 inches, and the length is 16 inches. 4. Let x represent the time each walked. The total distance is 3.5 miles. Rate # Time Distance Rate

Time

Distance

Justin

5

x

5x

Hope

2

x

2x

a

1 4

Justin’s Hope’s b 3.5 b a distance distance

5x 2x 3.5 7x 3.5 70x 35 1 35 x 70 2 1 It will take Justin and Hope hour to meet. 2 5. Let t represent the time the student traveled and let 1 t represent the time his mom traveled. They both 4 traveled the same distance.

Amount of solution

Percent

Amount of pure orange juice

Original solution

36

0.10

(0.10)(36)

Pure orange juice

x

1.00

1.00x

36 x

0.20

0.20(36 x)

Final solution a

Pure orange juice Pure orange juice b b a added in original solution a

Pure orange juice b in final solution

10.1021362 1.00x 0.20136 x2 36 10x 2136 x2 36 10x 72 2x 36 8x 72 8x 36 36 8x 8 8 x

9 36 8 2

The amount of pure orange juice that must be added is 4.5 quarts.


Chapter 3

537

2y 12,400 2 2

2. Let x represent the amount of 10% solution and let 6000 x represent the amount of 20% solution.

y 6200 Amount of solution

Percent

Amount of pure ethanol

10% solution

x

0.10

(0.10)(x)

20% solution

6000 x

0.20

0.20(6000 x)

16% solution a

6000

0.16

0.16(6000)

Pure ethanol Pure ethanol b a b in 10% solution in 20% solution a

If y 6200, then 8000 y 1800. The amount invested at 7% is $6200, and the amount invested at 5% is $1800. 4. Let x represent the amount invested at 6% and let x 2000 represent the amount invested at 9%. Amount invested

Percent

Interest earned

6% investment

x

0.06

(0.06)(x)

9% investment

x 2000

0.09

0.09(x 2000) 870

Pure ethanol b in 16% solution

a

10.1021x2 0.2016000 x2 0.16160002 10x 2016000 x2 16160002

Interest earned Interest earned b a b at 6% at 9% a

10x 120,000 20x 96,000

Total interest b earned

10.0621x2 0.091x 20002 870

10x 120,000 96,000 10x 24,000

6x 91x 20002 87,000

24,000 10x 10 10

6x 9x 18,000 87,000 15x 18,000 87,000

x 2400

15x 105,000

If x 2400, then 6000 x 3600. There should be 2400 gallons of 10% solution and 3600 gallons of 20% solution. 3. Let y represent the amount invested at 7% and let 8000 y represent the amount invested at 5%.

105,000 15x 15 15 x 7000 If x 7000, then x 2000 5000. Mrs. Stewart invested $7000 at 6% and $5000 at 9%. 5. Let y represent Kyle’s age now. We represent the unknown quantities as follows:

Amount invested

Percent

Interest earned

7% investment

y

0.07

(0.07)(y)

5% investment

8000 y

0.05

0.05(8000 y)

y 3 1 y 4:

524

Kyle’s age 2 a b in one year 3

Total investment a

8000

Interest earned Interest earned b b a at 7% at 5% a

Total interest b earned

10.0721y2 0.0518000 y2 524 7y 518000 y2 52,400 7y 40,000 5y 52,400 2y 40,000 52,400 2y 12,400

y:

Kyle’s present age

y 3:

Jennifer’s present age

y 1:

Kyle’s age in one year

y1

#

Jennifer’s age in one year a

Jennifer’s age b in one year

2 1y 42 3

2 31y 12 3 c 1y 42 d 3 3y 3 21y 42 3y 3 2y 8 y38 y5 If y 5, then y 3 8. Kyle’s present age is 5, and Jennifer’s present age is 8.

538


Chapter 4

CHAPTER 4 11. 12 54y 3 6 13y 22 4 106

Section 4.1

1. 18x 2x 62 13x 5x 102 2

2

18x 3x 2 12x 5x2 16 102 2

2

11x 3x 4 2

12 54y 36 3y 24 106 12 54y 34 3y4 106 12 14y 4 3y 102 12 1y 62

2. 16x 42 12x 32 15x 82

12 y 6

16x 2x 5x2 14 3 82

y 6

13x 7 3. 12x2 3x 42 15x3 7x 22 12x 62 15x3 2 12x2 2 13x 7x 2x2 14 2 62

12. Area of circle: pr2 Area of rectangle: 10r Sum of areas pr2 10r

5x3 2x2 2x 4. 19x2 8x 22 13x2 2x 12

Section 4.2

19x 8x 22 13x 2x 12

1. (a) y3 # y5 y35 y8

6x2 6x 3

2. (a) 14x2 216x3 2 4 # 6 # x2 # x3

2

2

19x2 3x2 2 18x 2x2 12 12 5. 16a 42 12a2 3a 12

(b) a2 # a a21 a3 24x5

16a 42 12a2 3a 12

2

2 6 2 6 (b) a a5bb a a4b2 b # # a5 # a4 # b # b2 3 7 3 7

12a2 2 16a 3a2 14 12

1

2a2 9a 5

4 a9b3 7

6. To subtract, add the opposite. 11x2 8x 2 1

3. (a) 1y3 2 2 y3 2 y6 #

11x2 8x 2

14x2 8x 3 1 4x2 8x 3

(b) 1m2 2 4 m2 4 m8 #

7x2 9x 5

4. (a) 12x3 2 4 122 4 1x3 2 4 16x12

7. To subtract, add the opposite. 8

5y3 2y2

1

5y3 2y2

8

6y 4 1 8y3

8y3

6y 4 3y3 2y2 6y 12

8. 16x 12 13x 22 15x 82 116x 12 113x 22 115x 82

(b) 13a3b5 2 2 132 2 1a3 2 2 1b5 2 2 9a6b10

5. (a) 15a3 2 2 12a4 2 3 152 2 1a3 2 2 122 3 1a4 2 3 251a6 2 1821a12 2 200a18

(b) 13m4n2 2 3 1mn2 2 132 3 1m4 2 3 1n2 2 3 112 2 1m2 2 1n2 2 271m2 12 1n6 21121m2 2 1n2 2

6x 1 3x 2 5x 8

16x 3x 5x2 11 2 82 4x 7 9. 1x2 3x 42 14x2 5x 62 12x 72 x 3x 4 4x 5x 6 2x 7 2

2

1x2 4x2 2 13x 5x 2x2 14 6 72 3x2 9 10. 10y 37y 1y 42 4

10y 37y y 44

27m14n8

6. (a) 14a3 2 1a2 3ab 5b2 2

14a3 21a2 2 14a3 2 13ab2 14a3 215b2 2 4a5 12a4b 20a3b2

(b) 12y215x2y 6xy y3 2

12y2 15x2y2 12y2 16xy2 12y2 1y3 2 10x2y2 12xy2 2y4

7. Area of floor plan:

12x212x2 4x2 14x21x2 4x2 12x212x2 4x2

10y 8y 4

Area of left side: Area of middle region: Area of right side:

2y 4

Total area 4x2 4x2 4x2 12x2.

10y 38y 4 4


Chapter 4

11. Let y represent the length of each side of the square corner piece to be removed.

Section 4.3

1. 1x 521y 22 x1y 22 51y 22

Then 10 2y represents the length of the cardboard box after the corner pieces have been removed.

x1y2 x122 51y2 5122 xy 2x 5y 10

And 6 2y represents the width of the cardboard box after the corner pieces have been removed. Volume lwh

2. 1x 421y z 32 x1y z 32 41y z 32 x1y2 x1z2 x132 41y2 41z2 4132

110 2y216 2y2 1y2

160 20y 12y 4y2 21y2

xy xz 3x 4y 4z 12

160 32y 4y2 2 1y2

3. 1x 621x 52 x1x 52 61x 52

60y 32y2 4y3 cubic inches

x2 5x 6x 30

Surface area of original piece of cardboard Area of four cut-out pieces

x2 11x 30 4. 1x 221x 82 x1x 82 21x 82

1102162 1421y2 1y2

x 8x 2x 16 2

60 4y2 square inches

x 6x 16 2

Section 4.4

5. 1x 32 1x2 2x 62

1. (a)

x1x2 2x 62 31x2 2x 62 x3 2x2 6x 3x2 6x 18 x3 2x2 3x2 6x 6x 18

(b)

x3 5x2 12x 18 6. 1x 2y215x 3xy 4y 2 2

2. (a)

2

x15x2 3xy 4y2 2 2y15x2 3xy 4y2 2 2

2

2

2

32b8 8b82 8b6 4b2 6a4b7 1a41b72 a3b5 6ab2 21m4n5 14m2n3 7m2n

5x 3x y 4xy 10x y 6xy 8y 3

3

14m2n3 21m4n5 2 7m n 7m2n

5x3 3x2y 10x2y 4xy2 6xy2 8y3

3m2n4 2m0n2

5x3 7x2y 2xy2 8y3

3m2n4 2n2

7. 1x 121x 62 x2 7x 6

(b)

8. 1x 321x 72 x2 4x 21

2a2 3a 6

1x 52 3 1x 521x 52 4 x1x2 10x 252 51x2 10x 252

Section 4.5 1.

x7 x 4 x2 11x 28 x2 4x x2 7x 28 x2 7x 28 x2 7x 20 2 1x 11x 282 1x 42 x 7

2.

3x 1 x 2 3x2 5x 2 3x2 6x 3x2 6x 2 3x26 x 2 3x26x 0

x3 10x2 25x 5x2 50x 125 x3 10x2 5x2 25x 50x 125 x3 15x2 75x 125 Special Product Patterns

1. 1x 82 2 1x2 2 21x2182 182 2 2. 1x 12 1x2 2 21x2112 112 2 x2 2x 1 3. 13x 52 13x 52 13x2 2 152 2 9x2 25

12a5 18a4 36a3 3 3 6a 6a 6a3

2a2 3a 6a0

10. 1x 52 3 1x 521x 521x 52

1x 521x2 10x 252

12a5 18a4 36a3 6a3

9. 14x 12 12x 32 8x2 10x 3

x2 16x 64

539

13x2 5x 22 1x 22 3x 1

540

3.


Chapter 4

2x 5 4x 3 8x 14x 15 8x2 6x 8x2 20x 15 8x2 20x 15 8x2 20x 0

(c)

2

(d) 125 2 1 2152 112 25 32

18x2 14x 152 14x 32 2x 5

4.

3. (a) z1 # z4 z1142 z5

4x 6 x 3 4x 6x 4 4x2 12x 6x 4 6x 18 6x 14

(c) a

2

(d) 12a5b2 215a3b2 122 152 1a53 2 1b21 2 10a2b3

4x 6x 4 14 4x 6 x3 x3

(b) 5,600,000 15.62 1106 2 (c) 9.54 19.542 1100 2 5. (a) 18.221107 2 82,000,000 (b) 13.4621104 2 0.000346 6. 150,000210.00000132 1521104 2 11.32 1106 2 15211.321104 2 1106 2 16.5211046 2

a4 a2 a a3 5a2 2a 3 a3 a2 a3 4a2 2a a3 4a2 4a a3 4a2 6a 3

16.521102 2 0.065

7.

6a 3 a2 5a2 2a 3 a4 2 2 a a a a

y6

1103 2

1102 2

200,000 8.

y2162 y26 y4

19002 10.00062

130,000210.022

5

k k55 k0 1 k5

2. (a) 35 # 38 358 33

14.22

#

122 1105 2

1 z3 (c) 2 z32 z5 5 z z (d)

18.42

122 1103122 2

m4 1 1. (a) 7 m47 m3 3 m m (b)

18.42 1103 2 8400 0.042 14.22 1102 2

Section 4.6

y2

10b3 a2

4. (a) 0.00000683 16.832 1106 2

a2 3a 9 3 a 3 a 0a2 0a 27 a3 3a2 a3 3a2 0a a3 3a2 9a a3 3a2 9a 27 a3 3a2 9a 27 a3 3a2 9a 20 a3 27 a2 3a 9 a3

6.

b2 a10

1m2 2 1 m2 1 m2 1 b 2 4 4 4 4 1 n n mn 1n 2

2

5.

1 z5

(b) 1a5b1 2 2 1a5 2 2 1b1 2 2 a10b2

8x2 14x 15 2x 5 4x 3

or

101 101142 1014 103 1000 104

132 1104 2 122 1102 2 192 1621102 2 1104 2

132 1221104 2 1102 2 1542 162

192

1 1 3 27 3

(b) 102 # 101 102 112 103

192 1102 2 162 1104 2

#

11024 2

#

11042 2

1102 2 110 2 2

192 11022 2 1 1 3 1000 10

192 1104 2 0.0009


Chapter 5

541

CHAPTER 5 Section 5.1

14a 2 # 7 # a # a b 17a5 7 # a # a # a # a # a

11. m1m 22 51m 22 0 1m 22 1m 52 0

2

1.

m 2 0 or m 5 0

The greatest common factor is 7 # a # a 7a2.

18m2n4 2 # 3 # 3 # m # m # n # n # n # n 2. 14m3n5 2 # 2 # m # m # m # n # n # n # n # n¥ 10m4n3 2 # 5 # m # m # m # m # n # n # n

The solution set is {5, 2}. 12. Let s represent the length of the side of the square. Use these formulas: A s2 Area of a square P 4s Perimeter of a square

The greatest common factor is 2 # m # m # n # n # n 2m2n3.

a

3. 15a2 21a6 3a2 152 3a2 17a4 2 3a2 15 7a4 2

a

Perimeter b of square

s2 12s 0 s1s 12s2 0

5. 48y8 16y6 24y4 8y4 16y4 2 8y4 12y2 2 8y4 132

s 0 or s 12 0

8y4 16y4 2y2 32

s 0 or

6. 8b 8b 8b 1b2 8b 112 2

#

s2 12s

2m3n14n m3 2

2

Area of b3 square

s2 314s2

4. 8m3n2 2m6n 2m3n14n2 2m3n1m3 2

3

m 5

m 2 or

s 12

Discard s 0, because length cannot be 0. The length is 12.

2

8b2 1b 12

7. (a) ab 5a 3b 15 a1b 52 31b 52 1b 521a 32

(b) xy 2x 4y 8 x1y 22 41y 22 1y 221x 42

1. 3y2 12 31y2 42 31y 221y 22 2. 32x3 2x 2x116x2 12

8. y2 7y 0

2x14x 12 14x 12

y1y 72 0

3. a4 81 1a2 92 1a2 92

y 0 or y 7 0 y 0 or

Section 5.2

1a2 92 1a 321a 32

y7

The solution set is {0, 7}. a 15a

9.

x2 64 0

1x 82 1x 82 0

a 15a 0 2

a1a 152 0

x80

a 0 or a 15 0 a 0 or

x2 64

4.

2

x 8 or

a 15


or x 8 0

4x2 9

5.

4x2 9 0

10. 5y2 2y 0 y15y 22 0

12x 32 12x 32 0

y 0 or 5y 2 0

2x 3 0

y 0 or y 0 or

x8


5y 2 y

2 5


or 2x 3 0

2x 3 or x

3 or 2

2x 3 x


3 2

542


Chapter 5

3x2 75 3x2 75 3 3

6.

2. m2 8m 15 We need two numbers with a sum of 8 and a product of 15.

x2 25

Product

1x 521x 52 0

3152 15 3 5 8

3152 15 3 152 8

or x 5 0

x50

x 5 or

The two numbers are 3 and 5.

x5

m2 8m 15 1m 321m 52

The solution set is {5, 5}. x3 49x 0

7.

3. y2 5y 24

x1x 492 0 2

We need two numbers with a sum of 5 and a product of 24.

x1x 721x 72 0 x 0 or x 7 0

or x 7 0

x 7 or

x 0 or

x7

The solution set is {7, 0, 7}. 8. Let y represent the length of the side of one square. Then 3y represents the length of the side of the other square. The sum of the areas of the squares is 250 square feet. a

Area of Area of b 250 b a second square first square y2 13y2 2 250

3182 24 3 182 5

132182 24 3 8 5

The two numbers are 3 and 8. y2 5y 24 1y 32 1y 82

4. y2 4y 12

Sum

11122 12 1 1122 11

10y2 250 10 10

2162 12 2 162 4

The two numbers are 6 and 2.

y2 25

y2 4y 12 1y 62 1y 22

y2 25 0

1y 521y 52 0

5. y2 7y 6

or y 5 0 y5

Discard y 5, because length cannot be negative. The length of the side of the first square is 5 feet, and the length of the side of the second square is 3(5) 15 feet.

Section 5.3 1. y2 9y 20 We need two numbers with a sum of 9 and a product of 20. Product

Sum

21122 24 2 1122 10

Product

10y2 250

y 5 or

Product

11242 24 1 1242 23


y2 9y2 250

y50

Sum

11152 15 1 15 16

x 25 0 2

Sum

11202 20 1 20 21 21102 20 2 10 12 4152 20 4 5 9 The two numbers are 4 and 5.

y2 9y 20 1y 421y 52

We need two numbers with a sum of 7 and a product of 6. Product

Sum

1162 6 1 6 7

112162 6 1 162 7 The two numbers are 6 and 1. y2 7y 6 1y 62 1y 12 6. a2 a 12 We need two numbers with a sum of 1 and a product of 12. Product 11122 12 2162 12 3142 12 132142 12

Sum 1 1122 11 2 162 4 3 142 1 3 4 1

The two numbers are 3 and 4. a2 a 12 1a 32 1a 42


We need two numbers with a sum of 7 and a product of 18. Product 11182 18 2192 18 3162 18 11182 18 2192 18 132162 18

Sum 1 18 19 2 9 11 369 1 1182 19 2 192 11 3 162 9

No two numbers will satisfy these conditions. y2 7y 18 is not factorable. 1y 1221y 22 0 y2

The solution set is {2, 12}. 9.

or a 4 0 a4


x2 18x 40 x2 18x 40 0

1x 2021x 22 0 x 20 0 or x 2 0 x 20 or

x 2

The solution set is {2, 20}. 11. Let n represent the first consecutive integer. Then n 1 represents the second consecutive integer. The product of the integers is 110. n1n 12 110 n2 n 110 0

1n 1121n 102 0 or n 10 0

n 11 or

n 10

If n 11, then n 1 10. If n 10, then n 1 11. The consecutive integers are 11 and 10 or 10 and 11. 12. Let z represent the width of the rectangular plot. Then z 2 represents the length. To solve, use A lw. 80 1z 22z 80 z2 2z

Discard z 10, because width cannot be negative. The width is 8 yards, and the length is z 2 10 yards. 13. Let x represent the length of one leg of a right triangle and let x 7 represent the other leg. To solve, use the Pythagorean theorem. a2 b2 c2

1x2 2 1x 72 2 1132 2

2x2 14x 120 0 1 1 12x2 14x 1202 102 2 2

x 12 0

or x 5 0 x5

x 12 or

Discard x 12, because length cannot be negative. If x 5, then x 7 12. The legs of the right triangle are 5 inches and 12 inches.

Section 5.4 1. 3y2 16y 5 The binomial will have the form (3y )(y There are only two factors of 5 that will give two possibilities. (3y 1)(y 5)

or

).

(3y 5)(y 1)

By checking we find that

3y2 16y 5 13y 12 1y 52 2. 4a2 8a 3

n2 n 110

n 11 0

z8

1x 1221x 52 0

1a 721a 42 0 a 7 or

z 10 or

x2 7x 60 0

a2 3a 28 0

a70

or z 8 0

2x2 14x 49 169

or y 2 0

y 12 or

0 1z 82 1z 102 z 10 0

x2 x2 14x 49 169

y2 14y 24 0

y 12 0

543

0 z2 2z 80

7. y2 7y 18

8.

Chapter 5

We find that 4a2 can be written as 4a # a or 2a # 2a. Because the last term is positive and the middle term is negative, we have two possible forms: (4a

)(a

)

or

(2a

)(2a

)

The factors of 3 are 3 and 1, so we have the following possibilities: (4a 1)(a 3) (4a 3)(a 1)

(2a 1)(2a 3)

By checking we find that 4a2 8a 3 (2a 1)(2a 3) 3. 6y2 27y 30 has a common factor of 3. So 6y2 27y 30 312y2 9y 102 . Now factor 2y2 9y 10.

544


Chapter 5

(b) 36x2 6x 1 16x2 2 216x2 112 112 2

We find that 2y2 can be written as 2y # y. The binomial will have the form (2y )(y ) because the last term is positive and the middle term is positive.

16x 1216x 12 16x 12 2

The possible factors of 10 are 1 # 10 and 2 # 5, which will give the following possibilities: (2y 1)(y 10) (2y 10)(y 1)

(c) 49m2 56mn 16n2

17m2 2 217m2 14n2 14n2 2 17m 4n217m 4n2

(2y 2)(y 5) (2y 5)(y 2)

17m 4n2 2


2. 3x2 6x 72 31x2 2x 242

2y 9y 10 12y 521y 22 2

Therefore, 6y2 27y 30 312y 521y 22 . 4. 6a2 3a 7

We find that 6a2 can be written as 6a # a or 3a # 2a.

Because the last term is positive and the middle term is negative, we have two possible forms: (6a

)(a

)

(3a

or

)(2a

31x 621x 42 3. 4x 100 41x2 252 2

4. 9x2 30x 25 13x2 2 213x2 152 152 2 13x 5213x 52 13x 52 2

5. x2 8x 24

)


The factors of 7 are 7 and 1, so we have the following possibilities: (6a 1)(a 7) (6a 7)(a 1)

Product

21122 24 2 1122 10 3182 24 3 182 5

None of these choices gives a middle term of 3a. Therefore, 6a2 3a 7 is not factorable. 5.

142162 24 4 162 2

2y2 7y 6 0

There are no numbers that satisfy those conditions. Therefore, x2 8x 24 is not factorable.

12y 321y 22 0 2y 3 0 or y 2 0 2y 3 or y 2 2y 3 or y 2 2 2 3 or y 2 y 2

6. 3x2 10x 8

We find that 3x2 can be written as 3x # x. Because the last term is negative and the middle term is negative, the two binomials will have different signs. The possible factors of 8 are 1 # 8 and 2 # 4, so we must test for both possibilities. Testing for 1 # 8:

3 The solution set is e 2, f . 2 6.

(3x 1)(x 8) (3x 1)(x 8) (3x 8)(x 1) (3x 8)(x 1) None of these gives 10x as the middle term.

2a 5a 12 0 2

12a 321a 42 0 2a 3 0

Testing for 2 # 4:

or a 4 0

2a 3 or 2a 3 or 2 2 3 a or 2

a4

(3x 2)(x 4) (3x 4)(x 2)

a4


(3x 2)(x 4) (3x 4)(x 2)

3x2 10x 8 13x 221x 42

a4

7. 4x2 100 41x2 252 41x 521x 52


8.

y2 4y y2 4y 0

Section 5.5 1. (a) a 10a 25 1a2 21a2152 152 2

Sum

11242 24 1 1242 23

(3a 1)(2a 7) (3a 7)(2a 1)

2

1a 521a 52

1a 52

2

y1y 42 0 2

y 0 or y 4 0 y 0 or

y4



y 6 0 or y 2 0

z1z 252 0 2

z1z 52 1z 52 0

y 6 or

z 0 or z 5 0 z 0 or

or z 5 0

z 5 or

z5

The solution set is {5, 0, 5}. 10.

14. Let x represent one number. Then 2x 1 represents the other number. Their product is 36. x12x 12 36

13y 421y 22 0

2x2 x 36 2x2 x 36 0

3y 4 0 or y 2 0

y

4 3

or

12x 92 1x 42 0

y 2

2x 9 or

x4

9 2

x4

x

15y 2215y 22 0

8.

or 5y 2 0

5y 2 or

5y 2

2 y or 5

2 y 5


If x 4, then 2x 1 2(4) 1 8 1 9. 9 Therefore, the numbers are and 8 or 4 and 9. 2 15. Let a represent the altitude and let a 4 represent the side. To solve, use A

12. 1x 12 1x 22 20

30

x2 3x 2 20 x2 3x 18 0

0 a2 4a 60

x3

0 1a 621a 102


a 6 0 or a 10 0

3y2 12y 36 0

13.

1 1a2 1a 42 2

60 a2 4a

or x 3 0

x 6 or

1 bh. 2

60 1a2 1a 42

1x 621x 32 0 x60

or

9 9 If x , then 2x 1 2 a b 1 9 1 2 2

25y2 20y 4 0

5y 2 0

or x 4 0

2x 9 0

y 2

4 The solution set is e 2, f . 3 11.

y 2


3y2 2y 8 0

3y 4 or

a 6 or

1 1 13y2 12y 362 102 3 3

a 10

Discard a 10, because length cannot be negative. If a 6, then a 4 10. The altitude is 6 inches, and the side is 10 inches.

y2 4y 12 0


2a1a 42 2a2 8a 2a 2 1a 421a 42 a4 a 16

2.

11x 32 1 x3 1x 321x 22 x2 x 5x 6

3. 4.

5.

2

1a 521a 42 a2 a 20 a4 2 12a 121a 52 2a 1 2a 11a 5 xy2 x2y x xy 2

545

1y 62 1y 22 0

z3 25z 0

9.

Chapter 6

xy1y x2 x1x y2

y y 1

6.

15a 22 1a 22 5a2 8a 4 3 3a b 12ab 3ab1a2 42

15a 22 1a 22

3ab1a 22 1a 22

5a 2 3ab1a 22

51x 22 5 5x 10 6x 12 61x 22 6

546

7.


514 y2 51121y 42 20 5y y4 y4 11y 42

6.

x2 4x 12 1 x2 4x 12 # 4x 6 2x 3 4x 6 2x 3 1

5 5 1

8.

Chapter 6

14 a213 a2 12 a a2 1a 421a 22 a2 2a 8

1a 22

21x 621x 22 12x 32 12x 32

a3 a2

7.

3x2 5x 2 3x2 5x 2 # 1 1x 12 2 3x x 2 3x2 x 2 x 1

Section 6.2 1.

a a # a4 2 b 1a 121a 42 a 5a 4

2.

#

1a 42

y2 4y 21 # 2 y 2 y 3y

1y 721y 32 2

y y7 y1y 32 2

1. (a)

2x 6 # x 5x 4 3x 12 x2 2x 3

21x 32 31x 42

#

1x 121x 42 1x 321x 12

21x 32 1x 12 1x 42

(b)

2 31x 42 1x 32 1x 12 3 y 25

4.

2

5.

#

1 x1

51y 62 3y

2x 4 x 1 3

x3 3

214y 12 3y

2

1y 521y 52

12y 321y 52

#

y12y 12 y5 2. (a)

y1y 521y 5212y 12 12y 321y 521y 52 2y 3

5a 15 a2 9 5a 15 # ab 4b 20b ab 4b 20b a2 9 51a 32 20b

#

b1a 42

1a 321a 32

3y

5y 30 8y 2 3y

13y 28 3y

8x 4 5x 4 9

3x x 9 3

4x 6 3x 8 2 x 5x 6 x 5x 6 2

3x 8 4x 6 x2 5x 6

20 # b1a 321a 32

1x 321x 22

a4 41a 32

1 x3

5 # b1a 321a 42 4

51y 62 214y 12

1

(b)

8x 4 15x 42 5x 4 8x 4 9 9 9

y12y 12

113x 221x 12

13x 221x 12 1x 12

2y y 2y 7y 15 y 5 2

1 x1

2x 4 1x 12 x1 2x 4 3 3 3

2

3.

13x 221x 12

#

Section 6.3 1

#

13x 221x 12

b

a b1a 12

1

21x 621x 22

1a 421a 22

113 a2

#

12x 32

11a 4213 a2

1x 621x 22 212x 32

11x 22


1 4 3 1 7 4 a b a b 7 3 7 3 3 7 12 7 12 7 19 21 21 21 21 4. LCD 2 # 2 # 3 # 5 60. 5 7 5 5 7 2 a b a b 12 30 12 5 30 2 25 14 25 14 39 60 60 60 60 13 3 # 13 # 3 20 20 3.

5. LCD 3 # 5 15. x4 2x 7 3 x4 5 2x 7 a b a b 5 3 5 3 3 5 312x 72 51x 42 15 15 312x 72 51x 42 15 6x 21 5x 20 11x 1 15 15 # 6. LCD 2 3 6. 5x 3 x1 5x 3 x1 3 a ba b 6 2 6 2 3 31x 12 5x 3 6 6 5x 3 31x 12 6 5x 3 3x 3 6 x 2x 2#x # 6 2 3 3

3 2 a4 3 a5 2 a b a b a5 a4 a5 a4 a4 a5

9. LCD n(n + 1). 4 4 14 n 1 14 n a a b b n n n1 n1 n1 n 141n 12 4n n1n 12 n1n 12 4n 14n 14 n1n 12 18n 14 n1n 12

21a 42

1a 521a 42

31a 52

1a 421a 52

21a 42 31a 52 1a 52 1a 42

2a 8 3a 15 a 7 1a 52 1a 42 1a 521a 42

Section 6.4 1.

2.

4 2 2 4 2 y y y1y 32 y 3y LCD y(y 3). 2 4 y3 2 4 a b y y y3 y1y 32 y1y 32 41y 32 4y 12 2 2 y1y 32 y1y 32 y1y 32 4y 14 y1y 32 7 8 8 7 2 a2 a2 1a 22 1a 32 a 5a 6 LCD (a 2)(a 3). 8 7 a2 1a 22 1a 32 a

5 1 5 3 1 2 a b a ba b 2x 3x 2x 3 3x 2 2 15 2 17 15 6x 6x 6x 6x 3 7 3 5 7 4 a b a ba b 8a 10a 8a 5 10a 4 15 28 15 28 13 13 40a 40a 40a 40a 40a

547

10. LCD (a 5)(a 4).

7. LCD 6x.

8. LCD 40a.

Chapter 6

3.

7 a3 8 ba b a2 a3 1a 22 1a 32 71a 32

1a 221a 32 71a 32 8 1a 221a 32

8 1a 22 1a 32

7a 21 8 1a 22 1a 32

7a 13 1a 221a 32

5 3 2 x2 4 x 6x 8 3 5 1x 221x 22 1x 221x 42 LCD (x 2)(x 2)(x 4). 5 3 1x 22 1x 22 1x 22 1x 42

3 x4 a b 1x 221x 22 x 4

5 x2 a b 1x 221x 42 x 2 31x 42

1x 221x 22 1x 42 51x 22 1x 221x 42 1x 22

548


Chapter 6

31x 42 51x 22

1x 221x 221x 42

3x 12 5x 10 1x 221x 221x 42

4y

6 5 y3 y1 y 2y 3 4y 6 5 1y 321y 12 y3 y1

2

6 9.

61y 12 4y 1y 321y 12 1y 321y 12

2 m

2 m

3 7 n

1y 321y 12

5y 21 1y 321y 12

3 2 3#2 8 3 7 3 # 16 3 # 16 6 5. # # 7 8 16 8 7 8 7 1 7 7 1 16

#

4 4 1 1 10 a b b 5 2 5 2 3 1 3 1 a b 10 a b 10 5 10 5 a

4 1 10 a b 10 a b 5 2 85 3 32 5 3 1 10 a b 10 a b 10 5

3b 2a2 a1b 5a2

a

2 b m

3 7b n 2mn m

3mn 7mn n

mn a 6

mn a

2 b m

3 7b n

2n13m 12 6mn 2n or 3m 7mn m13 7n2

(b) If m pounds of apples cost n dollars, then the price per pound can be expressed by dividing the price by the number of pounds. Therefore, the price per n pound for apples is . m (c) If Taylor washes b automobiles in a hours, to find his rate for washing cars, divide the number of cars by the number of hours. Therefore, Taylor’s rate b for washing cars is . a

LCM 2 # 5 10

4 1 5 2 10 3 1 10 10 5

or

10. (a) To express Selena’s rate in miles per hour, divide the number of miles by the number of hours. x Therefore, Selena’s jogging rate is expressed as . y

2 a 5 2 2 b 2#b 2b # # 6. a a 5 5 b a 5 5a b 1 4 5 2 7. 3 1 10 5

#

a6

6mn

1y 321y 12

3b 2a2 ab 5a2

mn mn

4y 61y 12 51y 32 4y 6y 6 5y 15 1y 321y 12

3 2 3 2 a2b a 2 b b 2 b b a a 5 5 1 1 a b a2b a b a a b b

LCM mn

3 7 n 6

51y 32

#

a

3 2 3a2b 2a2b a2b a 2b a2b a b b b a a2 2 1 5 5a2b 1a b a2b a b a2b a b a a b b

LCD (y 3)(y 1). 4y 6 5 1y 321y 12 y3 y1 4y y1 6 a b 1y 321y 12 y3 y1 y3 5 a b y1 y3

LCM a2b

3 2 2 b a 2b a 2 5 1 ab a b

8x 2 1x 221x 221x 42

4.

3 2 2 b a 8. 5 1 a b

Section 6.5 1.

x3 2x 7 5 2 3 6 The LCD is 6. 6a 6a

2x 7 5 x3 b 6a b 2 3 6

2x 7 x3 b 6a b5 2 3


31x 32 212x 72 5

xa

3x 9 4x 14 5 7x 23 5 7x 28 x4

xa

70 x b x122 84 x 70 x 2x 84 70 x 84 x 14

5 1 7 2. x x 4


The LCD is 4x; x 0.

6. Let x represent the smaller number. Then x 4 represents the larger number.

5 1 7 b 4x a b x x 4

x 2 x4 3

1 28x 5 4x a b 4x a b x x 4

The LCD is 3(x 4); x 4.

4x 20x 28 x 4

31x 42 a

x 2 b 31x 42 a b x4 3 # 3x1x 42 31x 42 2 x4 3

20 x 28 x8 The solution set is {8}.

3x 21x 42

4 10 3. x2 x4

3x 2x 8

The LCD is (x 2)(x 4); x 4, x 2. 1x 221x 42 a

4 10 b 1x 221x 42 a b x2 x4

41x 42 101x 22 4x 16 10x 20 6x 16 20 6x 36 x6 The solution set is {6}. 4.

5 x 3 x5 x5 The LCD is x 5; x 5. x 5 3b 1x 52 a b x5 x5 x1x 52 5 1x 52 a b 1x 52132 x5 x5 1x 52 a

51x 52 x5

x8 If x 8, then x 4 12. The smaller number is 8, and the larger number is 12. 7. If one angle has measure 30°, then the sum of the other two angles of the triangle is 180 30 150°. Let y represent the measure of the second angle and let 150 y represent the measure of the third angle. y 2 150 y 3 The LCD is 3(150 y); y 150. y 2 b 31150 y2 a b 31150 y2 a 150 y 3 3y1150 y2 31150 y2 # 2 150 y 3 3y 21150 y2 3y 300 2y 5y 300

31x 52 x

5 3x 15 x 3x 10 x 10 2x 5x But x 5, so there is no solution. The solution set is . 5.

549

70 x 84 2b x a b x x


4x a

Chapter 6

84 70 x 2 x x The LCD is x; x 0.

y 60 If y 60, then 150 y 90. The other two angles are 60° and 90°. 8. Let r represent the rate Aldrin rides his bike and let r 2 represent the rate Javier rides his bike. They both ride the same amount of time. Rate # Time Distance Time

Distance Rate

550


Rate Javier

r2

Aldrin

r

Time 24 r2

Chapter 6

Distance

1x 22 1x 22 c

11x 22 1x 22

24

18 r

18

1x 22 1x 22 a

x1x 22 x x2 4 x2 2x x x2 4

The LCD is r(r 2); r 0, r 2.

x2 x x2 4

24 18 b r1r 22 a b r r2

x 4 x4

18r1r 22 24r1r 22 r r2

The solution set is {4}. 3. y

181r 22 24r 18r 36 24r

1 5 y 2

The LCD is 2y; y 0.

36 6r

5 1 2y a y b 2y a b y 2

6r If r 6, then r 2 8. Aldrin’s rate is 6 mph, and Javier’s rate is 8 mph.

2y # 5 1 2y1y2 2y a b y 2 2y2 2 5y

Section 6.6

2y 5y 2 0 2

3 1 7 6x 2 3x 1 20

12y 121y 22 0

3 1 7 213x 12 3x 1 20

2y 1 0

or

y20

2y 1

or

y2

1 2

or

y2

1 The LCD is 20(3x 1); x . 3

y

2013x 12 c

7 3 1 d 2013x 12 a b 213x 12 3x 1 20

2013x 12 c

7 d 213x 12

2013x 12 a 14013x 12 213x 12

2013x 12 3 b 3x 1 20

6013x 12 3x 1

3x 1

70 60 3x 1 10 3x 1 9 3x

1 The solution set is e , 2 f . 2 1 4. Let x represent the number. Then represents its x reciprocal. x

1 50 x 7

The LCD is 7x; x 0. 7x a x

The solution set is {3}. x x 1 2 x2 x 4 x x 1 x2 1x 221x 22 The LCD is (x 2)(x 2); x 2, x 2.

50 1 b 7x a b x 7

7x # 50 1 7x1x2 7x a b x 7

3x

2.

x d 1x 221x 22

1x 22 1x 22

24 18 r r2

1.

x b x2

1x 221x 22 c

Aldrin’s Javier’s a b a b time time

r1r 22 a

x x d x2 1x 22 1x 22

7x2 7 50x 7x 50x 7 0 2

17x 12 1x 72 0 7x 1 0

or

x70

7x 1

or

x7

1 7

or

x7

x


3m 51m 102 110 3m 5m 50 110

1 1 . x 7

8m 50 110

1 If the number is , then its reciprocal is 7. If the num7 1 ber is 7, then its reciprocal is . 7 5. Let r represent the rate Kim jogs and let r 4 represent the rate Ryan rides his bicycle. Rate # Time Distance Time

Distance Rate

Rate

a

Kim

r

Ryan

r4

Time 16 r

8m 160 m 20 If m 20, then m 10 10. Therefore, Marlene worked 10 minutes making sandwiches. 7. Let y represent the number of minutes Dean and 1 Tasha worked together. Dean’s rate is of the job per 60 1 minute, and Tasha’s rate is of the job per minute. 90 Minutes

Distance

36 r4

16

60

Tasha

90

Together

y

The sum of individual rates must equal the rate working together. 1 1 1 y 60 90

16 36 1 r r4 The LCD is r(r 4); r 0, r 4. 36 16 r1r 42 a b r1r 42 a 1b r r4 161r 42 r1r 42 a

Dean

36 b 1r1r 42 r4

The LCD is 180y; y 0. 180y a 180y a

16r 64 36r r1r 42

1 1 1 b 180y a b y 60 90

180y 1 1 b 180y a b y 60 90 3y 2y 180

16r 64 36r r2 4r

5y 180

16r 64 32r r2

y 36

16r 64 r2

It should take them 36 minutes working together.

r2 16r 64 0

1r 821r 82 0 r 8 0 or r 8 0 r8

If r 8, then r 4 12. Kim’s rate is 8 mph, and Ryan’s rate is 12 mph. 6. Let m represent the number of minutes Derek spends making sandwiches. Then m 10 represents the number of minutes Marlene spends making sandwiches. Rate

Time

Rate/min 1 60 1 90 1 y

36

Kim’s Ryan’s b1 b a time time

r 8 or

551

Total sandwiches 110

1 1 7 1 If x , then 1 1 # 7. x 7 7 1 If x 7, then

Chapter 6

Quantity Rate Time

Derek

3

m

3m

Marlene

5

m 10

5(m 10)

8. Let c represent the number of minutes it takes Julio to clean the office. Then 3c represents the number of minutes it takes Martin to clean the office. Julio’s rate 1 1 is of the job per minute, and Martin’s rate is of c 3c the job per minute. Minutes Julio

c

Martin

3c

Together

30

Rate/min 1 c 1 3c 1 30

552


Chapter 7

The sum of individual rates must equal the rate working together. 1 1 1 c 3c 30

Time to do Working Time Fractional part entire job rate working of the job done David

16 hours

1 16

12

12 16

Carlos

h hours

1 h

5

5 h

The LCD is 30c; c 0. 1 1 1 30c a b 30c a b c 3c 30 1 30c 1 30c a b 30c a b c 3c 30 30 10 c 40 c If c 40, then 3c 120. Julio can clean the office in 40 minutes, and Martin can clean the office in 120 minutes. 9. Let h represent the number of hours it takes Carlos to 1 build the deck by himself. Carlos’s working rate is . h Because David can build the deck in 16 hours, his 1 working rate is . However, David actually worked 16 12 hours (7 alone, 5 with Carlos). a

Working b rate

#

a

Time Fractional part b a b working of job done

Adding fractional parts 1 whole job. a

Fractional part Fractional part b a b1 David does Carlos does

12 5 1 16 h The LCD is 16h; h 0. 16h a 16h a

12 5 b 16h112 16 h

5 12 b 16h a b 16h 16 h 12h 16152 16h 80 4h 20 h

Carlos can build the deck by himself in 20 hours.


y

1. The ordered pairs are: Naples (1, 4) Tallahassee (3, 1)

Tampa (2, 3) Orlando (1, 3) Miami (0, 2)

y = 3x − 2

y

Tampa

x Orlando

x Tallahassee Miami Naples

1 3. Graph y x 1. 3

2. Graph y 3x 2. x

Select values of x that are divisible by 3.

y

(x, y)

0

2

(0, 2)

y

(x, y)

1

1

(1, 1)

0

1

(0, 1)

2

4

(2, 4)

3

0

(3, 0)

3

7

(3, 7)

6

1

(6, 1)

1

5

3

2

(1, 5)

x

(3, 2)


Chapter 7

Section 7.2

y

1. Graph 2x 3y 12. y

1 x 3

To find the y intercept, set x 0. If x 0, then

1

2102 3y 12 3y 12

x

y4 To find the x intercept, set y 0. If y 0, then 2x 3102 12 2x 12 x6

4. Solve 2x 3y 9 for y.

To find a check point, let x 3. If x 3, then

2x 3y 9 3y 2x 9

2132 3y 12

3y 2x 9 3 3 3

6 3y 12 3y 6

2 y x3 3

y2 Three solutions are (0, 4), (6, 0), and (3, 2).

5. Solve 3x 5y 7 for x.

y

3x 5y 7

2x + 3y = 12

3x 5y 7 5y 7 3x 3 3 3

x

7 5 x y 3 3 6. Graph 4x 2y 8. 4x 2y 8 2y 4x 8

2. Graph 2x y 6.

2y 4x 8 2 2 2

To find the y intercept, set x 0. If x 0, then

y 2x 4

2102 y 6 y 6

x

y

0

4

1

6

2

8

1

2

2

0

y

y 6 To find the x intercept, set y 0. If y 0, then 2x 102 6

y = 2x + 4

2x 6 x

x3 To find a check point, let x 2. If x 2, then 2122 y 6 4y6

553

554


Chapter 7

y 2 y 2 Three solutions are (0, 6), (3, 0), and (2, 2).

4. Graph y 3x. Because the graph goes through (0, 0), we need to find more points.

x

y

y

y

0

0

2

6

2

6

(–2, 6) y = −3x

(3, 0) x

x (2, −2) 2x − y = 6 (0, −6)

3. Graph x 2y 5. To find the y intercept, set x 0. If x 0, then 102 2y 5 2y 5 5 y 2 To find the x intercept, set y 0. If y 0, then

(2, –6)

5. Graph x 2. For any values of y, x 2

x

y

2

0

2

2

2

4

y

x = −2 (−2, 2) (−2, 0)

x 2102 5

x

(−2, −2)

x5 To find a check point, let x 3. If x 3, then 132 2y 5 2y 2 y 1

5 Three solutions are a 0, b , (5, 0), and (3, 1). 2

y

x − 2y = 5

(0, – −5) 2

(5, 0) x (3, –1)

6. Graph y 4. For any values of x, y 4.

x

y

0

4

2

4

3

4

y (0, 4) (−2, 4)

(2, 4) y=4 x


7. (a) C m 5.

Chapter 7

555

y

m

1

3

5

10

C

6

8

10

15

(b) C m 5.

18 16 14 12 10 8 6 4 2

x

C

(10, 15) (5, 10) (3, 8) (1, 6)

0

C

m

5

2 4 6 8 10 12 14 16 18 m

(c) C ⬇ 7 when m 2.5. (d) C 2.5 5 7.5.

2. Graph y 3x. Graph y 3x as a solid line. Choose a test point other than (0, 0) because the line passes through (0, 0). Use (2, 2) as the test point, so y 3x becomes 2 3(2), which is equivalent to 2 6, which is a true statement. Because the test point does satisfy the inequality, the graph is the line along with the half-plane that does contain the test point. y

8. (a) D 1.5e. e

2

4.3

7.5

15.10

D

3

4.5

7.5

15.00

(b) D 1.5e. x 18 16 14 12 10 8 6 4 2 0

D (10, 15.00)

(5, 7.5) (3, 4.5) (2, 3)

D

1.5e

2 4 6 8 10 12 14 16 18 e

(c) D ⬇ 9 when e 6.

3. Graph y 2 Graph y 2 as a solid line. Choose (0, 0) as the test point, so y 2 becomes 0 2, which is a true statement. Because the test point does satisfy the inequality, the graph is the line along with the halfplane that does contain the test point. y

(d) D (1.5)(6) $9.

Section 7.3 1. Graph x 2y 6. Graph x 2y 6 as a dashed line. Choose (0, 0) as a test point, so x 2y 6 becomes (0) 2(0) 6, which is equivalent to 0 6, which is a false statement. Because the test point does not satisfy the inequality, the graph is the half-plane that does not contain the test point.

x

556


Chapter 7

4. Graph x 3.

y

Graph x 3 as a solid line. Choose (0, 0) as the test point, so x 3 becomes 0 3, which is a false statement. Because the test point does not satisfy the inequality, the graph is the line along with the halfplane that does not contain the test point.

(3, 1) (− 2, − 1)

x

y

x

3 4. Graph the line with m that passes through 2 (3, 4). 3 3 m 2 2 Plot (3, 4). Locate another point by moving 3 units down and 2 units right to obtain (1, 1). Locate a third point by again moving 3 units down and 2 units right to obtain (1, 2). y

Section 7.4 1. (a) Let (1, 2) be P1 and let (3, 8) be P2. m

y2 y1 82 6 3 3 x2 x1 31 2 1

(− 3, 4)

(−1, 1)

(b) Let (5, 3) be P1 and let (2, 4) be P2.

x

y2 y1 43 1 1 m x2 x1 25 3 3 (c) Let (4, 2) be P1 and let (1, 2) be P2. m

y2 y1 22 0 0 x2 x1 1 142 5

To find the y intercept: If x 0, then

To find the x intercept: If y 0, then

2102 y 4

2x 102 4

y 4

2x 4

y 4

x2

Use (0, 4) and (2, 0) to find the slope. m

0 142 y2 y1 4 2 x2 x1 2 102 2

2 that passes through 5 (2, 1). Plot (2, 1). Locate another point by moving 2 units up and 5 units right to obtain (3, 1). 2 2 We can locate another point using m . 5 5 From (2, 1) move 2 units down and 5 units left to obtain (7, 3).

3. Graph the line with m

2 . Therefore, let x 100 represent the unknown vertical distance.

5. A 2% grade means a slope of

2. Find the intercepts of 2x y 4.

x 2 1250 100 100x 2112502 100x 2500 x 25 The vertical distance is 25 feet.

Section 7.5 2 and 3 contains the point (4, 2). Let (x, y) represent any point on the line other than (4, 2). The slope of the line 2 connecting (4, 2) and (x, y) is . 3

1. Find the equation of a line that has a slope of


y2 2 x4 3 21x 42 31y 22 2x 8 3y 6 2x 8 3y 6 2x 3y 2 2. Find the equation of the line that contains (2, 3) and (1, 5). m

y2 y1 53 2 x2 x1 1 122 3

2 Use a slope of and either ordered pair to write the 3 equation. y5 2 x1 3 21x 12 31y 52 2x 2 3y 15 2x 2 3y 15 2x 3y 13 3. A y intercept of 4 means the line passes through 1 (0, 4). The slope is . 3 y 142 x0

1 3

Chapter 7

5. Use the slope-intercept form to write the equation of 2 the line with a slope of and a y intercept of 5. 3 m

2 3

b5

y mx b 2 y x5 3 6. To find the slope, solve for y. The slope is the numerical coefficient of x. 4x 2y 3 2y 4x 3 2y 4x 3 2 2 2 y 2x

3 2

The slope is 2. 3 7. Graph y x 2. 5 The y intercept is 2, which gives the point (0, 2). 3 Use m to move from (0, 2) three units up and 5 five units right to obtain (5, 1).

y4 1 x 3

y

31y 42 x

y

3y 12 x

3 x 5

2 (5, 1)

12 x 3y

x

4. Use the point-slope form to write the equation of the 5 line. The slope is , and the line contains (x, y), which 2 is represented by (1, 3).

(0,

2)

y2 y1 m1x2 x1 2 y3

5 3 x 112 4 2

5 y 3 1x 12 2 5 21y 32 2 c 1x 12 d 2 2y 6 51x 12 2y 6 5x 5 6 5x 2y 5 11 5x 2y

557

8. (a) To find the slope and y intercept, solve for y. 3x 2y 6 2y 3x 6 2y 3x 6 2 2 2 3 y x3 2 3 m b3 2

558


Chapter 8

From (0, 3) move 3 units down and 2 units right to obtain (2, 0).

y

y (0, 3)

x (1, − 2)

(0, − 3)

(2, 0) x

(c) To find the slope and y intercept, solve for y. y 4 m 0; y intercept is 4. Because the line has slope of 0, it is horizontal. From (0, 4) graph the horizontal line. (b) To find the slope and y intercept, solve for y.

y

y x 3 y x 3 1 1 1 yx3 1 m 1

x

b 3 (0, − 4)

(3, − 4)

From (0, 3) move 1 unit up and 1 unit right to obtain (1, 2).

CHAPTER 8 Section 8.1 1. a

2x 2y 25 b 5x 2y 17

Substitute 3 for x and 1 for y. 2x y 5

becomes

2132 112 5

5x 2y 17

becomes

5132 2112 17

Because the ordered pair does satisfy both equations, (3, 1) is a solution of the system. 2. Solve the system a xy2

y

(2, 4)

x 2y 2 b. x 2y 6

x

x 2y 6

x

y

x

y

0

2

0

3

2

0

6

0

1

3

6

6



3. Solve the system a

3x y 6 b. 3x y 0

3x y 6


3x y 0

y

x

y

0

6

0

0

2

0

1

3

1

3

1

3

559

2x 2y 5 b. 2x 3y 6

Graph the following lines as broken lines. xy5

x

Chapter 8

2x 3y 6

x

y

x

y

0

5

0

2

5

0

3

0

2

3

3

4

Use (0, 0) as the test point for x y 5. Substituting, we find 0 0 5, which is a true statement. Therefore, we shade the half-plane that contains the test point.

y

Use (0, 0) as the test point for 2x 3y 6. Substituting, we find 2102 3102 6, which is a false statement. Therefore, we shade the half-plane that does not contain the test point. y

x

x

The solution set is . 4. Solve the system a

2x 4y 8 b. 2x 2y 4

2x 4y 8

x 2y 4

Section 8.2 x

y

x

y

0

2

0

2

4

0

4

0

2

1

2

3

y

(0, 2) (4, 0)

x

2x y 6 b. 1x y 6 Add the equations.

1. Solve the system a 2x y 16 x y 16

3x y 12 x4 Substitute 4 for x in the first equation. 2142 y 6 8y6 y 2 The solution set is 514, 22 6. x 2y 17 b. 3x 2y 13 Add the equations.

2. Solve the system a x 2y 17 3x 2y 13 There are infinitely many solutions.

4x 2y 20 x 15

560


Chapter 8

Substitute 5 for x in the first equation. 152 2y 7 2y 2


Multiply the first equation by 3. Multiply the second equation by 2. Then add the equations.

y 1 The solution set is {(5, 1)}. yx7 3. Solve the system a b. x 2y 5

15x 6y 39 4x 6y 20 19x 6y 59 x

In the first equation, subtract x from both sides. Leave the second equation alone. Add the equations. x 1y 17

Substitute

x 2y 15

2a

3y 12 y 14 Substitute 4 for y in the first equation. 4x7

59 19

59 for x in the second original equation. 19

59 b 3y 10 19 118 3y 10 19 3y 10

3 x The solution set is {(3, 4)}. 4. Solve the system a

5x 2y 13 b. 2x 3y 10

3x 2y 3 b. 4x 2y 18

Leave the first equation alone. Multiply the second equation by 2. Then add the equations. 33x 2y 3 88x 2y 36 11x 2y 33 x13

118 19

3y

118 190 19 19

3y

72 19

72 1 1 13y2 a b a b 3 19 3 y

24 19

The solution set is e a

Substitute 3 for x in the first original equation.

59 24 , b f. 19 19

3132 2y 3 9 2y 3 2y 12 y 6 The solution set is {(3, 6)}. 3x 2y 14 b. 5. Solve the system a 4x 5y 13

7. Let x represent the smaller number and let y represent the larger number. The sum of two numbers is 31. x y 31 Twice the smaller number is subtracted from the larger number, with a result of 1. y 2x 1 x y 31 b. 2x y 31


Solve the system a

15x 10y 20

Multiply the first equation by 1. Leave the second equation alone.

8x 10y 26 23x 10y 46

x y 31

x 12

2x y 31

Substitute 2 for x in the first original equation. 3122 2y 4 6 2y 4

3x y 30 x 10 Substitute 10 for x in the first original equation to find y.

2y 2

10 y 31

y 1

y 21

The solution set is {(2, 1)}.

The smaller number is 10, and the larger number is 21.


8. Let x represent the cost of one pizza and let y represent the cost of one basket of wings. The cost of two pizzas and three baskets is $29. 2x 3y 29

Chapter 8

561

y 2x 5 y 2132 5 y 6 5 1 The solution set is {(3, 1)}.

The cost of five pizzas and two baskets is $45. 5x 2y 45

3. Solve the system

Multiply the first equation by 2. Multiply the second equation by 3. Add the equations. 4x 6y 58

Substitute

°

x 4y 22 ¢. 1 y x4 2

1 x 4 for y in the first original equation 2

15x 6y 5135

to find x.

11x 6y 2277

1 x 4 a x 4b 22 2

x 11 7 Substitute 7 for x in the first original equation to find y.

x 2x 16 22

2172 3y 29

3x 16 22

14 3y 29

3x 6

3y 15

x2

y5 The cost of one pizza is $7, and the cost of one basket of wings is $5.

Substitute 2 for x in the second original equation to find y. 1 y x4 2 y

Section 8.3 1. Solve the system a

x y 11 b. x 2y 10

Substitute y 11 for x in the second equation. 1y 112 2y 10 3y 11 10 3y 21 y 7

1 122 4 2

y145 The solution set is {(2, 5)}. 4. Solve the system a

2x 3y 4 b. 2x 2y 1

Solve the second equation for x. x y 1 x y 1

Substitute 7 for y in the first original equation to find x.

Substitute y 1 for x in the first original equation to find y.

x y 11

21y 12 3y 4 2y 2 3y 4

x 7 11

y24

x4

y6

The solution set is {(4, 7)}. 2. Solve the system a

4x 3y 15 b. y 2x 5

Substitute 2x 5 for y in the first original equation to find x. 4x 312x 52 15 4x 6x 15 15 10x 15 15 10x 30 x 3 Substitute 3 for x in the second original equation to find y.

Substitute 6 for y in the second original equation to find x. x y 1 x 6 1 x 1 6 7 The solution set is {(7, 6)}. 5. Solve the system a

2x 3y 18 b. 3x 2y 14

Solve the second equation for y. 3x 2y 14 2y 3x 14

562


Chapter 8

2y 3x 14 2 2 2 y

33x 10y 165

3x 7 2

x 115

3x Substitute 7 for y in the first original equation to 2 find x. 2x 3y 8 3x 2x 3 a 7b 8 2 9x 2x 21 8 2 4x 9x 42 16 13x 42 16 13x 26 x 2 Substitute 2 for x in the first original equation to find y. 2x 3y 8

Substitute 5 for x in the first original equation. 9x 5y 15 9152 5y 15 45 5y 15 5y 30 y6 The solution set is {(5, 6)}. 8. Solve the system a

y 3x 2 b. 3x y 4

Use substitution to solve. Substitute 3x 2 for y in the second original equation to find x. 3x 13x 22 4 3x 3x 2 4 2 4

2122 3y 8

The statement is false. Therefore, the solution set is .

4 3y 8


3y 12 y4 The solution set is {(2, 4)}. x 2y 6 6. Solve the system a b. 4x 3y 16 Use substitution to solve. Substitute 2y 6 for x in the second original equation to find y. 412y 62 3y 16 8y 24 3y 16 5y 24 16 5y 40 y8 Substitute 8 for y in the first original equation to find x. x 2y 6

1x 2y 14 b. 3x 6y 12

Multiply the first equation by 3. Leave the second equation alone. Then add the equations. 3x 6y 12 3x 6y 12 0 10 This is a true statement. Therefore, there are infinitely many solutions. 10. Let x represent the amount invested at 5% and let y represent the amount invested at 7%. Simoni invested $7000. x y 7000 The total interest earned was $440. 0.05x 0.07y 440 Multiply the second equation by 100. Multiply the first equation by 5. Then add the equations. 5x 5y 35,000

x 2182 6

5x 7y 44,000

x 16 6 10 The solution set is {(10, 8)}. 7. Solve the system a

18x 10y 130 15x 10y 135

9x 5y 15 b. 3x 2y 27

2y ,09000 y 1,4500 Substitute 4500 for y in the first original equation. x y 7000

Use elimination-by-addition to solve.

x 4500 7000


Simoni invested $2500 at 5% and $4500 at 7%.

x 2500


11. Let x represent the number of liters of 6% saline solution and let y represent the number of liters of 12% saline solution. The total number of liters is 30. x y 30 The total amount of saline is 10% of 30. 0.06x 0.12y 0.101302 Multiply the first equation by 6. Multiply the second equation by 100. Then add the equations. 6x 16y 180 26x 12y 300 6y 120 y 120 Substitute 20 for y in the first original equation. x y 30 x 20 30

12. Let w represent the width of the rectangle and let l represent the length of the rectangle. The length is 6 more than twice the width. l 6 2w The perimeter is 168 inches. Use P 2l 2w. 168 2l 2w Use substitution to solve. Substitute 6 2w for l in the second original equation. 168 216 2w2 2w 168 12 4w 2w 168 12 6w 156 6w 26 w Substitute 26 for w in the first original equation. l 6 2w l 6 21262 l 6 52 58 The length is 58 inches, and the width is 26 inches.

Section 8.4 3x y 1z 15 1. Solve the system ° y 2z 16 ¢ . 3z 12 Solve the third equation.

563

Substitute 4 for z in the second original equation. y 2z 6 y 2142 6 y86 y 2 Substitute 4 for z and 2 for y in the first original equation. 3x y z 15

3x 122 4 15 3x 6 15 3x 9 x3

The solution set is {(3, 2, 4)}. 2. Solve the system °

x 10 There are 10 liters of 6% saline solution and 20 liters of 12% saline solution.

Chapter 8

3x 4y 4z 14 y 4z 17 ¢ . 2y 3z 19

Multiply the second original equation by 2. Leave the third original equation alone. Then add the equations. 2y 8z 34 2y 3z 19 5z 15 z 13 Substitute 3 for z in the second original equation. y 4z 17 y 4132 17 y 12 17 y5 Substitute 3 for z and 5 for y in the first original equation. 3x 4y z 14 3x 4152 3 14 3x 20 3 14 3x 17 14 3x 3 x 1 The solution set is {(1, 5, 3)}. 1x 3y 2z 1 3. Solve the system ° 2x 1y 3z 5 3x 2y 1z 8

112 122 ¢ . 132

Multiply equation (3) by 3. Leave equation (2) alone. Then add the equations. 9x 6y 3z 24

3z 12

2x 2y 3z 25

z4

11x 5y 3z 29

(4)

564


Chapter 8

Multiply equation (3) by 2. Leave equation (1) alone. Then add the equations.


6x 4y 2z 16

7x 14y 122

6x 3y 2z 11

7x 35y 126

5x 1y 2 17 Solve the system a

31y 124

(5)

y 124

11x 5y 29 b. 5x 5y 17

Substitute 4 for y in equation (4) to find x.

11x 5y 29

(4)

7x 4y 2

5x y 17

(5)

7x 4142 2

Multiply equation (5) by 5. Leave equation (4) alone. Then add the equations. 11x 5y 29

7x 14 x 2

25x 5y 85 14x

7x 16 2

Substitute 2 for x and 4 for y in equation (1) to find z.

56 x 14

2x 3y z 9 2122 3142 z 9

Substitute 4 for x in equation (4) to find y. 11x 5y 29

4 12 z 9

11142 5y 29

8z9

44 5y 29

z1

5y 15

The solution set is {(2, 4, 1)}.

y 3 Substitute 4 for x and 3 for y in equation (1) to find z. x 3y 2z 1

142 3132 2z 1

2x 2y 2z 24 5. Solve the system ° 3x 4y 2z 25 2x 4y 2z 10


4 9 2z 1 5 2z 1

2x 4y 2z 8

2z 4

2x 4y 2z 10

z 2

0 2

The solution set is {(4,3,2)}. 2x 3y 2z 29 4. Solve the system ° 3x 2y 2z 16 5x 4y 3z 29

112 122 ¢ . 132


This is a false statement, which means the system is inconsistent. The solution set is . 2x y z 8 6. Solve the system ° 2x y z 5 2x y 5z 1

4x 6y 2z 18

Add equation (1) to equation (3).

3x 2y 2z 16

2x y 2z 8

7x 4y

12

(4)

Multiply equation (1) by 3. Leave equation (3) alone. Then add the equations. 6x 9y 3z 27

Solve the system a 7x 4y 2 x 5y 18

(5)

3x y 6z 9 (4) Add equation (2) to equation (3). x y 5z 1

(5)

7x 4y 12 b. x 5y 18 (4)

x y 5z 1

x y 1z 5

5x 4y 3z 29 x 5y 3z 18

112 122 ¢ . 132

2x y 4z 6

(5) 3x 6z 9 b. 2x 4z 6 (4) (5)

Solve the system a 3x 6z 9 2x 4z 6

112 122 ¢ . 132


1 Multiply equation (4) by . Multiply equation (5) by 3 1 . Then add the equations. 2 x 2z 3 x 2z 3

Chapter 9

565

Add equation (2) to equation (3). a b c 010 a b c 180 2a 2b c 190 (4) Multiply equation (1) by 2. Add to equation (4). 2a 2b 30

0 0 This is a true statement. Therefore, the system has infinitely many solutions. 7. Let a represent the measure of ⬔A, let b represent the measure of ⬔B, and let c represent the measure of ⬔C. The measure of ⬔A is 15 less than the measure of ⬔B.

2a 2b 190 4a 2b 160 a 1140

(5)

Substitute 40 for a in equation (1). a b 15 40 b 15 b 55 b 55

a b 15 The sum of the measures of ⬔A and ⬔B is 10 more than the measure of ⬔C.

Substitute 40 for a and 55 for b in equation (2).

a b 10 c

40 55 c 10

The sum of the measures of the angles of a triangle is 180°. a b c 180 a b c 150 Solve the system ° a b c 1 10 a b c 180

112 122 ¢ . 132

a b c 10 95 c 10 c 85 c 85 The measure of ⬔A is 40°, the measure of ⬔B is 55°, and the measure of ⬔C is 85°.


1. (a) 1289 17 because 1172 289 2

(b)

25 5 5 2 25 because a b A 81 9 9 81

(c) 10.04 0.2 because 10.22 2 0.04 2. (a) 264 4 because 142 64 3

(b)

3

27 27 3 3 3 because a b B 1000 10 10 1000 3

3 (c) 20.008 0.2 because 10.22 3 0.008

3. (a) 1425 ⬇ 20.616 (b) 1138 ⬇ 11.747 3 (c) 2 450 ⬇ 7.663 3 (d) 2 212 ⬇ 5.963

4. 6 15 12 13 2 15 4 13 6 15 2 15 12 13 4 13 16 22 15 112 42 13 4 15 8 13

5. 612 4 17 5 12 2 17 3 12 612 5 12 3 12 417 2 17 16 5 32 12 14 22 17 812 2 17 ⬇ 811.4142 212.6462 ⬇ 11.312 5.292 ⬇ 16.604 To the nearest tenth, 16.6. 6. Substitute 3.2 for L in the formula. Use 3.14 as an approximation for p. L T 2p A 32 3.2 T 213.142 6.28 10.10 1.9859 A 32 The period to the nearest tenth is 2.0 seconds. 7. Substitute 0.35 for f and 200 for D in the formula S 130Df. S 130Df 1301200210.352 12100 ⬇ 45.8275 To the nearest mile per hour, the car was traveling at approximately 46 mph.

566


Chapter 9

Section 9.2

3 3 9122 2 2 2132 2 2

1. (a) 124 14 # 6 1416 2 16

3

(b) 163 19 # 7 1917 3 17

3

1222

(c) 132 116 # 2 11612 4 12 2. (a) 120 14 # 5 1415 2 15 (b) 172

136 # 2

13612 6 12

(c) 6 150 6 125 # 2 6 12512 6152 12 30 12 3. (a) 216 28 # 2 2822 2 22 3

3

3

3

3

3

3

3

3

1.

7 17 17 A9 3 19

2.

3 3 3 3 81 1 81 1 271 3 313 3 A8 2 2 18

3

3

18 118 19 # 2 1912 3 12 12 A 81 9 9 9 3 181

4.

3 13 A5 15

5.

5 15 15 15 A 12 112 1413 213

4. (a) 212x2y 24x2 # 3y 24x2 13y 2x 13y (b) 275a3b2 225a2b2 # 3a 225a2b2 13a

3

3.

3

(c) 281 227 # 3 227 23 3 23 3

Section 9.3

3

(b) 240 28 # 5 2825 2 25 3

5ab 13a 3

3

(c) 216x y 28x y 4 3

3 3

# 2x

3

3 3 3

28x y 22x

3

2xy 22x

6. (a)

5. 2 145 4 15 2 1915 4 15 2132 15 4 15 6 15 4 15 10 15 6. 118 3 150 172 1912 3 12512 13612

4 4 1a 1a

3 3 13 13 5 4

12n 110mn 110mn 2 2n 12n 24n

110 5 130 5 130 130 10 2 110 1100

(d)

29a2 3a 3a 9a2 B 16b 116b 1161b 41b

#

3a 1b 2

42b

7 13 8.

#

1b 1b

3a 1b 4b

2 2 13 2 13 7 13 7 13 13 13 13 19

1 3 1913 13 5 4 1 3 132 13 13 5 4

1a 4 1a 4 1a 2 a 1a 2a

513 5 13 110 110

1 3 1 3 127 13 19 3 13 5 4 5 4

#

(c)

7. 713

12 12

13 13

#

13 15 62 12

#

15m 5m A 2n 12n

3 12 15 12 6 12

2 13 2 a7 b 13 3 3

23 23 13 13 3 3

2 12 190 19 # 10 A5 15

12 15 19110 15 15

a

3 3 b 13 5 4

110 110 1 3 110 3110 a 3b 110 5 5 125

a

12 15 b 13 20 20

16 110 5

3

15 115 115 5 15 125

115 115 115 2132 6 219

3 12 3152 12 6 12

3

#

(b)

16 42 15

7.

3

18 22 6 22

27 13 20

Section 9.4

3

3

8. 9 216 2 254 9 28 2 2 227 2 3

3

3

3

928 22 2 227 22

1. (a) 12114 12 14 128 1417 2 17 (b) 15132 12172 10121

Practice Your Skills Solutions 3 3 3 3 3 3 3 (c) 210 24 240 28 5 2825 2 25 3 3 3 3 3 (d) 12 25213 2502 6 2250 6 2125 22 3 3 6152 2 2 30 2 2

2. (a) 131 17 152 121 115 (b) 121 118 52 136 5 12 6 5 12

7.

(c) 1a1 1a 1b2 2a 1ab 2

3

3

3

120 3 12 1100 3 110 3110 9

1415 3 12 10 9

2 15 3 12 2 15 3 12 1

1 1a 32 1 1a 42

1 1a 42 1 1a 42

3

(d) 231 23 2182 29 254

29 227 # 2 3

3

3

3

3 3 2 9 3 22

118 142 16 114

8.

1912 142 16 114

5 6 15 8 13 6 15 4. (a) 1 15 621 15 62 125 615 6 15 36 5 36 31

16 3

1.

(c) 1 16 122 1 16 122

4

6.

5 16 10 136 2 16 136 2 16 2 16 4

316 10 6 64

316 4 2

1y 9

1 1y2 2 192 2

31 17 122 149 114 114 14

1a 12

1 1a2 2 1122 2 Because 1144 12, 144 is not the solution, so the solution set is .

1 17 122

1 17 1221 17 122

3.

13y 1 4

1 13y 12 2 142 2 3y 1 16 3y 15

31 17 122 72 31 17 122 5

#

a 144

1 17 122

31 17 122

12 12 110 3 110 3

1 16 22 1 16 22

Because 181 9, the solution set is {81}. 2.

62

1 16 22

y 81

136 112 112 14

1 16 22

Section 9.5

13

#

15 162 1 16 22

(b) 1 15 421 15 22 125 215 4 15 8

(b) 14 13214 132 16 413 413 19

a 71a 12 a 16

5 16 5 16 16 2 16 2

3 12 142 16 114

#

2a2 4 1a 4 1a 16

3

29 227 22

3 3 5. 17 12 1 17 122

2a2 4 1a 3 1a 12

3. (a) 1 16 122 1 13 172

567

1a 3 1 1a 42 1a 3 1a 4 1a 4 1 1a 42

a 1ab 3

Chapter 9

1 110 32 1 110 32

121 110 32

1 110 321 110 32

y5 Check: 13y 1 4 13152 1 ⱨ 4 115 1 ⱨ 4 116 ⱨ 4

44 The solution set is {5}.

568

4.


Chapter 9

0 y2 10y 24

22y 8 10

0 1y 621y 42

1 22y 82 2 1102 2 2y 8 100 2y 92

y40

y 6 or

y4

Check y 6.

y 46

12y 8 y 4

Check:

12162 8 ⱨ 162 4

22y 8 10 221462 82 ⱨ 10

112 8 ⱨ 2

292 8 ⱨ 10

14 ⱨ 2

2100 ⱨ 10

22 Check y 4.

10 10 Because y 46 does not check, there is no real number solution. So the solution set is . 5.

y 6 0 or

12y 8 y 4

12142 8 ⱨ 142 4

2 13x 6 24

18 8 ⱨ 0

13x 6 12

10 ⱨ 0

1 13x 62 2 1122 2

00

3x 6 144


3x 150

8. 2 1y y

x 50

1y y 2

1 1y2 2 1y 22 2

Check: 2 13x 6 24

y y2 4y 4

2 131502 6 ⱨ 24

0 y2 5y 4

0 1y 12 1y 42

21150 6 ⱨ 24 2 1144 ⱨ 24

y10

21122 ⱨ 24

y1

24 24

1 12y 52 1 14y 72

2 1112 ⱨ 112 2

21ⱨ1

2y 5 4y 7 2y 5 7 2y 12 y6

31 Check y 4.

2 2142 142 22ⱨ4 44

Check: 12y 5 14y 7 12162 5 ⱨ 14162 7 112 5 ⱨ 124 7 117 117 The solution set is {6}. 7.

y4

2 1y y

12y 5 14y 7 2

or

Check y 1.


or y 4 0

12y 8 y 4

1 12y 82 2 1y 42 2 2y 8 y2 8y 16

The solution set is {4}. 9. Use D D D

S2 . 30f

1552 2 3010.402 15521552 12

D ⬇ 252


Chapter 10

569

CHAPTER 10 8. 1a 52 2 9

Section 10.1

a 5 19

1. y2 7y 0

a 5 3

y1y 72 0 y0

or

y0

or

a53

y70 y 7

1x 72 1x 32 0

a 8

y 7 120 y 7 1415

x30

x 7 or

a 2 or 9. 1y 72 2 20

x2 4x 21 0

x 7 0 or

a 5 3



or

y 7 2 15

x 3

y 7 2 15


y 7 2 15 or

1a 521a 52 0 a 5 0 or

a50

a 5 or

a5

10. 315n 12 2 4 58 315n 12 2 54 15n 12 2 18


5n 1 118 5n 1 1912

n2 144

5n 1 312

n 144 0 2

1n 1221n 122 0 n 12 0

or

5n 1 3 12

n 12 0

n 12 or

or

1 312 5

or

n 12

n


5. y2 64 y 164 y 8

11.

5n 1 3 12 n

1 3 12 5

1 3 12 1 3 12 , f. 5 5

c

a5

b 12

The solution set is {8, 8}. 6. n2 12

or 5n 1 312

5n 1 312


Use the Pythagorean theorem to solve.

n 112

c2 a2 b2

n 112 1413

c2 152 2 1122 2

y 2 13

c2 25 144

The solution set is 52 23, 2236 .

c2 169 c 1169

7. 3y2 8

c 13

8 y2 3 y

y 7 215

The solution set is 57 2 25, 7 2 256 .

3. a 10x 25 0 2

4.

or y 7 2 15

Discard c 13, because length cannot be negative. The length of c is 13 inches. 18 13 124 1416 8 3 A3 13 13 19

y

2 16 3

2 16 216 The solution set is e , f. 3 3

12. Let x represent the length. 30 feet x 12 feet

570


Chapter 10

Use the Pythagorean theorem to solve. a b c 2

2

2

x2 1122 2 1302 2 x2 144 900 x2 756 x 1756 x ⬇ 27.5 Discard x 27.5, because length cannot be negative. The length of x is 27.5 feet. 13. Let a represent the length of each leg of the isosceles triangle. 12 feet

a

Section 10.2 1. y2 10y 6 0 y2 10y 6 6 y2 10y 25 6 25 1y 52 2 31

y 5 131 y 5 131

The solution set is 55 131, 5 1316. 2. y2 6y 2 0 y2 6y 2 2 y2 6y 9 2 9 1y 32 2 7

y 3 17 y 3 17

a

Use the Pythagorean theorem to solve. a2 b2 c2

a2 a2 1122 2 2a2 144 a2 72 a 172 c 13612 c 6 12

Discard 6 12, because length cannot be negative. The length is 6 12 feet. 14. In a 30-60 right triangle, the side opposite the 30° angle is equal in measure to one-half the length of the hypotenuse. The ladder is 16 feet in length, and the distance from the base of the ladder to the corner of the building is 8 feet. To solve for x, use the Pythagorean theorem. 30˚ 16 feet x 60˚ 8 feet

a2 b2 c2

182 2 x2 1162 2 64 x2 256 x2 192 x 1192 x ⬇ 13.9 Discard x 13.9, because length cannot be negative. The top of the ladder reaches 13.9 feet.

The solution set is 53 17, 3 176. 3. a2 5a 3 0 a2 5a 3 3 a2 5a

25 25 3 4 4

5 2 37 aa b 2 4 a

5 37 2 A4

a

137 5 2 2 a a

5 137 2 2

5 137 2


5 137 5 137 , f. 2 2

4. 3y2 6y 2 0 1 Multiply both sides by . 3 2 y2 2y 0 3 2 2 y2 2y 3 3 2 y2 2y 1 1 3 5 1y 12 2 3 y1

5 A3

y1

15 13

#

13 115 13 19


y1

115 3

y 1 y

3 19 72 2 3 181 x 2 3 9 x 2 3 9 3 9 or x x 2 2 12 6 or x x 2 2 or x 3 x 6 x

3 115 115 3 3 3

3 115 3


3 115 3 115 , f. 3 3

5. n2 n 12 0 n2 n

12

1 1 n2 n 12 4 4

The solution set is {6, 3}. x2 5x 3 0 Use the quadratic formula.

1 49 n 2 A4

a 1, b 5, c 3 x

1 7 n 2 2

x

1 7 n 2 2

n

8 2

n 4

1 7 or n 2 2 or

x2 5x 3

2.

1 2 49 an b 2 4

1 7 n 2 2

n

6 2

or n 3

b 2b2 4ac 2a

152 2152 2 4112 132 2112

x

5 125 12 2

x

5 137 2


n2 n 12 0

1n 421n 32 0 n40

5 137 5 137 , f. 2 2

3. 6y2 5y 4 0

or

n30

n 4 or

n3


Use the quadratic formula. a 6, b 5, c 4 y

6. y2 6y 22 20 y2 6y 22 22

y

y2 6y 29 22 9 1y 32 2 13

Any value of y will give a nonnegative value for 1y 32 2. Therefore, there are no real number solutions.

1. x2 3x 18 0

b 2b2 4ac 2a 152 2152 2 4162142 2162

y

5 125 96 12

y

5 1121 12

y

5 11 12

y

5 11 12

Section 10.3 Use the quadratic formula.

or y

5 11 12

a 1, b 3, c 18

y

16 12

or y

6 12

b 2b2 4ac 2a

y

4 3

or y

1 2

x x

Chapter 10

132 2132 2 41121182 2112


571

572


Chapter 10

5x 1 7

4. x2 4x 6 0 Use the quadratic formula. a 1, b 4, c 6

142 2142 4112162 2112

3. y

4 116 24 x 2

or

5x 6

8 5

or

x

2 3 y

4 140 2

Multiply both sides by y.

x x

4 14110 2

y2 3y 2 0

x

212 1102 4 2 110 2 2

a 1, b 3, c 2

y2 2 3y

Use the quadratic formula.

y

x 2 110

The solution set is 52 110, 2 1106 . y

5. y2 3y 11 0

132 2132 2 4112122 2112 3 19 8 2

b 2b2 4ac 2a

y

3 117 2

132 2132 2 41121112


a 1, b 3, c 11

y

b 2b2 4ac 2a

y


y

6 5


2

x

5x 8 x

b 2b2 4ac x 2a

or 5x 1 7

2112

y

3 19 44 2

y

3 135 2

3 117 3 117 , f. 2 2

x2 10x

4.

x2 10x 0 x1x 102 0

Because 135 is not a real number, there are no real number solutions.

x0

or x 10 0

x0

or

x 10

The solution set is {0, 10}. 5. t2 30t 216 0

Section 10.4 1. 3x2 3x 6 0


1 Multiply both sides by . 3 x x20 2

1x 22 1x 12 0 x 2 0 or

x10

x 2 or

x 1

The solution set is 51, 26. 2. 15x 12 2 49 5x 1 149 5x 1 7

a 1, b 30, c 216 t t t

b 2b2 4ac 2a 1302 21302 2 4112 12162 2112 1302 1900 864 2

t

30 136 2

t

30 6 2


t

30 6 2

30 6 2

or t

24 t 2

36 t 2

or

t 18

or t 12

The perimeter is 84 feet. 84 2l 2w

b 2b2 4ac y 2a

y

1 1 1842 12l 2w2 2 2

1102 21102 2 41121152 2112

360 lw

2

1102 4 110 2

42 l w The area is 360 square feet.

1102 1100 60

2

n 19.

3. Let w represent the width of the garden and let l represent the length.

a 1, b 10, c 15

1102 1160

n 19 0

The two consecutive whole numbers are 18 and 19.


y

or

Discard n 19, because it is not a whole number.

y2 10y 15 0

y

1n 1821n 192 0 n 18 or

y2 10y 15

y

573

n2 n 342 0

n 18 0


Chapter 10

10 116110 2 215 21102 2

y 5 2110

The solution set is 55 2 110, 5 21106.

Solve the system a

42 l w b. 360 lw

Solve 42 l w for l. 42 w l. Substitute 42 w for l in the second equation.

360 142 w2w 360 42w w2

0 w2 42w 360

Section 10.5 1. Let w represent the width of the rectangle and let l represent the length. The length is 3 inches more than the width. l3w The area is 88 square inches. 88 lw l3w Solve the system a b. 88 lw Use substitution to solve.

88 13 w2w

Multiply both sides by 1 0 w2 42w 360

0 1w 1221w 302 w 12 0

w 30 0

or

w 12 or

w 30

If w 12, then l 42 w 30. If w 30, then l 42 w 12. The dimensions of the garden are 30 feet by 12 feet. 4. Let x represent the first number and let y represent the second number.

88 3w w

2

0 w2 3w 88

Their sum is 10.

w 8 0 or

Their product is 6.

0 1w 821w 112

x y 10

w 11 0

w 8 or

w 11

Discard w 11, because length cannot be negative. If w 8, then l 3 8 11. The length is 11 inches, and the width is 8 inches. 2. Let n represent the first whole number and let n 1 represent the second consecutive whole number. Their product is 342.

xy 6

Solve the system a

x y 10 b. xy 6

Solve the first equation for y. y 10 x Substitute 10 x for y in the second equation. x110 x2 6

n1n 12 342

10x x2 6

n2 n 342

x2 10x 6 0

574


Chapter 10

Multiply both sides by 1. x2 10x 6 0

Substitute

1r 102 1t 12 90

Use the quadratic formula. a 1, b 10, c 6 x x x

b 2b2 4ac 2a 1102 21102 2 4112162

80

80 10t 10 90 t 70

10 1100 24 2

20

10 176 10 14119 2 2 1192 215 10 2119 x 2 2 x 5 119 The two numbers are 5 119 and 5 119. 5. Let r represent the rate for Mark’s car and let t represent the time Mark drove. Use Rate Time Distance. rt 80 Let r 10 represent the rate of John’s car and let t 1 represent the time John drove. Use Rate Time Distance. (r 10)(t 1) 90 Solve the system a

rt 80 b. 1r 1021t 12 90

Solve the first equation for r. r

80 a 10b 1t 12 90 t

2112

x

80 t

80 for r in the second equation and solve. t

80 10t 90 t 80 10t 0 t

Multiply both sides by t. 20t 80 10t2 0 Multiply both sides by

1 . 10

2t 8 t2 0 t2 2t 8 0

1t 42 1t 22 0 t40 t4

or t 2 0 or

t 2

Discard t 2, because time is nonnegative. If t 4, then r

80 80 20. t 4

If r 20, then r 10 20 10 30. The rate for Mark’s car is 20 mph, and the rate for John’s car is 30 mph.

Answers to Odd-Numbered Problems and All Chapter Review, Chapter Test, and Cumulative Review Problems

CHAPTER 1 Problem Set 1.1 (page 9)

21. Prime 23. Prime 25. Composite 27. Prime 29. Composite 31. 2 # 13 33. 22 # 32 35. 72 37. 23 # 7 39. 23 # 3 # 5 41. 33 # 5 43. 2 # 59 45. 3 # 67 47. 5 # 17 49. 32 # 13 51. 3 # 43 53. 4 55. 8 57. 9 59. 12 61. 18 63. 12 65. 24 67. 48 69. 140 71. 392 73. 168 75. 90 79. All other even numbers are divisible by 2. 81. 61 83. x 85. xy

Problem Set 1.2 (page 22)


1. Negative integers 3. Nonnegative integers 5. 8 7. 11 9. 15 11. 7 13. 31 15. 19 17. 9 19. 61 21. 18 23. 92 25. 5 27. 13 29. 12 31. 6 33. 1 35. 45 37. 29 39. 27 41. 65 43. 29 45. 11 47. 1 49. 8 51. 13 53. 35 55. 32 57. 2 59. 30 61. 56 63. 60 65. 126 67. 154 69. 225 71. 9 73. 7 75. 12 77. 14 79. Undefined 81. 4 83. 47 85. 69 87. 4 89. 31 91. 28 93. 10 95. 66 97. 7 99. 28 101. 15 103. 10 105. 77 107. 104 109. 17 14 3 111. 3 (2) (3) (5) 7 (7 under par) 113. 2 1 3 1 (2) 1 115. 800(19) 800(2) 800(4)(1) 13,600 117. 5 4(3) 7

1.

1. 52, 46 3. 51, 3, 5, . . .6 5. 7. 9. 16 11. 35 13. 51 15. 48 17. 72 19. 82 21. 22 23. 55 25. 60 27. 9 29. 66 31. 26 33. 2 35. 47 37. 21 39. 11 41. 15 43. 14 45. 79 47. 6 49. 74 51. 12 53. 187 55. 29 57. 884 59. 49 61. 9 63. 18 65. 55 67. 99 69. 72 71. 11 73. 48 75. 21 83. 36 12 (3 3) 6 # 2 85. 36 (12 3 3) 6 # 2

Problem Set 1.3 (page 31) 1. Distributive property 3. Associative property of addition 5. Commutative property of multiplication 7. Additive inverse property 9. Identity property of addition 11. Associative property of multiplication 13. 56 15. 7 17. 1800 19. 14,400 21. 3700 23. 5900 25. 338 27. 38 29. 7 31. 5x 33. 3m 35. 11y 37. 3x 2y 39. 16a 4b 41. 7xy 3x 43. 10x 5 45. 6xy 4 47. 6a 5b 49. 5ab 11a 51. 8x 36 53. 11x 28 55. 8x 44 57. 5a 29 59. 3m 29 61. 8y 6 63. 5 65. 40 67. 72 69. 18 71. 37 73. 74 75. 180 77. 34 79. 65 83. Equivalent 85. Not equivalent 87. Equivalent

Problem Set 1.4 (page 39) 1. True 3. False 5. True 7. True 9. True 11. False 13. True 15. False 17. True 19. False

15. 27. 41. 53. 65. 77. 85. 95. (e)

2 3

2 3

5 3

1 6

3 4

27 28

15 28 3 4 1 10 7 3 17. 19. 21. 23. 25. 10 3 4 21 5 10 5 5 2 1 2 27 29. 31. 33. 3 35. 37. 39. 7 9 3 2 3 8 9 37 59 19 1 43. 45. 47. 49. 51. 15 16 30 96 72 24 1 1 31 21 1 37 55. 57. 59. 61. 63. 3 6 7 4 4 30 1 12 1 7 67. 69. 71. 14 73. 68 75. 3 5 30 26 1 1 3 1 79. 2 cups 81. 1 cups 83. 16 yards 4 4 4 4 1 2 3 2 yards 87. 36 yards 89. 1 miles 8 3 4 (a) Larger (b) Larger (c) Smaller (d) Larger Larger (f) Smaller 3.

5.

7.

9.

11.

13.

Problem Set 1.6 (page 64) 1. Real, rational, integer, and negative 3. Real, irrational, and positive 5. Real, rational, noninteger, and positive 7. Real, rational, noninteger, and negative 9. C 11. A 13. B 15. I 17. D 19. 0.62 21. 1.45 23. 3.3 25. 7.5 27. 7.8 29. 0.9 31. 1.16 33. 0.272 35. 24.3 37. 44.8 39. 1.2 41. 7.4 43. 0.38 45. 7.2 47. 0.42 49. 0.76 51. 4.7 53. 4.3 55. 14.8 57. 1.3 59. 1.2x 61. 3n 63. 0.5t 65. 5.8x 2.8y 67. 0.1x 1.2 69. 3x 2.3 4 11 71. 73. 75. 17.3 77. 97.8 79. 2.2 81. 13.75 12 3 83. 0.6 85. $9910.00 87. 19.1 centimeters 575

576

Answers to Odd-Numbered Problems

89. 4.7 centimeters 91. $6.55 99. a. The denominator has only factors of 2. b. The denominator has a factor of 3. 7 11 13 17 9 3 c. , , , , , and 8 16 32 40 20 64

75.

7 8

76.

2 3

77.

7 12

78.

1 5

79.

13 36

80.

23 72

5 17 82. 83. 77.52 84. 31.4 85. 4.08 18 24 86. 26.05 87. 0.776 88. 2.287 89. 29.52 90. 11.556 91. 0.1521 92. 0.009 93. 0.1008 94. 0.04 95. 14.1 96. 1.582 97. 3.16 98. 5.32 99. 100 100. 720 101. 64 102. 81 103. 25 104. 0.16 105. 4 29 1 106. 14 107. 28 108. 25 109. 110. 12 2 111. 25 112. 33 113. 4 114. 16 115. 5x 4y 116. 3xy y 117. 2x 7 118. 7x 18 7 1 119. 3x 12 120. 2a 4 121. x y 8 10 1 1 122. x y 123. 2.2a 0.6b 124. 0.58ab 0.36bc 2 6 125. 22 126. 16 127. 37 128. 39 129. 2 1 5 130. 1 131. 132. 38 133. $168 134. yard 72 8 135. 3.377 inches 136. 297 euros 81.

Chapter 1 Review Problem Set (page 71)

1. 50, 1, 2, 3, 46 2. 51, 3, 5, 7, 96 3. 52, 3, 5, 76 4. 52, 4, 6, . . .6 5. 6. 7. 8. 9. Real, rational, integer, and negative 10. Real, rational, integer, and zero 11. Real, rational, noninteger, and negative 12. Real, rational, noninteger, and positive 13. Real, irrational, and positive 14. Real, irrational, and negative 15. 3 16. 25 17. 13 18. 12 19. 22 20. 6 21. 25 22. 39 23. 24 24. 10 25. 5 26. 15 27. 1 28. 2 29. 45 30. 11 31. 56 32. 12 33. 34 34. 45 35. 156 36. 252 37. 6 38. 13 39. 48 40. 7 41. 175°F 42. 20,602 feet 43. 2(6) 4 3(8) 1 31 44. $3444 45. Commutative Property of Addition 46. Distributive property 47. Prime 48. Composite 49. Composite 50. Composite 51. Composite 52. Composite 53. 23 # 3 54. 32 # 7 55. 23 # 3 # 5 56. 33 # 5 57. 22 # 3 # 7 58. 26 59. 18 60. 12 3 5 6 3 61. 180 62. 60 63. 64. 65. 66. 10 16 7 4 12 16 2 20 7 67. 18 68. 69. 70. 71. 72. 5 15 3 9 5 1 12 73. 74. 4 7

Chapter 1 Test (page 73) 1. 11 2. 54 3. 26 4. 25 5. 20 6. 6°F 7. Commutative Property of Addition 8. 12x 6y 9. 13x 21

10. 2a2 11b2

11. 5x2 7x 5

12.

13. Prime 14. 23 # 32 # 5 15. 12 16. 72 5 2 2 17. 18. 19. 20. 2.58 21. 2.85 22. 37 9 5 24 23. 4 24. 14 teaspoons 25. $200.31

7 9

CHAPTER 2 Problem Set 2.1 (page 81) 1. 586

3. 566

13. 5236 21. 50.276 31. 536 41. 5246

5. 596

15. 576

7. 566 9. 5136 4 17 17. e f 19. e f 12 15

23. 53.56 33. e

13 f 2

43. 5356

25. 586 35. 51446 45. e

3 f 10

11. 5486

27. 5176 37. 5246 47. e

9 f 10

29. e

37 f 5

39. 5156 1 49. e f 2

1 27 5 7 51. e f 53. e f 55. e f 57. e f 3 32 14 5 1 3 59. e f 61. e f 63. 50.36 65. 596 67. 556 12 20

5. Two times the quantity, l plus w 7. The quotient of A divided by w 9. The quantity, a plus b, divided by 2 11. Two more than three times y 13. l w 15. ab d 17. 19. lwh 21. y x 23. xy 2 25. 7 y2 t xy 27. 29. 10 x 31. 10(n 2) 33. xy 7 4 35. xy 12 37. 35 n 39. n 45 41. y 10 d d d 43. 2x 3 45. 10d 25q 47. 49. 51. p t 12 53. n 1

55. n 2

63. 8w 65. 3l 4

57. 3y 2 67. 48f 72

59. 36y 12f 69. 2w 2

61.

71. 9s2

Problem Set 2.3 (page 95) Problem Set 2.2 (page 87) Answers may vary somewhat for Problems 1–11. 1. The difference of a and b 3. One-third of the product of B and h

1. 546 13. 566

3. 566

5. 586

15. 516

7. 5116

17. 556

9. e

17 f 6

19. 566

19 f 2 11 21. e f 2 11. e

f 3


23. 526 33. 41. 51. 63.

25. e

10 25 f 27. 5186 29. e f 31. 576 7 4 24 5 4 12 e f 35. e f 37. e f 39. e f 7 2 17 5 9 43. 22 45. $18 47. 35 years old 49. $7.25 6 53. 5 55. 11 57. 8 59. 3 61. $300 4 meters 65. 341 million 67. 1.25 hours

7. 5x 0x 26 −2

9. 5x 0x 16

−1

11. 5x 0x 26

2

Problem Set 2.4 (page 101) 5 9. e f 2 13 5 11. 596 13. 526 15. 536 17. e f 19. e f 2 3 13 16 1 21. 5176 23. e f 25. e f 27. 526 29. e f 2 3 3 19 31. e f 33. 17 35. 35 and 37 37. 36, 38, and 40 10 3 39. 41. 6 43. 32° and 58° 45. 50° and 130° 2 47. 65° and 75° 49. $42 51. $9 per hour 53. 150 men and 450 women 55. $91 57. $145 1. 556

3. 586

8 5. e f 5

7. 5116

13. 5x 0x 36 15. 5x 0x 16 19. 5x 0x 16 21. 5x 0 x 06 25. 31. 37. 43. 49. 55. 65.

17. 5x 0 x 46 23. 5x 0x 206 10 5x 0 x 96 27. 5x 0x 96 29. e x 0x f 3 24 5x 0x 86 33. 5n 0n 86 35. e n 0n f 7 5n 0n 76 39. 5x 0 x 56 41. 5x 0 x 66 8 5 5x 0x 216 45. e x 0x f 47. e x 0x f 3 4 5x 0x 16 51. 5t 0t 46 53. 5x 0x 146 3 1 9 e x 0x f 57. e t 0t f 59. e x 0x f 2 4 4 {All reals} 67. 69. {All reals} 71.

Problem Set 2.7 (page 127) Problem Set 2.5 (page 110) 17 29 f 9. e f 4 3 5106 13. 5446 15. 5266 17. 536 19. 516 22 1 526 23. 5166 25. e f 27. e f 29. 5576 3 6 7 27 3 e f 33. 526 35. 536 37. e f 39. e f 5 10 28 18 24 51 e f 43. e f 45. 556 47. 506 49. e f 5 7 10 5126 53. 5156 55. 57. {All reals} {All reals} 61. 63. 65. 7 and 8 14, 15, and 16 69. 6 and 11 71. 48 73. 12 and 18 8 feet and 12 feet 77. 15 nickels, 20 quarters 40 nickels, 80 dimes, and 90 quarters 8 dimes and 10 quarters 4 crabs, 12 fish, and 6 plants 85. 30° 20°, 50°, and 110° 89. 40° Any three consecutive integers

1. 516 11. 21. 31. 41. 51. 59. 67. 75. 79. 81. 83. 87. 97.

3. 5106

5. 596

7. e

1. 5x 0x 26

10 f 3 11 7. 5n 0 n 116 9. 5t 0t 116 11. e x 0 x f 5 5 3 13. e x 0x f 15. 5x 0x 86 17. e n 0n f 2 2 5 19. 5y 0y 36 21. e x 0 x f 23. 5x 0x 86 2 17 25. 5x 0x 216 27. 5x 0 x 66 29. e n 0n f 2 4 9 31. 5n 0 n 426 33. e n 0n f 35. e x 0 x f 2 3 37. 5n 0n 46 39. 5t 0 t 3006 41. 5x 0x 506 33 43. 5x 0 x 06 45. 5x 0 x 646 47. e n 0n f 5 16 49. e x 0x f 3 51. −1

53.

Problem Set 2.6 (page 119) 1. 5x 0 x 26 3. 5x 0 x 36

3. 5x 0 x 16

2

−2

55.

5. e x 0x

1

−2

2

57.

−2

2

59. 3

5. 5x 0 x 26

−4

61.

∅

63. 2

0

2

577

578


65.

51.

All reals

1

67. All numbers greater than 7 69. 15 inches 71. 158 or better 73. Better than 90 75. More than 250 sales 77. 77 or less

All reals

53.

3

54.

1


55. n 5

1. 536

59.

2. 516

3 3. e f 4

4. 596

5. 546

6. 5126

7. e 12. 16. 21. 25. 27.

40 9 15 f 8. e f 9. e f 10. 5326 11. 5216 3 4 8 11 5 2 e f 13. e f 14. 576 15. e f 4 21 41 8 19 1 e f 17. e f 18. 5606 19. 5106 20. e f 7 2 5 22. {All reals} 23. {All reals} 24. {All reals} 26. 5x 0 x 16

28. 5x 0 x 26

1

2

29. 5x 0 x 06

30. 5x 0 x 36

Chapter 2 Test (page 136) 1. 526

2. 536 3. 596 4. 556 5. 5536 6. 5186 5 35 11 7. e f 8. e f 9. 5126 10. e f 11. 5226 2 18 5 31 12. e f 13. 5x 0x 56 14. 5x 0x 16 2 23 15. 5x 0 x 96 16. 5x 0x 06 17. e x 0x f 2 18. 5n 0n 126 19. −2

4

20. 1

31. 5x 0 x 46 32. 5x 0x 46 33. 5x 0 x 136 11 34. e x 0 x f 35. 5y 0 y 246 36. 5x 0x 106 2 26 2 37. 5x 0 x 356 38. e x 0x f 39. e n 0n f 5 11 9 43 40. 5n 0 n 336 41. e x 0 x f 42. e x 0 x f 2 3 5 43. 5n 0 n 26 44. e n 0n f 45. 5n 0 n 1206 11 180 46. e n 0 n f 13 47. 3

2

2

5

49.

3

21. $72.00 per hour 22. 15 meters, 25 meters, and 30 meters 23. 96 or better 24. 17 nickels, 33 dimes, and 53 quarters 25. 60°, 30°, 90°

Cumulative Practice Test (Chapters 1–2) (page 138) 5 19 5. 20 6. 12 90 9 1 7. 0.09 8. 12 9. 36 10. x y 11. 3x 40 24 28 12. 22x 16 13. 16x 1 14. 5x 22 15. 5166 17 16. 5356 17. e f 18. 19. 5x 0x 166 11 20. 5x 0x 16 21. 22 # 3 # 52 22.

1. 3

2. 128

3. 2.4

1

3

50.

64. 69. 72. 74.

56. 5 n 57. 10(x 2) 58. 10x 2 d x 3 60. 61. x2 9 62. (x 9)2 63. x3 y3 r xy 4 65. 24 66. 7 67. 33 68. 8 16 and 24 70. 18 71. 8 nickels and 22 dimes 8 nickels, 25 dimes, and 50 quarters 73. 52° 700 miles 75. 89 or better 76. 88 or better

0

3

48.

4

52.

4.

3

23. 9 on Friday and 33 on Saturday 24. 67 or fewer people 25. 11 and 13

1

CHAPTER 3 Problem Set 3.1 (page 146) 1. 596

3. 5106

13. 5286

5. e

15. 5346

15 f 2

7. 5226

17. 566

9. 546

8 19. e f 5

11. 566

21. 576

9 53 23. e f 25. e f 27. 5506 29. 51206 2 2 1 9 31. e f 33. 15 feet by 19 feet 35. 330 miles 7 2 37. 60 centimeters 39. 7.5 pounds


1 41. 33 pounds 43. 90,000 3 49. $150,000

51. 55%

45. 137.5 grams 47. $300

53. 60%

2 55. 16 % 3

1 57. 37 % 59. 150% 61. 240% 63. 2.66 65. 42 2 67. 80% 69. 60 71. 115% 73. 90 79. 81. 83. All real numbers except 0


1. 51.116 3. 56.66 5. 50.486 7. 5806 9. 536 11. 5506 13. 5706 15. 52006 17. 54506 19. 51506 21. 522006 23. $50 25. $3600 27. $20.80 29. 30% 31. $8.50 33. $12.40 35. $1000 37. 40% 39. 8% 41. $4166.67 43. $3400 45. $633.33 49. Yes, if the profit is figured as a percent of the selling price. 51. Yes 53. 51.6256 55. 53506 57. 50.066 59. 515.46

579

Problem Set 3.5 (page 179) 1. 12.5 milliliters 3. 15 centiliters 1 1 5. 7 quarts of the 30% solution and 2 quarts of the 2 2 50% solution 7. 12 gallons 9. 16.25% 11. $1500 at 3% and $2250 at 5% 13. $500 at 4% and $700 at 6% 15. $4000 17. $600 at 6% and $1700 at 8% 19. $3000 at 8% and $2400 at 10% 21. Pam is 9 years old and Bill is 18 years old 23. Tom is 30 years old and his father is 70 years old 25. Dylan is 45 years old and Tobin is 15 years old 27. Maria is 3 years old and Mario is 10 years old 29. 32 years old

Chapter 3 Review Problem Set (page 184) 1. 20

2. 21

3. 6 1 3

4. 8

5.

23 5

6.

1 2

7. 20

8. 12


9. 34

1. 7 3. 500 5. 20 7. 48 9. 9 11. 46 centimeters 13. 15 inches 15. 504 square feet 17. $6 19. 7 inches 1 21. 150p square centimeters 23. p square yards 4 25. S 324p square inches and V 972p cubic inches 27. V 1152p cubic feet and S 416p square feet V 3V 29. 12 inches 31. 8 feet 33. h 35. B B h P 2l 3V 5 37. w 39. h 2 41. C 1F 322 2 9 pr 9 7y A 2pr2 9x 13 43. h 45. x 47. y 2pr 3 6 11y 14 y 4 3 49. x 51. x 53. y x 2 3 2 2y 22 ax c 55. y 57. x 59. y mx b b 5 65. 125.6 square centimeters 67. 245 square centimeters 69. 65 cubic inches

13. 300 miles 14. $4000 15. 25 pounds 3 2 16. 9 feet and 14 feet 17. $300 and $450 5 5 18. 2750 females and 1650 males 19. 65% 20. 32% 21. 87.5% 22. 12.5% 23. 175% 24. 18.75% 2 1 25. 66 % 26. 233 % 27. 60% 28. 16% 29. 40 3 3 30. 560 31. 600 32. 8.4 33. 556 34. 53.56 35. 556 36. 58006 37. 5166 38. 5736 39. $52 40. $60 41. 40% 42. 35% 43. $8 44. $36.25 45. $110 46. $1.60 a pound 47. $2720 48. 4 years 49. $3000 50. 8.2% 51. w 6 52. C 25 53. r 0.0625 or r 6.25% 54. t 2.5 55. 77 square inches 56. 6 centimeters 57. 354 yards 2A V V 58. 15 feet 59. w 60. h 2 61. b lh h pr 2x 12 3V 62. B 63. y 64. y 3x 2 h 5 2y 5 4y 7 65. x 66. x 6 3 67. 6 meters by 17 meters 68. 15 centimeters by 40 centimeters 1 69. 29 yards by 10 yards 70. 34° and 99° 71. 1 hours 2 72. 58 miles per hour and 65 miles per hour 73. 5 hours 1 74. 3 hours 75. 20 liters 76. 18 gallons 77. 26% 2 2 78. 4 liters 79. 2 cups 80. 25 millimeters 9 81. $675 at 3% and $1425 at 5% 82. $3000 at 5% and $4500 at 6% 83. $1000 at 8% and $3000 at 9% 84. $1000 85. Shane is 20 years old and Caleb is 12 years old 86. Melinda is 22 years old and her mother is 42 years old

Problem Set 3.4 (page 172) 1 1. e 8 f 3. 5166 5. 5256 7. 576 9. 5246 11. 546 3 1 1 13. 12 years 15. 33 years 2 3 17. The width is 14 inches and the length is 42 inches. 19. The width is 12 centimeters and the length is 34 centimeters. 21. 80 square inches 23. 24 feet, 31 feet, and 45 feet 25. 6 centimeters, 19 centimeters, 21 centimeters 27. 12 centimeters 29. 7 centimeters 31. 9 hours 1 33. 2 hours 35. 55 miles per hour 2 37. 64 and 72 miles per hour 39. 60 miles

10.

11. 14 feet by 17 feet 12. 130 miles

580


87. Kaitlin is 12 years old and Nikki is 4 years old 88. Jessie is 18 years old and Annilee is 12 years old

15.

Chapter 3 Test (page 187)

21. 81

1. 5226

2. e

6. e

7. 51006

11. 14. 16. 21. 24.

12 f 5

17 f 18

3. 5776 8. 5706

4 4. e f 3

9. 52506

5. 5146

11 f 2 y8 9x 47 5F 160 12. x 13. y C 9 2 4 64p square centimeters 15. 576 square inches 14 yards 17. 125% 18. 70 19. $189 20. $52 40% 22. 875 women 23. 10 hours 4 centiliters 25. 11.1 years 10. e

7 8

16.

5 6

17. 0.28

22. 64

1 4

23. 0.064 24.

26.

1 512

27. 546

31. e

25 f 2

4 32. e f 7

28. 526

35. {All reals} 36. 5116 39. 5x 0x 76

18.

33. e

3 37. e f 2

42. 5x 0 x 216

43. 5t 0t 1006

Cumulative Review Problem Set Chapters 1–3 (page 189)

45. 5n 0n 186

46. e x 0x

1. 16x 2. 4a 6 3. 12x 27 4. 5x 1 1 1 5. 9n 8 6. 14n 5 7. x 8. n 9. 0.1x 4 10 10. 0.7x 0.2 11. 65 12. 51 13. 20 14. 32

47. 49. 51. 53.

1 32

29.

34 f 3

40. 5x 0 x 66

19. 5 25.

20.

1 4

25 36

30. 586

34. 52006 38. 506

41. e n 0n

7 f 5

44. 5x 0x 16)

5 f 3 $15,000 48. 45° and 135° 8 nickels and 17 dimes 50. 130 or higher 12 feet and 18 feet 52. $40 per pair 45 miles per hour and 50 miles per hour 54. 5 liters



1. C 3. B 5. A 7. 6x 2 9. 8x 11 11. 4y 10 13. 2x2 2x 23 15. 17x 19 17. 6x2 5x 4 19. 5n 6 21. 7x2 13x 7 23. 5x 5 25. 2x 5 27. 3x 7 29. 2x2 15x 6 31. 5n2 2n 3 33. 3x3 2x2 11 35. 9x 2 37. 2a 15 39. 2x2 5 41. 6x3 12x2 5 43. 4x3 8x2 13x 12 45. x 11 47. 3x 14 49. x2 13x 12 51. x2 11x 8 53. 10a 3b 55. n2 2n 17 57. 8x 1 59. 5n 1 61. 2a 6 63. 11x 6 65. 6x 7 67. 5n 7 69. 8x 6 71. 20x2

1. xy 3x 2y 6 3. xy x 4y 4 5. xy 6x 5y 30 7. xy xz x 2y 2z 2 9. 6xy 2x 9y 3 11. x2 10x 21 13. x2 5x 24 15. x2 6x 7 17. n2 10n 24 19. 3n2 19n 6 21. 15x2 29x 14 23. x3 7x2 21x 27 25. x3 3x2 10x 24 27. 2x3 7x2 22x 35 29. 8a3 14a2 23a 9 31. 3a3 2a2 8a 5 33. x4 7x3 17x2 23x 12 35. x4 3x3 34x2 33x 63 37. x2 11x 18 39. x2 4x 12 41. x2 8x 33 43. n2 7n 12 45. n2 18n 72 47. y2 4y 21 49. y2 19y 84 51. x2 2x 35 53. x2 6x 112 55. a2 a 90 57. 2a2 13a 6 59. 5x2 33x 14 61. 6x2 11x 7 63. 12a2 7a 12 65. 12n2 28n 15 67. 14x2 13x 12 69. 45 19x 2x2 71. 8x2 22x 15 73. 9x2 9x 4 75. 72n2 5n 12 77. 27 21x 2x2 79. 20x2 7x 6 81. x3 6x2 12x 8 83. x3 9x2 27x 27 85. 8n3 12n2 6n 1 87. 27n3 54n2 36n 8 89. x2 14x 49 91. 25x2 4 93. x2 2x 1 95. 9x2 42x 49 97. 4x2 12x 9 99. 4x2 9y2 101. 1 10n 25n2 103. 9x2 24xy 16y2 105. 9 24y 16y2 107. 1 49n2 109. 16a2 56ab 49b2 111. x2 16xy 64y2 113. 25x2 121y2 115. 64x3 x 117. 32x3 2xy2

Problem Set 4.2 (page 204) 1. 45x2 3. 21x3 5. 6x2y2 7. 14x3y 9. 48a3b3 11. 5x4y 13. 104a3b2c2 15. 30x6 17. 56x2y3 2 2 19. 6a2b3 21. 72c3d3 23. x3y5 25. a2b5 5 9 27. 0.28x8 29. 6.4a4b2 31. 4x8 33. 9a4b6 35. 27x6 37. 64x12 39. 81x8y10 41. 16x8y4 43. 81a12b8 45. x12y6 47. 18x5 49. 432x5 51. 25x7y8 53. a12b5c9 55. 16x11y17 57. 15x2 10x 59. 18x3 6x2 61. 28x3 16x 63. 2x3 8x2 12x 65. 18a3 30a2 42a 67. 28x3y 7x2y 35xy 69. 9x3y 2x2y 6xy 71. 13x 22y 73. 2x 9y 75. 4x3 3x2 14x 77. x 14 79. 7x 12 81. 7x 5 83. 3px2 89. x7n 91. x6n1 93. x6n3 95. 20x10n 97. 12x7n


119. (a) x2 8x 16 (b) x2 10x 25 (c) x2 12x 36 (d) 9x2 12x 4 (e) x2 2x 1 (f) x2 6x 9 123. 4x3 56x2 196x; 196 4x2

Problem Set 4.4 (page 218) 1. x8 3. 2x2 5. 8n4 7. 8 9. 13xy2 11. 7ab2 13. 18xy4 15. 32x5y2 17. 8x5y4 19. 1 21. 14ab2c4 23. 16yz4 25. 4x2 6x3 27. 3x3 8x 29. 7n3 9 31. 5x4 8x3 12x 33. 4n5 8n2 13 35. 5a6 8a2 37. 3xy 5y 39. 8ab 10a2b3 41. 3bc 13b2c4 43. 9xy2 12x2y3 45. 3x4 5x2 7 47. 3a2 7a 13b 49. 1 5xy2 7xy5

Problem Set 4.5 (page 223) 1. x 12 3. x 2 5. x 8 with a remainder of 4 7. x 4 with a remainder of 7 9. 5n 4 11. 8y 3 with a remainder of 2 13. 4x 7 15. 3x 2 with a remainder of 6 17. 2x2 3x 4 19. 5n2 4n 3 21. n2 6n 4 23. x2 3x 9 25. 9x2 12x 16 27. 3n 8 with a remainder of 17 29. 3t 2 with a remainder of 6 31. 3n2 n 4 33. 4x2 5x 5 with a remainder of 3 35. x 4 with a remainder of 5x 1 37. 2x 12 with a remainder of 49x 5 39. x3 2x2 4x 8

Problem Set 4.6 (page 231) 1.

1 9

3.

1 64

5.

2 3

7. 16

9. 1

11.

27 8

13.

1 4

1 27 1 17. 19. 21. 27 23. 1000 9 64 8 1 1 25. or 0.001 27. 18 29. 144 31. x5 33. 2 1000 n 27 1 15 35. 5 37. 8x 39. 4 41. 3 43. 96 45. x10 x y a 1 3 1 47. 4 49. 2n2 51. 4 53. 4 55. x6 57. 4 n x x 1 1 8 1 81 x2 59. 3 4 61. 6 3 63. 6 65. 67. 8 69. 6 25 xy a xy n 16n 2 x2y y 1 x 71. 73. 2 75. a4b8 77. 6 79. x 81. 2 x 8x3 y x4 83. 85. 4.211010 2 87. 4.31103 2 89. 8.91104 2 4 91. 2.51102 2 93. 1.11107 2 95. 8000 97. 52,100 99. 11,400,000 101. 0.07 103. 0.000987 105. 0.00000864 107. 0.84 109. 450 111. 4,000,000 113. 0.0000002 115. 0.3 117. 0.000007 119. $875 121. 358 feet

15.

581

8x2 18 11. 11x 8 12. 9x2 8x 20 2y2 54y 18 14. 13a 30 15. 27a 7 n 2 17. 5n2 2n 18. 17n2 14n 16 35x6 20. 54x8 21. 24x3y5 22. 6a4b9 8a6b9 24. 9x2y4 25. 16x2y6 26. 125a6b3 35x2 15x 28. 24x3 3x2 3x3 5x2 3x 2 30. 2x3 17x2 26x 24 x3 3x2 8x 12 32. 2x3 7x2 10x 7 x4 x3 2x2 7x 5 n4 5n3 11n2 30n 4 35. x2 17x 72 x2 10x 24 37. x2 8x 12 x2 11x 30 39. x2 5x 14 40. x2 6x 40 2x2 11x 5 42. 3x2 14x 8 5x2 14x 3 44. 4x2 17x 15 14x2 x 3 46. 20a2 3a 56 30n2 19n 5 48. 12n2 13n 4 a3 15a2 75a 125 50. a3 18a2 108a 216 y2 16y 64 52. m2 8m 16 9a2 30a 25 54. 4x2 28x 49 55. 4n2 1 16n2 25 57. 600 4x2 square feet 4x3 60x2 216x cubic inches 59. 12x3y3 7a3b4 61. 3x2y 9x4 62. 10a4b9 13a3b7 14x2 10x 8 64. 4a2 5a 1 x 4, R 21 66. 7x 6 67. 2x2 x 4, R 4 1 68. 3x 5, R 4 69. 13 70. 25 71. 72. 1 16 5 16 1 11 73. 1 74. 9 75. 76. 77. 8 78. 79. 9 4 18 4 9y3 1 1 1 80. 81. 3 82. 12x3 83. x2 84. 2 85. 2 25 x x x 4 4 x2 b6 1 6 86. 2 2 87. 8a 88. 89. 90. 4 91. n y 2x xy a n3 1 92. 93. 94. 12b 95. 610 96. 56,000 8 9n4 97. 0.08 98. 0.00092 99. 1921103 2 100. (4.7)(10) 101. 14.721102 2 102. 12.12110 4 2 103. 0.48 104. 4.2 105. 2000 106. 0.00000002

10. 13. 16. 19. 23. 27. 29. 31. 33. 34. 36. 38. 41. 43. 45. 47. 49. 51. 53. 56. 58. 60. 63. 65.

Chapter 4 Test (page 239) 1. 2x2 2x 5 2. 3x2 6x 20 3. 13x 2 4. 28x3y5 5. 12x5y5 6. x2 7x 18 7. n2 7n 98 8. 40a2 59a 21 9. 9x2 42xy 49y2 10. 2x3 2x2 19x 21 11. 81x2 25y2 12. 15x2 68x 77 13. 8x2y4 14. 7x 9y 15. x2 4x 5 16. 4x2 x 6 27 24 5 x3 x6 17. 18. 1 19. 16 20. 2 21. 22. 10 8 16 4 x y 23. 12.72110 4 2 24. 9,200,000 25. 0.006

Cumulative Review Problem Set Chapters 1– 4 (page 240)


1. 130

1. C 2. F 3. E 4. D 5. B 6. A 7. 8x2 13x 2 8. 3y2 11y 9 9. 3x2 2x 9

8. 16

2. 1 9. 36

3. 27 10. 1

3 4

4. 16 11. 0

5. 81 12.

13 40

6. 32 13.

2 13

7.

3 2

14. 5

582


1 16. 33 17. 15x3y7 18. 12ab7 8x6y15 20. 6x2y 15xy2 21. 15x2 11x 2 21x2 25x 4 23. 2x2 7x 6 24. 49 4y2 3x3 7x2 2x 8 26. 2x3 3x2 13x 20 4n2 12n 9 28. 1 6n 12n2 8n3 2x4 x3 4x2 42x 36 30. 4x2y2 31. 14ab2 7y 8x2 9x3y3 33. 2x2 4x 7 xy2 2 6 z2 34. x2 6x 4 35. 36. 37. 38. 2 4 x x 3 xy 39. 0.12 40. 0.0000000018 41. 200 42. 5116 3 43. 516 44. 5486 45. e f 46. 596 47. 5136 7 15. 19. 22. 25. 27. 29. 32.

48. e

9 f 14

49. 55006

50. 5x 0 x 16

51. 5x 0 x 06

52. e x 0 x 55. 58. 59. 61. 62.

4 12 f 53. 5x 0x 26 54. e x 0 x f 5 7 5x 0 x 3006 56. 3 57. 40 8 dimes and 10 quarters $700 at 8% and $800 at 9% 60. 3 gallons 1 3 hours 2 The length is 15 meters and the width is 7 meters.

CHAPTER 5 Problem Set 5.1 (page 248) 2

2 2

1. 6y 3. 12xy 5. 14ab 7. 2x 9. 8a b 11. 412x 3y2 13. 7y12x 32 15. 9x12x 52 17. 6xy12y 5x2 19. 12a2b13 5ab3 2 21. xy2 116y 25x2 23. 818ab 9cd2 25. 9a2b1b3 32 27. 4x4y113y 15x2 2 29. 8x2y15y 12 31. 3x14 5y 7x2 33. x12x2 3x 42 35. 4y2 111y3 6y 52 37. 7ab12ab2 5b 7a2 2 39. 1y 121x z2 41. 1b 421a c2 43. 1x 321x 62 45. 1x 1212x 32 47. 1x y215 b2 49. 1x y21b c2 51. 1a b21c 12 53. 1x 521x 122 55. 1x 221x 82 57. 12x 121x 52 59. 12n 1213n 42 61. 50, 86 63. 51, 06 3 3 65. 50, 56 67. e 0, f 69. e , 0 f 71. 55, 06 2 7 3 5 73. e 0, f 75. 50, 76 77. 50, 136 79. e , 0 f 2 2 4 81. 55, 46 83. 54, 66 85. 0 or 9 87. 20 units 89. p 91. The square is 3 inches by 3 inches and the rectangle is 3 inches by 6 inches. 95. (a) $116 (b) $254 (c) $750 (d) $2000 c c 97. x 0 or x 2 99. y 1ab b

Problem Set 5.2 (page 254) 1. (x 1)(x 1) 3. (x 10)(x 10) 5. (x 2y)(x 2y) 7. (3x y)(3x y) 9. (6a 5b)(6a 5b) 11. (1 2n)(1 2n) 13. 5(x 2)(x 2) 15. 8(x2 4) 17. 2(x 3y)(x 3y) 19. x(x 5)(x 5) 21. Not factorable 23. 9x(5x 4y) 25. 4(3 x)(3 x) 27. 4a2(a2 4) 29. (x 3)(x 3)(x2 9) 31. x2(x2 1) 33. 3x(x2 16) 35. 5x(1 2x)(1 2x) 37. 4(x 4)(x 4) 39. 3xy(5x 2y)(5x 2y) 41. (2x 3y)(2x 3y)(4x2 9y2) 43. (3 x)(3 x)(9 x2) 45. 53, 36 47. 52, 26 2 2 4 4 49. e , f 51. 511, 116 53. e , f 55. 55, 56 3 3 5 5

57. 54, 0, 46 63. 69. 75. 79. 87. 89. 91. 95. 97.

59. 54, 0, 46

1 1 61. e , f 3 3 1 9 9 1 510, 0, 106 65. e , f 67. e , 0, f 8 8 2 2 7 or 7 71. 4, 0, or 4 73. 3 inches and 15 inches The length is 20 centimeters and the width is 8 centimeters. 77. 4 meters and 8 meters 5 centimeters 85. (x 2)(x2 2x 4) (n 4)(n2 4n 16) (3a 4b)(9a2 12ab 16b2) (1 3a)(1 3a 9a2) 93. (2x y)(4x2 2xy y2) (3x 2y)(9x2 6xy 4y2) (5x 2y)(25x2 10xy 4y2) 99. (4 x)(16 4x x2)

Problem Set 5.3 (page 263) 1. (x 4)(x 6) 3. (x 5)(x 8) 5. (x 2)(x 9) 7. (n 4)(n 7) 9. (n 3)(n 9) 11. (n 4)(n 10) 13. Not factorable 15. (x 6)(x 12) 17. (x 6)(x 11) 19. ( y 8)( y 9) 21. (x 5)(x 16) 23. (x 6)(x 12) 25. Not factorable 27. (x 2y)(x 5y) 29. (a 4b)(a 8b) 31. 57, 36 33. 53, 66 35. 52, 56 37. 59, 46 39. 54, 106 41. 58, 76 43. 52, 146 45. 512, 16 47. 52, 86 49. 56, 46 51. 7 and 8 or 7 and 8 53. 12 and 14 55. 4, 3, 2, and 1 or 7, 8, 9, and 10 57. 4 and 7 or 0 and 3 59. The length is 9 inches and the width is 6 inches. 61. 9 centimeters by 6 centimeters 63. 7 rows 65. 8 feet, 15 feet, and 17 feet 67. 6 inches and 8 inches 73. (x a 5)(x a 8) 75. (x a 3)(x a 9)

Problem Set 5.4 (page 269) 1. (3x 1)(x 2) 3. (3x 2)(2x 5) 5. (4x 1)(x 6) 7. (4x 5)(3x 4) 9. (5y 2)( y 7) 11. 2(2n 3)(n 8) 13. Not factorable 15. (6x 1)(3x 7) 17. 3(7x 2)(x 4) 19. (2x 3)(4x 7) 21. (3t 2)(3t 7) 23. (12y 5)( y 7)


25. Not factorable 27. (7x 3)(2x 7) 2 1 1 29. (4x 3)(5x 4) 31. e6, f 33. e , f 2 3 4 3 3 1 2 7 5 35. e , 8 f 37. e , f 39. e7, f 41. e , f 3 5 3 6 8 2 5 1 2 4 2 5 5 2 43. e , f 45. e , f 47. e , f 49. e , f 3 3 5 2 2 3 4 4

Problem Set 5.5 (page 277) 1. (x 2)2 3. (x 5)2 5. (3n 2)2 7. (4a 1)2 9. (2 9x)2 11. (4x 3y)2 13. (2x 1)(x 8) 15. 2x(x 6)(x 6) 17. (n 12)(n 5) 19. Not factorable 21. 8(x2 9) 23. (3x 5)2 25. 5(x 2)(3x 7) 27. (4x 3)(6x 5) 29. (x 5)( y 8) 31. (5x y)(4x 7y) 33. 3(2x 3)(4x 9) 35. 6(2x2 x 5) 37. 5(x 2)(x 2)(x2 4) 39. (x 6y)2 41. 50, 56 1 3 2 11 43. 53, 126 45. 52, 0, 26 47. e , f 49. e , f 3 2 3 4 2 6 51. 53, 16 53. 50, 66 55. 54, 0, 66 57. e , f 5 5 10 4 59. 512, 166 61. e , 1 f 63. 50, 66 65. e f 3 3 4 5 5 67. 55, 06 69. e , f 71. 5 and 3 or and 12 3 8 4 1 43 24 73. 1 and 1 or and 2 75. and or 4 and 9 2 5 5 77. 6 rows and 9 chairs per row 79. One square is 6 feet by 6 feet and the other one is 18 feet by 18 feet. 81. 11 centimeters long and 5 centimeters wide 83. The side is 17 inches long and the altitude to that side is 6 inches long. 85. 6 inches and 12 inches

Chapter 5 Review Problem Set (page 282) 1. 2x y 2. 3xy 3. 4ab 4. 12m n 5. 3x (5x 7) 6. 4a3(6 5a3) 7. 5x2y(2x2 10xy y2) 8. 4m2n(3 5mn 6m2n2) 9. (a 3)(b 4) 10. (2x 5)(y 6) 11. (x 2 y 2)(x y) 12. (2m 3n)(4m 1) 13. (5x 4)(5x 4) 14. 2(3x 5)(3x 5) 15. (x 2 4)(x 2)(x 2) 16. x(x 7)(x 7) 17. (x 3)(x 8) 18. (x 12)(x 1) 19. (x 8)(x 3) 20. (x 10)(x 2) 21. (2x 3)(x 1) 22. (5x 2)(x 3) 23. (4x 3)(x 2) 24. (3x 2)(x 4) 25. (3x 2)2 26. (2x 5)2 2

2

2

3 2

2

2

27. 30. 33. 36. 38. 40. 42. 44. 46. 48. 51. 53. 57. 61. 65. 69. 73. 74. 75. 76. 78. 80. 81. 83.

(4x 1)2 28. (x 6y)2 29. (x 2)(x 7) 3x(x 7) 31. (3x 2)(3x 2) 32. (2x 1)(2x 5) (5x 6)2 34. n(n 5)(n 8) 35. ( y 12)( y 1) 3xy( y 2x) 37. (x 1)(x 1)(x2 1) (6n 5)(3n 1) 39. Not factorable (4x 7)(x 1) 41. 3(n 6)(n 5) x(x y)(x y) 43. (2x y)(x 2y) 2(n 4)(2n 5) 45. (x y)(5 a) (7t 4)(3t 1) 47. 2x(x 1)(x 1) 3x(x 6)(x 6) 49. (4x 5)2 50. (y 3)(x 2) (5x y)(3x 2y) 52. n2(2n 1)(3n 1) 5 8 1 56, 26 54. 50, 116 55. e4, f 56. e , f 2 3 3 5 2 9 52, 26 58. e f 59. 51, 0, 16 60. e , f 4 4 7 3 7 57, 46 62. 55, 56 63. e6, f 64. e , 1 f 5 2 3 52, 0, 26 66. 58, 126 67. e5, f 68. 52, 36 4 7 5 3 4 5 e , f 70. 59, 66 71. e5, f 72. e , f 3 2 2 3 2 8 19 and or 4 and 7 3 3 The length is 8 centimeters and the width is 2 centimeters. A 2-by-2-inch square and a 10-by-10-inch square 13 8 by 15 by 17 77. and 12 or 2 and 13 6 7, 9, and 11 79. 4 shelves A 5-by-5-yard square and a 5-by-40-yard rectangle 18 and 17 or 17 and 18 82. 6 units 2 meters and 7 meters 84. 9 and 11

Chapter 5 Test (page 284) (x 5)(x 2) 2. (x 3)(x 8) 3. 2x(x 1)(x 1) (x 9)(x 12) 5. 3(2n 1)(3n 2) (x y)(a 2b) 7. (4x 3)(x 5) 8. 6(x2 4) 2x(5x 6)(3x 4) 10. (7 2x)(4 3x) 11. 53, 36 5 2 12. 56, 16 13. 50, 86 14. e , f 15. 56, 26 2 3 1 2 16. 512, 4, 06 17. 55, 96 18. e12, f 19. e4, f 3 3 7 20. 55, 0, 56 21. e f 22. 14 inches 5 23. 12 centimeters 24. 16 chairs per row 25. 6, 5, 4 or 4, 5, 6

1. 4. 6. 9.

CHAPTER 6 Problem Set 6.1 (page 289) 1. 13.

3x 7y

3.

y x2

3xy 5 2x 3y 15. 3

3y 8

5.

7.

3y 2

4xy 2b2 11. 9 9z x1 19. 3x

9.

4x x2 17. x7

583

x12x 5y2 xy n 2n 1 23. 25. 27. x 21x 4y2 n1 n3 2x 7 3 x9 2a 1 29. 31. 33. 35. 3x 4 41x 12 x3 3a 1 21.

584


x 3y x3 n7 2n 3 39. 41. 43. x 2x y 8 n1 y 12 1x 45. 47. 1 49. 3 51. 4x 53. x y 14 y3 2x x9 1 55. 57. 59. 63. 4 5x x6 2 y5 x1 1 1 4 65. 67. 6 69. 3 2 71. 3 x5 xy a x 37.

13. 19. 31. 39.

xy 1 11. x 5 2x1x 42 a 15. 17. x9 15a 2213a 12 5y1x 82 3x2y 4 17 10b3 4 21. 23. 2a 25. 27. 29. 3 19 27 2 5b 3y12x y2 s 3a 5 33. 35. 37. 17 14 21x y2 xy y2 x1 4t 1 x6 41. 43. 2 45. x 314t 32 3 y 11 y21x y2

1 6

9 14 51x 62 3.

5.

2x2 7y

7.

3n2 10

9.

Problem Set 6.3 (page 301) 1. 13. 23. 33. 43. 53. 61. 69. 75. 79.

2 1 4 x4 3 5. 7. 2 9. 11. n x 3x x1 x 5t 2 3a 8 5n 4 1 15. 17. 19. 21. n 1 4 3 4 3x 5 9 1 25. 27. 3 29. 31. a 2 7x 4 7x 11y 3 13x 3n 47x 35. 37. 39. 41. x6 8 4 12 21 14x 13n 7x 14 4x 18n 11 45. 47. 49. 51. 15 24 10 9 12 9n 14 7x 23x 18 19 55. 57. 59. 18 24 60 24x 1 20x 33 25 10x 35 63. 65. 67. 18y 12x x1x 52 48x2 n 3 2n 16 6 71. 73. n1n 12 n1n 42 x12x 12 10x 12 6x 21 77. 1x 42 1x 32 1x 221x 12 x7 4 6 n3 85. 87. 89. 12x 1213x 12 x3 a1 2n 1

17 x

29. 35. 41. 51.


25.

3.

Problem Set 6.4 (page 311) 3x 8 5x 3 8n 50 4 3. 5. 7. x1x 42 x1x 22 n1n 62 n1 5x 7 5x 17 4 9. 11. 13. 2x1x 12 1x 421x 42 x1 11a 6 12 15. 17. a1a 221a 22 x1x 621x 62 2x 17 n8 19x 19. 21. 23. 31n 421n 42 613x 22 151x 12

59. 71.

5x 6 x2 10x 20 27. 1x 32 1x 42 1x 32 1x 221x 42 1x 52 ab 8x 14 15x 4 31. 33. ab 1x 521x 52 x1x 22 1x 22 4x 4 2x 6 3 37. 39. 1x 52 1x 22 13x 521x 42 x4 2y 3x y 1 10 28 43. 45. 47. 49. 2 3 27 3x 5y x x2 4y 6 2x 9 24x x2 2x 53. 55. 57. 3 4x 10x 42 6x 4 7xy 3x2 m d 2x 3 k 34 47 61. 63. 65. 67. 69. r n 4x 1 40 l l 96 n2 n 1 3x2 4x 2 73. 75. b n1 4x 2

Problem Set 6.5 (page 318) 1. 5126

3. e

2 f 21

5. 546

7. 516

9. 536

11. 5386

13. e 23. 33. 43. 51. 59.

13 5 7 f 15. 526 17. 566 19. e f 21. e f 8 18 10 1 e f 25. 586 27. 5376 29. 31. 5396 3 40 5 2 12 e f 35. 566 37. e f 39. e f 41. 4 3 7 48 10 45. 48° and 72° 47. 60° 49. 15 miles per hour 50 miles per hour for Dave and 54 miles per hour for Kent 57. All real numbers except 0 All real numbers except 2 and 3

Problem Set 6.6 (page 327) 1. 526 11. 19. 27. 37. 39. 43. 45. 49.

1.

3. 556

9 5. e f 2

7. e

49 f 10

2 9. e f 3 4 3 11 13 e f 13. e f 15. e f 17. e f 3 2 3 4 5 5 1 e1, f 21. e , 4 f 23. e , 6 f 25. 556 8 4 2 1 5 2 5216 29. 526 31. 58, 16 33. or 4 35. or 2 5 2 17 miles per hour for Tom and 20 miles per hour for Celia 16 miles per hour for the trip out and 12 miles per hour for the return trip 41. 30 minutes 90 minutes for Mike and 180 minutes for Barry 9 hours 47. 4 minutes 10 oysters per minute for Tchaika and 15 oysters per minute for Pachena 53. 506

Chapter 6 Review Problem Set (page 333) 1.

7x2 9y2

6. 5 y

2.

x x3 7. 6

4a 3 3n 5 4. 5. (x 4) n2 5a 2 x5 3x 8. 9. 10. x(x 3) x1 8 3.


11. 15. 20. 23. 25. 28. 32. 38. 43. 47. 50. 52.

x1x y2 2a 1 n7 12. 13. 14. x1 3 a6 n2 8xy3 7b3 5 22x 19 16. 17. 2n 3 18. 19. 3 y2 7 20 43x 3 10n 7 a 8 21. 22. n1n 12 1a 421a 22 12x2 5x 4 5x 9 24. 4x1x 32 1x 521x 521x 22 3y2 4x 6x 37 12x 23 26. 27. 1x 721x 32 1x 721x 42 4xy 5y2 2 2y xy x 6x 6 3n x 29. 30. 31. y 3xy 5x 5n 2m x 12 p d 40 20 2 33. 34. 35. e f 36. e f 37. 566 g m k 17 9 61 28 3 e f 39. 596 40. e f 41. e f 42. 516 60 3 4 1 1 576 44. e f 45. e , 2 f 46. 55, 106 7 2 2 1 e f 48. 516 49. Becky 2 hours, Nancy 8 hours 5 3 36 1 or 2 51. 72 Todd’s rate is 15 miles per hour and Lanette’s rate is 22 miles per hour. 53. 8 miles per hour 54. 60 minutes 31x 32

Chapter 6 Test (page 335) 8x2y x 2n 1 2x 3 2. 3. 4. 5. 2x2y2 9 x6 3n 4 x5 1x 42 2 x2 6x 5 9n 4 6. 7. 8. 9. x3 x1x 72 24 30 41 15x 2n 6 5x 18 10. 11. 12. 18x n1n 12 4x1x 62 5x 11 13x 43 13. 14. 1x 421x 82 12x 5213x 421x 62 4 19 15. 556 16. e f 17. e , 3 f 18. 56, 86 16 3 1.

585

1 3 19. e , 2 f 20. 5236 21. 526 22. e f 5 2 2 23. or 3 24. 14 miles per hour 25. 12 minutes 3

Cumulative Review Problem Set Chapters 1– 6 (page 337) 5 17 2. 6 3. 4. 0.6 5. 20 6. 0 7. 2 8. 3 2 12 1 3 12 1 9 9. 10. 11. 1 12. 13. 14. 27 2 7 16 4 2y 7x 2 4 1 19 15. 16. 17. 18. 19. 25 27 10x 3x 1x 621x 42 3x 4 x 12 x7 20. 2 21. 22. 3y 1x 421x 32 x 1x 42 23. 35x5y5 24. 81a2b6 25. 15n4 18n3 6n2 26. 15x2 17x 4 27. 4x2 20x 25 28. 2x3 x2 7x 2 29. x4 x3 6x2 x 3 30. 6x2 11x 7 31. 3xy 6x3y3 32. 7x 4 33. 3x1x2 5x 92 34. 1x 102 1x 102 35. 15x 22 1x 42 36. 14x 7212x 92 37. 1n 1621n 92 38. 1x y2 1n 22 39. 3x1x 121x 12 40. 2x1x 92 1x 62 16 41. 16x 52 2 42. 13x y2 1x 2y2 43. e f 3 1 44. 511, 06 45. e f 46. 5156 47. 51, 16 14 11 1 48. 56, 16 49. 526 50. e f 51. 51, 26 52. e f 12 18 7 1 1 16 53. e , f 54. e , 8 f 55. 59, 26 56. e f 2 3 2 5 1 57. e f 58. 6 inches, 8 inches, 10 inches 59. 75 2 60. 100 milliliters 61. 6 feet by 8 feet 62. 2.5 hours 63. 7.5 centimeters 64. 27 gallons 65. 92 or better 7 66. 51, 2, 36 67. All real numbers greater than 2

1.

CHAPTER 7 Problem Set 7.1 (page 347) 1. Yes

5. Yes

3. No y

(2, 7)

7. Yes

y

y

y

(1, 4) (0, 3)

(2, 1) x

x

(3, 0) (4, 1)

(0, 2) (3, 3)

x (3, 1)

(4, 0) x (2, 1) (6, 5)

(3, 6)

586


9. No

11. y

Sat.

(1, 0)

43.

45.

Thu. x

Wed.

x

Mon. Fri.

(2, 4)

y (2, 6) (1, 5)

41.

Tue.

(2, 4)

13.

39. y

y

15.

y (3, 3)

(0, 4) (1, 3)

(0, 2)

(2, 1)

(6, 0) x

x

x

(0, 3) (1, 5)

y

17.

(2, 1)

(0, 0)

47.

y

19.

(1, −15 ) x

x

(3, 2)

49.

(1, −35 ) 3 13 21. y x 23. x 3y 9 7 7 1 7 2 5 27. x y 29. y x 3 3 3 3 31.

14 1 25. y x 5 5

y

53.

y

55.

(0, 0) (2, 6)

(2, 6)

x (2, 4)

(0, 2)

33.

(2, 4)

x

57.

59.

y

y

(5, 8)

(2, 8)

35.

y

37.

y (0, 4) (0, 0)

( 4–3 , 0)

(3, 0)

(1, 0) x

(0, 4)

x

x (2, 8)

(1, 8)

x

Answers to Odd-Numbered Problems y (2, 16)

61. (2, 16)

17.

19.

21.

23.

25.

27.

29.

31.

33.

35.

37.

39.

(1, 1)

(1, 1)

x

(0, 0)

Problem Set 7.2 (page 357) y

1. (2, 3)

y

3.

(0, 2)

(2, 2)

(4, 0) x

(1, 0)

(0, 2)

y

5.

x

y

7. (4, 3)

(5, 0) x (2, 3)

9.

x

(2, 0) (0, 3)

(0, 5)

11.

y

(2, 0) x (1, 3) (0, 6)

13.

15.

587

588


41.

43.

y

5.

y

7.

(2, 0)

x

(3, 0) x

(0, −4)

y

9.

45.

(0, −4)

y

11.

(1, 2)

(− 2, 2)

(0, 0) x

(0, 0)

x

y

13.

y

15.

47. Hemoglobin A1c, h 6.0 6.5 7.0 8.0 8.5 9.0 10.0 Blood Glucose, G 120 135 150 180 195 210 240

(2, 0)

(4, 0)

(0, −1)

49. $41.76; $43.20; $44.40; $46.50; $46.92

x

x

(0, −2)

51. (a) C 0 5 10 15 20 5 10 15 20 25 F 32 41 50 59 68 23 14 5 4 13 (b)

53.

F

(0, 5)

C

y

19.

(0, 4)

(1, 8)

75 50 25

y

17.

c

(4, 0)

x

x

(−4, 0)

30 10 10 30

(0, − 3) t

y

21. y

59.

23.

y

y

61.

(0, 4)

(0, 0)

(0, 1) (1, 0)

(2, 0) x

(−1, 0) (0, −1)

x

x

(−2, 0)

(2, 0) (2, 0)

(1, 2)

x

(0, −4)

Problem Set 7.3 (page 364) y

1.

3.

y

25.

27.

(5, 3) (0, 3)

(0, 1)

(5, 0)

(2, 0)

(1, 0) x

y

y

x

x (3, 1)

(0, 1)

x


41. m 2, b 5

y

29.

589

y x

(0, 0) (4, 0) x (0, −5) (1, −7)

Problem Set 7.4 (page 372) 1. 15. 23. 35. 45. 59. 61.

3 4

10 3 3 9. 11. 0 13. 3 4 2 db bd Undefined 17. 1 19. 21. 4 or ac ca 6 25 –31. Answers will vary. 33. Negative 3 5 Positive 37. Zero 39. Negative 41. 43. 2 4 1 2 6 47. 2 49. 0 51. 53. 55. 3 57. 4 5 5 5 2 3 y y 63. 3.

7 5

5.

6 5

7.

(5, 0) x (0, −3)

y

4 45. m , b 2 9

(0, 2) x

(− 4.5, 0)

(−2, 3)

(3, 1) x

65.

y

3 43. m , b 3 5

y

x

67.

y

3 47. m , b 4 4

y

x (4, −1)

(0, 5)

(0, − 4)

x (2, −2) x

49. m

y

2 , b 1 11 (−5.5, 0)

69. 5.1%

71. 32 cm

73. 1.0 feet

x

(0, −1)

Problem Set 7.5 (page 381) 1. 2x 3y 5 3. x 2y 7 5. x 3y 20 7. y 7 9. 4x 9y 0 11. 3x y 16 13. 7x 5y 1 15. 3x 2y 5 17. x y 1 3 19. 5x 3y 0 21. 4x 7y 28 23. y x 2 5 1 5 25. y 2x 1 27. y x 4 29. y x 6 2 2 5 1 31. y x 33. y x 5 35. y 3x 1 9 2 5 37. y 2x 6 39. y 3

9 51. m , b 0 7

y (0, 0) x

(7, − 9)

590


57. Perpendicular 59. Intersecting lines that are not perpendicular 61. Parallel 63. 2x 3y 1

y

15.

y

16.

(1, 3) (3, 1)


(0, −1)

1. x intercept is 6, y intercept is 2 2. x intercept is 4, y intercept is 2 3. x intercept is 5, y intercept is 4 1 4. x intercept is , y intercept is 1 2 5. x intercept is 2, y intercept is 6 6. x intercept is 4, y intercept is 8 7. 8.

(0, 0) x

x

y

17.

y

18.

(0, 2) (3, 2) x

x (2, 3) (0, 3)

y

19. 9.

y

10.

y

20. (4, 3)

y

(1, 3)

(4, 0) x

x

(1, 0)

(0, 3) (0, 0) x

x

(4, 0)

21.

(2, 4)

14

16

10

12 8

6

11.

y

12.

s

22.

C

4

2

p

e 2

(1, 0)

x

y

y

14.

16

12

12

8

8

4

4

x (0, − 4)

8 12 16 20

P

y

25.

8

12 16

C

16

2 3 4 5 6 7 8 9

26.

(−2, 0)

(4, 0)

4

8 10

24.

4

13.

6

C

23. (0, 2)

4

P

y

(1, 0) (0, − 1)

x

(1, 0)

x

x (0, 2)

(0, 3)


27.

y

y

28.

y

39.

591

y

40. (3, 2)

(0, 1)

(3, 0) (0, 1)

29.

x

y

41. 3

( 52 , 0) x

y

x

(2, 4)

43.

44.

3 2

45. 0

46. 1

49. 50 feet

16 ; No 52. 7500 feet 75 54. 2x y 7 55. y 4x 2

50. 42.9 centimeters 51.

58. 3x 2y 9

x

x

y

42. 3

2 61. m , b 2 5

(0, 1)

(3, 0)

(3, 0) x

53. 3x 4y 11 3 56. y x 57. 7x 2y 12 5 1 59. y x 60. y 0x 5 4

y

32.

(0, 3)

33.

1 3 5 47. Undefined 48. 2

y (0, 5)

(1, 1)

31.

(0, 0) x

x

30.

(0, 1)

(0, 2)

(4, 0)

y

34.

1 62. m , b 1 3

(3, 3) (0, 0) x

(0, 0)

x

(1, 3)

35.

y

y

36.

1 63. m , b 1 2

(2, 1)

(0, 1) (2, 0)

(0, 0) (0, 0)

37.

x

x (2, 2)

y

(2, 3)

(4, 3)

(4, 0)

x

64. m 3, b 2

y

38.

y

y

x

(2, 0)

x

x (0, −2) (1, −5)

592


65. m 2, b 4

4. m 3, b 0

y

y

(0, 0) x

x

(1, −2)

(1, −3)

(0, − 4)

3 6. 0 7. 8 8. 5 9. 17, 62, 111, 72 or others 2 10. 13, 22, 14, 52 or others 11. (a) A (b) C (c) B (d) D 12. 4.6% 13. 2 14. 4 y 15. 16. 5.

y

3 66. m , b 3 4

(4, 0) x (0, −3)

(3, 2)

(0, 0)

x


17.

y

18.

y

5 1. m , b 5 3

(2, 3) (2, 0)

(−,32 0)

(0, − 3)

y

19. 2. m 2, b 4

x

x

y

20.

(0, 4)

y

(3, 4) x

(2, 0)

x

x

(0, − 4)

(1, −2) (0, − 4)

21.

22.

y

C

10 y

1 3. m , b 2 2

8 (3, 0)

6 (0, 1)

4 2

x (0, − 2)

x

t

(2, −3)

2 4 6 8 10

23. 3x 5y 20

24. 4x 9y 34

25. 3x 2y 0


593


1. No 3. Yes 5. Yes 7. Yes 9. No 11. 512, 126 13. 512, 126 15. 17. 5 10, 026 19. 511, 126 21. Infinitely many 23. 511, 326 25. 513, 226 27. 512, 426 29. 512, 326 y y 31. 33. (0, 4)

(0, 3) (1, 0)

(1, 0)

(3, 0)

(4, 0) x

(0, −2)

x (0, −4)

y

35.

y

37. (−1, 3)

(0, 3)

x

x

(−4, 0)

(0, 1) (2, 0)

x

(0, −2)

3 12 33. e a , b f 5 5

1 37. e a , 4b f 2

5 39. Infinitely many solutions 41. e a 5, b f 2 10 1 43. 5112, 42 6 45. e a , b f 47. 10 and 24 11 11 49. 42 women 51. $5000 at 5% and $8000 at 6% 53. $150 at 8% and $400 at 9% 55. 20 liters of 40% alcohol; 10 liters of 70% alcohol 57. 6 quarts of 10% juice; 4 quarts of 5% juice 59. Length 29 meters; width 8 meters 61. 60 five-dollar bills; 40 ten-dollar bills 63. 90 nickels and 30 pennies 69. 510, 02 6 71.

1. 513, 1, 22 6 (2, 0) (0, −2)

y

39.

31. 5110, 1226


(−3, 0)

(−1, 1)

6 6 29. e a , b f 7 7 5 35. e a , 6b f 2

7. 512, 3, 52 6

13. 513, 0, 22 6

3. 5 11, 3, 426

9. 5 13, 2, 426 15. 5 11, 4, 22 6

5. 5 15, 3, 126

11. 511, 3, 12 6

17. 10 quarters, 8 dimes, 2 nickels 19. ⬔A 120°, ⬔B 24°, ⬔C 36° 21. Plumber $50 per hour; apprentice $20 per hour; laborer $10 per hour 23. Peaches $1.29 per pound; cherries $0.99 per pound; pears $0.69 per pound 25. Helmet $250; jacket $350; gloves $50 29. Infinitely many solutions 31.

Chapter 8 Review Problem Set (page 431) Problem Set 8.2 (page 405)

1. 516, 82 6 3. 515, 426 5. 5 16, 1226 7. 515, 226 11 9 2 2 9. e a , b f 11. e a , b f 13. 514, 526 4 8 3 3 3 18 5 15. 514, 126 17. e a , 3b f 19. e a , b f 2 71 71 21. 51250, 50026 23. 51100, 20026 25. 9 and 21 27. 17 and 4 29. 8 and 15 31. 6 and 12 33. $0.45 per lemon and $0.50 per apple 35. 22 hardcover and 14 softcover 37. 35 double rooms and 15 single 41. 5112, 2426 43.

1. Yes 2. No 3. No 4. Yes 5. 511, 42 6 6. 512, 426 7. 8. 9. Infinitely many 10. Infinitely many y y 11. 12. (0, 5)

(0, 4) (5, 0) (2, 0)

(1, 0) x

x

(1, 0) (0, 3)

y

13.

y

14.

(0, 3)

Problem Set 8.3 (page 415) 1. 515, 926

3. 514, 1026

3 1 4 5. e a , 4b f 7. e a , b f 2 3 3 9. 5118, 2426 11. 5110, 1526 13. 15. 512, 62 6 6 9 9 7 17. 516, 1326 19. e a , b f 21. e a , b f 8 12 31 31 23. 51100, 40026 25. 513, 1026 27. 512, 626

(0, 2) (4, 0)

(1, 0)

x

(0, 1)

(4, 0) (2, 0)

(0, 6)

x

594

Answers to Odd-Numbered Problems y

15.

y

16.

y

8.

(1, 0) (0, 1)

(1, 0) (0, 2)

(5, 0) (0, 3)

x

1. No 2. Yes 3. Yes 7. Infinitely many

4. 512, 326

(3, 0)

(0, 3)

( 3 , 0) 2 x

(2, 0)

(4, 0)

5. 510, 326

(0, 2) x

x

(0, 1)

(0, 1)

(0, 3)

17. 517, 1326 18. 5110, 2526 19. 516, 226 20. 515, 426 21. 5 13, 126 22. 5 10, 226 17 5 23. 5 116, 526 24. e a , b f 25. 519, 626 26. 16 16 27. 517, 12 6 28. 5 13, 826 29. Elimination-by-addition 30. Elimination-by-addition 31. Substitution 32. Substitution 33. 510, 2, 526 34. 512, 1, 326 35. 511, 1, 426 36. 5 14, 0, 126 37. 38 and 75 38. $250 at 6% and $300 at 8% 39. 18 nickels and 25 dimes 40. Length of 19 inches and width of 6 inches 41. 32 and 58 42. 50 and 130

43. $3.25 for a cheeseburger and $2.50 for a milkshake 44. $1.59 for orange juice and $0.99 for water


y

9.

6.

y

10.

(0, 3) (0, 2) (6, 0) x

(1, 0)

11. 514, 62 6 12. 513, 426 13. 512, 12 6 14. 512, 626 1 15. e a , 3b f 16. Infinitely many 17. 2 18. 514, 3, 22 6 19. 5 11, 3, 526 20. 5 12, 0, 426 21. 25 and 71 22. Paper $3.49, notebook $2.29 23. 13 inches 24. 3 liters 25. $2500 at 5% and $3000 at 6%

CHAPTER 9 Problem Set 9.1 (page 442) 3. 8 5. 11 7. 60 9. 40 11. 80 13. 18 5 17. 0.4 19. 3 21. 2 23. 9 25. 6 3 4.36 29. 7.07 31. 8.66 33. 9.75 35. 4.64 4.33 39. 66 41. 34 43. 97 45. 20 47. 32 3 3 3 21 12 51. 8 17 53. 111 2 55. 31 7 21 5 12 2 13 59. 917 2 110 61. 17.3 13.4 65. 1.4 67. 26.5 69. 4.1 71. 2.1 29.8 75. 1.6 seconds; 2.1 seconds; 2.2 seconds 2.2 seconds; 2.8 seconds; 18.2 seconds

1. 7 15. 27. 37. 49. 57. 63. 73. 77.

Problem Set 9.2 (page 449) 3 1. 216 3. 3 12 5. 313 7. 2110 9. 3 1 2 3 11. 4 15 13. 3 113 15. 2412 17. 615 3 19. 10 15 21. 3216 23. 312 25. 13 2 1 3 27. 2 15 29. 14 31. xy 1y 33. x 12y 2 3 35. 2x 12 37. 3a 13ab 39. 4x 2 xy2 41. 3x2y 17 2 5x 3 43. 12x 13 45. 36x3 12x 47. 16xy 49. 1 2x 3 8 26 3 51. a4 53. 33 12 55. 315 57. 18 1 2 3

59. 217 61. 913 63. 215 65. 1512 55 15 69. (a) 912 (c) 5 111

Problem Set 9.3 (page 456) 1. 13. 23. 33. 41. 49. 57.

1 5 5 119 7. 9. 11. 8 4 16 5 3 212 51 3 13 16 110 15. 17. 19. 21. 7 6 3 2 4 16 110 213 29. 31. 17 25. 3 27. 6 9 12 137 2 115 416 121 35. 37. 39. 5 9 8 3 2 13 512x 13x 31x 43. 45. 47. x x x 2x x1xy 110xy 115xy 31x 51. 53. 55. 5y 9y 2y x2 31xy 41x 2213 19110 59. 61. 63. 3 5 x4 2y2

4 5

3. 3

5.

14 16 69. 613 3 3 3 21 5 1 14 77. 79. 5 4

65. 315 67. 3

75.

1 63 3

71. 7115


69. 566

Problem Set 9.4 (page 463) 3

1. 135 3. 4 13 5. 512 7. 31 2 9. 416 3 11. 15 121 13. 10 1 3 15. 72 17. 4016 19. 2415 3 21. 16 110 23. 213 5 16 25. 2 1 20 27. 6 12 4 16 29. 416 8 115 31. 56 15 12 33. 9 2 16 35. 312 2 16 4 13 6 37. 15 39. 15 41. 51 43. x 1y 45. 5x 47. 12a1b 49. x 16 2 13xy 51. x 21x 15 53. x 49 3 12 12 55. 57. 416 8 59. 15 13 14 41x 8 x 31x 61. 16 12 63. 65. x4 x9 6 212 313 16 a 7 1a 10 67. 69. a 25 7

71. 5286 72. 5326 73. 11 74. 75. 546 76. b r 77. 55, 46 78. 5106 2 79. 536 80. 516 81. 2.5 seconds, 4.7 seconds, 5.6 seconds c2 82. b 22a

1.

12.

19.

1. 5496

23.

13. 23. 33. 41. 47.

9 7. e f 4

4 9. e f 11. 5146 9 7 21 15. e f 17. 5366 19. e f 21. 566 3 4 526 25. 526 27. 53, 26 29. 566 31. 5126 1 5256 35. 536 37. e f 39. 526 2 56 feet; 106 feet; 148 feet 43. 3.2 feet; 5.1 feet; 7.3 feet 5186 49. 54, 126 51. 5576 53. 51106 55. 52526

1. 8

37. 42. 47. 51. 55. 59. 61. 63. 66.

3. 40

2 5. 3

7 6. 6

7. 3

8. 5

10.

3. 2.8

8.

12 3

4. 5.6

5. 2.1

6. 315

5 12 16 110 10. 11. 2 3 4 115xy 3 5xy12x 13. 14. 3xy13x 15. 4 16 5y 3 241 10 17. 1212 12 13 18. 1 5115 312 13 20. 4 16 21. 556 22. 5 536 24. 52, 16 25. 2.1 seconds

1. 64

9.

21. 23. 26. 30. (c) 33. 35. 37. 41.

2. 64

3. 144

4. 8

5.

2 3

13 7. 9 9 x 17 12. 1x 22 1x 32 6.

15 4x 5y 1 8x 41 14. 15. 16. 60x4y4 2 x4 1x 621x 32 8 3a3 1 18. 19. 20. 27x2 30x 8 2 b x 3n4 5x2 12x 7 22. 6x3 x2 13x 4 3x3y6 4y3 24. 2x2 2x 3 25. y x 4.67 gallons 27. 25% 28. 108 29. 130 feet 314 square inches 31. (a) 18.521104 2 (b) 1921104 2 11.0421106 2 (d) 15.32 1107 2 32. 2x12x 5213x 42 312x 32 12x 32 34. 1y 32 1x 22 15 x2 16 5x2 36. 41x 121x 121x2 12 6 15 17x 22 13x 42 38. 817 39. 315 40. 5 516 2 1ab 42. 6y2 12xy 43. 44. 48 18 5

8. 49

17.

2 11. 12.57 12. 35.36 13. 29.33 5 15.81 15. 8.49 16. 9.24 17. 215 18. 4 12 3 3 10 12 20. 4 15 21. 21 3 22. 51 2 23. 2x13 3 2 2 5x 12y 25. 6ab15a 26. 4a 13 27. 2y 2 x 3 3x 1x 29. 1812 30. 1213 31. 7 12x 3 3 2 3 2 15y 33. x 1 x 34. 92 x 35. 2 36. 1 5 114 x12x 3 12a 6 17 3 110 38. 39. 40. 41. 3 2b 7 5 4 110xy 5 16 13 2 1x 43. 44. 45. 2 46. x 3 5y 6 3 1xy 16 2 12 21x 48. 49. 50. 3 3 5 4y2 17 16 16 110 5215 31115 52. 53. 54. 3 5 5 5 3 6 12 56. 18 12 57. 40 58. 10 1 28 3 1 6 2 60. 6110 12 115 3 121 115 135 62. 24 716 5117 152 4 5 142 64. 18 15 65. 2 3 142 4 115 3 12 16 68. 3 12 2 13 67. 6 23

9. 4 14. 19. 24. 28. 32.

2. 7

9 4. 5

2. 0.05

Cumulative Review Problem Set Chapters 1– 9 (page 480)

13.


8 7

3 7. 121 2


70. 5856


16.

3. 5186

595

9. 29

10. 49

11.

45. 216 36 16 46. 11 47. 413 4 12 21315 162 3715 48. 49. 12 50. 13 12 51.

52.

y (0, 4)

(3, 5)

x

596

Answers to Odd-Numbered Problems y

53.

y

54.

59. 2

(0, 3)

x

x

(0, 0)

67. 71.

55.

76.

y

56.

81. (3, 0) x

92. y

y

58.

61. 2x 3y 8 62. 4x 3y 13

3 65. 511, 226 66. 5 12, 42 6 2 1 23 516, 122 6 68. e a , 3b f 69. 5176 70. e f 2 5 19 47 556 72. e f 73. e f 74. 55006 75. 5536 10 7 2 e f 77. 56, 26 78. 52, 26 79. 5296 80. 5276 3 2 1 5n 0 n 56 82. e n 0n f 83. e x 0x f 4 5

84. 5n 0n 66 87. 90.

(0, −6)

57.

4 7

63. x 4y 12 64.

(1, 2)

(5, 1)

60.

95.

85. e x 0 x

5 16 f 86. e x 0 x f 16 3 65° and 115° 88. 13 and 37 89. 7, 9 7 nickels, 15 dimes, and 25 quarters 91. $3780 1 $48 93. 4 hours 94. 12.5 milliliters 2 91 or better 96. More than 11 97. 12 minutes

(0, 3) (−4, 0)

(−2, 0) x

(0, −2)

x

CHAPTER 10 Problem Set 10.1 (page 491) 1. 515, 06 11. 19. 25. 29. 33. 39. 45. 49. 51. 55. 59. 63. 65. 67. 69.

3. 50, 126 5. 50, 56 7. 51, 86 9. 52, 76 5 1 2 5 1 56, 16 13. e , f 15. e , f 17. e f 3 2 5 6 2 5 5 58, 86 21. e , f 23. 54, 46 3 3 5114, 1146 27. No real number solutions 54 12, 4 126 31. 5312, 3126 3 12 3 12 5 12 512 e , f 35. e , f 37. 51, 36 2 2 4 4 5 58, 26 41. e , 3 f 43. 56 15, 6 156 3 3 2 15 3 2 15 51 2 12, 1 2 126 47. e , f 2 2 No real number solutions 5 2110 5 2 110 3 5 e , f 53. e , f 3 3 7 7 58 512, 8 5126 57. 512 inches 2 17 meters 61. 2 111 feet a 4 inches and b 413 inches c 12 feet and b 613 feet a 413 meters and c 8 13 meters a 10 inches and c 10 12 inches

912 meters 73. 8.2 feet 75. 30 meters 2 77. 35 meters 81. 5.2 centimeters 83. 10.8 centimeters 71. a b


54 117, 4 1176 55 123, 5 1236 5. 52 212, 2 2126 No real number solutions 51 312, 1 3126 1 113 1 113 11. e , f 2 2 5 133 5 133 13. e , f 2 2 4 122 4 122 15. e , f 2 2 6 142 6 142 17. e , f 3 3 2 12 2 12 19. e , f 2 2 21. No real number solutions 9 165 9 165 23. e , f 2 2 1. 3. 7. 9.


3 117 3 117 , f 4 4 1 17 1 17 27. e , f 29. 514, 126 3 3 31. 511, 156 33. 56, 26 35. 59, 36 5 3 1 37. 55, 86 39. e , 4 f 41. e , f 2 2 2 25. e

45. e

b 2b2 4ac b 2b2 4ac , f 2a 2a


1. 51, 66 3. 59, 46 5. 51 16, 1 166 5 133 5 133 7. e , f 9. No real number solutions 2 2 7 11. 52 12, 2 126 13. 50, 66 15. e 0, f 2 2 1 17. 516, 186 19. 510, 86 21. 526 23. e , f 3 2 2 1 5 25. e1, f 27. e , f 5 4 3 5 173 5 173 29. e , f 4 4 2 17 2 17 3 31. e , f 33. e f 3 3 4 2 15 2 15 35. e , f 2 2 9 157 9 157 37. e , f 12 12 1 133 1 133 39. e , f 4 4 41. No real number solutions 5 1137 5 1137 43. e , f 14 14 1 185 1 185 45. e , f 47. 514, 96 6 6 53. 52.52, 7.526 55. 58.10, 2.106 57. 53.55, 1.226 59. 51.33, 3.586 61. 51.95, 2.156

2 41. e1, f 3 45. 55, 36

1. 17 and 18 3. 19 and 25 5. 3 15 and 3 15 4 5 7. or 9. 12, 14, and 16 11. 8 or 8 5 4 13. 8 meters by 12 meters 15. 15 centimeters by 25 centimeters 17. 27 feet by 78 feet 19. 15 rows and 20 seats per row 21. 7 feet by 9 feet 23. 6 inches and 8 inches 1 25. 1 inches 27. 30 students 2 29. 24 inches and 32 inches 31. 18 miles per hour

Chapter 10 Review Problem Set (page 518) 1. 56, 46

4. 54, 06 1 2 110 5. 53, 136 6. 512, 46 7. 54, 46 8. e f 3 9. 57 1696 10. 54 2110 6 11. 51 2 126

1 9. e f 3

11. 50, 26

7 15. e1, f 5 21. 27. 31. 35.

13 1 , f 5 5

5. 51, 26

8 7. e 0, f 3

13. 57 2117, 7 2 1176

5 1 17. e , f 19. 54 123, 4 1236 3 5 3 133 3 133 512, 76 23. e , f 25. 51, 46 4 4 No real number solutions 29. 516, 306 1 113 1 113 3 4 e , f 33. e , f 2 2 4 3 3 13 3 13 513, 16 37. 511, 176 39. e , f 2 2

2. 55, 86

3. 57, 26

3 129 7 173 f 14. e f 2 2 1 113 2 110 e f 16. e f 17. 56, 16 3 2 2 1 54 113, 4 1136 19. e , f 20. 50, 176 7 3 7 161 7 161 54, 16 22. 511, 156 23. e , f 6 6 1 e f 25. No real number solutions 26. 55, 16 2 52 12, 2 126 28. 52 312, 2 3 126 5315, 3156 30. 53, 96 31. 50, 16 52 113, 2 1136 33. 520, 246 8 52 2 12, 2 2 126 35. e , 1 f 5 1 173 1 173 1 e , f 37. e , 4 f 12 12 2 4 e , 1 f 39. 30 yards 40. 4 yards and 6 yards 3 1613 612 inches 42. centimeters 3 9 inches by 12 inches 44. 18 and 19 45. 7, 9, and 11 15 meters and 315 meters 47. 10 meters Jay’s rate was 45 mph and Jean’s rate was 48 mph; or 1 1 Jay’s rate was 7 mph and Jean’s rate was 10 mph 2 2

12. 55 1296 13. e 15. 18. 21. 24. 27. 29. 32. 34.

Problem Set 10.4 (page 508) 3. e

5 1193 5 1193 , f 6 6


36. 1. 59, 56

43. e

597

38. 41. 43. 46. 48.

Chapter 10 Test (page 520) 1. 2113 inches 2. 13 meters 3. 7 inches 5 4. 4 13 centimeters 5. e3, f 6. 54, 46 3

598


1 3 7. e , f 2 4

8. e

3 129 3 129 , f 2 2

9. 51 110, 1 1106 10. No real number solutions 11. 512, 26 12. e

3 141 3 141 , f 4 4

1 14. e , 2 f 5

15. 513, 156

13. e

1 157 1 157 , f 2 2

3 16. e , 4 f 4

3 1 313 1 3 13 f 19. e , f 7 4 4 20. No real number solutions 21. 15 seats per row 22. 14 miles per hour 23. 15 and 17 24. 9 feet 25. Length 18 inches, width 10 inches 18. e1,

17. e 0,

1 f 6

Index

A Abscissa, 341 Absolute value of a number, 13 Addition: of algebraic fractions, 296 of decimals, 58 of fractions, 46 of integers, 14 of polynomials, 193 of radicals, 439 of rational expressions, 296 of rational numbers, 46 Addition method of solving systems, 398 Addition property of equality, 77 Addition property of inequality, 115 Additive inverse, 12 Additive inverse property, 26, 58 Algebraic equation, 76 Algebraic expression, 2 Algebraic fraction, 286 Algebraic inequality, 113 Analytic geometry, 339 Approximation of square roots, 439 Associative property: of addition, 25, 57 of multiplication, 25, 57 Axes of a coordinate system, 340

B Base of a power, 6 Binomial, 193

C Cartesian coordinate system, 341 Circle, 157 Circumference, 157 Common binomial factor, 244 Commutative property: of addition, 25, 57 of multiplication, 25, 57 Complementary angles, 100 Completing the square, 494 Complex fraction, 306 Composite number, 34 Compound inequality, 123 Cone, right circular, 158 Conjugates, 460 Consecutive number problems, 109

Consistent system of equations, 394 Contradiction, 107 Coordinate geometry, 339 Coordinates of a point, 57, 341 Cross products, 140 Cube root, 438 Cylinder, right circular, 158

D Decimals: basic operations, 58 nonrepeating, 56 repeating, 56 terminating, 56 Declaring the variable, 92 Degree: of a monomial, 192 of a polynomial, 192 Denominator: common, 46 least common, 47 rationalizing a, 453 Dependent equations, 394 Descartes, René, 339 Difference of two cubes, 256 Difference of two squares, 251 Discount, 150 Distributive property, 26, 58 Divisibility, rules of, 35, 36 Division: of algebraic fractions, 293 of decimals, 60 definition of, 33 of fractions, 45 of integers, 18 of monomials, 215 of polynomials, 219 of radicals, 451 of rational expressions, 293 of rational numbers, 45 Division property of equality, 79 Division property of inequality, 117 Domain of a relation, 86

E Element of a set, 3 Elimination-by-addition method, 398 Empty set, 3 I-1

I-2

Index

English system of measure, 86 Equal sets, 3 Equations: dependent, 394 equivalent, 76 first-degree in one variable, 76 first-degree in two variables, 342, 350 linear, 76, 342, 350 quadratic, 191, 484 radical, 464 rational, 313 root of, 76 second-degree in one variable, 191, 484 solution of, 76 Equivalent equations, 76 Equivalent fractions, 47 Equivalent inequalities, 115 Evaluating algebraic expressions, 7 Exponents: definition of, 6 integers as, 224 negative, 225 positive integers as, 6, 199 properties of, 200, 201, 202, 224, 225, 226 zero as an, 225 Extraneous solution, 465

F Factoring: common factor, 243 complete, 244, 251 composite numbers, 35 difference of cubes, 256 difference of squares, 251 by grouping, 244 sum of cubes, 256 trinomials, 256 Factors, 34 First-degree equations, 76 FOIL, 209 Formulas, 156 Fractional equation, 76, 313 Fractions: addition of, 46 algebraic, 286 complex, 306 division of, 45 equivalent, 47 fundamental principle of, 43, 286 least common denominator of, 47 lowest terms of, 43 multiplication of, 42, 291 reduced form of, 43 subtraction of, 46 Fundamental principle of fractions, 43, 286

G Geometric formulas, 157 Graphing method of solving systems, 393 Graph of an equation, 342 Graph of an inequality, 115, 360 Graph of a solution set, 114 Greatest common factor, 36, 242 Greatest common monomial factor, 243

H Half-plane, 360 Hypotenuse, 262, 488

I Identity, 107 Identity property: for addition, 25, 57 for multiplication, 25, 58 Inconsistent equations, 394 Index of a radical, 438 Inequalities: equivalent, 115 graphs of, 115, 360 open sentences as, 113 properties of, 115, 117 solutions of, 113 systems of, 395 Integers, 12 Intercepts, 351 Interest, simple, 152 Intersection of sets, 123 Irrational number, 56 Isosceles right triangle, 490

L Least common denominator, 47 Least common multiple, 37 Like terms, 28 Linear equations: definition of, 342, 350 graph of, 342 slope-intercept form of, 378 systems of, 391, 393 Linear inequalities of two variables, 360 Literal factors, 28 Lowest terms, 43

M Metric system of measure, 86 Monomial, 192 Multiplication: of algebraic fractions, 291 of decimals, 59

Index

of fractions, 42 of integers, 17 of monomials, 199 of polynomials, 206 of radicals, 458 of rational expressions, 291 of rational numbers, 42 Multiplication property of equality, 79 Multiplication property of inequality, 117 Multiplication property of negative one, 26, 58 Multiplication property of zero, 26, 58 Multiplicative inverse property, 58

N Null set, 3, 76 Number line, 12, 57, 340 Numbers: absolute value, 13 composite, 34 integers, 12 irrational, 56 negative, 12 positive, 12 prime, 34 rational, 41 real, 56 whole, 3 Numerical coefficient, 28 Numerical expression, 1, 76, 113 Numerical statement, 76

O Open sentence, 113 Operations, order of, 6 Opposite of an integer, 12 Ordered pairs, 341 Ordered triples, 417 Ordinate, 341 Origin, 340

P Parabola, 348 Parallelogram, 157 Percent, 143 Perfect squares, 437 Perfect square trinomial, 270, 494 Point-slope form, 377 Polynomials: addition of, 193 completely factored form of, 251 definition of, 192 degree of, 192 division of, 219 equations, 191

multiplication of, 206 subtraction of, 194 Power of a base, 6 Prime factor, 34 Prime number, 34 Principal cube root, 438 Principal square root, 437 Prism, 157 Problem-solving suggestions, 98, 168, 174, 509 Profit, 151 Properties of equality, 77 Proportion, 140 Pyramid, 158 Pythagorean theorem, 262, 488

Q Quadrant, 340 Quadratic equations, 191, 484 Quadratic formula, 500

R Radicals: addition of, 439 definition of, 436 division of, 451, 460 equation, 464 multiplication of, 458 simplest form of, 444, 452 subtraction of, 439 Radical sign, 436 Radicand, 436 Ratio, 140 Rational approximation, 439 Rational equation, 313 Rational expressions: addition of, 296 definition of, 286 division of, 293 multiplication of, 291 simplifying, 286 subtraction of, 296 Rationalizing the denominator, 453 Rational number, 41 Real number, 56 Real number line, 12, 57 Reciprocal, 42, 58, 293, 321 Rectangle, 157 Rectangular coordinate system, 341 Rectangular prism, 158 Reflexive property of equality, 77 Right triangle, 262 Right triangle 30 – 60 , 490 Root of an equation, 76 Root of a number, 438

I-3

I-4

Index

S Scientific notation, 228 Second-degree equation in one variable, 484 Set, 2 Set-builder notation, 114 Similar terms, 28 Simplest radical form, 444, 452 Simplifying algebraic fractions, 286 Simplifying numerical expressions, 6 Simplifying radicals, 444, 452 Simplifying rational expressions, 286 Slope, 365 Slope-intercept form, 378 Solution of an equation, 76 Solution of an inequality, 113 Solution set, 76, 113 Special product patterns, 210 Sphere, 157 Square root, 436 Square-root property, 486 Standard form of a quadratic equation, 484 Statement of inequality, 113 Substitution method of solving systems, 406 Substitution property of equality, 77 Subtraction: of algebraic fractions, 296 of decimals, 58 of fractions, 46 of integers, 15 of polynomials, 194 of radicals, 439 of rational expressions, 296 of rational numbers, 46 Subtraction property of equality, 77 Subtraction property of inequality, 115 Suggestions for solving word problems, 98, 168, 174, 509 Sum of two cubes, 256 Supplementary angles, 100 Symmetric property of equality, 77

Systems of linear equations, 391, 393 Systems of linear inequalities, 395 Systems of three equations, 417

T Term, 28 Transitive property of equality, 77 Translating English phrases, 82 Trapezoid, 157 Triangle, 157 Trinomial, 193

U Union of sets, 124

V Value, absolute, 13 Variable, 1

W Whole numbers, 3

X x intercept, 351

Y y intercept, 351

Z Zero: addition property of, 14 as an exponent, 225 multiplication property of, 26, 58

Some Approximate Comparisons Between English and Metric Systems

(the symbol ⬇ means “is approximately equal to”) Length

1 inch ⬇ 2.5 centimeters 1 foot ⬇ 30.5 centimeters 1 yard ⬇ 0.9 of a meter 1 mile ⬇ 1.6 kilometers

1 centimeter ⬇ 0.4 of an inch 1 centimeter ⬇ 0.03 of a foot 1 meter ⬇ 1.1 yards 1 kilometer ⬇ 0.6 of a mile

Volume

1 pint ⬇ 0.5 of a liter 1 quart ⬇ 0.9 of a liter 1 gallon ⬇ 3.8 liters

1 liter ⬇ 2.1 pints 1 liter ⬇ 1.1 quarts 1 liter ⬇ 0.3 of a gallon

Mass

1 ounce ⬇ 28.35 grams 1 pound ⬇ 453.6 grams 1 pound ⬇ 0.45 of a kilogram

Simple Interest

I Prt I P A r t

A P Prt simple interest principal amount rate of interest time (in years)

1 gram ⬇ 0.04 of an ounce 1 gram ⬇ 0.002 of a pound 1 kilogram ⬇ 2.2 pounds

Temperature

5 1F 322 9 9 F C 32 5 F Fahrenheit degrees C Celsius degrees C

Distance-Rate-Time

r

d rt d r t

d t

distance rate time

t

d r

Subsets of the Real Numbers

51, 2, 3, p 6 50, 1, 2, 3, p 6 5 p , 3, 2, 1, 0, 1, 2, 3, p 6 a e ` a and b are integers and b 0 f b 5x 冟x is real number and x is not rational6

Natural numbers Whole numbers Integers Rational numbers Irrational numbers Properties of Real Numbers

For real numbers a, b, and c, the following properties are true. Addition

Multiplication

Commutative Property Associative Property

abba 1a b2 c a 1b c2

Inverse Property

a 1a2 0

ab ba 1ab2c a1bc2 1 aa b 1 a a # 1a

a0a

Identity Property Distributive Property

a1b c2 ab ac

Inequalities Statement

x a

Set Notation

Graph

5x 冟x a6

x a

5x 冟x a6 5x 冟x a6

0

xa

5x 冟x a6

0

xa a x b

5x 冟 a x b6

0

axb

5x 冟 a x b6

0

1ab2 n a nbn

2ab 2a 2b

a n an a b n b b

a 2a n Bb 2b

n

n

n

1q, a2

a

b

3 a, q 2

a

1q, a 4

]

bn bnm bm n

b

[

0

1bn 2 m bn m

a

)

Properties of Exponents and Radicals

# bm bnm

1a, q 2

( 0

bn

Interval Notation

n

a

(

)

a

b

[

]

a

b

Multiplication Patterns

1a, b2 3 a, b4 Factoring Patterns

1a b2 2 a2 2ab b2

a2 b2 1a b21a b2

1a b2 2 a2 2ab b2

a3 b3 1a b21a2 ab b2 2

1a b21a b2 a2 b2

a3 b3 1a b21a2 ab b2 2

1a b2 3 a3 3a2b 3ab2 b3 1a b2 3 a3 3a2b 3ab2 b3

Beginning Algebra (Available 2010 Titles Enhanced Web Assign)

Intermediate Algebra (Available 2010 Titles Enhanced Web Assign)

Prealgebra , Fourth Edition (Available 2011 Titles Enhanced Web Assign)

Single Variable Calculus (Available 2011 Titles Enhanced Web Assign)

Trigonometry, (8th Edition) (Available 2011 Titles Enhanced Web Assign)

Finite Mathematics and Applied Calculus, (5th Edition) (Available 2011 Titles Enhanced Web Assign)

Intermediate Algebra: An Applied Approach (Available Titles Coursemate) , Eighth Edition