Algebra (Curriculum Binders (Reproducibles))

Math Binder 2 Algebra By Michael Buckley Three Watson Irvine, CA 92618-2767 Web site: www.sdlback.com Development and...

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Math Binder 2

Algebra By Michael Buckley

Three Watson Irvine, CA 92618-2767 Web site: www.sdlback.com Development and Production: Frishco Ltd. and Pearl Production

ISBN 1-59905-023-4 Copyright © 2006 by Saddleback Educational Publishing. All rights reserved. No part of this book may be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without the written permission of the publisher, with the exception below. Pages labeled with the statement Saddleback Educational Publishing ©2006 are intended for reproduction. Saddleback Educational Publishing grants to individual purchasers of this book the right to make sufficient copies of reproducible pages for use by all students of a single teacher. This permission is limited to a single teacher, and does not apply to entire schools or school systems. Printed in the United States of America 10 09 08 07 06 05 9 8 7 6 5 4 3 2 1

Table of Contents Classifying Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Order of Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Writing a Variable Expression – Addition & Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Writing a Variable Expression – Subtraction & Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Evaluating Variable Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Simplifying Variable Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Adding Integers Using Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Subtracting Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Multiplying Integers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Dividing Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Distributive Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Exponents . . . . . 12 Negative Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Scientific Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Fractional Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Writing an Equation from a Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Writing an Equation from a Word Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Solving One-Step Equations by Adding or Subtracting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Solving One-Step Equations by Multiplying or Dividing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Solving Two-Step Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 Solving Multi-Step Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Solving Equations with Variables on Both Sides. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Identifying a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Writing a Function Rule from a Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Types of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Plotting Points on a Coordinate Plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Finding Solutions of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 Graphing a Linear Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 Direct Variation 29 Inverse Variation 30 Slope of a Line . 31 Slope Intercept Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Point Slope Form I. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Point Slope Form II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Parallel Lines. . . 35 Perpendicular Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 Using Reciprocals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 Solving Equations that Contain Decimals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Solving Equations that Contain Fractions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 Graphing Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 Writing Inequalities from a Graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 Solving One-Step Inequalities by Adding or Subtracting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 Solving One-Step Inequalities by Multiplying or Dividing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 Solving Two-Step Inequalities. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 The Pythagorean Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 Irrational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 Simplifying Radical Expressions by Multiplying Two Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . 47 Simplifying Radical Expressions by Removing Perfect Squares . . . . . . . . . . . . . . . . . . . . . . . . . 48 Simplifying Radical Expressions with Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Solving a Radical Equation by Isolating the Radical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 Estimating Square Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 Estimating Cube Roots and Higher Power Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 Multiplying a Polynomial by a Monomial. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 Multiplying Binomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 Squaring a Binomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 Adding Polynomials. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 Subtracting Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 Multiplying a Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 Factoring a Binomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 Finding the Greatest Common Factor for Variable Terms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 Factoring a Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 Factoring Trinomials in the Form x2 + bx +c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 Factoring Trinomials in the Form ax2 + bx +c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 The Difference of Two Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 Solving Systems of Equations by Graphing. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 Solving Systems of Equations by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 Solving Systems of Equations by Elimination. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 Solving Linear Systems by Multiplying . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 Solving Quadratic Equations Using Square Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 The Quadratic Formula. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 Using the Discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 Zero-Product Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 Solving a Quadratic Equation by Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Solving a Quadratic Equation by Completing the Square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 Evaluating Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 Exponential Growth Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 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Exponential Decay Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 Raising a Power to a Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 Raising a Product to a Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 Raising a Quotient to a Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 Dividing Powers with the Same Base . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 Simplifying Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 Multiplying Rational Expressions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 Dividing Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 Finding the LCD of a Rational Expression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 Adding Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 Subtracting Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 Finding Trigonometric Ratios. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 Using Trigonometric Ratios to Find a Missing Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 Theoretical Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 Experimental Probability. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 Mean . . . . . . . . . 92 Median . . . . . . . 93 Permutations . . 94 Combinations . . 95 Matrices. . . . . . . 96 Matrix Addition 97 Matrix Subtraction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 Scalar Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 Matrix Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

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Classifying Numbers Numbers can be classified into two categories. One category is rational numbers. The other category is irrational numbers. Rational Numbers include: 1. Natural numbers–numbers used for counting (e.g. 1, 2, 3, 4, etc.) 2. Whole numbers–natural numbers and 0 (e.g. 0, 1, 2, 3, 4, etc.) 3. Integers–whole numbers and their opposites (e.g. -11, -2, 0, 2, 11) 4. All integers plus fractions that result in a decimal or repeating decimal (e.g. _12 , −_13 ) Irrational Numbers include: 1. Numbers that cannot be expressed as fractions where the numerator and

denominator are both fractions.

Example

__

1 1 Classify the following numbers: −10, √5 , __ , 0, 1 __ , 6, 0.23 3 2 __ Step 1 Which numbers are irrational? √5

Step 2 Which numbers are rational?

−10, _13 , 0, 1 _12, 6, 0.23

Step 3 Which rational numbers are integers?

−10, 0, 6

Step 4 Which rational numbers are whole numbers?

0, 6

Step 5 Which rational numbers are natural numbers? 6

Practice Classify each number. Include as many categories as is appropriate. __

___

1. −3, −√2 , −_34, 5 _23, 0, √11 , 2, 5, 1.35

___

Which numbers are irrational?

√11 ,

Which numbers are rational?

−3, −_34 , 2, 5,

Which rational numbers are integers?

−3,

,

,

Which rational numbers are whole numbers?

,

Which rational numbers are natural numbers?

,5

__

,

, ,5 ___

1 Use the following list for items 2−6: −4.2, −3, √3 , −__ , −1, 0, 0.34, 1.5, 4, —√10 . 3

2. Which numbers are irrational numbers? 3. Which numbers are rational numbers? 4. Which numbers are integers? 5. Which numbers are whole numbers? Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Order of Operations Suppose you were given the following expression: 3 × 2 + 4 × 5 = ? Is the answer 50 or is it 26? To solve an expression with several operations, you need to perform your calculations in a certain order. The order of operations lists the sequence of operations in an expression. 1. Parentheses: simplify any operations in parentheses. 2. Exponents: simplify any terms with exponents. 3. Multiply and Divide: do all multiplication and division from left to right. 4. Addition and Subtraction: do all addition and subtraction from left to right.

Example Simplify. (9 − 3) + 42 × 5 Step 1 Parentheses

(9 − 3) + 42 × 5 = 6 + 42 × 5

Step 2 Exponents

6 + 42 × 5 = 6 + 16 × 5

Step 3 Multiplication and division

6 + 16 × 5 = 6 + 80

Step 4 Addition and subtraction

6 + 80 = 86

Practice Simplify each expression.

1. (5 × 3) × 4 + 24 − 3 Parentheses

(5 × 3) × 4 + 24 − 3 =

Exponents

15 × 4 + 24 − 3 =

Multiplication and division

15 × 4 + 16 − 3 =

Addition and subtraction

60 + 16 − 3 = 76 − 3 =

2. 6 − 22 × 4 + (2 − 2) 3. 3 (5 − 2) + 2 − 42 4. 24 − 6 + 32 + 4 5. 3 × 4 − (−22) + (6 − 4) 6. 14 + 42 × 6 −(22 + 2) 7. 36 − 22 + 4 (3 + 7)

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Writing a Variable Expression–Addition & Multiplication There are times when you are given a math situation in words and must create an expression that models the situation. When you need to create an expression, look carefully at the words. Certain words will tell you which operation to use. Key Math Operation Words Addition

Multiplication

sum

product

more than

times

plus

multiplied

increases

twice (times 2)

Rules for Writing an Expression 1. Write out the written phrase. 2. Replace each word with a variable, number, or operation symbol.

Example Use the following written phrase to create an expression: four times a number plus five. Step 1 Write out the written phrase. Step 2 Replace each word with a variable,

number or operational symbol.

Four times number plus five 4 × 4c + 5

c

+

5

OR

Practice Use the written phrase to create an expression.

1. Five times a number minus fifteen. Write out the written phrase.

Five times a number minus fifteen.

Replace each word with a variable, number or operational symbol. 2. Twelve more than twice a number.

6. Ten plus the product of fifty and a number.

3. Five times a number times six.

7. The product of a number and seven, plus the product of six and another number.

4. The sum of nine plus a number plus six.

8. The sum of twelve and a number plus six.

5. Eighteen more than the product of a number and twenty. Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Writing a Variable Expression–Subtraction & Division There are times when you are given a math situation in words and you must create an expression that models the situation. When you need to create an expression, look carefully at the words. Certain words will tell you which operation to use. Key Math Operation Words Subtraction

Division

difference

quotient

less than

divided by

minus

divided equally

It is important to note that when writing an expression you must follow the order of the numbers and the variables. The order is important–numbers or variables in the wrong order will yield an incorrect result. Rules for Writing an Expression 1. Write out the written phrase. 2. Replace each word with a variable, number, or operation symbol.

Example Use the following written phrase to create an expression: two less than a number divided by two. Step 1 Write out the written phrase.

Number divided by two less two.

Step 2 Replace each word with a variable,

x

÷

2

−

2

number or operational symbol.

Practice Use the written phrase to create an expression.

1. The difference of thirteen and twelve divided by a number. Write out the written phrase.

Thirteen minus twelve divided by a number.

Replace each word with a variable, number or operational symbol. 2. The difference of nine and a number divided by 5. 3. Five minus a number divided by 2. 4. A number minus sixteen divided by 4. 5. Eighteen minus a number minus six. 6. Fourteen divided by a number minus 3.

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Evaluating Variable Expressions As you know, a variable is a letter used to represent one or more numbers. When you know the number or numbers that represent a variable, you can put the number in the expression in the place of the variable. This process is known as substitution. When you replace the variable with a number and you perform all the operations in the expression, you evaluate the expression. Rules for Evaluating an Expression 1. Write the expression. 2. Replace the variable or variables with the number or numbers given. 3. Perform any operation to end up with one number.

Example Evaluate. 4x – 2 ÷ y, for x = 3 and y = 2 Step 1 Write the expression.

4x – 2 ÷ y

Step 2 Replace the variable with the number

Replace x with 3 and y with 2: 4 (3) − 2 ÷ 2

given.

Step 3 Perform any operations to end up with Multiply, divide then subtract:

a single number.

4 (3) − 2 ÷ 2 = 12 −2 ÷ 2 = 12 − 1 = 11

Practice Evaluate each expression when x = 5 and y = 3.

1. 3x − 2y + 4 Write the expression.

3x − 2y + 4

Replace the variable with the number given.

Replace x with

and y with

:

Perform any operations to end up with a single number. 2. x + 5y − 4 3. x2 + 11 − y 4. 12 + xy ÷ (9 − 6) Evaluate the expression when x = 2 and y = 4.

5. 4y + 23 − x 6. (x + y)2 + 9 7. 4(x + y) − y Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Simplifying Variable Expressions Once you have identified like terms, you can combine the like terms by using addition or subtraction. When you combine like terms you add the numbers that come before the variables, but you do not change the variable or any exponents. Rules for Combining Like Terms 1. Identify the like terms, including the sign before each term. 2. Combine (add or subtract) the number in front of the variable

of the terms that are like terms.

Example Simplify: 3x + 2 + 2x + 1 + 3x2 Step 1 Identify the like terms, including the

sign before each term.

3x, 2x 2, 1 3x2

Step 2 Combine the number in front of the

3x + 2x = 5x variable of the terms that are like terms. 2 + 1 = 3 If there is a variable without a number, 3x2 = 3x2 a “1” is assumed to be in front of the 3x2+ 5x + 3 number.

Practice Simplify

1. 2x + 4 + 3x2 + 7 − 2x2 + 6x Identify the like terms, including the sign before each term.

2x, , −2x2 4,

Combine the number in front of the variable of the terms that are like terms. If there is a variable without a number, a “1” is assumed to be in front of the number.

2x +

= – 2x2 =

4+

=

2. 2x + 6xy + y + 2xy 3. 4x + 4y2 − x 4. 8x − 5 + 14x + 12 5. −21x2 + 6x3 + 4x2 − 16 6. 4x2y − 3xy2 − 2x2y + 2x − 3 Algebra

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Adding Integers Using Absolute Value When you add integers using absolute value, you can use the following rules. Two numbers

Add or subtract absolute value

Sign

Example

postive + positive

add

positive

17 + 4 = 21

negative + negative

add

negative

—17 + (—4) = —21

postive + negative or negative + positive

subtract

sign of number with the larger absolute value

17 + (—4) = 13 —17 + 4 = —13

Example Add.

a. −21 + (−6)

Step 1 Find the absolute value of each number | −21 | = 21

|−6 | = 6

b. −21 + 6

| −21 | = 21 |6|=6

Step 2 What are the signs?

Negative and negative Negative and positive

Step 3 Do you add or subtract the absolute

Add

Subtract

Step 4 Solve

21 + 6 = 27

21 − 6 = 15

Step 5 What is the sign?

Adding two negatives 21 has a larger = a negative absolute value = −21 + (−6) = −27 negative −21 + 6 = −15

value?

Practice Add.

Find the absolute value of each number

1. −27 + 19

2. 45 + (−55)

| −27 | =

| 45 | =

| 19 | =

| −55 | =

27 − 19 =

55 − 45 =

−27 + 19 =

45 − 55 =

Do you add or subtract the absolute value? Solve What is the sign?

3. 12 + (− 8)

6. −36 + 14

4. 42 + 16

7. 17 + 17

5. −24 + (−10)

8. −61 + 21


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Subtracting Integers Consider the following examples of adding and subtracting integers. Adding

Subtracting

−8 + (−8) = −16

−8 − 8 = −16

10 + (−12) = -2

10 − 12 = −2

In the first instance, subtracting 8 is like adding −8; so, too, subtracting 12 is like adding −12. Rule for Subtracting Integers

Change the second number to its opposite, change the minus sign to a plus sign, and add using the rules for adding integers.

Example Subtract

a. −19 − (−4)

Step 1 Change the second number to its

−4 becomes 4

opposite Step 2 Change the minus sign to a plus

−19 + 4

Step 3 Find the absolute value of each number. |−19 | = 19

|4|=4 Step 4 Do you add or subtract?

Negative plus a positive = subtract

Step 5 Solve

19 has a larger absolute value, so the answer is negative, 19 − 4 = 15 −19 + (−4) = −15

Practice For each of the following subtraction sentences determine the sign of the answer.

1. −16 − (−21)

2. −13 − 27

3. 15 − (−36)

Subtract

4. −9 − (−17) Change the second number to its opposite

−17 becomes

Change the minus sign to a plus

−9

Find the absolute value of each number.

| −9 | =

Do you add or subtract?

A negative plus a positive,

17 | 17 | =

17 is larger, so the answer is 17 − 9 =

Solve

; −9 − (−17) =

5. −8 − 13

7. −9 − 5

9. 6 − (−12)

6. −6 − (−2)

8. 12 − 4

10. −21 − (−21)

; . .

Algebra

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Multiplying Integers Use the examples in the table below to discover the rules for multiplying integers. Example

Signs of Numbers

7 × 5 = 35

positive times

−7 × (−5) = 35

negative times

Signs of Answer

−7 × 5 = −35

times positive

7 × (−5) = −35

times

Example Multiply.

a. 8 × (−6)

b. −4 × (−12)

Step 1 What are the signs of the two numbers? Positive and Negative Negative and Negative Step 2 What is the sign of the answer?

Negative

Positive

Step 3 Multiply the absolute value of the two

8 × 6 = 48

4 × 12 = 48

8 × (−6) = −48

−4 × (−12) = 48

numbers. Step 4 Solve.

Practice For each of the following multiplication sentences determine the sign of the answer.

1. 4 × (−8)

3. −12 × 6

2. −9 × (−6)

4. 13 × 3

Multiply.

5. −7 × 6 What are the signs of the two numbers?

and

What is the sign of the answer? Multiply the absolute value of the two numbers.

7×6=

Solve.

−7 × 6 =

6. 10 × (−4) =

10. −5 × 12 =

7. −12 × (−8) =

11. 17 × (−2) =

8. 8 × (−9) =

12. 16 × 4 =

9. −7 × 7 =

13. −11 × (−3) =


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Dividing Integers Use the examples in the table below to discover the rules for dividing integers. Example

Signs of Numbers

30 ÷ 5 = 6

positive divided by

30 ÷ (−5) = -6

positive divided by

−30 ÷ 5 = -6

negative divided by

−30 ÷ (−5) = 6

negative divided by

Signs of Answer

Example Divide.

a. 45 ÷ (−5)

b. −72 ÷ (−9)

Step 1 What are the signs of the two numbers? Positive and Negative Negative and Negative Step 2 What is the sign of the answer?

Negative

Positive

Step 3 Divide the absolute value of the two

45 ÷ 5 = 9

72 ÷ 9 = 8

45 ÷ (−5) = −9

−72 ÷ (−9) = 8

numbers. Step 4 Solve.

Practice For each of the following division sentences, determine the sign of the answer.

1. 24 ÷ (−6)

3. 60 ÷ 4

2. −32 ÷ (−4)

4. −64 ÷ 8

Divide.

5. −54 ÷ 9 What are the signs of the two numbers?

and

What is the sign of the answer? Divide the absolute value of the two numbers.

54 ÷ 9 =

Solve.

−54 ÷ 9 =

6. −100 ÷ (−5) =

10. 27 ÷ 3 =

7. 88 ÷ (−8) =

11. 144 ÷ (−12) =

8. 64 ÷ (−4) =

12. −55 ÷ (−5) =

9. −63 ÷ (−9) =

13. −96 ÷ 6 =

Algebra

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Distributive Property Suppose you have the expression 2 × (4 + 5) [or 2(4 + 5)]. In this expression you are using two operations, multiplication and addition. You can rewrite the expression using the Distributive Property. When you use the Distributive Property, you distribute the number outside the parentheses to each number inside the parentheses. Rules for the Distributive Property 1. Multiply the number outside the parentheses by each number inside

the parentheses. 2. Place the operation symbol inside the parentheses between the two

multiplication expressions. 3. Simplify using order of operations.

Example Simplify using the Distributive Property: 4(5x + 2) Step 1 Multiply the number outside the

4 (5x + 2) = (4 • 5x) __ (4 • 2)

parentheses by each number inside the parentheses. Step 2 Place the operation symbol inside

(4 • 5x) __ (4 • 2) = (4 • 5x) + (4 • 2)

the parentheses between the two multiplication expressions. Step 3 Simplify using the order of operations.

(4 • 5x) + (4 • 2) = 20x + 8

Practice Use the Distributive Property to simplify each expression.

1. 5(x − 4) Multiply the number outside the parentheses by each number inside the parentheses.

5(x − 4) = (5 • x) __

Place the operation symbol inside the parentheses between the two multiplication expressions.

(5 • x) __ (5 • x)

Simplify using the order of operations.

(5 • x)

=

=

2. 2(3 − 4x) 3. −3(4x2 − 2x) 4. 7(−x + 5) 5. −4(−x2 − 9) 6. 6(x + 6) 7. −3(−2x2 − 4x) Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Exponents You can show the repeated multiplication of the same number using exponents. In an expression such as 43, the “4” is known as the base, and the “3” is the exponent. Rules for Working with Exponents

To solve an expression with an exponent: Multiply the base by itself the number of times equal to the exponent. To write an expression using an exponent: Count the number of times a number is multiplied by itself; that amount is your exponent. The number being multiplied is the base.

Example Solve. 64. Then write 7 × 7 × 7 × 7 × 7 using an exponent. Step 1 Multiply the base by itself the number

of times equal to the exponent. Step 2 Count the number of times a number

is multiplied by itself, that amount is your exponent. Step 3 The number being multiplied is the

base.

The exponent is 4 so you multiply 6 by itself 4 times; 64 = 6 × 6 × 6 × 6 = 1296. 7 is multiplied by itself 5 times; the exponent is 5. 7 is being multiplied by itself, so 7 is the base. 7 × 7 × 7 × 7 × 7 = 75.

Practice 1. Solve the following expression, 43. Multiply the base by itself the number of times equal to the exponent.

4×

=

2. Write the expression. 2 × 2 × 2 × 2 × 2 × 2 using an exponent. Count the number of times a number is multiplied by itself, that amount is your exponent.

is multiplied by itself

times.

The number being multiplied is the base. Solve the following expressions.

Write the following expressions using an exponent.

3. 54 =

6. 1 × 1 × 1 × 1 × 1 =

4. 113 =

7. 12 × 12 =

5. 95 =

8. 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 =

Algebra

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Negative Exponents When you have a negative exponent, you treat the negative exponent as follows: 1 x−2 = __ 2 x

In other words, a power with a negative exponent is treated as a fraction with 1 in the numerator and the power with a positive exponent as the denominator. Rules for Working with Negative Exponents 1. Identify the base and the exponent. 2. Create a fraction with 1 as the numerator and the

power as the denominator with a positive exponent. 3. Evaluate the power by multiplying the base by itself the number of times equal to the exponent.

Example Simplify 5−3. Step 1 Identify the base and the exponent.

exponent

5−3 base

Step 2 Create a fraction with 1 as the

numerator and the power as the denominator with a positive exponent. Step 3 Evaluate the power by multiplying

the base by itself the number of times equal to the exponent.

1 5−3 = __ 3 5

1 _______ 1 1 __ = = ___ 125 53 5 × 5 × 5

Practice Simplify the following.

1. 2−5 Identify the base and the exponent.

2−5

Create a fraction with 1 as the numerator and the power as the denominator with a positive exponent.

2−5 = ____ 0))0

Evaluate the power by multiplying the base by itself the number of times equal to the exponent.

00 00 00 ___ _____________ ____ 000 = 000000000000000 = 0000

0)

2. 8−1

4. (−5)−3

6. a−2

3. 3−2

5. (−3)−4

7. 4x−3


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Scientific Notation A shorthand way to write a large number or small number is to use scientific notation. 3400 → 3.4 × 103

0.00923 → 9.23 × 10−3

As you can see, a number in scientific notation is made of a number between 1 and 10 multiplied by 10 raised to a power. Rules for using Scientific Notation 1. Move the decimal point to the left or right to get a number between 1 and 10. 2. Multiply that number by 10 with an exponent. 3. The exponent is equal to the number of places the decimal point moved. 4. The exponent is positive if the decimal point is moved to the left; negative if

moved to the right.

Example Write 376,700 in scientific notation. Step 1 Move the decimal point to the left or

right to get a number between 1 and 10. Step 2 Multiply the number by 10 with an exponent. Step 3 The exponent is equal to the number of places the decimal point moved. Step 4 The exponent is positive if the decimal point is moved to the left; negative if moved to the right.

376,700 (5 decimal places): 3.767 3.767 × 10? The decimal point moved 5 places: 3.767 × 105 The decimal point moved to the left: 3.767 × 105

Practice Write each number in scientific notation.

1. 0.0404 Move the decimal point to the left or right to get a number between 1 and 10. Multiply the number by 10 with an exponent. The exponent is equal to the number of places the decimal point moved. The exponent is positive if the decimal point is moved to the left; negative if moved to the right.

0.0404 (

decimal places): 4.04

4.04 × 10? 4.04 × 10 The decimal point moved to the right: 4.04 × 10

2. 1,243

5. 0.0042

3. 10,045

6. 0.00075

4. 1,423,000

7. 0.0000303 Algebra

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Fractional Exponents When an exponent is expressed as a fraction, the numerator tells you the power the number is raised to, and the denominator tells you the root you take. The order in which you perform these operations does not matter. raise to the x power

x _

ay

cube 5

3 _

52

then take the square root

take the y root of ax

Rules for Working with Fractional Exponents 1. Identify the power the number is raised to; identify

the root you will find. 2. Raise the number to the identified power. 3. Take the identified root of the answer from Rule #2.

Example 4 __

Solve. 7 3

raise to the 4th power

4 _

Step 1 Identify the power the number is raised 7 3

to; identify the root you will find. Step 2 Raise the number to the identified

power. Step 3 Take the identified root of the answer

from Rule #2.

take the cube root

Raise 7 to the 4th power: 74 = 7 x 7 x 7 x 7 = 2401 take the cube root of 2401 _____ 3 √2401 = 13.4

Practice Solve. 5 _

1. 12 2

raise to the

5 _

Identify the power the number is raised to; identify the root you will find.

12 2

Raise the number to the identified power.

Raise 12 to the

Take the identified root of the answer from Rule #2.

Take the

power

take the

12

power:

= of the answer from

Rule #2: 4 _

3 _

2. 8 4

4. –4 3 2 _

3. 10 5

3 _

5. 4 9


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Writing an Equation from a Table You can use the data in a table to create a linear equation. You can use the data to find the slope of the line. Once you know the slope, you can use one ordered pair and the point-slope form to create your equation. Rules for Writing an Equation from a Table 1. Choose two sets of ordered pairs. Find the slope. 2. Use one of the ordered pairs as the x and y

coordinates in the point-slope form of an equation. 3. Place the values into the point-slope form.

Example Write an equation to model the data in the following table. x

3

5

6

9

y

—2

2

4

10

Step 1 Choose two sets of ordered pairs. Find

the slope.

y2 − y1 ______ 2 − (−2) _ 4 _____ x2 – x1 = 5−3 = 2 = 2

The slope (m) = 2.

Step 2 Use one of the ordered pairs as the

x and y coordinates in the point-slope form of an equation. Step 3 Place the values into the point-slope

form.

(3, −2) → 3 = x-coordinate; −2 = y-coordinate y − y1 = m(x − x1) → y − (−2) = 2(x − 3)

Practice Write an equation to model the data in each table.

1.

2.

x

—5

—1

1

7

y

—9

—7

—6

—3

y −y

–7 – (–9)

Choose two sets of ordered pairs. Find the slope.

2 1 _______ slope = _____ x − x = –1 – (–5) =

Use one of the ordered pairs as the x and y coordinates in the point-slope form of an equation.

(−5, −9) →

Place the values into the point-slope form.

y − y1 = m(x − x1) → y −

x

4

3

0

—2

y

8

6

0

—4

2

1

= x-coordinate; = y-coordinate =

(x − 3.

x

—10

—4

—1

3

y

12

0

—6

—14

Algebra

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Writing an Equation from a Word Problem To translate a word problem into an equation, you will write phrases from the problem as mathematical expressions. Replace words that mean equals with an equal sign. Use other key words to determine operations within each expression. Rules for Writing an Equation from a Word Problem 1. Read the problem. Write the phrases from the problem as a word-based

mathematical sentence. 2. Identify unknown values—assign a variable to each unknown. 3. Replace each phrase with a mathematical expression. Connect the expression

with operation symbols and equal signs.

Example DVDs sell for $15 each. Write an equation for the total cost of a given number of DVDs. Step 1 Read the problem. Write the phrases Total cost = 15 times the number of DVDs.

from the problem as a word-based mathematical sentence.

Step 2 Identify unknown values—assign a

variable to each unknown. Step 3 Replace each phrase with a

mathematical expression. Connect the expressions with operation symbols and equal sign.

Total cost is variable c. The number of DVDs bought is variable n. Total cost is 15 times number of DVDs c = 15 × n c = 15 × n

Practice Write an equation for each situation.

1. DVDs bought on-line cost $12 each, plus a shipping fee of $5. Write an equation for the total cost of a given number of DVDs. Read the problem. Write the phrases from the problem as a word-based mathematical sentence. Identify unknown values—assign a variable to each unknown. Replace each phrase with a mathematical expression. Connect the expressions with operation symbols and equal signs.

is 12

the number of

DVDs plus

. is variable c. is variable is 12

. number of

DVDs + $5 c

12

+5

2. Bowling costs $5 per game, plus $3 rental fee for shoes. Write an equation for the total cost of bowling a certain number of games. 3. You are buying food for your dog and cat. Dog food costs $0.35 per can; cat food is $0.25 per can. Write an equation for the total cost of buying a certain number of cans of food. Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Solving One-Step Equations by Adding or Subtracting When given an algebraic equation, you are asked to get the variable by itself on one side of the equal sign. To do so you must “undo” any operations that are on the same side of the equal sign as the variable. When you have the variable by itself, you have solved the equation. Rules for Isolating a Variable using Addition or Subtraction 1. Identify the operation and number on the same side of the equal sign or the

variable. 2. Perform the opposite operation of that number on each side of the equation.

Example Solve. x + 24 = 36 Step 1 Identify the operation and number

on the same side of the equation as the variable.

“+ 24” is on the same side of the equation as the variable.

Step 2 Perform the opposite operation of that You will “−24” on each side.

number on each side of the equation.

x + 24 − 24 = 36 − 24 x + 0 = 12 or x = 12

Practice Solve.

1. 172 = x − 125 Identify the operation and number on the same side of the equation as the variable.

is on the same side of the equation as the variable.

Perform the opposite operation of that number on each side of the equation.

You will 172

to each side of the equation. = x − 125

297

=x−

or

297 = x 2. x − 24 = 72 3. 87 = x + 16 4. 144 = x − 63 5. x + 21 = 21 6. x − 3 = 104 7. 264 + x = 475

Algebra

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Solving One-Step Equations by Multiplying or Dividing When given an algebraic equation, you are asked to get the variable by itself on one side of the equal sign. To do so you must “undo” any operations that are on the same side of the equal sign as the variable. When you have the variable by itself, you have solved the equation. Rules for Isolating a Variable using Multiplication and Subtraction 1. Identify the operation and number on the same side of the equal sign or the

variable. 2. Perform the opposite operation of that number on each side of the equation.

Example Solve. 3x = 36 Step 1 Identify the operation and number

You are multiplying the variable 3 times.

on the same side of the equation as the variable. Step 2 Perform the opposite operation of that The opposite of multiplication is division;

number on each side of the equation.

so divide each side by 3. 3x ÷ 3 = 36 ÷ 3 x = 12 or x = 12

Practice Solve.

1. 24 = _8x Identify the operation and number on the same side of the equation as the variable.

The variable is

by 8.

Perform the opposite operation of that number on each side of the equation.

The opposite of is multiplication, so multiply each side by 8. 24 × 8 = _8x × = =x

2. 5x = 45 3. 16x = 32

5. _7x = 84

x 6. __ 10 = 240

4. 14x = 140 Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Solving Two-Step Equations When solving an equation, the goal is to get the variable by itself on one side of the equal sign. There are times when solving an equation requires two steps. The first step is usually either addition or subtraction. The second step is usually multiplication or division. Rules for Solving Two-Step Equations 1. Write the equation. 2. Isolate the term with the variable using addition or subtraction. 3. Isolate the variable using multiplication or division.

Example Solve. 4x − 16 = 24 Step 1 Write the equation.

4x − 16 = 24

Step 2 Isolate the term with the variable using 4x − 16 + 16 = 24 + 16

addition or subtraction.

4x − 0 4x = 40

=

40

4x ÷ 4 = 40 ÷ 4 x = 10

or

Step 3 Isolate the variable using multiplication 4x = 40

or division.

x = 10

Practice Solve.

1. 7x + 7 = 70 Write the equation.

7x + 7 = 70

Isolate the term with the variable using addition or subtraction.

7x + 7 7x +

= 70 0

=

7x = Isolate the variable using multiplication or division.

7x = 7x x

= 63 =

2. _4x + 10 = 17

5. 24 = 5x − 96

3. _7x − 3 = 21

6. _2x − 17 = 3

4. 6x − 11 = 37

7. 7x − 9 = 68

or

x=

Algebra

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Solving Multi-Step Equations Multi-step equations are the equations that require more than two operations to solve (to isolate the variable). As you solve a multi-step equation, keep in mind the goal is to get the variable on one side of the equation by using inverse operations. Rules for Solving Multi–Step Equations 1. Write the equation. 2. Combine like terms, if necessary. 3. Isolate the term with the variable. 4. Isolate the variable.

Example Solve. 2x + x + 12 = 78 Step 1 Write the equation.

2x + x + 12 = 78

Step 2 Combine like terms.

2x + x + 12 = 78 3x + 12 = 78

2x and x are like terms Step 3 Isolate the terms with the variable.

3x + 12 − 12 = 78 − 12 3x = 66

Step 4 Isolate the variable.

3x ÷ 3 = 66 ÷ 3 x = 22

Practice Solve.

1. 4x + 3x + 12 = 61 Write the equation.

4x + 3x + 12 = 61

Combine like terms.

+ 12 = 61

Isolate the terms with the variable.

+ 12

= 61

= ÷7=

Isolate the variable.

÷7

x= 2. 5x − 5 − 2x = 22

5. 3x + 6 + x = 90

3. 15 = 4x − 2x + 1

6. 72 + 4 − 14x = 34

4. 3x + 4 − (−4x) = 39

7. 26 = 10 − 26 + 3x


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Solving Equations with Variables on Both Sides To solve an equation with variables on both sides of the equal sign, you will need to add or subtract all the terms with the variable to get all the variables on one side of the equation. Rules for Solving an Equation with Variables on Both Sides 1. Use addition or subtraction to get all the variables on one side of the equation. 2. Use addition or subtraction to combine all numbers without a variable. 3. Use multiplication or division to solve for the variable.

Example Solve for x, 2x + 1 = 43 − 4x Step 1 Use addition or subtraction to get

all the variables on one side of the equation.

4x + 2x + 1 = 43 − 4x + 4x 6x + 1 = 43 − 0

Step 2 Use addition or subtraction to combine 6x + 1 = 43

all numbers without a variable. Step 3 Use multiplication or division to solve

for the variable.

6x + 1 − 1 = 43 − 1 6x = 42 6x = 42 6x ÷ 6 = 42 ÷ 6 x=7

Practice Solve each equation:

1. 6x − 4 = x + 16 Use addition or subtraction to get all the variables on one side of the equation.

6x – x − 4 = x − x + 16

Use addition or subtraction to combine all numbers without a variable.

5x − 4 = 16

− 4 = 0 + 16

5x − 4 + 4 = 16 + 4 5x + 0 =

Use multiplication or division to solve for the variable.

5x = 20 5x

= 20

x= 2. −4x − 4 = 3x + 10

5. 3 + 4x = 3x + 6

3. 8 − x = 3x − 16

6. 5x − 3 = 2x + 12

4. 2x − 5 = 4x + 7 Algebra

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Identifying a Function A function is defined as a relationship between two numbers: one number is the input and the other number is the output. In a function, for each input there is only one output. If an input has more than one output then the relationship is not a function. In a function the input is known as the domain. The output is known as the range. Rules for Identifying a Function 1. Identify the input values–this is the domain. 2. Identify the output values–this is the range. 3. Make sure that for each input there is only one output.

Example Decide whether the table below represents a function. x

—2

—1

0

1

2

y

—4

—2

0

2

4

Step 1 Identify the input values.

The top row contains the input values.

Step 2 Identify the output values.

The bottom row contains the output values.

Step 3 Make sure that each input value has

None of the input values repeat, so the table represents a function.

only one output.

Practice Decide whether each table represents a function.

1.

Hours Envelopes stuffed

3

4

4

5

6

6

18

16

17

22

25

28

Identify the input values.

The input values are

Identify the output values.

The output values are

Make sure that for each input there is only one output.

For the inputs 4 and there is more than one output. For input 4 the outputs are 16 and 17; for input and

2.

Dollars Tip

3.

10

20

30

40

50

1.50

3.00

4.50

6.00

7.50

x

4

4

9

9

16

y

2

—2

3

—3

4

4.

the outputs are 25

.

Month

1

2

3

4

5

Inches of Rain

3.5

4.1

5.4

3.5

1.25


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Writing a Function Rule from a Table When given a table of values, you can analyze the table to see if there is a pattern. If you uncover the pattern, you can write a rule for a function. Rules for Writing a Function Rule from a Table 1. Compare the value of x to the value of y in an ordered pair. Find out

what is done to get from x to y. 2. Check another ordered pair. Find out what is done to get from X to Y.

If it is not the same as in Step 1, then consider another pattern. 3. Write a function rule from the pattern.

Example Write a function rule for the table below. x

1

2

3

4

y

6

7

8

9

Step 1 Compare the value of x to the value of

y in the ordered pair. Find out what is done to get from x to y.

In the first ordered pair, (1, 6), you add 5 to get from the x–value (1) to 6.

Step 2 Check another ordered pair. Find out

In the second ordered pair, (2, 7) you add 5 what is done to get from x to y. If it is to get from the x–value (2) to 7. not the same as in Step 1, then consider another pattern.

Step 3 Write a function rule from the pattern.

y=x+5

Practice Write a function rule for each table.

1.

2.

x

2

4

6

8

y

6

12

18

24

Compare the value of x to the value of y in an ordered pair. Find out what is done to get from x to y.

In the first ordered pair, (2, 6), you

Check another ordered pair. Find out what is done to get from x to y. If it is not the same as Step 1, then consider another pattern.

In the second ordered pair, (4, 12) you

Write a function rule from the pattern.

y=

x

2

4

6

8

y

0

2

4

6

to get from the x-value (2) to 6. to get from the x-value (4) to 12.

Algebra

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Types of Functions As you know, a function is defined as a relationship between two numbers–an input number and an output number. There are many types of functions. A linear function is a function whose graph is a straight line. Another function, a quadratic function, is a function whose graph is a U–shaped line. By looking carefully at an equation you can determine the type of function. Types of Function

General Form

Description

Example

1. Linear

y = ax + b

Equation with variables but no exponents.

y = 2x + 2

2. Quadratic

y = ax2 + bx + c

Equation with a squared term.

y = x2 + 3x – 2

3. Exponential

y = ax

Equation with a variable as an exponent.

y = (2)x

4. Rational

y = __ax + b + c

Equation with a variable in the denominator.

5 y = — __ x +4+2

Example

a Classify the following function as linear, quadratic, exponential, or rational: __ +3=6 4 Step 1 Does any variable have a 2 as an No

exponent? Step 2 Is the variable an exponent?

No

Step 3 Is the variable in the denominator?

No

Step 4 Identify the function.

It is a linear function.

Practice Classify each function as linear, quadratic, exponential, or rational.

1. y = _1x − 1 Does the variable have a 2 as an exponent? Is the variable an exponent? Is the variable in the denominator? Identify the function. x2

It is a

3. y = 4x +2

5 1 _ 5. y = __ 6x × 2 6. y = 3(0.5)x

1 4. y = __ 2x + 3

7. y = 3x + 22 − 1

2. y = __ 2 + 4x − 3

function.


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Plotting Points on a Coordinate Plane A point on a coordinate plane is defined by its location on the x-axis and on the y-axis. An ordered pair gives the location of a point. Ordered Pair (x, y) (x-coordinate, y-coordinate)

Symbol of Coordinate

x

y

+

right

up

−

left

down

Rules for Plotting Points on a Coordinate Plane 1. Move across the x-axis the number of units of the x-coordinate. 2. Move up or down from the y-axis the number of units of the y-coordinate.

Example Graph the following point: (5, −3) Step 1 Move across the x-axis the number of

Move across the x-axis 5 units to the right.

units of the x-coordinate. Step 2 Move up or down from the x-axis the

Move down from the x-axis 3 units.

number of units of the y-coordinate.

Practice Graph the following points.

1. (−7, −4) Move across the x-axis Move across the x-axis the number of units units to the . of the x-coordinate. Move up or down from the x-axis the number of units of the y-coordinate.

Move x-axis

1.

from the units.

2. (6, 1) 3. (−7, 0)

C

4. (0, −5) Give the coordinates for each point.

A

5. A 6. B

D

B

7. C 8. D Algebra

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Finding Solutions of Linear Equations An equation with two variables whose solution on a coordinate plane is a straight line is a linear equation. In many instances, the equation may have more than one solution. While it is impossible to find every solution, you can find several solutions. One way to find the solution is by using an input/output table. Rules for Finding Solutions of a Linear Equation 1. Create an input/output table. 2. Select several values for x. 3. Substitute each value of x into the equation. 4. Solve the equation for y (the y-coordinate).

Example List four solutions of y = 4x + 3 Step 1 Create an input/output table. Step 2 List several values for x. Step 3 Substitute each value of x into

the equation. Step 4 Solve the equation for y.

x

y = 4x + 3

y

−1

y = 4 (—1) + 3

—1

0

y = 4 (0) + 3

3

1

y = 4 (1) + 3

7

2

y = 4 (2) + 3

11

x

y = 3x − 2

y

−2

y = 3 (—2) − 2

0

y = 3 (—0) − 2

The solutions for y = 4x + 3 are (−1, −1), (0, 3), (1, 7), and (2, 11).

Practice List four solutions for each equation.

1. y = 3x − 2 Create an input/output table. List several values for x. Substitute each value of x into the equation. Solve the equation for y. The solutions for y = 3x − 2 are

y = 3 (1) − 2 y = 3 (2) − 2

.

2. y = 5x − 4 3. y = _2x + 1 4. y = −x + 3 5. y = 4x – 2 6. y = _3x − 1 7. y = x + 1 Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Graphing a Linear Equation When you find the solution of an equation, you are finding two values,—one for x and one for y—that make the equation true. Each set of values is known as an ordered pair. You can use the ordered pairs to plot points on a coordinate plane. If the solution (ordered pairs) makes a line, then you have a linear equation. Rules for Graphing a Linear Equation 1. Create an input/output table. 2. Select several values for x. 3. Substitute the values for x into the equation and solve for y. 4. Plot each solution on the coordinate plane. Draw a line so

it goes through each point.

Example Graph the following equation y = x + 3 Step 1 Create an input/output table. x

y=x+3

y

(x, y)

−2

y = —2 + 3

1

(−2,1)

0

y=0+3

3

(0, 3)

1

y=1+3

4

(1, 4)

2

y=2+3

5

(2, 5)

x

y = 3x − 1

y

(x, y)

Create an input/output table.

2

y = 3(2) − 1

5

(2, 5)

List several values for x.

0

y = 3(0) − 1

—1

(0, −1)

y = 3(1) − 1

2

(1, 2)

y = 3(—2) − 1

—7

(—2, −7)

Step 2 List several values for x. Step 3 Substitute each value of x into the

equation. Solve the equation for y. Step 4 Plot each solution on a coordinate

plane. Draw a line so it goes through each point.

Practice Graph the following equations.

1. y = 3x − 1

Substitute each value of x into the equation. Solve the equation for y. Plot each solution on a coordinate plane. Draw a line so it goes through each point. 2. y = −x + 3 3. y = −2x − 1 4. y = 4x + 5 5. y = _3x + 1 6. y = 2x + 3

Algebra

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Direct Variation When two variables have a constant ratio, they show a direct variation. In a direct variation, when one variable increases, the other variable increases. Similarly, when one variable decreases, the other variable decreases. Rules for Direct Variation 1. Examine the data. When one variable changes does the other variable change

in the same direction? 2. Does the ratio of the two variables result in the same number? 3. Does the graph go through the origin (0,0)? 4. If the answer is “yes” to all questions, then the data shows a direct variation.

Example Tell whether the data shows a direct variation. Time (hours) Distance (miles)

1

2

3

4

40

80

120

160

Step 1 Examine the data. When one variable

Yes, as hours increase, the distance increases.

changes does the other variable change in the same direction? distance 80 Yes, the ratio of ______ is 40 (e.g. __ 2 = 40) time

Step 2 Does the ratio of the two variables

result in the same number? Step 3 Does the graph go through the origin?

Yes, a graph would go through the origin (0, 0)

Step 4 If the answer is yes to all questions,

Each question was answered “yes;” the data shows a direct variation.

then the data shows a direct variation.

Practice Tell whether the data shows a direct variation.

1.

2.

Time (hours)

0.5

2.0

3.5

5.0

Distance (miles)

30

40

55

80

Examine the data. When one variable changes does the other variable change in the same direction?

As the time increases, the distance

Does the ratio of the two variables result in the same number?

distance The ratio of ______ time number.

Does the graph go through the origin?

The graph

If the answer is yes to all three questions, then the data shows a direct variation.

The answer to all three questions

Drop Height (cm) Bounce (cm)

. the same go through the origin.

“yes.” The data variation.

10

20

30

40

50

9

18

27

36

45

show a direct


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Inverse Variation When you have a set of data in which one variable increases while the other decreases and the product of the variable is the same value, you have an inverse variation. Rules for an Inverse Variation 1. Examine the data. Does one variable increase while the other one decreases? 2. Is the product of xy a constant value? 3. If the answers to 1 and 2 are “yes,” the data shows an inverse variation.

Example Does the data in the table represent an inverse variation? x

2

4

6

8

y

12

6

4

3

Step 1 Examine the data. Does one variable

Yes, as the value for x increases, the value for increase while the other one decreases? y decreases.

Step 2 Is the product of xy a constant value?

Yes, the product of xy is the same value, 24. For example, 2 × 12 = 24; 4 × 6 = 24.

Step 3 If the answer to 1 and 2 is “yes,” the data The answer to both is yes. The data shows an

inverse variation.

shows an inverse variation.

Practice Does the data in the tables below represent an inverse variation?

1.

x

9

6

3

1

y

4

6

12

36

Examine the data. Does one variable increase while the other one decreases?

As the value for x decreases the value for y

Is the product of xy a constant value?

The product of xy

. a constant. For

example, 9 × 4 = If the answer to 1 and 2 is “yes,” the data shows an inverse variation.

;6×6=

.

The answer to question #1 is

, the

answer to question #2 is

, the data

show an inverse variation. 2.

3.

x

2

4

6

8

10

y

4

8

12

16

20

x

4

12

20

30

y

15

5

3

2

4.

x

18

12

9

3

y

4

6

8

24

Algebra

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Slope of a Line If you look at the graph of a linear equation, you will see it forms a straight line. You may have noticed that most lines have a “slant” to them. The slope of a line is a measure of the steepness of a line. The slope of a line is the ratio of the vertical change (the number of units of change along the y-axis) to horizontal change (the number of units of change along the x-axis). To find the slope of a line, you pick any two points on the line. You then subtract the y-coordinates and subtract the x-coordinates. Suppose a line passes through two points, for example (2, 3) and (4, 2). You make one set of coordinates (x1, y1) and the other set, (x2, y2). vertical change

y −y

2 1 slope = _____________ = _____ horizontal change x − x 2

1

Example Find the slope of a line that passes through (6, 4) and (2, 1). Step 1 Make one set of coordinates (x1, y1)

(x1, y1) (3, 5)

Step 2 Use the equation for slope, place the

2 1 1−4 ____ slope = _____ x −x = 2−6

and the other set, (x2, y2)

numbers into the formula. Step 3 Solve.

(x2, y2) (2, 1) y −y 2

1

3 1−4 _ slope = ____ 2 − 6= 4

The slope is _34 .

Practice Find the slope of the line passing through each set of points.

1. (3, 5) and (6, 8) Make one set of coordinates (x1, y1) and the other set, (x2, y2)

(x1, y1)

(x2, y2)

Use the equation for slope; place the numbers into the formula.

−5 2 1 ________ slope = _____ x −x = −3

Solve.

− 5 ______ slope = ________ − 3 = 0000 =

(3, 5) y −y 2

The slope is

1

.

2. (−3, 1) and (4, 7) 3. (5, 2) and (7, 4) 4. (−5, 2) and (4, −1) 5. (−3, −2) and (−2, −1) 6. (−3, 0) and (3, −4) 7. (0, 4) and (−8, −2) Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Slope Intercept Form Looking at an equation can tell you certain pieces of information about the graph of that equation. An equation written with y isolated on one side of the equal sign and x on the other side of the equation is in slope–intercept form. An equation in slope-intercept form is written as: y = mx + b

y-intercept

slope

The y-intercept is the point on the y-axis through which the line passes.

Example Find the slope and the y-intercept of the line y = 4x − 2 Step 1 Find the number in front of the x-term. y = mx + b

Be sure to include the negative sign if necessary. This is the slope. Step 2 Find the term without a variable.

This number is the y-coordinate of where the line crosses the y-axis. Be sure to include a negative if necessary.

y = 4x − 2 m = slope = 4 y = mx + b y = 4x − 2 b = y–intercept = −2

Practice Find the slope and y-intercept for each equation.

1. y = −3x + 7 Find the number in front of the x-term. Be sure to include the negative sign if necessary. This is the slope.

y = mx + b y = −3x + 7

Find the term without a variable. This number is the y-coordinate of where the line crosses the y-axis. Be sure to include a negative if necessary.

y = mx + b y = −3x + 7

m = slope =

b = y–intercept =

2. y = _13x − 3 3. 2y = 2x + 2 4. y = _34x − 13 5. 3y = −2x + 9 Algebra

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Point Slope Form I There are instances in which you are given the slope and an ordered pair. For example, you may know that the slope of a line is −2 and the graph of the equation passes through (−2, 1). You can use the point-slope form of a linear equation to write an equation of the line. y-coordinate

Point–slope form: y − y1 = m (x − x1) y-coordinate

slope

Rules for Using the Point–Slope Form 1. Identify the slope m. 2. From the ordered pair identify the x-coordinate and the y-coordinate. 3. Use the point–slope form to write the equation: y − y1 = m (x − x1)

Example Write the equation of the line that has a slope of 3 and passes through the point (2, 5). Step 1 Identify the slope.

The slope (m) is 3.

Step 2 From the ordered pair, identify the

The ordered pair is (2, 5) The x-coordinate is 2; the y-coordinate is 5.

x-coordinate and the y-coordinate. Step 3 Use the point–slope form to write

the equation.

y − y1 = m(x − x1) y − 5 = 3 (x − 2)

Practice Write the equation of the line.

1. Slope = 6, (−3, −1) Identify the slope (m).

The slope is

.

From the ordered pair, identify the x-coordinate and the y-coordinate.

The ordered pair is (−3, −1) The x-coordinate is the y-coordinate is

Use the point–slope form to write the equation.

; .

y − y1 = m(x − x1)

2. slope = −_12, (7, 1) 3. slope = 2, (−3, −3) 4. slope = _23, (4, −5) 5. slope = −3, (−1, 3)


33

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Point- Slope Form ll When you are given the slope of a line and an ordered pair identifying a point on the graph of the line, you can use the point–slope form. You can also use the point-slope form when given two sets of ordered pairs. To use the two ordered pairs, you first will need to use the ordered pairs to find the slope. Rules for Using Point-Slope Form Using Two Points y2 − y1 vertical change 1. Use the formula for slope (slope = _____________ = _____ x − x ) to find the slope. horizontal change

2

1

2. Use one set of ordered pairs for the x-coordinate and y–coordinate. 3. Use point-slope form to write the equation.

Example Write the equation of the line that passes through (−3, −3) and (1, 5). y −y

2 1 Step 1 Use the formula for slope (_____ x −x )

to find the slope.

2

1

Step 2 Use one set of ordered pairs for the

x-coordinate and the y-coordinate. Step 3 Use point–slope form to write the

equation.

y2 − y1 ______ 5 − (−3) _ 8 Slope = _____ x2 − x1 = 1 − (−3) = 4 = 2

Use the ordered pair (1, 5) The x-coordinate is 1; the y-coordinate is 5. y − y1 = m (x − x1) y − 5 = 2(x − 1)

Practice Use the point–slope form to write an equation.

1. (−2, −2), (0, −4) y −y

2 1 Use the formula for slope ( _____ x2 − x1 ) to find the slope.

Use one set of ordered pairs for the x-coordinate and the y-coordinate.

y2 − y1 _______ −4 − (−2) __ −2 Slope = _____ x2 − x1 = 0 − (−2) = 2 = −1

Use the ordered pair (−2, −2) The x-coordinate is the y-coordinate is

Use point–slope form to write the equation.

; .

y − y1 = m (x − x1)

2. (0, 1), (2, 2) 3. (−6, 4), (3, −5) 4. (2, 6), (0, 0) 5. (−1, −4), (5, 2) 6. (6, 0), (3, −2) Algebra

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Parallel Lines Parallel Lines are lines in the same plane that do not intersect. The equation of line A is y = 2x + 3 The equation of line B is y = 2x − 1 As you can see both lines have the same slope, but a different y–intercept. y = mx + b m = slope b = y-intercept y = 2x + 3 2 3 y = 2x − 1 2 −1

� �

Rules for Parallel Lines 1. Write all equations in slope-intercept form. 2. Identify the slope of each line. 3. If the slopes are equal the lines are parallel.

Example Are the graphs of y = x + 4 and 6y – 3x = 6 parallel? Step 1 Write all equations in slope-intercept

form. Step 2 Identify the slope of each line.

y = _12x + 4: is in slope-intercept form. 6y − 3x = 6 → y = 1 + x = 1 + x y = x + 4: slope = _12 y = 1 + x: slope = _12

Step 3 If the slopes are equal the lines are

The slopes are equal so the lines are parallel.

parallel.

Practice For each set of equations, determine if graphs of the equations are parallel.

1. y = 3x + 12 and 6y = −3x − 6 Write all equations in slope-intercept form.

y = 3x + 12: is in slope-intercept form. 6y − 3x = −6: is not in slope-intercept form. 6y − 3x = −6 → y =

Identify the slope of each line.

y = 3x + 12: m = 6y − 3x = − 6: m =

If the slopes are equal the lines are parallel.

The slopes

equal.

The lines

parallel.

2. y =–_14x + 5 and 12y + 3x = 24 3. 8x + 4y = 8 and Y = −2x + 4 4. y = 2x + 6 and −2x + 2y = 12 5. y = –_14x + 12 and 8x + 6y = 9 Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Perpendicular Lines Perpendicular lines are lines that intersect to form right angles. The equation of line A is y = 2x − 1. The equation of line B is y = –_12x + 4. As you can see the slope of one line is the opposite (negative) reciprocal of the other line. y = mx + b m = slope b = y-intercept y = 2x − 1 2 −1 1 1 y = –_2x + 4 −_2 4

�

�

Rules for Perpendicular Lines 1. Identify the slope of the known line. 2. Write the reciprocal of the slope. 3. Give the new slope a sign opposite to the slope of the first line. This is the slope

of the new line. 4. Use the slope-intercept form to create the equation of a line perpendicular to the given line.

Example Write an equation of the line that has a y–intercept of 2 and is perpendicular to y = 3x + 5 Step 1 Identify the slope of the known line.

y = 3x + 5; slope = m = 3

Step 2 Write the reciprocal of the slope. This

m = 3, the reciprocal is _13.

is the slope of the new line. Step 3 Give the new slope a sign opposite to

the slope of the first line. Step 4 Use the slope-intercept form to create

the equation of a line perpendicular to the given line.

The slope of the first line is positive; make the new slope negative: −_13 y = mx + b = −_13x + 2

Practice Write an equation of the line with the given y-intercept that is perpendicular to the given equation.

1. y = −_12x + 2; new y-intercept: −3 Identify the slope of the known line.

y = −_12x + 2; slope = m = −_12

Write the reciprocal of the slope. This is the slope of the new line.

m = −_12; the reciprocal of _12 is

Give the new slope a sign opposite to the slope of the first line.

The slope of the first line is

Use the slope-intercept form to create the equation of the line perpendicular to the given line.

line

:

y = mx + b =

x+

; make the slope of the new

2. y = _34x + 5; new y-intercept: 4 3. 2y = 4x + 2; new y-intercept: 3 36

Algebra


Name

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Using Reciprocals Two numbers are reciprocals if their product is 1. For example, the reciprocal of _34 is _43. As you can see, in a reciprocal the numerators and denominators are switched. When the product of two numbers is −1, then one number is the negative reciprocal of the other. 3 _ 4 _ 4×3=1

1 4 _ _ 4 × − 1 = −1

Rules for Finding the Reciprocal of a Number 1. Start with the original number—identify the numerator and denominator. 2. The numerator of the first is the denominator of the reciprocal; making the

denominator of the first number the numerator of the reciprocal. 3. If you are making a negative reciprocal change the sign from what it was in the first number.

Example

7 Write the reciprocal and negative reciprocal of __ . 8 7 __________ numerator _ Step 1 Start with the original number— 8=

identify the numerator and denominator.

Step 2 The numerator of the first is the

denominator of the reciprocal; make the denominator of the first number the numerator of the reciprocal.

denominator

7 8 _ _ 8→7 8 7 _ _ 7 is the reciprocal of 8 .

Step 3 If you are making a negative reciprocal The sign in front of _78 is understood to be

change the sign from what it was in the positive. Change the sign in front of _8 to a 7 first number. 7 _ negative. The negative reciprocal of 8 is − _87.

Practice

Write the reciprocal and negative reciprocal of each number.

1. −_25 2 __________ numerator _ Start with the original number— 5 = denominator identify the numerator and denominator. The numerator of the first is the denominator of the reciprocal; make the denominator of the first number the numerator of the reciprocal.

− _25 →

If you are making a negative reciprocal change the sign from what it was in the first number.

The sign in front of − _25 is

2. _17 3. _53

Change the sign in front of

. to

. The negative reciprocal of − _52 is 5 4. __ 13 5. _98

.


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Solving Equations That Contain Decimals Not all equations you encounter will have integers in all parts. You may solve equations that contain decimals, such as; 1.2x + 7.33 = 14.2

or

0.32x − 12.2 = 8.75

Rules for Solving an Equation That Contains Decimals 1. Count the number of decimal places in each number. 2. Find the most numbers of decimal places. 3. Multiply all numbers by a multiple of 10 with a

number of “0” equal to most number of decimal places. 4. Solve for the variable.

Example Solve. 1.5x + 2.1 = 17.1 Step 1 Count the number of decimal places

in each number. Step 2 Find the most number of decimal

1.5 → one place 2.1 → one place 17.1 → one place The most number of decimal places is one.

places. Step 3 Multiply all numbers by a multiple of

10 with a number of “0” equal to the most number of decimal places. Step 4 Solve for the variable.

Practice

The multiple of 10 with one “ 0” is 10. (10)(1.5x) + (10)(2.1) = (10)(17.1) 15x + 21 = 171 15x + 21 −21 = 171 − 21 15x = 150 x = 10

Solve.

1. 0.3x + 16.25 = 22.65 Count the number of decimal places in each number.

0.3 = one place 16.25 =

places

22.65 =

places

Find the most number of decimal places.

The most number of decimal places is

Multiply all numbers by a multiple of 10 with a number of “0” equal to the most number of decimal places.

The multiple of 10 you use is (100) (0.3x) +

Solve for the variable.

. .

(16.25) = (22.65)

x+

=

x=

−

x= x= 2. 2.5x − 2 = 15

3. 1.2x + 10.25 = 14.05 Algebra

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Solving Equations that Contain Fractions Not all equations you encounter will have integers in all parts. You may solve equations that contain fractions, such as; 4x __ 2x __ 5 + 3 =5

or

x + _2x = _78

Rules for Solving Equations that Contain Fractions 1. Place each fraction in front of each variable. 2. Rewrite the fractions so each has the same denominator. 3. Combine like terms. 4. Isolate the variable by multiplying each side by the reciprocal of the fraction.

Example 3x 2x Solve. __ + __ =4 3 4

Step 1 Place each fraction in front of each

variable. Step 2 Rewrite the fraction so each has the

same denominator.

3 2 _ _ 3x + 4x = 4

12 is the common denominator. 9 8 __ __ 12 x + 12 x = 4

9 8 __ __ 12 x + 12 x = 4 17 __ 12 x = 4 17 12 __ 12 __ Step 4 Isolate the variable by multiplying each (__ 17 ) ( 12 )x = 4( 17 ) side by the reciprocal of the fraction. 48 14 __ x = __ 17 = 2 17

Step 3 Combine like terms.

Practice Solve. 3x 1. _2x + __ 5 = 10

Place each fraction in front of each variable.

1 _ 2x +

x = 10

Rewrite the fraction so that each has the same denominator.

5 __ 10 +

x = 10

Combine like terms.

5 __ 10 +

x = 10 x = 10

Isolate the variable by multiplying each side by the reciprocal of the fraction. 2. _2x − _3x = 6 3. _5x + _2x = 7

x = 10 x= 4. _6x − _23 = _16 5. 4 _9x − _3x = 12


39

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Graphing Linear Inequalities When you find the solution of an inequality, you are finding two values—one for x and one for y—that makes the inequality true. Each set of values is known as an ordered pair. You can use the ordered pairs to plot points on a coordinate plane. Rules for Graphing a Linear Inequality 1. Create an input/output table. 2. Select several values for x. Substitute the values for x into the inequality. Solve for y. 3. Plot each solution on the coordinate plane. 4. Draw a line so it goes through each point. If the inequality is  or  the line is a

dashed line, if the inequality is  or , then the line is a solid line. 5. Select a point on either side of the line. Plug the values into the inequality. Shade the

side of the line where the test point is true.

Example Graph the following equation y < 3x + 4. Step 1 Create an input/output table. Step 2 List several values for x. Step 3 Substitute each value of x into the

inequality. Solve the inequality for y. Step 4 Plot each solution on a coordinate

x

y < 3x + 4

y

(x, y)

−2

y < 3(−2) + 4

—2

(—2, —2)

0

y < 3(0) + 4

4

(0, 4)

1

y < 3(1) + 4

7

(1, 7)

2

y < 3(2) + 4

10

(2, 10)

plane. Draw a line so it goes through each point. Step 5 Select a point on either side of the

line. Shade the side of the line where the test point is true.

Select two points (3, 3) and (−2, 1). The point (3, 3) is true, so shade the area on this side.

Practice Graph the following equations.

1. y  4x − 6 Create an input/output table.

x

y  4x − 6

List several values for x.

2

y  4(2) − 6

0

y  4(0) − 6

Substitute each value of x into the inequality. Solve the inequality for y. Plot each solution on a coordinate plane. Draw a line so it goes through each point. Select a point on either side of the line. Shade the side of the line where the test point is true. 2. y  5x − 2

y  4(

)−6

y  4(

)−6

y

(x, y)

Select two points: (3, 3) and (−2, 1). The point (3, 3) is , so shade the area on this side. 3. y  −2x + 3 Algebra

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Writing Inequalities from a Graph There are four inequality symbols. The symbol of each type and its graph are summarized below. Inequality Symbol

Meaning

Example



less than

x9



less than or equal to

x9



greater than

x9



greater than or equal to

x9

Graph 0

9

0

9

0

9

0

9

As you may have noticed, the inequality symbol points to the number with the lesser value. For example, you can say “7 is less than 9” or 7  9. Rules for Writing an Inequality from Its Graph (the number). 1. Identify the point on the number line. Write x 2. Look at the direction the graph points to and if you have an open or closed circle. Point

Direction of graph

Symbol

to the left to the left

•

 

Point

Direction of graph

Symbol

•

to the right to the right

 

Example Write an inequality for the graph.

-1

Step 1 Identify the point on the number line.

The point is located at: −1;

Write x

(the number).

Step 2 Look at the direction the graph points

to and if you have an open or closed circle.

0

x

−1.

The graph points to the right, with a closed circle: x  −1

Practice Write an inequality for each graph.

1. Identify the point on the number line. Write x

0

6

The point is located at :

.

The graph points to the

, with a (n)

(the number).

Look at the direction the graph points to and if you have an open or closed circle.

circle. x

2.

-5

0

3.

0

3


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Solving One-Step Inequalities by Adding or Subtracting When given an algebraic inequality, you are asked to get the variable by itself on one side of the inequality symbol to isolate the variable. To do so, you must “undo” any operations that are on the same side of the equal sign as the variable. When you have the variable by itself, you have solved the inequality. Rules for Isolating a Variable using Addition or Subtraction 1. Identify the operation and number on the same side of the inequality symbol

as the variable. 2. Perform the opposite operation of that number on each side of the inequality.

Example Solve. x − 10 > 21 Step 1 Identify the operation and number on

the same side of the inequality symbol as the variable.

“− 10” is on the same side of the inequality as the variable.

Step 2 Perform the opposite operation of that You will “+ 10” to each side.

number on each side of the inequality.

x − 10 + 10  21 + 10 x − 0  31 or x  31

Practice Solve.

1. 44  16 + x is on the same side of the inequality

Identify the operation and number on the same side of the inequality symbol as the variable.

as the variable.

Perform the opposite operation of that number on each side of the inequality.

You will

on each side. 44  16 

+x + x or

x

2. x + 12  10 3. 15 + x  23 4. 41  x − 12 5. 17  5 + x 6. 27  x − 10 7. 17 + x  -5

Algebra

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Solving One-Step Inequalities by Multiplying or Dividing When given an algebraic inequality, you are asked to get the variable by itself on one side of the inequality symbol. To do so you must “undo” any operations that are on the same side of the inequality symbol as the variable and thus isolate the variable. When you have the variable by itself, you have solved the inequality. Rules for Isolating a Variable using Multiplication and Subtraction 1. Identify the operation and number on the same side of the inequality

symbol as the variable. 2. Perform the opposite operation of that number on each side of the

inequality symbol. 3. If you multiply or divide each side of the inequality by a negative

number, switch the inequality symbol.

Example Solve. 10x  240 Step 1 Identify the operation and number on

You are multiplying the variable times 10.

the same side of the inequality symbol as the variable. Step 2 Perform the opposite operation of that The opposite of multiplication is division; so

number on each side of the inequality symbol.

divide each side by 10. 10x ÷ 10  240 ÷ 10 x  24

Practice Solve.

1. − _3x  9 Identify the operation and number on the same side of the inequality symbol as the variable.

The variable is

by −3.

Perform the opposite operation of that number on each side of the inequality symbol.

The opposite of is multiplication; so multiply each side by

If you multiply or divide each side of the inequality by a negative number, switch the inequality symbol.

−_3x

.

9

x? x

2. 3x  24

4. −4x  −36

6. _5x  −10

3. 2x  −16

5. _2x  14

7. − _4x  12


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Solving Two-Step Inequalities When solving an inequality, the goal is to get the variable by itself on one side of the inequality symbol. There are times when solving an inequality requires two steps. The first step is usually either addition or subtraction. The second step is usually multiplication or division. Rules for Solving Two–Step Equations 1. Isolate the term with the variable using addition or subtraction. 2. Isolate the variable using multiplication or division. 3. If you multiply or divide each side of the inequality by a negative

number, switch the inequality symbol.

Example Solve. 3x + 5 < 26 Step 1 Isolate the term with the variable using 3x + 5  26

addition or subtraction.

3x + 5 − 5  26 − 5 3x + 0  21 or 3x  21

Step 2 Isolate the variable using multiplication 3x  21

or division.

3x ÷ 3  21 ÷ 3 x7

Practice Solve.

1. −2x − 7  17 Write the equation.

−2x − 7  17

Isolate the term with the variable using addition or subtraction.

−2x − 7

Isolate the variable using multiplication or division.

−2x  24

−2x

−2x

 17 

= −2x  24

? 24

x? x 2. 3x + 2  − 10

5. _5x + 10  −3

3. 4 − 2x  20

6. _5x − 3  9

4. −5x − 7  18

7. −_4x + 4  7

Algebra

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Date

Name

The Pythagorean Theorem A right triangle is a triangle with one 90º angle (also known as a right angle). In a right triangle the sides next to the right angle are the legs. The side opposite the right angle is the hypotenuse. hypotenuse (c)

leg (a)

leg (b)

In a right triangle there is a relationship between the legs and the hypotenuse. This relationship (the Pythagorean theorem) says that a2 + b2 = c2 Rules for Using the Pythagorean Theorem 1. Identify the legs and the hypotenuse. 2. Plug the numbers into the Pythagorean theorem and square the numbers. 3. If the unknown side is a leg, solve the equation for the unknown leg. 4. If the unknown side is the hypotenuse, add the squares of the two legs and

then find the square root.

Example Find the unknown length in a right triangle if a = 5 and c = 13. Step 1 Identify the legs and the hypotenuse.

a = 5 is a leg; c = 13 is the hypotenuse.

Step 2 Plug the numbers into the Pythagorean 52 + b2 = 132

theorem. Square the numbers. Step 3 If the unknown side is a leg, solve the

equation for the unknown leg.

25 + b2 = 169 25 − 25 + b2 = __ 169 − 25 ____ 2 b = 144 √b2 = √144 = 12

Practice Find the unknown length in each right triangle.

1. b = 15, a = 8 Identify the legs and hypotenuse.

a = 8 is a leg. b = 15 is

Plug the numbers into the Pythagorean theorem. Square the numbers.

82 +

64 +

=

If the unknown side is the hypotenuse, add the squares of the two legs and then find the square root.

64 +

=

.

2=

=

2. b = 8; c = 12

4. a = 20; b = 15

3. a = 3; b = 4

5. a = 200; c = 250

=


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Irrational Numbers Rational numbers are defined as a number that can be written as the ration of two integers. When written as a decimal, a rational number either ends or repeats. 2 _ 3 = 0.666 9 (a repeating decimal)

23 __ 2 = 11.5 (an ending, or terminating, decimal)

A number that cannot be written as a ratio of two integers is known as an irrational number. Instead, an irrational number written as a decimal continues without repeating. __ __

√_17 = 0.377964473

√5 = 2.23606797

Rules for Identifying Rational and Irrational Numbers 1. Look at the number under the square root sign. Ask if the number is a perfect square. 2. Find the square root. 3. Look at the result. If the result is a terminating decimal or repeating decimal, the

number is rational. If the decimal does not terminate or repeat, it is irrational.

Example Is the expression a rational or irrational?

___

___

a. √81

b. √10

___

a. √81 = 9 is a perfect square. root sign. Ask if the number is a perfect So the 81 is rational. ___ square. b. √10 is not a perfect square.

Step 1 Look at the number under the square

___

Step 2 Find the square root.

b. √10 = 3.16227766

Step 3 Look at the result. If the result is a

b. 3.16227766 ___does not terminate and does not repeat. √10 is irrational.

terminating decimal or repeating decimal, the number is rational. If the decimal does not terminate or repeat it is irrational.

Practice Is each expression rational or irrational?

Look at the number under the square root sign. Ask if the number is a perfect square.

__

__

a. √7

a perfect square.

____

b. √100 =

; it is a perfect square.

So √100 is

.

____

__

Find the square root.

a. √7 =

Look at the result. If the result is a terminating decimal or repeating decimal, the number is rational. If the decimal does not terminate or repeat, it is irrational.

a.

____

__

does not terminate and

does not repeat. √7 is

.

__

1 _ 3 √___

2. √1.21

4.

3. √25

5. √11

___

____

1. a. √7 b. √100

Algebra

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Simplifying Radical Expressions by Multiplying Two Radicals A radical expression contains a number or expression under a square root sign. The following expressions are radical expressions. ____

__

√121

√2

___

_____

√4x2

√x + 5

You can simplify a radical expression by finding and removing perfect squares. When simplifying a radical expression in which radicals are multiplied, you multiply the values and expressions under the square root symbol as you would any expression. Rules for Simplifying Radical Expressions by Multiplying Two Radicals 1. Place both numbers under the square root symbol under a single square

root symbol, separated by a multiplication sign. 2. Multiply. 3. Simplify the radical by looking for a perfect square. Follow the rules for simplifying a radical expression by removing a perfect square.

Example ____

_____

Simplify. √8x2 × √18x5

Step 1 Place both numbers under the square

root symbol under a single square root symbol, separated by a multiplication sign. Step 2 Multiply. Step 3 Simplify the radical by looking for a

perfect square. Follow the rules for simplifying a radical expression by removing a perfect square.

___

_________

____

√8x2 × √18x5 = √8x2 × 18x5 _________

______________

_____

8x2 × 18x5 = √(8 × 18)(x2 × x5) = √144x7 √_____ ____

144x7 144 is a perfect square: √144 = 12 √_____ __ __ √144x6 × √x = 12x3√x

Practice Simplify. ___

___

1. √3x × √6x

2. 3.

___

___

__________

Place both numbers under the square root symbol under a single square root symbol, separated by a multiplication sign.

√3x × √6x = √3x ×

Multiply.

√3x ×

Simplify the radical by looking for a perfect square. Follow the rules for simplifying a radical expression by removing a perfect square. ___ ___

√

__________ ______

√5x3 × √8x2

4.

√2x5 × √2x4

5.

___

___

___________

=√

______

= √

___

___

×√

______

=√

=

____

5x4 × √10x √___ ___ 6 2x × √ √8x2


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Simplifying Radical Expressions by Removing Perfect Squares A radical expression contains a number or expression under a square root sign. The following expressions are radical expressions. _____ ___ ____ __ √121 √4x2 √2 √x + 5 You can simplify a radical expression by finding and removing perfect squares. Rules for Simplifying Radical Expressions by Removing Perfect Squares 1. Look at the number expression under the square root sign. Find two

factors of the number or expression: one factor must be a perfect square. 2. Rewrite the radical expression as the product of two square roots of the

factor found in #1. 3. Find the square root of the perfect square. Place the result next to the

square root symbol of the other factor.

Example ___ Simplify. √50

Step 1 Look at the number or expression under The factors of 50 are: 1 and 50; 2 and 25;

the square root sign. Find two factors 5 and 10; 25 is a perfect square; so use the of the number or expression: one of factors 2 and 25. the factors must be a perfect square. ___ ___ __ √50 = √25 × √2 Step 2 Rewrite the radical expression as the product of two square roots of the factor found in # 1. ___ __ __ 25 × 2 = 5 2 √ √ √ Step 3 Find the square root of the perfect square. Place the result next to the square root symbol of the other factor.

Practice Simplify. ___

1. √27

Look at the number or expression under the square root sign. Find two factors of the number or expression: one of the factors must be a perfect square.

The factors of 27 are 1 and 3 and

.

___

_________

√27 = √3 ×

Find the square root of the perfect square. Place the result next to the square root symbol of the other factor.

√3 × √

____

__

___

__

= √3 ×

__

___

= √3 × √

=

___

2. √500 =

4. √48 =

3. √80 =

5. √75 =

___

is a perfect square,

so use 3 and 9.

Rewrite the radical expression as the product of two square roots of the factor found in #1.

;

___

Algebra

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Simplifying Radical Expressions with Variables A radical expression contains a number or expression under a square root sign. The following expressions are radical expressions. _____ ___ ____ __ √121 √4x2 √2 √x + 5 You can simplify a radical expression by finding and removing perfect squares. Rules for Simplifying Radical Expressions with Variables 1. Find two factors (including variables) of the expression under the square root symbol;

one of the factors must be a perfect square. 2. Rewrite the radical expression as a product of two square roots of the factor found in #1. 3. Find the square root of the prefect square. Place the result next to the square root of the

symbol of the other factor.

Example _____ Simplify. √27a5

Step 1 Find two factors (including variables)

Look for a number that is a perfect square; of the expression under the square root with the perfect square place the variable symbol; one of the factors must be a raised to an even power. The factors of 27a5 perfect square. are 3a and 9a4. 4 × 3a = (9 × 3)(a4 × a) = 27a5 9a____ _______ ___

Step 2 Rewrite the radical expression as a

product of two square roots of the factor found in #1. Step 3 Find the square root of the perfect

square. Place the result next to the square root symbol of the other factor.

___

√27a5 = √9a4 × 3a = √9a4 × √3a __

Find divide___ the exponent by 2. ___ √ 9 and ___ 4 2 √9a × √3a = 3a √3a

Practice Simplify.

1.

____

√32x7

Find two factors (including variables) of the expression under the square root symbol; one of the factors must be a perfect square.

2. 3.

Look for a number that is a perfect square; with the perfect square place the variable raised to an even power. The factors of 32x7 are ____ __________

Rewrite the radical expression as a product of two square roots of the factor found in #1.

√32x7 = √

Find the square root of the perfect square. Place the result next to the square root symbol of the other factor.

√16x6 × √2x = 4x3√2x

____

√____ √48x5 50x2

____

4. 5.

___

and 2x.

___________

× 2x = √

___

× √2x

___

____

63x6 √_____ √180x9


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Solving a Radical Equation by Isolating the Radical A radical equation is an equation that has a variable under the square root symbol. To solve a radical equation, you can use many of the methods you have learned to solve any equation. The difference with radical equations is that you must get rid of the square root symbol. Rules for Solving a Radical Equation by Isolating the Radical 1. Isolate the radical on one side of the equation. 2. Square both sides to remove the square root symbol.

Example __

Solve. √x − 5 = 1 Step 1 Isolate the radical on one side of the

equation. Step 2 Square both sides to remove the square

root symbol.

__ √x − 5 + 5 = 1 + 5 __ √x = 6 __ √x = 6 __ (√x )2 = 62

x = 36

Practice Solve.

_____

1. √x − 4 + 1 = 9 Isolate the radical on one side of the equation.

_____

√x − 4 + 1 = 9 _____

√x − 4 + 1

=9

_____

√x − 4 = Square both sides to remove the square root symbol.

_____

(√x − 4 )

=

2

=

+

x−4= x−4 x=

__

2. √x + 9 = 14 ___

3. √2x − 4 = 10 _____

4. √x + 3 + 1 = 8 ______

5. √3x − 2 +2 = 15 ___

6. √4x = 4

Algebra

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Estimating Square Roots Almost all of the numbers you work with are rational numbers. A rational number is a number that can be written as a ratio of two integers. When written as a decimal a rational number either ends or repeats. Below are two rational numbers written as a ratio and a decimal. 19 9.5 = __ 2

0.33 3 = _13

You can use perfect square to help you estimate the square root of a number that is not a perfect square. Rules for Estimating Square Roots 1. Look at the number and find a perfect square that is less than the number. 2. Look at the number and find a perfect square that is greater than the number. 3. The square root of the number is between the two perfect squares.

Example

___

Between what two integers is √10 ? Step 1 Look at the number and find a perfect

__

The perfect square less than 10 is 9 (√9 = 3).

square that is less than the number. Step 2 Look at the number and find a perfect

The ___perfect square greater than 10 is 16. square that is greater than the number. (√16 = 4).

Step 3 The square root of the number is

___

√10 is between 3 and 4.

between the two perfect squares.

Practice Between what two integers is the square root of each number? ___

1. √22

Look at the number and find a perfect square that is less than the number.

The perfect square less than 22 is 16

Look at the number and find a perfect square that is greater than the number.

The perfect square that is greater than 22 is

The square root of the number is between the two perfect squares.

___

(√16 =

). ___

___

.(√

=

√22 is between

) and

.

2. 5 3. 34 4. 29 5. 52 6. 130 Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Estimating Cube Roots and Higher Power Roots Powers above 3 are called higher powers. Roots above 3 are higher power roots. You write cube roots and higher power roots in a manner similar to square roots. 3 __ 4 __ Cube root: √ Fourth Root: √ __ The number outside the √ symbol tells you the root you are finding. Rules for Finding or Estimating Cube or Higher Power Roots 1. Look at the number outside the symbol to determine the root to calculate. 2. If the number under the symbol is a perfect power, find the root. 3. If the number under the symbol is not a perfect power, find a perfect power

less than the number. 4. Then find a perfect power greater than the number. 5. The estimated root is between the two perfect roots.

Example

3

__

Find the root: √8

3

__

Step 1 Look at the number outside the symbol √ 8

—determine the root to calculate.

Step 2 Determine if the number under the

symbol is a perfect power. If it is a perfect power, find the root.

The “3” outside the symbol means you find the cube root. 8 is a__perfect cube (23 = 8) 3 So, √8 = 2

Practice Find each root. 3

____

1. √100

Look at the number outside the symbol —determine the root to calculate.

3

____

√100

The

outside the symbol means find the root.

Determine if the number under the symbol is a perfect power. If it is a perfect power, find the root.

100

If the number under the symbol is not a perfect power, find a perfect power less than the number.

The perfect cube less than 100 is

Then find a perfect power greater than the number.

100 is

The estimated root is between the two perfect roots. 5

_____

5

___

2. √5000 3. √32

3

____

4

____

4. √168 5. √250

3

(

.

= 64). The perfect cube greater than .

3

( 3

a perfect cube.

= 125)

____

√100 is between

and 4

____

3

____

.

6. √625 7. √216

Algebra

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Multiplying a Polynomial by a Monomial When you multiply a polynomial by a monomial, you multiply each term of the polynomial by the monomial. When you multiply one term by another, you multiply the numbers in front of each term. You also apply the rules for multiplying exponents by adding exponents. (4x4)(3x2) = 12x6

(2x5)(7 x2y) = 14x7y

Rules for Multiplying a Polynomial by a Monomial 1. Multiply each term in the polynomial by the monomial. 2. Multiply the numbers in front of the variables. (Remember that

“1” is understood to be in front of a variable with no number.) 3. Add the exponents of variables of the same letter. (Remember that “1” is understood to be the exponent of a variable with no exponent.)

Example Multiply. 5x (2x3 + x2 − 3) Step 1 Multiply each term in the polynomial

by the monomial.

(5x) (2x3) = 10x4 (5x)(x2) = 5x3 (5x)(−3) = −15x 10x4 + 5x3 − 15x

Practice Multiply.

1. −x4 (4x2 + _2x + 10) Multiply each term in the polynomial by the monomial.

(−x4) (4x2) = (−x4)

=

(−x4)

= x5

5

x 4 6 __ 4 −4x6 + (−__ 2 ) + (−10x ) = −4x − 2 − 10x

2. 2x3 (−x2 − 4x + 2) 3. x4 (3x4 + 7x2 − 10) 4. −3x2 (−2x3 − 5x2 + _4x )


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Multiplying Binomials When you multiply two binomials, you will multiply each term in the first binomial by each term in the second binomial. One way to remember to multiply each term is to remember to FOIL. This tells you which terms to multiply. Rules for the FOIL Method F: The first term in each binomial. O: The outer terms in each binomial. I: The inner terms in each binomial. L: The last term in each binomial.

Combine like terms.

Example Multiply. (5x − 3)(4x + 2)

Apply the FOIL Method

(5x − 3)(4x +2)

F: the first terms

(5x) (4x) = 20x2

O: the outer terms

(5x)(2) = 10x

I: the inner terms

(−3)(4x) = −12x

L: the last terms

(−3)(2) = −6 20x2 + 10x − 12x − 6 = 20x2 − 2x − 6

Combine like terms.

Practice Multiply.

1. (2x + 1)(6x + 3) Apply the FOIL Method

(2x + 1)(6x + 3)

F: the first terms

(2x)(6x) =

O: the outer terms

(2x)(3) =

I: the inner terms

(1)

L: the last terms

= =

Combine like terms. 2. (3x + 2)(2x − 5) 3. (6x − 4)(2x − 3) 4. (2x +2)(x − 5) 5. (3x − 4)(2x +1)

Algebra

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Squaring a Binomial When you square a binomial, you can apply the FOIL method to find the product. You can also apply the following rules as a short cut. Rules for Squaring a Binomial 1. Square the first term. 2. Find 2 times the product of the two terms; use the same

operation sign as the one between the two terms. 3. Square the last term.

Example Solve. (x + 3)2 Step 1 Square the first term.

x is the first term (x × x) = x2

Step 2 Find 2 times the product of the two

2(x × 3) = 2(3x) = 6x Use the plus sign.

terms; use the same operation sign as the one between the two terms. Step 3 Square the last term.

32 = 9 (x + 3)2 = x2 + 6x + 9

Practice Solve.

1. (5x − 2)2 Square the first term.

5x is the first term (5x × 5x) =

Find 2 times the product of the two terms; use the same operation sign as the one between the two terms.

2(5x × 2) = 2(10x) =

Square the last term.

−22 = 4

Use the

sign.

(5x − 2)2 =

2. (x + 4)2 3. (x − 8)2 4. (2x + 6)2 5. (4x − 4)2 6. (6x + 12)2


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Adding Polynomials When you add two polynomials, you do so by adding like terms. Terms are like terms if they have the same variable raised to the same power. Like Terms

Not Like Terms

2x2, −4x2

5x4, 4x5

Rules for Adding Polynomials 1. Write each polynomial in standard form. 2. Line up like terms. 3. Add the numbers in front of each variable. (Remember

“1” is understood to be in front of a variable with no number).

Example Add. (4x2 + 2x − 5) + (3x4 − 3x + 5x2) Step 1 Write each polynomial in standard form. (4x2 + 2x − 5) + (3x4 + 5x2 − 3x)

4x2 + 2x −5 3x4 + 5x2 − 3x

Step 2 Line up like terms. Step 3 Add the numbers in front of each

variable.

4x2 + 2x − 5 +3x4 + 5x2 − 3x 3x4 + 9x2 − 1x − 5

Practice Add.

1. (16x3 + 5 − 2x2) + (5 + 3x3 + x2) Write each polynomial in standard form.

(16x3 − 2x2 + 5) +

Line up like terms.

16x3 − 2x2 +5

Add the numbers in front of each variable.

16x3 − 2x2 + 5 +

2. (−2x3 + 5 − 3x) + (2x3 + 2x − 3x2) 3. (15 + 4x2 + x3) + (5x3 − 10 − x2) 4. (5xy + 3 − 2x2y) + (14 + 5x2y + xy) 5. (7 − 5y3 + 3y4) + (2y2 − y4 + 12y3) Algebra

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Subtracting Polynomials When you subtract two polynomials, you do so by subtracting like terms. Terms are like terms if they have the same variable raised to the same power. Like Terms

Not Like Terms

3x3, −x3

5x, x5

Rules for Subtracting Polynomials 1. Write polynomials in standard form. 2. Line up like terms. 3. Change the sign of each term in the second polynomial. 4. Add the numbers in front of each variable. (Remember,

“1” is understood to be in front of a variable with no number).

Example Subtract. (2x2 + 5x3 + 1) − (5 + 5x2 + 3x3) Step 1 Write polynomials in standard form. Step 2 Line up like terms. Step 3 Change the sign of each term in the

second polynomial. Step 4 Add the numbers in front of each

variable

(5x3 + 2x2 + 1) − (3x3 + 5x2 + 5) 5x3 + 2x2 + 1 −(3x3 + 5x2 + 5) 5x3 + 2x2 +1 −3x3 − 5x2 − 5 5x3 + 2x2 + 1 + (−3x3 − 5x2 − 5) 2x3 − 3x2 − 4

Practice Subtract.

1. (x + 5 + 4x2) − (10 + 3x2 – 5x) Write polynomials in standard form. Line up like terms. Change the sign of each term in the second polynomial. Add the numbers in front of each variable

(4x2 + x + 5) – 4x2 + x + 5 −

4x2 + x + 5 4x2 + x + 5

+

2. (x2 + 5 − 4x3) − (−10 − 2x2 + 5x3) 3. (−4 + x3 − 3x2) − (5x + 5x2 + 14) 4. (−3x2 +10 −2x) − (9 + 4x2 − 10x) 5. (−4x3 + 12 − x) – (−5x + 10x2 − 3) Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Multiplying a Polynomial When you multiply a polynomial by a monomial, you multiply each term of the polynomial by the monomial. When you multiply one term by another, you multiply the numbers in front of each term. You also apply the rules for multiplying exponents by adding exponents. (5x2)(2x3) = 10x5

(2x3)(3xy) = 6x4y

Rules for Multiplying a Polynomial by a Monomial

Multiply each term in the polynomial by the monomial: 1. Multiply the numbers in front of the variables. (Remember that “1” is understood to be in front of a variable with no number) 2. Add the exponents of variables of the same letter (Remember that “1” is understood to be the exponent of a variable with no exponent.)

Example Multiply. 4x(2x2 + x − 6) Step 1 Multiply each term in the polynomial

by the monomial.

(4x)(2x2) = 8x3 (4x)(x) = 4x2 (4x)(−6) = −24x 8x3 + 4x2 − 24x

Practice Multiply.

1. 3x3(4x2 + 2x − 1) Multiply each term in the polynomial by the monomial.

(3x3)(4x2) = (3x3)(2x) = (3x3)(−1) =

2. 7x2(−3x3 + 2) 3. –2x4 (−4x2 + x − 3) 4. x2(3x5 + 6x3 + x) 5. 5x4(−4x4 − 2x3 + 5)

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Factoring a Binomial By applying the Distributive Property in reverse, you can factor out a common factor. 20 + 15 = (5 × 4) + (5 × 3) = 5 (4 + 3) Rules for Factoring Out the Greatest Common Factor: Factoring Binomials 1. Find the greatest common factor of all the terms. 2. Determine the terms that when multiplied by the greatest common factor

will result in each original term. 3. Rewrite the expression with the greatest common factor outside the

parentheses and the terms you found in Step 2 inside the parentheses.

Example Factor out the greatest common factor: 5x2 + 10x Step 1 Find the greatest common factor of all

the terms.

Step 2 Determine the terms that when

multiplied by the greatest common factor will result in each original term. Step 3 Rewrite the expression with the

List the factors of 5x2 and 10x. 5x2: 1, 5, x2 10x: 1, 2, 5, 10, x The greatest common factor is 5x. (5x) (x) = 5x2 (5x) (2) = 10x 5x (x + 2)

greatest common factor outside the parentheses and the terms you found in Step 2 inside the parentheses.

Practice Factor out the greatest common factor.

1. 10x4 − 15x3 Find the greatest common factor of all the terms.

Determine the terms that when multiplied by the greatest common factor will result in each original term. Rewrite the expression with the greatest common factor outside the parentheses and the terms you found in Step 2 inside the parentheses. 2. 27x4 − 9x5

3. 36x3 + 24x

List all the factors of 10x4 and −15x3. 10x4: 1, 2, 5, 10, x4 −15x3: 1, 3, 5, 15, x3 The greatest common factor is ( (

.

) = 10x4 ) = −15x3

)( )( (

)

4. 5x2y5 + 15xy7


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Finding the Greatest Common Factor for Variable Terms The greatest common factor can also be found for two or more variable terms. To find the greatest common factor for variable terms, you apply the rules you have learned for finding the greatest common factor of two or more numbers. Rules for Finding the Greatest Common Factor for Variable Terms 1. List all the factors of the coefficients (the number in front of the variable). 2. From the lists, identify the greatest common factor; this is the coefficient-

part of the greatest common factor. 3. List the variables with their exponents. 4. From the list, select the variable with the lowest exponent.

Example Find the greatest common factor of 30x3, 40x6, and 50x7. Step 1 List all the factors of the coefficients

(the number in front of the variable). Step 2 From the list, identify the greatest

common factor; this is the coefficientpart of the greatest common factor.

30: 1, 2, 3, 5, 6, 10, 15, 30 40: 1, 2, 4, 5, 8, 10, 20, 40 50: 1, 2, 5, 10, 25, 50 Of the factors listed, 10 is the greatest common factor.

Step 3 List the variables with their exponents.

The variable part of each term: x3, x6, x7.

Step 4 From the list, select the variable with

The variable with the lowest exponent is x3. The greatest common factor is 10x3.

the lowest exponent.

Practice Find the greatest common factor for each list of terms.

1. 12m3n2 and 18m5n4 List all the factors of the coefficients (the number in front of the variable). From the list, identify the greatest common factor; this is the coefficientpart of the greatest common factor. List the variables with their exponents. From the list, select the variable with the lowest exponent.

12: 1, 2, 3, 4, 6, 12 18: Of the factors listed,

is the greatest

common factor. m3n2 = m3 × n2

m5n4 =

×

The variable with the lowest exponents are m3 and . The greatest common factor is

2. 6x4 and 8x6

4. 12x8 and 20x9

3. 9m2 and 3m5

5. 4x4y3 and 6xy2

m3

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Factoring a Polynomial By applying the Distributive Property in reverse, you can factor out a common factor. 20 + 15 = (5 × 4) + (5 × 3) = 5 (4 + 3) Rules for Factoring Out the Greatest Common Factor: Factoring Polynomials 1. Find the greatest common factor for all of the terms. 2. Determine the terms that when multiplied by the greatest common factor will

result in each original term. 3. Rewrite the expression with the greatest common factor outside the parentheses

and the terms found in Step 2 inside the parentheses.

Example Factor out the greatest common factor. 20x5 + 10x6 − 15x4 Step 1 Find the greatest common factor for

all of the terms.

Step 2 Determine the terms that when

multiplied by the greatest common factor will result in each original term. Step 3 Rewrite the expression with the greatest common factor outside the parentheses and the terms found in Step 2 inside the parentheses.

List the factors of 20x5, 10x6, and −15x4. 20x5 = 1, 2, 4, 5, 10, 20, x5 10x6 = 1, 2, 5, 10, x6 −15x4 = 1, 3, 5, 15, x4 The greatest common factor is 5x4. (5x4) (4x) = 20x5; (5x4)(2x2) = 10x6 (5x4)(−3) = −15x4 5x4 (4x + 2x2 − 3)

Practice Factor out the greatest common factor.

1. 13x8 + 26x4 − 39x2 List all the factors of 13x8, 26x4, and −39x2. 13x8: 26x4: −39x2: The greatest common factor is 13x2. 13x2 (x6) = 13x8 13x2 ( ) = 26x4 Determine the terms that when multiplied by the greatest common factor 13x2 ( ) = −39x2 will result in each original term. 13x2 (x6 ) Rewrite the expression with the greatest common factor outside the parentheses and the terms found in Step 2 inside the parentheses. Find the greatest common factor for all of the terms.

2. 8x3 − 24x2 − 80x

4. x2 + 2x5 − 9x7

3. 3x4 + 30x3 + 48x2

5. 3x5 − 6x4y − 45x3y2


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Factoring Trinomials in the Form x2 + bx + c You can also factor a polynomial. When you factor a polynomial you look for pairs of expressions whose product (when they are multiplied) is the original polynomial. (x + 2) (x + 4) = x2 + 6x + 8 Rules for Factoring a Trinomial in the Form x2 + bx + c 1. Create a table. The left column lists the factors of c. The right

column is the sum of the factors in column 1. 2. Choose the pair of factors in the right column whose sum equals b. 3. Create two expressions of “x +” the factors.

Example Factor: x2 + 9x + 18 Step 1 Create a table. The left column lists

the factors of c. The right column is the sum of the factors in column 1.

Step 2 Choose the pair of factors in the right

column whose sum equals b. Step 3 Create two expressions of “x +” the factors.

FACTORS OF 18

SUM OF FACTORS

1 and 18

1 + 18 = 19

2 and 9

2 + 9 = 11

3 and 6

3+6=9

The last pair of factors, 3 and 6, have a sum (9) that equals the value of b. The factors are: (x + 3)(x + 6).

Practice Factor.

1. x2 − 8x + 12 Create a table. The left column lists the factors of c. The right column is the sum of the factors in column 1.

FACTORS OF 12

SUM OF FACTORS

−1 and −12

−1 + −12 = −13

Note that since b is negative, we are using negative numbers for the factors of c.

−2 and −6

−2 + −6 = −8

−3 and −4

−3 + −4 = −7

Choose the pair of factors in the right column whose sum equals b. Create two expressions of “x +” the factors.

The pair of factors

and

have a

sum that equals the value of b, . The factors are: (x + (−2))(x + (−6)) or (x − 2)(x − 6)

2. x2 + 4x +3

4. x2 + 8x + 15

3. x2 + 9x + 8

5. x2 − 15x + 36 Algebra

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Factoring Trinomials in the Form ax2 + bx + c You can use FOIL to multiply binomials, creating a trinomial. You can also use FOIL to “undo” a trinomial, creating two binomials. Rules for Factoring Trinomials in the Form ax2 + bx + c 1. Create a FOIL table. 2. In the “F” column place factors that result in a. In the “L”

column place factors that result in c. 3. In the “O “and “I ”columns, try different combinations of the factors from Step 2 by adding the products. The combination resulting in b shows the placement within the binomials.

Example Factor. 6x2 + 23x + 7 Step 1 Create a FOIL table. Step 2 In the “F” column, place factors that

result in a. In the “L” column, place factors that result in c.

6x2

23x

7

F

O

+

I

=

?

L

1×6

1×7 1×1

+ +

1×6 7×6

= =

13 43

1×7

2×3

2×7 2×1

+ +

1×3 7×3

= =

17 23

1×7

Step 3 In the “O + I” column, try different

combinations of the factors from Step 2 by adding the products. The combination resulting in b shows the placement within the binomials.

Value for b

The outer terms are 2 and 1. The inner terms are 7 and 3. 6x2 + 23x + 7 = (2x + 7)(3x + 1)

Practice Factor.

1. 5x2 − 14x − 3 Create a FOIL table. In the “F” column, place factors that result in a. In the “L” column, place factors that result in c. In the “O + I” column, try different combinations of the factors from Step 2 by adding the products. The combination resulting in b shows the placement within the binomials.

Value for b

5x2

–14x

–3

F

O

+

I

=

?

L

5×1

5×1

+ +

1×1

= =

2

1 × (−3)

5 × (–1) 5 × (3)

+ +

(–1) × 1

= =

5x2 − 14x − 3 = ( =

2. 2x2 + 8x + 8

5. 2x2 − 8x + 6

3. 2x2 − 3x − 9

6. 2x2 − 7x − 4

x+

)(

x+

)

4. 7x2 + 50x + 7 Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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The Difference of Two Squares The difference of squares involves multiplying two binomials with the same two terms. One binomial is the sum of the terms—for example, (a2 + b). The other binomial is the difference of the terms—for example, (a2 – b). As you can see, the two terms are a2 and b. When you multiply the two squares, the product follows a pattern. Rules for the Difference of Squares 1. Square the first term. 2. Square the second term. 3. Place a minus sign between the two squared terms.

Example Solve. (x2 + 4)(x2 − 4) Step 1 Square the first term.

(x2)2 = (x2)(x2) = x4 Remember when you multiply terms with exponents, you add the exponents.

Step 2 Square the second term.

(4)2 = 4 × 4 = 16

Step 3 Place a minus sign between the

Result of the first squaring: x4 Result of the second squaring: 16 x4 − 16

two squared terms.

Practice Solve.

1. (x4 + y)(x4 − y) Square the first term.

(x4)2 = (x4)(x4) =

Square the second term.

(y)2 = (y)(y) =

Place a minus sign between the two squared terms.

The result of the first squaring is The result of the second squaring is

. .

− 2. (x3 + y2)(x3 − y2) 3. (x5 + 5)(x5 − 5) 4. (2x3 + 4)(2x3− 4) 5. (4x2 + 3)(4x2 − 3) 6. (3x4 + y2)(3x4 − y2)

Algebra

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Solving Systems of Equations by Graphing A system of equations is a set of two or more equations that contain two or more variables. A solution of a system of equations is a set of values that are solutions to all the equations. One method used in solving a system of equations is to use a graph.

Steps for Solving Systems of Equations by Graphing Example: y = x + 1; 2x + y = 4 Step 1 Find two solutions

for the first equation. Step 2 Find two solutions

for the second equation. Step 3 Plot the points from

each equation and draw a line for each equation. Step 4 Locate the point where

x

y=x+1

y

(x, y)

−2

y = (–2) + 1

−1

(−2, −1)

0

y=0 +1

1

(0, 1)

x

2x + y = 4

y

(x, y)

2

2(2) + y = 4

0

(2, 0)

0

2(0) + y = 4

4

(0, 4)

the two lines cross. The lines cross at (1, 2).

Practice

y = 4x − 7

2

y = 4(2) − 7

0

y = 4(0) − 7

x

x+y=8

3

3+y=8

0

0+y=8

Step 4 The lines cross at

1 + x = y

y

(x, y)

y

(x, y)

=4

Step 2

x

+y

1. y = 4x − 7; x + y = 8 Step 1

2x

Solve the system of equations using graphing for the following.

Step 3

.

Locate the point where the two lines cross.

2. x + 2y = 7; x + y = 4

4. x + 4y = −6; 2x + 3y = −1

3. x + y = 2; x = y

5. y + 2x = 5; 2y − 5x = 10


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Solving Systems of Equations by Substitution A system of equations is a set of two or more equations that contain two or more variables. A solution of a system of equations is a set of values that are solutions to all the equations. One method used in solving a system of equations is by substitution. Rules for Solving Systems of Equations by Substitution 1. Make sure each equation has y isolated. 2. Set the parts of equations with the x-term equal to each other. 3. Rearrange the resulting equation so x is on one side. 4. Isolate x. 5. Substitute the value of x into one of the original equations.

Example Solve the system of equations: y = 6 − x; x − 2 = y Step 1 Make sure each equation has y isolated. y is already isolated. Step 2 Set the parts of the equations with the

x-term equal to each other. Step 3 Rearrange the resulting equation so x

is on one side. Step 4 Isolate x. Step 5 Substitute the value of x into one of

the original equations.

y=6−x x−2=y 6 − x = x− 2 6−x+x=x+x−2 6 = 2x − 2 6 + 2 = 2x − 2 + 2 8 = 2x 8 ÷ 2 = 2x ÷ 2

4=x

y=6−x y=6−4=2 The solution is (4, 2).

Practice Solve the system of equations.

1. y = 5 − x; y = x − 1 Make sure each equation has y isolated.

y is already isolated.

Set the parts of the equations with the x-term equal to each other.

5−x=

Rearrange the resulting equation so x is on one side.

5−x 5= 5

Isolate x. Substitute the value of x into one of the original equations.

= =

6=

6 ÷ 2 = 2x ÷ 2

y=x−1

y = 3 − 1= 2

The solution is

x=3

.

2. y = x + 2; y = 2x − 1

4. y = 3x; y = −0.5x + 7

3. y = 8 − 2x; y = x + 5

5. y = 4x − 5; y = 10 − 2x Algebra

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Solving Systems of Equations by Elimination One method of solving a system of equations is by elimination. When you use this method, you add or subtract the equations to remove, or eliminate, a variable. Rules for Solving Systems of Equations by Elimination 1. Line up the equations by like terms. 2. Look for a variable to eliminate. 3. Eliminate the variable and solve for the remaining variable. 4. Solve for the eliminated variable by using one of the

equations and substituting the solution from step 3.

Example Solve: 2x + 3y = 10

x − 3y = 2

Step 1 Line up the equation by like terms. Step 2 Look for a variable to eliminate. Step 3 Eliminate the variable and solve for

the remaining variable. Step 4 Solve for the eliminated variable

by using one of the equations and substituting the solution from Step 3.

Practice

2x + 3y = 10 x − 3y = 2 You can eliminate 3y by adding the two equations since 3y + (−3y) = 0 2x + 3y = 10 + x − 3y = 2 3x + 0 = 12 3x = 12 x=4 2x + 3y = 10 2(4) + 3y = 10 8 + 3y = 10 3y = 2 y = _23 The solution is (4, _23).

Solve

1. 4x + 3y = 12; −4x + y = 4 Line up the equations by like terms. Look for a variable to eliminate. Eliminate the variable and solve for the remaining variable.

Solve for the eliminated variable by using one of the equations and substituting the solution from Step 3.

−4x + 3y = 12 −4x + y = 4 You can eliminate equations. 4x + 3y = 12 −4x + y = 4 0x + 4x + 3y = 12 4x + 3

by adding the two

y= = 12

4x +

= 12

4x =

x=

The solution is (0,4).

2. 2x + 4y =10, −2x – 2y = 4

4. 4x + 3y = 10, x − 3y = 10

3. 3x + y = 13, 2x − y = 2

5. 5x + 3y = 12, −x − 3y = 0


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Solving Linear Systems by Multiplying You can multiply all terms in one equation in order to solve a system of equations by elimination. Rules for Solving Systems by Multiplying 1. Identify a variable in one equation that is a multiple of the

same variable in the other equation. 2. Multiply all terms in one equation to eliminate one variable. 3. Eliminate one variable and solve for the remaining variable. 4. Solve for the eliminated variable by substituting the solution

in Step 3 into one of the equations.

Example Solve. 2x + 3y = 10; −4x + 2y = 4 Step 1 Identify a variable in one equation that 2x + 3y = 10

−4x + 2y = 4 is a multiple of the same variable in the You can multiply the first equation by 2. other equation. Step 2 Multiply all terms in one equation to 2(2x + 3y = 10) = 4x + 6y = 20 eliminate one variable. Step 3 Eliminate one variable and solve for the 4x + 6y = 20 remaining variable. +(−4x + 2y = 4) 0x + 8y = 24 y=3 2x + 3y = 10 2x + 3(3) = 10 Step 4 Solve for the eliminated variable by substituting the solution in Step 3 2x + 9 = 10 2x =1 x = _12 into one of the equations. The solution is (_12, 3).

Practice Solve.

1. −2x + 10y = 20; 5x − 5y = 6 Identify a variable in one equation that is a multiple of the same variable in the other equation. Multiply all terms in one equation to eliminate one variable. Eliminate one variable and solve for the remaining variable.

Solve for the eliminated variable by substituting the solution in Step 3 into one of the equations.

−2x + 10y = 20

5x − 5y = 6

You can multiply the second equation by

.

(5x − 5y = 6) = −2x + 10y = 20

−2x + 10y =20 + 10y = 20

x= −2 10y = 28

+ 10y = 20 y=

The solution is 2. 3x + 10y = 25; −2x + 20y =10

4. x + 8y = 18; −4x − 4y = −2

3. 12x + y = 30; −6x + 3y = −36

5. 6x − 8y = 14; 2x + 4y = 8 Algebra

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Solving Quadratic Equations Using Square Roots The following are examples of quadratic equations: 2x2 + 3x − 4 = 0

x2 + 4 = 0

4x2 − 2x = 0

When you solve a quadratic equation, you are finding the points where the graph of the equation crosses the x-axis. In many quadratic equations, the graph crosses the x-axis at two locations. When solving a quadratic equation, look to see if there is an x-term (for example, 3x). If the equation does not have an x-term, then check to see if you can solve it using square roots. Rules for Solving a Quadratic Equation Using Square Roots 1. Get the x2 term on one side of the equation. 2. Isolate x2. 3. Find the square roots. Remember the solution of a square

root is both a positive number and a negative number.

Example Solve: 2x2 − 32 = 0 Step 1 Get the x2 term on one side of the

equation. Step 2 Isolate x2.

2x2 − 32 + 32 = 0 + 32 2x2 = 32 2x2 ÷ 2 = 32 ÷ 2 x2 = 16 __

Step 3 Find the square roots. Remember the

___

√x2 = √16

solution of a square root is both a x = 4, x = −4 positive number and a negative number.

Practice Solve.

1. 3x2 − 25 = 50 Get the x2 term on one side of the equation.

3x2 − 25 + 25 = 50 + 25

Isolate x2.

3x2 ÷ 3 = 75 ÷ 3

3x2 = x2 = __

Find the square roots. Remember the solution of a square root is both a positive number and a negative number.

___

√x2 = √00 x=

,x=

2. x2 = 49

4. x2 − 25 = 0

3. 2x2 −2 = 6

5. x2 + 15 = 115


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The Quadratic Formula When you solve a quadratic equation, you are finding the points where the graph of the equation crosses the x-axis. The graph of a quadratic equation is U–shaped and crosses the x-axis at two points. Therefore, in finding the solution, you are finding the x-coordinate. The y-coordinates are always 0. One way to solve a quadratic equation is to use the quadratic formula. To use the quadratic formula, your equation must be in the form of ax2 + bx + c = 0. _______ ± √b – 4ac __________ The quadratic formula is x = –b 2a 2

Example Use the quadratic formula to solve the following quadratic equation: x2 + 5x − 50 = 0 Step 1 Identify a, b, and c.

a=1

Step 2 Plug the values for a, b, and c into the

± √b – 4ac __________ x = –b 2a

quadratic formula.

b= 5 c = −50 _______ 2

___________

–5 ± √52–4(1)(–50) x = ______________ 2(1) ________ ____ –5 ± √25 + 200 _______ –5± √225 ___________ x= = 2 2 –5 ± 15 ______ x= 2 10 −20 ___ x = __ 2 or x = 2

Step 3 Solve.

x = 5 or x = −10

Practice Solve using the quadratic formula.

1. x2 + 3x – 4 = 0 Identify a, b, and c. Plug the values for a, b, and c into the quadratic formula.

a=

b=

_______

±√ 4ac __________ x = –b 2a

____________________

2 − 4( − ±√ ) )( x = __________________________ 2

____________

Solve.

c=

b2 –

______

−3 ± √000000000000 −3 ± √ + x = _______________ = __________ 2 2 −3 ± 0000 ________ x= 2 0 _ x = 2 or x = _02

x=

or x =

2. x2 + 15x + 26 = 0

4. 2x2 − 10x + 12 = 0

3. x2 – 6x − 72 = 0

5. 3x2 − 12x − 15 = 0

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Using the Discriminant The solution to a quadratic equation is where the graph of the equation crosses the x-axis. The graph of a quadratic equation can cross the x-axis at two points, one point, or no point. You can find out the number of solutions a quadratic equation has by using the discriminant. The discriminant is the b2 − 4ac part of the quadratic formula. Rules for using the Discriminant 1. Identify a, b, and c in a quadratic equation. 2. Plug the numbers for a, b, and c into b2 − 4ac. 3. Solve. If the result is positive: There are two solutions.

If the result is 0: There is one solution. If the result is negative: There are no solutions.

Example Find the number of solutions for 3x2 – 5x − 1 = 0. Step 1 Identify a, b, and c in the quadratic 3x2 − 5x − 1 = 0

equation. Step 2 Plug the numbers for a, b, and c into

b2 − 4ac. Step 3 Solve.

a=3 b = −5 c = −1 2 2 b − 4ac = (−5) − 4 (3)(−1) (−5)2 − 4 (3) (−1) = 25 − (−12) 25 + 12 = 37 The result is positive so, there are two solutions for the equation.

Practice Find the number of solutions for the following equations.

1. x2 + 3x + 7 = 0 Identify a, b, and c in the quadratic equation. Plug the numbers for a, b, and c into b2 − 4ac. Solve.

x2 + 3x + 7 = 0 a=1 b= b2 − 4ac = 2

2

c= − 4(1)

− 4 (1)

Since the result is

=

−

=

; there are

solutions. 2.

x2

+ 2x + 1 = 0

3. 2x2 − 7x + 4 = 0 4. x2 − 5 = 0 5. 2x2 − 12x + 18 = 0 6. 3x2 − 9x + 12 = 0 Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Zero-Product Property You can solve a quadratic equation in a number of ways. One method is to factor the quadratic equation that has been written in standard form. Once you have factored the equation, you can use the zero-product property to solve the equation. Rules for Using the Zero–Product Property 1. Set each factor equal to zero. 2. Solve each factor/equation.

Example Solve. (x + 3)(2x − 1) = 0 Step 1 Set each factor equal to zero. Step 2 Solve each factor/equation.

(x + 3)(2x − 1) = 0 (x+3) = 0 x+3−3=0−3 x=−3

2x − 1 = 0 2x − 1 + 1 = 0 + 1 2x ÷ 2 = 1 ÷ 2 x = _12

Practice Solve.

1. (x + 5)(x − 2) =0 Set each factor equal to zero.

(x + 5)(x − 2) = 0 (x + 5) = 0

Solve each factor/equation.

=0

x+5−5=0−5 x = −5

2. (x − 1)(x + 8) = 0

x=

,x=

3. (2x − 7)(x − 3) = 0

x=

,x=

4. (6x + 5)(x + 4) = 0

x=

,x=

5. (x − 2)(x − 3) = 0

x=

,x=

=0 x=

Algebra

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Solving a Quadratic Equation by Factoring You have learned to factor an equation in the form ax2 + bx + c. By applying this method of factoring and by using the Zero-Product Property, you can solve a quadratic equation. Rules for Solving a Quadratic Equation by Factoring 1. Be sure the equation is in the form ax2 + bx + c = 0. Set up a

FOIL table to help factor the equation. 2. Use the FOIL table to identify the numbers in each binomial. Write the factored form of the original equation. 3. Set each binomial equal to 0 and solve for the two values of x.

Example Solve. 2x2 + 5x + 2 = 0 Step 1 Be sure the equation is in the form

2x2

ax2

+ bx + c = 0. Set up a FOIL table to help factor the equation.

Step 2 Use the FOIL table to identify the

numbers in each binomial. Write the factored form of the original equation. Step 3 Set each binomial equal to 0 and solve for the two values of x.

Practice

+5x

+2

F

O

+

I

=

?

L

2×1

2×1 2×2

+ +

2×1 1×1

= =

4 5

1×2

The outer terms are 2 and 2; the inner terms are 1 and 1. 2x2 + 5x + 2 = (2x + 1)(x + 2) = 0 (2x + 1)(x + 2) = 0 2x + 1 = 0 x+2=0 1 _ x = −2 x = −2

Solve.

1. 2x2 − 5x + 2 = 0 Be sure the equation is in the form ax2 + bx + c = 0. Set up a FOIL table to help factor the equation.

2x2

5x

F 2×1

Use the FOIL table to identify the numbers in each binomial. Write the factored form of the original equation. Set each binomial equal to 0 and solve for the two values of x.

O 2× 2×

+ + +

2 I ×1 ×1

=

?

= =

−4 −5

L

The outer terms are 2 and –2; the inner terms are 1 and –1. 2x2 – 5x + 2 = ( )( )=0 (2x − 1)(x − 2) = 0 2x − 1 = 0 x−2=0 x= x=

2. 6x2 − 23x + 7 = 0 3. 2x2 + x − 3 = 0 4. 3x2 − 7x − 6 = 0 Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Solving a Quadratic Equation by Completing the Square An equation such as x2 + 6x + 7 = 0 is not easy to solve. However, a method called Completing the Square is a way to solve a quadratic equation of this type. Rules for Solving a Quadratic Equation by Completing the Square 1. Identify the coefficient in front of the variable that is not squared. 2. Take _12 of the coefficient in Step 1 and then square that number. 3. Add the result to both sides of the equation. 4. Factor the expression on the left side; add the terms on the right side. 5. Take the square root of each side. 6. Isolate the variable.

Example Solve. x2 − 8x = 5 Step 1 Identify the coefficient in front of the

variable that is not squared. Step 2 Take _12 of the coefficient in Step 1 and square that number. Step 3 Add the result to both sides of the equation. Step 4 Factor the expression on the left side; add the terms on the right side. Step 5 Take the square root of each side. Step 6 Isolate the variable.

The coefficient in front of the variable that is not squared is −8. Half of −8 is −4; (−4)2 = 16. x2 − 8x + 16 = 5 + 16 x2 − 8x + 16 = 5 + 16 (x_______ − 4)2 = 21 ___ ___ = √21 → x − 4 = √21 √(x − 4)2___ x = 4 ± √21

Practice Solve.

1. x2 + 2x = 5 Identify the coefficient in front of the variable that is not squared. Take _12 of the coefficient in Step 1 and

square that number. Add the result to both sides of the equation.

The coefficient in front of the variable that is not squared is Half of is x2 + 2x +

. ;

2

=

.

=5+

Factor the expression on the left side; add x2 + 2x + =5+ the terms on the right side. 2= ____ ___ ___ Take the square root of each side. = √00 √000 = √00 →__ Isolate the variable. x= ±√ 2. x2 − 4x = −3

4. x2 + 2x = 5

3. x2 − 2x = 8

5. x2 + 4x = −1 Algebra

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Evaluating Exponential Functions An exponential function is a function in which a variable is an exponent. In general terms, an exponential function takes the form y = abx. In this general form a is a constant, b is greater than 0 and not equal to 1, and x is a real number. The following are examples of exponential functions: y = 0.23 × 3x

y = −1.5 × 0.75x

y = 3 × 0.4x−2

Rules for Evaluating Exponential Functions 1. For each value of x, plug the number into the variable exponent. 2. Follow order of operations by simplifying the power. 3. Multiply or divide from left to right.

Example Evaluate. y = 2 × 0.75x, for x = 2 Step 1 For each value of x, plug the number

into the variable exponent. Step 2 Follow order of operations by

simplifying the power. Step 3 Multiply or divide from left to right.

y = 2 × 0.75x → replace x with 2 y = 2 × 0.752 y = 2 × 0.752 → (0.752 = 0.5625) y = 2 × 0.5625 y = 2 × 0.5625 = 1.125

Practice Evaluate

1. y = −3 × 3x, for x = 4 For each value of x, plug the number into y = −3 × 3x → replace x with the variable exponent. y = −3 × 3 Follow order of operations by simplifying y = −3 × 3 the power. y = −3 × Multiply or divide from left to right.

y = −3 ×

→3

=

=

2. y = 4 × (−3)x, for x = 3 3. y = 12 × 0.25x, for x = 4 4. y = 3x, for x = −3 5. y = 4 × 2x, for x = −5 6. y = 1.5 × 4x, for x = −2 7. y = 3.2 × 2x, for x = −4


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Exponential Growth Functions You can model exponential growth using the following formula: time period

}

y = C(1 + r)t initial amount

(1 + r) is the growth factor r = growth rate

Rules for Writing and Evaluating an Exponential Growth Function 1. Identify C, the initial amount. Identify r, the growth rate. Identify t, the time. 2. Plug C, r, and t into the formula for exponential growth. 3. Evaluate the equation; the result is the amount after a certain period of time.

Example A savings account starts with a balance of $200.00. Interest on the account is 6% each year. What is the balance after 10 years? Step 1 Identify C; the initial amount.

Identify r; the growth rate. Identify t, the time period.

C, the initial amount is $200.00 r, the growth rate or 6% or 0.06. t, the time period is 10 years.

Step 2 Plug C, r, and t into the formula for

y = 200(1 + 0.06)10

exponential growth. Step 3 Evaluate the equation; the result is the amount after a certain period of time.

y = 200(1 + 0.06)10 = 200(1.06)10 y = 200(1.79) = $358

Practice Solve.

1. A savings account starts with a balance of $500.00 Interest on the account is 10% each year. What is the balance after 5 years? Identify C; the initial amount. Identify r; the growth rate. Identify t, the time period.

C, the initial amount is

.

r, the growth rate, is 10% or t, the time period is

.

years.

Plug C, r, and t into the formula for exponential growth.

y=

(1 +

)

Evaluate the equation; the result is the amount after a certain period of time.

y=

(1 +

)

y=

(

=

(

)

)=

2. A population of bacteria has a growth rate of 2% per hour. You start with 50 bacteria. How many bacteria are there after 20 hours? 3. An organism’s weight increases at a growth rate of 5% each day. If the initial weight is 0.75 grams, what is the weight after 14 days? Algebra

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Exponential Decay Functions You can model exponential decay using the following formula: time period

}

y = C(1 – r)t

(1 − r) is the decay factor

initial amount

r = decay rate

Rules for Writing and Evaluating an Exponential Decay Function 1. Identify C, the initial amount. Identify r, the growth rate. Identify t, the time. 2. Plug C, r, and t into the formula for exponential decay. 3. Evaluate the equation; the result is the amount after a certain period of time.

Example A car was bought for $15,000.00. The value of the car decreases in value by 10% each year. What is the value of the car after 5 years? Step 1 Identify C; the initial amount.

Identify r; the decay rate. Identify t, the time period. Step 2 Plug C, r, and t into the formula for

C, the initial amount, is $15,000.00. r, the decay rate, is 10% or 0.10. t, the time period, is 5 years. y = 15,000(1 – 0.10)5

exponential decay. Step 3 Evaluate the equation; the result is the

amount after a certain period of time.

y = 15,000 (1 − 0.10)5 = 15,000(0.90)5 = 8,857.35

Practice 1. A copy machine is bought for $2,000. The value of the copier decreases at a rate of 25% each year. What is the value of the copier after 4 years? Identify C; the initial amount. Identify r; the decay rate. Identify t, the time period.

C, the initial amount is

.

r, the decay rate, is 25% or t, the time period, is

. years.

Plug C, r, and t into the formula for exponential decay.

y=

(1 –

)

Evaluate the equation; the result is the amount after a certain period of time.

y=

(1 –

)

=

=

2. A business has a profit of $50,000. Profits decrease by 2.5% each year. What is the profit in the 10th year? 3. A truck is bought for $25,000. The value of the truck decreases at a rate of 10.5% per year. What is the value of the truck after 3 years?, 6 years?, 12 years? Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Raising a Power to a Power You can show the repeated multiplication of the same number by using exponents. In an expression such as 43, “4” is known as the base and “3” is known as the exponent. When you raise a power to a power, you take a power (the base and its exponent) and you apply an exponent to the power. The following are examples of raising a power to a power: (53)2 = 53 × 2 = 56

(x4)2 = x4 × 2 = x8

Rules for Raising a Power to a Power 1. Multiply the two exponents.

The product is the new exponent. 2. Do not change the base. 3. Simplify if needed.

Example Simplify. (x5)3 exponents

Step 1 Multiply the two exponents.

The product is the new exponent.

(x5)3

→ 5 × 3 = 15 base

Step 2 Do not change the base.

x15

Step 3 Simplify if needed.

The expression is simplified.

Practice Simplify.

1. (x6)6 Multiply the two exponents. The product is the new exponent.

(x6)6

Do not change the base.

x

Simplify if needed.

The expression

→6×

2. (x5)4

5. (x−2)−4

3. (x2)−3

6. (x−4)−2

4. (x5)5

7. (x−6)−3

=

simplified.

Algebra

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Raising a Product to a Power You can show the repeated multiplication of the same number by using exponents. In an expression such as 43, “4” is known as the base and “3” is known as the exponent. A product is an expression with a number, a variable, and (sometimes) an exponent. When a product is raised to a power, you apply the exponent (the power) to the number, the variable, and (if necessary) the exponent in the expression. Rules for Raising a Product to a Power 1. Distribute the power to each factor. 2. Simplify the number raised to the power. 3. Multiply the exponent to the right of the

variable by the power. 4. Simplify.

Example Simplify: (4x3)4 Step 1 Distribute the power to each factor.

(4x3)4 (4x3)4 = 44(x3)4

Step 2 Simplify the number raised to the

44 = 256

power. Step 3 Multiply the exponent to the right of

(x3)4 = x3 × 4 = x12

the variable by the power. Step 4 Simplify.

256x12

Practice Simplify.

1. (2x3)5 Distribute the power to each factor.

(2x3)5 (2x3)5 = 2

Simplify the number raised to the power.

2

Multiply the exponent to the right of the variable by the power.

(x3)

Simplify.

(x3)

= =x x

2. (5x3)2 3. (4x4)3 4. (3x−4)3 5. (4x2)−3 Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Raising a Quotient to a Power Suppose you raise the following quotient to the 4th power. 4

x (_xy )4 = _xy × _xy × _xy × _xy = __ 4 y

As you can see, you apply the power to the numerator and denominator. Rules for Raising a Quotient to a Power 1. Apply the power to the numerator and

the denominator. 2. Multiply any exponents. 3. Simplify.

Example

x3 2 Simplify. (__ 2) y

Step 1 Apply the power to the numerator

and the denominator. Step 2 Multiply any exponents. Step 3 Simplify.

3

(x3)4 (y )

x 4 ____ (__ 2) = ( 2 4 ) y

3×4 (x3)4 (y ) y 3×4 12 x x (___ ) = (___ ) y2×4 y8

x ___ (____ 2 4 ) = ( 2×4 )

Practice Simplify. x5 3 1. (__ 3) y

5

(x5) (y )

Apply the power to the numerator and the denominator.

x 3 ______) (__ 3) = ( 3

Multiply any exponents.

x _______ (____ ) 3 3) = ( 3 ×

Simplify.

y

5× (x5)3 (y ) y 5×3 x____ x ( 3 × 3 ) = (_____ ) y y

x4 5 2. (__ 2)

y x6 –2 3. (___ ) y–2 3 2 4. (__ ) x4 x5 3 5. (__ 4)

Algebra

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Dividing Powers with the Same Base You can show the repeated multiplication of the same number by using exponents. In an expression, such as 43, “4” is known as the base and “3” is known as the exponent. What happens when you divide powers with the same base? Suppose you have the following problem: 75 ____________ __ = 7 × 7 ×7 ×7 ×7 7 × 7 = 73 72

As you can see, like terms (the factors 7) can be eliminated when they are in the numerator and the denominator. Another way to look at a problem in which you divide powers with the same base is to subtract exponents and keep the base the same. Rules for Dividing Powers with the Same Base 1. Subtract the exponent in the denominator from

the exponent in the numerator. 2. Keep the base the same.

Example Simplify. Step 1 Subtract the exponent in the

denominator from the exponent in the numerator. Step 2 Keep the base the same.

x7 __ = x7−3 x3

x7−3 = x4

Practice Simplify x10 1. ___ 3 x

Subtract the exponent in the denominator from the exponent in the numerator. Keep the base the same.

x10 ___ =x x3

x

=x

x4 2. __ 8 = x

x7 3. ___ 12 = x

2 5

xy 4. ___ 5 3 = xy

4 6

xy 5. ____ 2 –3 = xy


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Simplifying Rational Expressions The following expressions are examples of rational expressions: 4 _ x

5 ___ x+3

x2+2 ____ y2–1

x ___ y–2

As you can see, a rational expression has a variable in the denominator. You may have noticed that a rational expression looks like a fraction. Like a fraction, a rational expression is in simplest form if the numerator and denominator have no common factors other than 1. Rules for Simplifying a Rational Expression 1. Factor the numerator and denominator. 2. Divide out common factors. 3. Simplify.

Example

10x2 Simplify. ____ 4 15x

2

(5x )(2) 10x2 Step 1 Factor the numerator and denominator. ____4 = ________ 2 2

Use guess and test to find two factors that create each expression.

Step 2 Divide out common factors.

A factor in the numerator cancels out the same factor in the denominator.

15x

(5x )(3x )

(5x2)(2) 10x2 ________ ____ = 15x4 (5x2)(3x2)

2 ___ 3x2

Step 3 Simplify.

Practice Simplify. 18x3 1. ____ 3x+6

3

(6x ) 18x3 __________ Factor the numerator and denominator. _____ 3x + 6 = (x + 2) Use guess and test to find two factors that create each expression.

Divide out common factors. A factor in the numerator cancels out the same factor in the denominator. Simplify.

( )(6x3) ____________ ( )(x + 2)

(

)

6x+15 2. _____ 18

3x+12 5. _____ 2x+8

5x4 3. ____ 7

20x+5x 6. ______ 10+5x

35x

4x+12 4. _____ 2 2x

Algebra

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Multiplying Rational Expressions When multiplying a fraction, you multiply the numerators and multiply the denominators. You then simplify the product as needed. 3 1____ ×3 _ 3 1 _ _ 2 × 4 = 2×4 = 8

When multiplying rational expressions you follow the same rules. Rules for Multiplying Rational Expressions 1. Multiply the numerators. 2. Multiply the denominators. 3. Express the resulting rational expression

in simplest form.

Example

5 4 __ Multiply. __ 2 × 3 x

x

Step 1 Multiply the numerators.

4×5 20 4 __ __ × 5 = ______ = ______ x2 x3 (x2)(x3) (x2)(x3)

Step 2 Multiply the denominators.

20 20 ______ = __ (x2)(x3) x5

(Multiplying like variables with exponents means adding the exponents.) Step 3 Express the resulting rational

expression in simplest form.

20 __ is in the simplest form. x5

Practice Multiply. x2 5x __ 1. ___ x–1 × 4 Multiply the numerators.

Multiply the denominators. Express the resulting rational expression in simplest form.

(x2)(5x) ______ x2 5x _______ ____ __ × = = _______ 4 x−1 (x − 1)(4) (x − 1)(4) ______ 00000000 _______ = _______ (x − 1)(4) 00000000 00000000 _______ in the simplest form. 00000000

3x2 ___ 6 2. ___ 2 × 4

5x x+4 ____ 2x2 3. ___ x × 2x−3 4x−1 4. _23 × ____ 4

x 3x x+1 ___ 5. ___ x−2 × 3 6x2 5−2x3 _____ 6. ____ × 2x+2 8x Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Dividing Rational Expressions When dividing fractions you flip the second fraction in the expression and then follow the rules for multiplying fractions. 3 1____ ×3 _ 3 1 _ 2 _ 1 _ _ 2 ÷ 3 = 2 × 2 = 2×2 = 4

When multiplying rational expressions you follow the same rules. Rules for Dividing Rational Expressions 1. Flip the second expression and change the division sign to

a multiplication sign. 2. Multiply the numerators. 3. Multiply the denominators. 4. Express the resulting rational expression in simplest form.

Example

−3 2 Divide. x_____ ÷ _____ x+1 x2 Step 1 Flip the second expression and change

the division sign to a multiplication sign. Step 2 Multiply the numerators.

Leave the numerator in factored form.

x − 3 ____ x − 3 ____ ____ ÷ x +2 1 = ____ × x +2 1 x2 x2 (x − 3)(x + 1) x − 3 ____ ____ × x +2 1 = __________ 2 x (x2)(2)

Step 3 Multiply the denominators.

(x − 3)(x + 1) __________ (x − 3)(x + 1) __________ = (x2)(2) 2x2

Step 4 Express the resulting rational

(x − 3)(x + 1) __________ is in the simplest form. 2x2

expressions in simplest form.

Practice Divide x–2 __ 2 1. ___ x+2 ÷ 2 x

Flip the second expression and change the division sign to a multiplication sign.

x − 2 __ x−2 2 ____ ____ x + 2 ÷ x2 = x + 2 ×

Multiply the numerators. Leave the numerator in factored form.

x−2 ____ x+2 ×

(x − 2)(x2)

= ________ = (x + 2)(2)

Multiply the denominators. Express the resulting rational expressions in simplest form.

is in the simplest form.

3x2 ____ −5x4 2. ___ 2 ÷ 4

– 1 5x +6 _____ 4. x____ x – 3 ÷ x2

+ 2 x____ −1 3. x____ x+3 ÷ 2

+ 6 ____ 8x2 _____ 5. 2x ÷ 4x x+1 Algebra

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Finding the LCD of a Rational Expression You can find the least common denominator of rational expressions in the same way you find the least common denominator of fractions. Rules for Finding the Least Common Denominator (LCD) of a Rational Expression 1. Factor denominators into prime factors. 2. List the number of times each factor appears in the denominators. 3. Multiply the factors from Rule 2 to get the least common denominator. 4. Multiply numerator and denominator by a factor that will result in an

equivalent expression with the LCD.

Example Find the LCD for the following pair of rational expressions. Then rewrite each 3 1 expression with the LCD. __ and ___ 8x 10x Step 1 Factor denominators into prime

8x = 2 × 2 × 2 x

10x = 2 × 5 x

factors. Step 2 List the maximum number of times

each factor appears in each of the denominators.

2 appears three times, 5 appears once, and x appears once.

Step 3 Multiply the factors from Step 2 to get

2 × 2 × 2 × 5 × x = 40x

the least common denominator. Step 4 Multiply numerator and denominator by a factor that will result in an equivalent expression with the LCD.

5 ___ 3 1 4 __ _ _ 8x : multiply by 5 ; 10x : multiply by 4 ´ 5 ___ 3 1 _ 5 4 ___ 12 ___ ___ _ 8x × 5 = 40x 10x × 4 = 40x

Practice Find the LCD of the following pair of rational expressions. Then rewrite each expression with the LCD. 4 ____ 5 1. ___ 3x, 2 x –4x

x2 − 4x = x × (

Factor each denominator into prime factors.

3x = 3 × x

List the maximum number of times each factor appears in each of the denominators.

3 appears

,

x appears

,

(x − 4) appears

Multiply the factors from Step 2 to get the least common denominator.

3×

Multiply numerator and denominator by a factor that will result in an equivalent expression with the LCD.

4 __ 3x × 5 _____ × x2−4x 5 ___ 4. ____ , 8 12x2 8x3 6 ____ 5. ____ , 9 4x2y3 3x4y

2 __ 10 2. __ 5x , 6x 12 __ 7 3. ___ 15x , 4x

)

. = ________

= ________ (3x)(x − 4) ________ = ________ (3x)(x − 4)


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Adding Rational Expressions When adding a fraction, the denominators of each fraction (the bottom number) must be the same. If the denominators are not the same, you must find the least common denominator (LCD). Addition of rational expressions follows the same rules as addition of fractions. Rules for Adding Rational Expressions 1. Find the least common denominator (LCD). 2. Rewrite each expression as an equivalent expression with the LCD

as the denominator. 3. Add the numerators. The LCD is the denominator of the answer. 4. Express the answer in simplest terms.

Example

2x 4 Add. ____ + ___ x–1 x2–1 Step 1 Find the least common denominator.

Step 2 Rewrite each expression as an

equivalent expression with the LCD as the denominator. Step 3 Add the numerators. The LCD is the

denominator of the answer. Step 4 Express the answer in simplest terms.

(x2 − 1) = (x − 1)(x + 1) x − 1 = (x − 1) The LCD = (x + 1)(x − 1) 2x 2x ____ = __________ x2 – 1 (x – 1)(x + 1) x + 1 __________ 4x + 4 4 4 ____ ____ ____ x – 1 = x – 1 × x + 1 = (x + 1)(x – 1) 2x 4x + 4 6x + 4 __________ + __________ = __________ (x – 1)(x + 1) (x + 1)(x – 1) (x + 1)(x – 1) 6x + 4 __________ is in simplest terms. (x + 1)(x – 1)

Practice Add. 3 6 ___ 1. __ 4x + 2 7x

Find the least common denominator.

4x = 2 × 2 × x 7x2 = The LCD is 2 × 2

Rewrite each expression as an equivalent expression with the LCD as the denominator.

3 3 7x __ __ __ 4x = 4x × 7x =

Add the numerators. The LCD is the denominator of the answer. Express the answer in simplest terms. 1 3 __ 2. ___ x+2 + 2

x (–4) x ____ 3. ___ x+3 + x2–9

6 6 ___ = ___ × 2 7x 7x2

+

= = in simplest terms.

4 2 ___ 4. ___ x+3 + x–2 3y 4 5. ___ + __ 2 2x 2x Algebra

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Subtracting Rational Expressions When subtracting fractions, the denominators of each fraction (the bottom number) must be the same. If the denominators are not the same, you must find the least common denominator (LCD). Subtraction of rational expressions follows the same rules as subtraction of fractions. Rules for Subtracting Rational Expressions 1. Find the least common denominator (LCD). 2. Rewrite each expression as an equivalent expression with the LCD as

the denominator. 3. Subtract the numerators. The LCD is the denominator of the answer. 4. Express the answer in simplest terms.

Example

3 8 Subtract. ___ − __ x x–2 Step 1 Find the least common denominator

(LCD). Step 2 Rewrite each expression as an

equivalent expression with the LCD as the denominator. Step 3 Subtract the numerators. The LCD is

the denominator of the answer. Step 4 Express the answer in simplest terms.

List the factors of each denominator. x−2=x−2 x=x So the LCD is x(x − 2) 8 8 8x ___ ___ _x ______ x–2 = x–2 × x = (x)(x–2) 3 _ 3 ___ x–2 ______ 3x–6 _ x = x × x–2 = (x)(x–2) 8x 3x–6 8x–3x+6 ______ ______ − ______ = _______ = 5x+6 (x)(x–2) (x)(x–2) (x)(x–2) (x)(x–2)

The answer is in simplest terms.

Practice Subtract. 7x 6 1. ____ − ___ x2–1 x–1 Find the least common denominator (LCD).

Rewrite each expression as an equivalent expression with the LCD as the denominator. Subtract the numerators. The LCD is the denominator of the answer. Express the answer in simplest terms. 4 2 _ 2. ____ x−1 − x (–8) x ____ 3. ____ x−3 − 2 Algebra

x –9

x2 − 1 = x − 1 = (x − 1) The LCD is 7x _____ already has the LCD x2 − 1 7x 7x _____ = (__________ ). (x − 1)(x + 1) x2 − 1 6 x+1 ___ ____ x–1 × x + 1 = 7x __________ – (x – 1)(x + 1) 7x − 6x + 6 __________ = (x − 1)(x + 1)

The answer

=

in simplest terms.

− x _____ 4 4. 5____ x − x2 + 2x + 3 2x +2 _____ 5. x____ 3x − 4x


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Finding Trigonometric Ratios As you know, the sides of a right triangle exhibit a special relationship known as the Pythagorean theorem. The sides of a right triangle exhibit other special properties. The ratios of different sides of right triangles are called trigonometric ratios. A hypotenuse leg adjacent to A B

C

leg opposite to A

There are 3 basic trigonometric ratios – sine, cosine, and tangent. These ratios are based on the length of two of the sides in a right triangle. length of the leg opposite A opposite ________ sine of A= ___________________ = sin A = length of the hypotenuse hypotenuse length of the leg adjacent A adjacent ________ cosine of A = ___________________ = cos A = length of the hypotenuse hypotenuse length of the leg opposite A opposite ______ tangent of A = ___________________ = tan A = length of the leg adjacent A adjacent

Example Find sin A, cos A, and tan A.

3

5 A

4

length of the leg opposite A

Step 1 Find the sin A

sine of A = ___________________ = _35 length of the hypotenuse

Step 2 Find the cos A

cosine of A = ___________________ = _45 length of the hypotenuse

Step 3 Find the tan A

tangent of A = ___________________ = _3 length of the leg adjacent A 4

length of the leg adjacent A


B

Practice

5

1. Find sin A, cos A, and tan A of Triangle 1.

Triangle 1

Y

13 12

Triangle 2

6 A

10 8

X


Find the sin A

sine of A = ___________________ = length of the hypotenuse

Find the cos A

cosine of A = ___________________ = length of the hypotenuse

Find the tan A

tangent of A = ___________________ = length of the leg adjacent A

length of the leg adjacent A


2. Use Triangle 1 above to find sin B, cos B, tan B. 3. Find the sin X, cos X tan X of Triangle 2. Simplify the ratios. 4. Find the sin Y, cos Y, and tan Y of Triangle 2. Simplify the ratios. Algebra

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Using Trigonometric Ratios to Find a Missing Length As you know the sides of a right triangle exhibit a special relationship known as the Pythagorean theorem. The sides of a right triangle exhibit other special properties. The ratios of different sides of right triangles are called trigonometric ratios. Rules for Finding a Missing Side Length 1. Identify the location of the sides in relation to the given angles. 2. Determine the trigonometric ratio to use. 3. Plug the numbers into the formula for the selected trigonometric ratio. 4. Solve for the unknown. Use your calculator to find the value of the angle.

Example

45˚

16

Find the value of x in the right triangle to the right. x

Step 1 Identify the location of the sides in

relation to the given angles. Step 2 Determine the trigonometric ratio

to use. Step 3 Plug the numbers into the formula for

The missing side is opposite the angle. You know the measure of the hypotenuse. The trigonometric ratio using the opposite side and the hypotenuse is sine. length of the leg opposite A

x sin A = ___________________ = sin 45˚ = __ 16 length of the hypotenuse

the selected trigonometric ratio. x Step 4 Solve for the unknown. Use your sin 45˚ = __ 16 = sin 45˚(16) = x calculator to find the value of the angle. sin 0.707(16) = x = 11.31

Practice

Find the length of the unknown side.

1.

12 x

25˚

Identify the location of the sides in relation to the given angle.

The missing side is

the angle.

Determine the trigonometric ratio to use.

The trigonometric ratio that uses the

You know the measure of the

.

and the the

is

length of the leg adjacent to A

Plug the numbers into the formula for the selected trigonometric ratio.

cosine of A = _____________________ = length of the hypotenuse cos

=

Solve for the unknown. Use your calculator to find the value of the angle.

cos

=

cos

= cos

=x

=x=

2. angle 45˚, side adjacent to angle is unknown, side opposite angle = 10 3. angle 60˚, side adjacent to angle = 15, side opposite angle is unknown Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Theoretical Probability The probability of an event tells you how likely it is that the event will happen. In situations in which each outcome is equally likely, you can find the theoretical probability. number of favorable outcomes theoretical probability = ________________________ total number of possible outcomes

Rules for Finding the Theoretical Probability of an Event 1. Count the total number of outcomes. 2. Count the total number of favorable outcomes. 3. Plug the outcomes into the formula for theoretical probability. 4. Express the probability as a fraction in simplest terms.

Example What is the probability of tossing a “2” on a number cube? Step 1 Count the total number of outcomes. There are 6 possible outcomes, one for each

face of the cube. A number cube has only one number “2.” outcomes. The number of favorable outcomes is 1. Step 3 Plug the outcomes into the formula for theoretical probability = number of favorable outcomes ________________________ theoretical probability. = _1 total number of possible outcomes 6 Step 2 Count the number of favorable

Step 4 Express the probability as a fraction in

simplest terms.

The probability of tossing a “2” is _16 . 1 _ 6 is in simplest terms.

Practice Find the theoretical probability.

1. The names of the days of the week are placed on slips of paper. What is the theoretical probability of picking a slip of paper that has a day of the week starting with an “S”? There are possible outcomes: Count the total number of outcomes. , , , , , , . Count the number of favorable outcomes.

There are

Plug the outcomes into the formula for theoretical probability. Express the probability as a fraction in simplest terms.

theoretical probability = The fraction

favorable outcomes, and .

in the simplest terms.

The names of each month are placed on slips of paper.

2. What is the probability of picking a slip with a month that begins with a “M”? 3. What is the probability of picking a slip with a month that ends with a “R”? Algebra

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Experimental Probability The probability of an event tells you how likely it is that the event will happen. There are situations in which the probability of an event occurring is not known without performing an experiment. Rules for Finding Experimental Probability 1. Identify the number of times the event occurs. 2. Identify the number of times the experiment is done. 3. Plug the numbers into the formula for experimental probability. 4. Express the probability as a fraction in simplest terms or as a decimal.

Example A basketball player made 122 shots out of 250 attempts. What is the probability of the player making a shot? Step 1 Identify the number of times the event The number of times the event occurs (shots

occurs. Step 2 Identify the number of times the

experiment is done.

made) is 122. The number of times the experiment was done (number of attempts, or shots) is 250.

Step 3 Plug the numbers into the formula for theoretical probability =

experimental probability. Step 4 Express the probability as a fraction in

simplest terms or as a decimal.

number of times event occurs 122 ________________________ = ___ number of times experiment done 250 61 122 ___ ___ 250 = 125 = 0.488

Practice Find the probability of the event occurring.

1. A 6–sided number cube comes up “6” 44 times out of 250 throws. Identify the number of times the event occurs.

The number of times the event occurs:

Identify the number of times the experiment is done.

The number of times the experiment was

Plug the numbers into the formula for experimental probability.

theoretical probability =

done:

Express the probability as a fraction in simplest terms or as a decimal.

.

.

=

=

=

2. A number cube is rolled 500 times. A 1 or 2 is rolled 188 times. 3. A baseball player gets 215 hits in 625 times at bat. 4. A coin is tossed 1000 times. Heads is tossed 495 times. Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Mean In statistics, there are several ways to use a single number to represent a data set. One method is to calculate the mean. The mean of a set of data is the sum of the numbers divided by how many numbers there are in the data set. The mean is often referred to as the average of a set of data. Rules for Finding the Mean 1. Add all the numbers in the data set. 2. Divide that result by how many

numbers are in the data set.

Example Find the mean of this data set: 25, 30, 30, 47, 28 Step 1 Add all the numbers in the data set.

25 + 30 + 30 + 47 + 28 = 160

Step 2 Divide the result by how many

There are 5 numbers in the data set. 160 ___ 5 = 32

numbers are in the data set.

Practice Find the mean of the following sets.

1. 56, 44, 63, 58, 51, 59 Add all the numbers in the data set.

56 + 44 + 63 + 58 + 51 + 59 =

Divide the result by how many numbers are in the data set.

There are

numbers in the data set.

=

2. 18, 17, 20, 26, 24 3. 43, 36, 38, 38, 63 4. 67, 50, 65, 49, 66, 63 5. 45, 47, 34, 36, 38 6. 18, 35, 28, 15, 36, 10 7. 141, 154, 148, 172, 161, 155

Algebra

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Median In statistics, there are several ways to use a single number to represent a set of data. One method is to find the median value of a data set. In fact, you may have seen news reports about housing in your area, which report the median price of a home. The median is the middle number when the numbers are written in order. Rules for Finding the Median 1. Place the numbers in order from least to greatest. 2. Count the number of items in the data set. 3a. If there are an odd number of data points, the median

is the middle item. 3b. If there is an even number of items, the median is the average of the two middle numbers.

Example Find the median of this set: 10, 2, 14, 12, 17, 6, 15, 9, 19 Step 1 Place the numbers in order from least

2, 6, 9, 10, 12, 14, 15, 17, 19

to greatest. Step 2 Count the number of data points.

There are nine data points.

Step 3 If there is an odd number of data, you

Nine is an odd number. 2, 6, 9, 10, 12, 14, 15, 17, 19 The middle value is 12.

find the middle value.

Practice Find the median of the following data sets.

1. 72, 83, 80, 79, 89, 84 Place the numbers in order from least to greatest. Count the number of data points.

There are

If there is an even number of items then find the average of the middle two numbers.

There is an

data points. number of values.

The two middle numbers are so,

; .

2. 17, 23, 32, 19, 21 3. 67, 62, 55, 49 4. 17, 10, 9, 21, 17, 4 Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Permutations When you can find the number of possible outcomes in a situation when order does matter, you are finding the number of permutations. The formula for finding the number of permutations of a certain number of things (n) taken a certain number at a time (r) is: n = number of things from which to choose. nPr

n! = _____ (n–r)!

r = number of things chosen at a time. ! = factorial–multiplying all whole numbers from that number to 1 (e.g. 3! = 3 × 2 × 1 = 6). Note that 0! = 1.

Example A softball team fields 10 players. How many different batting orders can there be? Step 1 What is n (the number of things from

There are 10 players to place in the order.

which to chose)? Step 2 What is r (number of things taken at

All ten are taken at a time.

a time)? Step 3 Plug the numbers into the equation

for combinations.

10P10

10! = ______ (10–10)! 10! = ___ 0!

3,628,800 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 __________________________ = _______ = 1 1

3,628,800

Practice 1. Eight students run for class office. The student getting the most votes is class president; the student with the second highest total votes is vice-president. The student with the third most votes is class secretary. How many possible ways can the students finish in the vote? What is n (the number of things from which to chose)?

There are office.

What is r (number of things taken at a time)?

Only

Plug the numbers into the equation for combinations.

8P3

students running for class can be elected.

8! = ______ 0000000 8! = _____ 00000 40,320

8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 _____ = ____________________ = 000 = 000

2. There are 8 runners in a road race. How many different ways can the runners finish 1st, 2nd and 3rd? 3. You must read 5 books over the summer. How many choices do you have in picking out the first two books? 4. Your class has 30 students. Everyone is eligible for class president and vice-president. How many different pairs are possible? Algebra

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Combinations When you are finding the number of possible outcomes in a situation when order does not matter, you are finding the number of combinations. The formula for finding the number of combinations of a certain number of things (n) taken a certain number at a time (r) is: n = number of things from which to choose. n! ______ nCr = r!(n–r)!

r = number of things chosen at a time. ! = factorial–multiplying all whole numbers from that number to 1 (e.g. 3! = 3 × 2 × 1 = 6)

Example Five people volunteer to work the school dance. Only two can collect tickets. How many different combinations of people can collect tickets? Step 1 What is n (the number of toppings

There are 5 people from which to choose.

from which to chose)? Step 2 What is r (number of things chosen at

There are 2 people chosen at a time.

a time)? 5! Step 3 Plug the numbers into the equation for 5C2 = ______ 2!(5–2)!

combinations.

Practice

5! = ____ 2!(3)! 5×4×3×2×1 120 = _____________ = ___ 12 = 10 (2 × 1)(3 × 2 × 1)

1. Ten students run for student council. Only 3 are elected. How many ways can the students be elected? What is n (the number of people from which to chose)?

There are

people from which to choose.

What is r (number of things chosen at a time)?

There are

people chosen at a time.

Plug the numbers into the equation for combinations.

10C3

10! = _______ 00000000 10! = ______ 000000 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = __________________________ 00000 3,628,800

= _______ = 120 000 2. An ice cream shop offers 6 kinds of toppings. You can choose any 2 at no cost. How many different topping combinations (at no charge) are there? 3. A book club offers 3 free books with a membership. You have 7 books from which to choose. How many different pairs of books are there? Algebra Saddleback Educational Publishing ©2006 • (888) 735-2225 • www.sdlback.com

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Name

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Matrices A matrix is a rectangular arrangement of numbers in rows and columns. You can arrange data in a chart or in a matrix: Public

Private

Elementary

29,300

2,900

29,300

2,900

Middle School

23,200

2,100

23,200

2,100

High School

15,400

1,200

15,400

1,200

Rules for Creating and Reading a Matrix 1. Create a matrix with dimensions that match the data table. 2. Transfer each data item from the table to its corresponding

position in the matrix.

Example Place the data from the table into a matrix. 1 10 2

2 20 4

3 30 6

4 40 8

Step 1 Create a matrix with dimensions that

match the data table. Step 2 Transfer each data item from the table to its corresponding position in the matrix.

Practice

The data in the table is in 3 rows and 4 columns; create a 3 × 4 matrix. 1

2

3

4

10

20

30

40

2

4

6

8

Place the data from the table into a matrix.

1. 25

50

75

100

125

150

200

180

160

140

120

100

Create a matrix with dimensions that match the data table.

The data in the table is in

Transfer each data item from the table to its corresponding position in the matrix.

matrix.

columns; create a

rows and ×

2. Northeast 1,720 2,889

North central 2,646 3,963

South 2,306 2,889

West 2,692 2,808

Algebra

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Name

Date

Matrix Addition When adding matrices, you add the corresponding elements in each matrix. corresponding elements

–2 4

0 2

+

3

–1

4

7

Rules for Matrix Addition 1. Combine corresponding elements in each matrix to form one large

matrix. Place a plus sign between each corresponding element. 2. Add the corresponding elements.

Example Add.

–4

2

–10

7

+

5

–9

9

–3

Step 1 Combine corresponding elements

in each matrix to form one large matrix. Place a plus sign between each corresponding element. Step 2 Add the corresponding elements.

–4

2

−10

7

+

5

−9

9

−3

–4 + 5 2 + (–9) (–10) + 9 7 + (–3)

=

–4 + 5 2+ (–9)

=

(–10) + 9 7 + (–3)

1

–7

–1

4

Practice Add.

1.

–5

8

3

–3

+ 11

–1

5

–7

Combine corresponding elements in each matrix to form one large matrix. Place a plus sign between each corresponding element.

–5

8

3

–3

+

11

–1

5

–7

Add the corresponding elements. 2. 3. 4.

2

–9

–4

3

–1

5

–4

7

–9

3

2 17

+

+ 5

4

16

9

11

1

2

+ 7

2

–4

20

=

= =

=

=

–12


97

Name

Date

Matrix Subtraction When subtracting matrices, you subtract the corresponding elements in each matrix. corresponding elements

–2

0

4

2

–

3

–1

–4

7

Rules for Matrix Subtraction 1. Combine corresponding elements in each

matrix to form one large matrix; place a minus sign between corresponding elements. 2. Subtract corresponding elements.

Example Subtract.

–2

5

0

–2

–

–4

6

8

5

Step 1 Combine corresponding elements

in each matrix to form one large matrix; place a minus sign between corresponding elements. Step 2 Subtract corresponding elements.

–2

5

–

0 −2

–4

6

8

5

–2–(–4)

5–6

0–8

–2–5

=

=

–2–(–4)

5–6

0–8

–2–5

2

–1

–8

–7

Practice Subtract. 3

3

–4

–1

1.

–

6

–2

8

–2

Combine corresponding elements in each matrix to form one large matrix; place a minus sign between corresponding elements. Subtract corresponding elements. 2. 3. 4.

–3

5

0

–4

9

–12

15

4

7

–5

6

12

–8

–4

–

–

–5

9

10

3

– –9

0

–2

–9

2

4

–3

5

–1

–3

3

3

–4

–1

–

6

–2

8

–2

=

=

Algebra

98


Name

Date

Scalar Multiplication A matrix is a rectangular arrangement of numbers in rows and columns. You can think of a matrix as a way to organize data, similar to the way data is displayed in a table. You can multiply all elements of a matrix by a single number known as a scalar. A scalar increases the value of each element by the same proportion. Rules for Scalar Multiplication 1. Create an expanded matrix in which each element is

multiplied by the scalar. 2. Find the product of each element times each element.

Example Solve.

2

–6

4

7

–3

Step 1 Create an expanded matrix in which

each element is multiplied by the scalar. 2

–6 7

4 –3

–6 × 2 7×2

Step 2 Find the product of each element times

each element.

=

–6 × 2 7×2

4×2 –3 × 2

=

4×2 –3 × 2 –12 14

8 –6

Practice Solve.

1. 5

11 –5

–9 6

–4 3

Create an expanded matrix in which each element is multiplied by the scalar. Find the product of each element times each element

2. –3

2 9 –11

3. 4

5 8

4. –6

–8 0

5

11 –5

–9 6

–4 3

11 × 5 –5 × 5

= =

11 × 5 –5 × 5 55 –25

16 –2 6 –12 –2

–4 2

1 –9


99

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Date

Matrix Multiplication When multiplying matrices, you multiply the elements of a row by the corresponding elements in a column of the second matrix. You then add the products. 3

6

5

5

3

8

5

×

2

=

2

3×5

+

6×2

+

5×2

5×5

+

3×2

+

8×2

Rules for Matrix Multiplication 1. Identify the elements to be multiplied. For example, circle the first column of

the first matrix; circle the top number in each column of the second matrix. In the same manner, use squares for the second column, and triangles for the third. 2. Create a large matrix showing the multiplication of the elements. 3. Find the products of each element times each element. Add the products in each row. The number of columns in the answer is the same as the number of columns in the second matrix.

Example

3

5

Multiply.

2

–3

×

2 9

Step 1 Identify the elements to be multiplied. Step 2 Create a large matrix showing the

multiplication of the elements. Step 3 Find the products of each element

times each element. Add the products in each row.

Practice Multiply.

1.

2

–4

3

7

×

1

9

3

2

Identify the elements to be multiplied. Create a large matrix showing the multiplication of the elements.

2.

Find the products of each element times each element. Add the products in each row. 7 6 3 8 × 4 9 2 4 2

3

5

2

–3

3

5

2

–3

3×2 +

2

×

9 2

×

9

5×9

=

2×2 + –3×9

2

–4

3

7

2

–4

3

7

× ×

3×2 +

=

2×2 + –3×9

6 + 45 4 + –27

1

9

3

2

1

9

3

2

5×9

=

51 −23

=

= 3.

2

5

4

6

×

9

4

2

3

Algebra

100


Answer Key PAGE 1__ Classifying Numbers 1. –√ 2 ,

0, 1.35, 5_23, 2, 5, 0, 0, 2, 2 ___ __ 2. –√ 10 , √ 3 3. −4.2, −3, −1, −_13 , 0, 0.34, 1.5, 4 4. −3, −1, 0, 4 5. 0, 4

PAGE 2 Order of Operations 1. 15 × 4 +24 − 3

15 × 4 + 16 – 3 60 + 16 – 3 73 2. −10 3. −5 4. 23 5. 10 6. 104 7. 72 PAGE 3 Writing a Variable

Expression—Addition & Multiplication 1. 5 × c − 15 2. 12 + 2x 3. 5x • 6 4. 9 + x + 6 5. 18 + (20t) 6. 10 + 50x 7. 7x + 6y 8. 12 + (x + 6)

PAGE 4 Writing a Variable

Expression—Subtraction & Division 1. 13 − 12 ÷ x 2. 9 − x ÷ 5 3. 5 − x ÷ 2 4. x −16 ÷ 4 5. 18 − x − 6 6. 14 ÷ n − 3

PAGE 5 Evaluating Variable

Expressions

1. 5, 3, 3(5) − 2 (3) + 4,

15 − 6 + 4 = 13 2. 16 3. 33 4. 17 5. 22 6. 45 7. 20

PAGE 6 Simplifying Variable

Expressions 1. 6x, 3x2, 7, 6x, 8x, 3x2, x2, 7, 11, x2 + 8x + 11 2. 2x + y + 8xy 3. 3x + 4y2 4. 22x + 7 5. 6x3 − 17x2 − 16 6. 2x2y − 3xy2 + 2x − 3

PAGE 7 Adding Integers Using

Absolute Value

1. 27, 19,

subtract, 8, negative, −8 2. 45, 55, subtract, 10 negative, −10 3. 4 4. 58 5. −34 6. −22 7. 34 8. −40 PAGE 8 Subtracting Integers 1. positive 2. negative 3. positive 4. 17

PAGE 10 Dividing Integers

positive, positive negative, negative positive, negative negative, positive 1. negative 2. positive 3. positive 4. negative 5. negative, positive negative 6 −6 6. 20 7. −11 8. −16 9. 7 10. 9 11. −12 12. 11 13. −16 PAGE 11 Distributive Property 1. (5 • 4)

(5 • 4), − (5 • 4) − (5 • 4), 5x − 20 2. 6 − 8x 3. –12x2 + 6x 5. 4x2 + 36 6. 6x + 36 7. 6x2+ 12x

+, 9, 17 subtract, positive 8, 8 5. −21 6. −4 7. −14 8. 8 9. 18 10. 0

PAGE 12 Exponents 1. 4 × 4, 64 2. 2, 6, 2 × 2 × 2 × 2 × 2 × 2 =26 3. 5 × 5 × 5 × 5 = 625 4. 11 × 11 × 11 = 1331 5. 9 × 9 × 9 × 9 × 9 = 59049 6. 15 7. 122 8. 48

PAGE 9 Multiplying Integers

PAGE 13 Negative Exponents 1. exponent, base

positive, positive negative, positive negative, negative positive, negative, negative 1. negative 2. positive 3. negative 4. positive 5. negative, positive negative 42 −42 6. −40 7. 96 8. −72 9. −49 10. −60 11. −34 12. 64 13. 33

2. 3. 4. 5. 6. 7.

1 __ 25 1 1 1 ____________ __ , , __ 25 2 × 2 × 2 × 2 × 2 32 1 _ 8 1 _ 9 1 −___ 125 1 __ 81 1 __ a2 1 4 4__ = __ x3 x3


101

PAGE 14 Scientific Notation 1. 2

2 −2 2. 1.243×103 3. 1.0045×104 4. 1.423×106 5. 4.2×10−3 6. 7.5×10−4 7. 3.03×10−5 PAGE 15 Fractional Exponents 1. 5th, square root

5th, 5, 12 × 12 × 12 × 12 × 12 = 248,832 _______ square root √248,832 = 498.8 2. 4.76 3. 2.51 4. 6.35 5. 1.59

PAGE 16 Writing an Equation

from a Table

1. _24 = _12

PAGE 17 Writing an Equation

from a Word Problem 1. total cost, times, $5 shipping total cost, number of DVDs, n Total cost, times, =, ×, n 2. c = 5n + 3 3. t = 0.35d + 0.25c

PAGE 18 Solving One-Step

2. 3. 4. 5. 6. 7.

Step Equations by Multiplying or Dividing 1. divided, x division, 8, 192, __ 8 , 192 2. x = 9 3. x = 2 4. x =10 5. x = 588 6. x =2400

PAGE 20 Solving Two-Step

Equations

1. − 7, − 7, 63, 63

63, ÷ 7, ÷ 7, 9, 9 2. x = 28 3. x = 168 4. x = 8 5. x = 24 6. x = 40 7. x =11 PAGE 21 Solving Multi-Step 1. 7x

−5, −9 (−9) = _12 (x − (−5)) 2. y – 8 = 2 (x − 4) 3. y − 12 = −2 (x − (−10)

1. −125

PAGE 19 Solving One-

Equations by Adding or Subtracting

+ 125, + 125, + 125, 0 x = 96 x = 71 x = 207 x=0 x = 107 x = 211

Equations

7x, −12, −12, 7x, 49 7x, 49, 7 2. x = 9 3. x = 7 4. x = 5 5. x = 21 6. x = 3 7. x = 14 PAGE 22 Solving Equations 1. 5x

with Variables on Both Sides

20 ÷ 5, ÷ 5, 4 2. x = −2 3. x = 6 4. x = −6 5. x = 3 6. x = 5 PAGE 23 Identifying a Function 1. 3, 4, 4, 5, 6, 6

18, 16, 17, 22, 25, 28 6, 6, 28 2. yes 3. no 4. yes PAGE 24 Writing a Function Rule

from a Table

1. multiply by 3

multiply by 3 3x 2. y = x − 2

PAGE 25 Types of Functions 1. No

No Yes rational 2. quadratic 3. linear 4. rational 5. rational 6. exponential 7. linear PAGE 26 Plotting Points on a 1. 7, left 2. 3. 4. 5. 6. 7. 8.

Coordinate Plane

down, 4, check graph check graph check graph check graph (2, 2) (5, -4) (−3, 7) (−9, −5)

PAGE 27 Finding Solutions of

Linear Equations

1.

x

y = 3x − 2

y

−2

y = 3 (—2) − 2

–8

0

y = 3 (—0) − 2

–2

1

y = 3 (1) − 2

1

2

y = 3 (2) − 2

4

(−2, −8), (0, −2), (1, 1), and (2, 4) answers may vary, possible answers 2. (0, −4), (−1, −9), (1, 1), (2, 6) 3. (−2, 0), (0, 1), (2, 2), (4, 3) 4. (−2, 5), (−1, 4), (0, 3), (1, 2) 5. (−1, −6), (0, −2), (1, 2), (2, 6) 6. (−3, −2), (0, −1), (3, 0), (6, 1) 7. (−1, 0), (0, 1), (1, 2), (2, 3) PAGE 28 Graphing a Linear

Equation

1. x

y = 3x − 1

y

(x, y)

2

y = 3(2) − 1

5

(2, 5)

0

y = 3(0) − 1

—1

(0, −1)

1

y = 3(1) − 1

2

(1, 2)

–2

y = 3(—2) − 1

—7

(—2, −7)

2. 3. 4. 5. 6.

check graphs, possible answers (−1, 4),(0, 3),(1, 2),(2, 1) (−1, 1),(0, −1),(1, −3),(2, −5) (−1, 1),(0, 5),(1, 9),(2, 13) (−3, 0),(0, 1),(3, 2),(6, 3) (−1, 1),(0, 3),(1, 5),(2, 7)

Algebra

102


PAGE 29 Direct Variation 1. increases

is not does was not, does not 2. no PAGE 30 Inverse Variation 1. increases

is, 36, 36 yes, yes, does 2. no 3. yes 4. yes PAGE 31 Slope of a Line 1. (6, 8) 8 _ 6 3 8 _ _ 6 , 3 , 1, 1 2. _67

3. 1

4. –_13 5. 1

−2 6. __ 3

7. _34

PAGE 32 Slope Intercept Form 1. −3

7 2. slope = _13 , y–int. =−3 3. slope = 1, y-int. = 1 4. slope = _34 , y-int. = −13 5. slope = –_23 , y-int. = 3 PAGE 33 Point Slope Form I 1. 6

−3−1 y – (−1) = 6 (x − (−3)) 2. y − 1 = –_12 (x − 7) 3. y− (−3) = 2(x − (−3)) 4. y − (−5) = _23 (x−4) 5. y − 3 = −3 (x + 1) PAGE 34 Point Slope Form II 1. −2, −2

y − (−2) = −1(x − (−2)) 2. y − 2 = _12 (x − 2) 3. y − 4 = −1(x − (−6)) 4. y − 6 = 3(x − 2) 5. y − 2 = 1(x − 5) 6. y − 0 = _23 (x −6)

PAGE 35 Parallel Lines 1. _12 x − 1

3, _12

are not, are not 2. m = –_14 for both equations; the graphs are parallel 3. m = −2 for both equations; the graphs are parallel. 4. m = 2 and −1; the graphs are not parallel. 5. m = –_14 and –_43 , the graphs are not parallel. PAGE 36 Perpendicular Lines 1. 2

negative, positive, 2, 2, −3 2. y = −_43 x + 4 3. y = −_12 x+3

PAGE 37 Using Reciprocals 1. –_52

–5 5 _ negative, __ 2 , positive, 2

2. 3. 4. 5.

7 7 _ _ 1 or 7 and − 1 or −7 3 3 _ _ 5 and − 5 13 13 __ __ 5 and − 5 8 8 _ _ 9 and − 9

PAGE 38 Solving Equations That 1. 2, 2

Contain Decimals

2 100, 100, 100, 30, 1625, 2265 30, 2265, 1625 30, 640 21.33 2. x = 6.8 3. x = 3.1666 PAGE 39 Solving Equations that

Contain Fractions

1. _35 6 __ 10 6 __ 11 __ 10 , 10 10 ___ 100 10 __ 11 __ 1 __ __ 11 ,10 , 11 , 11 =9 11 2. x = 36 3. x = 10 4. x = 5 5. x = 108

PAGE 40 Graphing Linear

Inequalities

1. true 2. check graph 3. check graph

PAGE 41 Writing Inequalities 1. 6, 6

from a Graph

left, open, 3

PAGE 42 Solving One-Step 1. + 16 2. 3. 4. 5. 6. 7.

Inequalities by Adding or Subtracting

−16, −16 +, −16, 28, 0, 28 x < −2 x>8 53 > x 12 < x 37 > x x < −22

PAGE 43 Solving One-Step

Inequalities by Multiplying or Dividing 1. divided division, −3 × − 3, ×−3, −27, >, −27 2. x  8 3. x > −8 4. x  9 5. x > 28 6. x  −50 7. x < −48

PAGE 44 Solving Two-Step

Inequalities

1. + 7, + 7, − 0, 24 2. 3. 4. 5. 6. 7.

÷ −2, ÷ −2, −12, −8 x < −5 x < −65 x < 60 x  −12

PAGE 45 The Pythagorean 1. a leg

Theorem

15, c2, 225, c2 2, 289, c2, 17, c 225, c___ 2. a = √ 80 3. c = 5 4. c = 25 5. b = 150


103

PAGE 46 Irrational Numbers 1. is not, 10, rational

2.64575131 2.64575131, irrational 2. 1.1 – rational 3. 5 – rational 4. 0.57735026 – irrational 5. 3.31662479 – irrational PAGE 47 Simplifying Radical

1. 6x

Expressions by Multiplying Two Radicals

1. 4

Roots

PAGE 48 Simplifying Radical

Expressions by Removing Perfect Squares 1. 27, 9, 9 9, 9 __ 9, 3,__3√3 2. 10√ __5 3. 4√__ 5 4. 4√__ 3 5. 5√ 3

PAGE 49 Simplifying Radical

Expressions with Variables

16x6, 16x6 __ 2. 5x√ 2 ___ 3. 4x2 √ 3x __ 4. 3x3√ 7 ___ 5. 6x4 √ 5x PAGE 50 Solving a Radical

Equation by Isolating the Radical 1. −1, −1, 8 2, 8, 64, +4, 64, 4, 68 2. x = 25 3. x = 98 4. x = 46 5. x = 57 6. x = 4

PAGE 57 Subtracting

Polynomials

1. (3x2 − 5x + 10)

25, 25, 5 4, 5 2. between 2 and 3 3. between 5 and 6 4. between 5 and 6 5. between 7 and 8 6. between 11 and 12 PAGE 52 Estimating Cube Roots

6x, (3×6)(x×x), 18x2 __ 2, 9x2, 2, 3x√ 2 18x ____ ____ 2. √___ 40x5 = 2x2 √10x __ 3. √____ 4x9 = 2x4√x ___ 4. √____ 50x5 = 5x2 √2x 5. √ 16x8 = 4x4

1. 16x6

PAGE 51 Estimating Square

and Higher Power Roots 1. 3, cube is not 4, 4, 5, 5, 4, 5 2. between 5 and 6 3. 2 4. between 5 and 6 5. between 3 and 4 6. 5 7. 6

PAGE 53 Multiplying a

Polynomial by a Monomial x5 __ 6 1. −4x , 2 , 10, −10x4 2. −2x5 − 8x4 + 4x3 3. 3x8 + 7x6 − 10x4 3

3x 4. 6x5 + 15x4 − ___ 4

PAGE 54 Multiplying Binomials 1. 12x2

6x (6x), 6x (1)(3), 3 12x2 + 6x + 6x + 3 = 12x2 + 12x + 3 2. 6x2 − 11x – 10 3. 12x2 − 26x +12 4. 2x2 − 8x − 10 5. 6x2 − 4x − 4 PAGE 55 Squaring a Binomial 1. 25x2

20x, minus 25x2 − 20x + 4 2. x2 + 8x + 16 3. x2 − 16x + 64 4. 4x2 + 24x + 36 5. 16x2 − 32x + 16 6. 36x2 + 144x + 144 PAGE 56 Adding Polynomials 1. (3x3 + x2+ 5)

3x3 + x2 + 5 3x3 + x2 + 5, 19x3 − x2 + 10 2. −3x2 − x + 5 3. 6x3 + 3x2 + 5 4. 3x2y + 6xy + 17 5. 2y4 + 7y3 + 2y2 + 7

(3x2 − 5x +10) −3x2 + 5x − 10 (−3x2 + 5x − 10), x2 + 6x − 5 2. −9x3 + 3x2 + 15 3. x3 − 8x2 − 5x − 18 4. −7x2 + 8x + 1 5. −4x3 − 10x2 + 4x + 15 PAGE 58 Multiplying a 1. 2. 3. 4. 5.

Polynomial 12x5, 6x4, −3x3, 12x5 + 6x4 − 3x3 −21x5 + 14x2 8x6 − 2x5 + 6x4 3x7 + 6x5 + x3 −20x8 − 10x7 + 25x4

PAGE 59 Factoring a Binomial 1. 5x3

5x3, 2x, 5x3, −3 5x3, 2x − 3 2. 9x4 (3 − x) 3. 12x(3x2 + 2) 4. 5xy5(x + 3y2)

PAGE 60 Finding the Greatest

Common Factor for Variable Terms 1. 1, 2, 3, 6, 9, 18 6 m5, n4, n2 6, n2 2. 2x4 3. 3m2 4. 4x8 5. 2xy2

PAGE 61 Factoring a Polynomial 1. 1, 13, x8; 1, 2, 13, 26, x4; 1, 3, 13,

39, x2 2x2, −3 + 2x2 − 3 2. 8x (x2 − 3x −10) 3. 3x2(x2 + 10x + 16) 4. x2 (1 + 2x3 − 9x5) 5. 3x3 (x2 − 2xy − 15y2)

PAGE 62 Factoring Trinomials in 1. 2. 3. 4. 5.

the Form x2 + bx + c −2, −6, −8 (x + 3)(x + 1) (x + 1)(x + 8) (x + 5)(x + 3) (x − 3)(x − 12)

PAGE 63 Factoring Trinomials in

the Form ax2 + bx + c

1. −3 × 1, 5x − 3, −14

5 × 1, 3 × 1, −2, −1 × 3, 14 5, 1, x, −3, (5x + 1)(x −3) 2. (2x + 4)(x + 2)

Algebra

104


3. 4. 5. 6.

(x − 3)(2x + 3) (7x + 1)(x + 7) (2x − 2)(x − 3) (2x + 1)(x − 4)

PAGE 64 The Difference of 1. x8

Two Squares

y2 x8, y2, x8, y2 2. x6 − y4 3. x10 − 25 4. 4x6 − 16 5. 16x4 − 9 6. 9x8 − y4 PAGE 65 Solving Systems of

Equations by Graphing

1. 1, (2, 1), −7, (0, −7)

5, (3, 5), 8, (0, 8) (3, 5) 2. (5, 1) 3. (1, 1) 4. (−2, −1) 5. (0, 5) PAGE 66 Solving Systems 1. x − 1

of Equations by Substitution

+ x, x − 1, + x, 2x − 1 + 1, 2x − 1, + 1, 2x (3, 2) 2. (3, 5) 3. (1, 6) 4. (2, 6) 5. (_52 , 5) PAGE 67 Solving Systems 1. 4x

of Equations by Elimination

4y = 16, 4 (4), 12, 0, 0 2. (–9, 7) 3. (3, 4) 4. (4, –2) 5. (3, –1) PAGE 68 Solving Linear 1. 2

Systems by Multiplying

2, 10x − 10y = 12 10x − 10y = 12, 8x + 0y = 32, 4 (4), −8, 2.8, (4, 2.8) 2. (5, 1) 3. (3, −6) 4. (–2, 2.5) 5. (3, 0.5)

PAGE 69 Solving Quadratic 1. 75

Equations Using Square Roots

25 25, 5, −5 2. x = 7, x = −7 3. x = 2, x = −2 4. x = 5, x = −5 5. x = 10, x = −10 PAGE 70 The Quadratic Formula 1. a = 1 b = 3 c = −4

3, 3, 1, –4, 1 9, 16, 25 5, 2, −8, 1, −4 2. x = –7.22, x = −22.78 3. x = 59.27, x = 84.73 4. x = 9, x = 11 5. x = 4.65, x = 19.35 PAGE 71 Using the Discriminant 1. 3, 7

3, (7) 3, (7), 9, 28, −19, negative, no 2. the result is 0; there is one solution 3. the result is 17; there are two solutions 4. the result is 20; there are two solutions 5. the result is 0; there is one solution 6. the result is -63; there are no solutions PAGE 72 Zero-Product Property 1. (x − 2)

x − 2 + 2, + 2, 2 2. −8, 1 3. _72 , 3 4. –_56 , −4 5. 2, 3 PAGE 73 Solving a Quadratic

Equation by Factoring 1. −1, −2, −1 × −2 −2, −1 2x − 1, x − 2 1 _ 2, 2 2. (3x − 1)(2x - 7) =0; x = _13 , _72 3. (2x+ 3)(x − 1) = 0; x = −_32 , 1

4. (x − 3)(3x + 2) = 0; x = 3, −_23

PAGE 74 Solving a Quadratic 1. 2

Equation by Completing the Square

2, 1, 1, 1 1, 1 1, 1, (x + 1), 6 (x + 1)2, 6, (x + 1), 6 −1, 6 __ 2. x = 2 ± √__ 1 = 3 or 1 3. x = 1 ± √ 9 __= 4 or −2 4. x = −1 ± √__ 6 5. x = −2 ± √ 3 PAGE 75 Evaluating Exponential 1. 4, 4

Functions

4, 4, 81, 81 81, −243 2. −108 3. 0.046875 1 4. __ 27

4 4 5. __ = __ = _1 = 0.125 25 32 8 6. 0.09375 3.2 7. ___ 16 = 0.2

PAGE 76 Exponential Growth

Functions

1. $500.00, 0.10, 5

500, 0.10, 5 500, 0.10, 5, 500, 1.1, 5 500, 1.61, $805 2. 74.50 3. 1.41 PAGE 77 Exponential Decay

Functions

1. $2000, 0.25, 4

2000, 0.25, 4 2000, 0.25, 4, 2000, 0.75, 4, 632.81 2. 38,816.48 3. 17,922.93, 12,849.27, 6,604.14 PAGE 78 Raising a Power to a

Power

1. exponents, 6, 36, base

36 is 2. x20 3. x−6 4. x25 5. x8 6. x8 7. x18


105

PAGE 79 Raising a Product to a 1. 5, 5

Power

x2 __ x2 1. __ 2, 2

5, 32 5, 15 32, 15 2. 25x6 3. 64x12 27 4. 27x−12 = ___ x12 1 5. 4−3(x−6) = ____ 6

is

12x2

6x 2. _____4 = ____ 2

PAGE 80 Raising a Quotient to a

Power

1. 3, 3, 3, 3, 15, 9

Rational Expression

y x−12 ____ 3. ___ = 121 4 y4 x y 9 4. __ x8 15 ___ 5. x64

1. once, once, once

x × (x − 4), 3x (x − 4) 3 x−4 _ ___ x−4 , 4x−16, 3 , 15

PAGE 81 Dividing Powers with

the Same Base

10-3, 7 2. x−4 3. x−5 4. x−3y2 5. x2y9

3. 4. 5. 6.

Expressions

2x+5 ____ 6 1 ___ 7x3 2x+6 ____ x2 3x + 12 ______ 2x + 8 4+x ____ 2+x

2. 3. 4. 5.

28x 28x 21x ____ + 24 24 21x ____ , , ______ 28x2 28x2 28x2 21x + 24 ______ , is 28x2 2 x________ + 3x + 6 (x2)(x + 2) x2 − 3x − 4 __________ (x + 3)(x − 3) 6x − 2 __________ (x + 3)(x − 2) 3xy +4 ______ 2x2

PAGE 87 Subtracting Rational

Expressions

1. (x − 1)(x + 1), (x − 1)(x + 1) 6x + 6 __________ (x − 1)(x + 1) 6x + 6 7x − 6x − 6 __________ , __________ , (x − 1)(x + 1) (x − 1)(x + 1) x−6 __________ (x − 1)(x + 1)

18x2 ____ 10x4 2x3+8x2 ______ 2x2−3x 8x−2 ____ 3x4 2+x 3x2+3x x_____ _____ 3x−6 = x − 2 30x2−12x5

pressions 1. 7 × x × x , × 7 × x × x = 28x2 21x _ 4 ____ 24 ____ 2, 4, 2

5.

_____ , is 4x − 4

is 2x + 2 2. ______ 6x4

− _______ 6. ________ = 15x 8x + 8 16x2+16x

PAGE 89 Using Trigonometric

Ratios to Find a Missing Length 1. adjacent, hypotenuse adjacent, hypotenuse, cosine x 25°, __ 12 x 25°, __ 12 , 12, 0.906, 12, 10.87 2. x = 1 3. x = 25.98

PAGE 90 Theoretical Probability 1. 7, Monday, Tuesday, Wednesday,

2 _ 7

4.

5x3

3. _35 , _45 , _34 4. _45 , _35 , _43

10x ____ 24 4. ____ 3 3,

3.

Expressions

Ratios

5 5 __ 12 __ 1. __ 13 , 13 , 12 5 __ 12 12 __ 2. __ 13 , 13 , 5

Thursday, Friday, Saturday and Sunday 2, Saturday and Sunday

2.

5x3, 5x3, 5x3

PAGE 88 Finding Trigonometric

105 48 ___ 3. ___ 60x , 60x

PAGE 86 Adding Rational Ex

PAGE 83 Multiply Rational 1.

50 12 ___ 2. ___ 30x , 30x

24x 24x 2 18x2 _____ 5. _____ , 36y 12x4y3 12x4y3

PAGE 82 Simplifying Rational

2.

–10x −5x 2x+4 3. ________ (x+3)(x–1) x3–x2 _________ 4. (x–3)(5x+6) (2x + 6)(x + 1) 5. ___________ 32x3

PAGE 85 Finding the LCD of a

x20 2. ___ 10

1. 3, 3

Expressions

3 2 x3–2x2 _____ _______ , x –2x (x+2)(2) 2x+4

64x

1. 10-3

PAGE 84 Dividing Rational

x(x − 1) x2 + 3x + 8 3. __________ (x − 3)(x + 3) 2 + 3x + 6 _________ 4. −x x(x + 2) + 6 −x +3 ______ _____ 5. −2x 12x = 6x

is 1 2 _ 2. __ 12 = 6 1 4 _ 3. __ 12 = 3

PAGE 91 Experimental

Probability

1. 44, 250,

number of times event occurs ________________________ number of times experiment done 44 ___ 250 44 ___ 22 ___ 250 , 125 , 0.176

2. = 0.376 3. = 0.344 4. = 0.495

PAGE 92 Mean 1. 2. 3. 4. 5. 6. 7.

331 331, 6, ___ 6 , 55.2 21 43.6 60 40 23.7 155.2

PAGE 93 Median 1. 72, 79, 80, 83, 84, 89

6, even, 72, 79, 80, 83, 88, 89, 80 and 83, (80 + 83) ÷ 2 = 81.5 2. 21 3. 58.5 4. 13.5

Algebra

106


PAGE 94 Permutations 1. 8

PAGE 98 Matrix Subtraction 1. 3−6

3 (8–3), 5!, 5 × 4 × 3 × 2 × 1, 120, 336 2. 56 3. 20 4. 870

3−6

2.

(3 × 2 × 1)(7 × 6 × 5 × 4 × 3 × 2 × 1), 30,240 2. 15 3. 35

3.

PAGE 96 Matrices 1. 2, 6, 2, 6 25

50

75

100

125

150

200

180

160

140

120

100

1,720

2,646

2,306

2,692

2,889

3,963

2,889

2,808

2.

−3

5

2 + (−12)

18 + (−8)

−12

1

3 + 21

27 + 14

−4

−10

10

−10

−7

24

41

7

−16

18

70

−1

8

−2

79

15

12

28

23

−6

5

48

34

2.

3.

2.

−6

6 −18

20

−48

32

−8

8 + (−1)

6

7

3 + 5 −3 + (−7)

8

−10

7

5

14

0

7

3

9

−13

23

,

−48

33

6

3×9+7×2

−9 × 5, −4 × 5, −45, −20 6 × 5, 3 × 5, 30, 15 −27

,

,

2

3 + (5) −3 + (−7)

3.

4.

4.

3 − (−2)

−4 − 8 −1 − (−2)

8 + (−1)

−5 + 11

3×1+7×3

PAGE 99 Scalar Multiplication 1. −9 × 5, −4 × 5, 6 × 5, 3 × 5

PAGE 97 Matrix Addition 1. −5 + 11

2 × 1 + (−4) × 3) 2 × 9 + (−4) × 2

3 − (−2)

−4 − 8 −1 − (−2)

PAGE 95 Combinations 1. 10, 3, 3!(10 − 3)!, 3!7!,

2.

PAGE 100 Matrix Multiplication 1.

3.

4.

48

24

−6

0

−12

54

7 5


107

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