Algebra (Curriculum Binders (Reproducibles))

AlgebraBy Michael Buckley

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.

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Math Binder 2

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

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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. 1 _ 2 , − 1 _ 3 )

Irrational Numbers include: 1. Numbers that cannot be expressed as fractions where the numerator and

denominator are both fractions.

ExampleClassify the following numbers: −10, √

__ 5 , 1 __ 3 , 0, 1 1 __ 2 , 6, 0.23

Step 1 Which numbers are irrational? √__

5

Step 2 Which numbers are rational? −10, 1 _ 3 , 0, 1 1 _ 2 , 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 , − 3 _ 4 , 5 2 _ 3 , 0, √___

11 , 2, 5, 1.35

Which numbers are irrational? √___

11 ,

Which numbers are rational? −3, − 3 _ 4 , 2, 5, , ,

Which rational numbers are integers? −3, , ,

Which rational numbers are whole numbers? , , 5

Which rational numbers are natural numbers? , 5

Use the following list for items 2−6: −4.2, −3, √__

3 , − 1 __ 3 , −1, 0, 0.34, 1.5, 4, — √___

10 .

2. Which numbers are irrational numbers?

3. Which numbers are rational numbers?

4. Which numbers are integers?

5. Which numbers are whole numbers?

Name Date

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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.

ExampleSimplify. (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

PracticeSimplify 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.

ExampleUse the following written phrase to create an expression: four times a number plus five.

Step 1 Write out the written phrase. Four times number plus five

Step 2 Replace each word with a variable, number or operational symbol.

4 × c + 5 OR 4c + 5

PracticeUse 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.

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

Key Math Operation Words

Addition Multiplication

sum product

more than times

plus multiplied

increases twice (times 2)

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Algebra


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.

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.

ExampleUse 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, number or operational symbol.

x ÷ 2 − 2

PracticeUse 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.

Key Math Operation Words

Subtraction Division

difference quotient

less than divided by

minus divided equally

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

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Algebra


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.

ExampleEvaluate. 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 given.

Replace x with 3 and y with 2:

4 (3) − 2 ÷ 2

Step 3 Perform any operations to end up with a single number.

Multiply, divide then subtract:

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

PracticeEvaluate 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

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Algebra


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.

ExampleSimplify: 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 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.

3x + 2x = 5x

2 + 1 = 3

3x2 = 3x2

3x2+ 5x + 3

PracticeSimplify

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

Name Date

Algebra


Adding Integers Using Absolute Value

When you add integers using absolute value, you can use the following rules.

Two numbersAdd or subtract absolute value Sign Example

postive + positive add positive 17 + 4 = 21

negative + negative add negative —17 + (—4) = —21

postive + negativeor

negative + positivesubtract sign of number with the

larger absolute value17 + (—4) = 13—17 + 4 = —13

ExampleAdd. a. −21 + (−6) b. −21 + 6

Step 1 Find the absolute value of each number | −21 | = 21 |−6 | = 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 value?

Add Subtract

Step 4 Solve 21 + 6 = 27 21 − 6 = 15

Step 5 What is the sign? Adding two negatives = a negative −21 + (−6) = −27

21 has a larger absolute value = negative −21 + 6 = −15

PracticeAdd. 1. −27 + 19 2. 45 + (−55)

Find the absolute value of each number | −27 | =

| 19 | =

| 45 | =

| −55 | =

Do you add or subtract the absolute value?

Solve 27 − 19 = 55 − 45 =

What is the sign?

−27 + 19 = 45 − 55 =

3. 12 + (− 8) 6. −36 + 14

4. 42 + 16 7. 17 + 17

5. −24 + (−10) 8. −61 + 21

Name Date

Algebra


Subtracting Integers

Consider the following examples of adding and subtracting integers.

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

ExampleSubtract a. −19 − (−4)

Step 1 Change the second number to its opposite

−4 becomes 4

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

PracticeFor 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 17

Find the absolute value of each number. | −9 | = | 17 | =

Do you add or subtract? A negative plus a positive, ;

17 is larger, so the answer is .

Solve 17 − 9 = ; −9 − (−17) = .

5. −8 − 13 7. −9 − 5 9. 6 − (−12)

6. −6 − (−2) 8. 12 − 4 10. −21 − (−21)

Adding Subtracting

−8 + (−8) = −16 −8 − 8 = −16

10 + (−12) = -2 10 − 12 = −2

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

Name Date

Algebra


Multiplying Integers

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

Example Signs of Numbers Signs of Answer

7 × 5 = 35 positive times

−7 × (−5) = 35 negative times

−7 × 5 = −35 times positive

7 × (−5) = −35 times

ExampleMultiply. 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 numbers.

8 × 6 = 48 4 × 12 = 48

Step 4 Solve. 8 × (−6) = −48 −4 × (−12) = 48

PracticeFor 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) =

Name Date

Algebra


Dividing Integers

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

Example Signs of Numbers Signs of Answer

30 ÷ 5 = 6 positive divided by

30 ÷ (−5) = -6 positive divided by

−30 ÷ 5 = -6 negative divided by

−30 ÷ (−5) = 6 negative divided by

ExampleDivide. 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 numbers.

45 ÷ 5 = 9 72 ÷ 9 = 8

Step 4 Solve. 45 ÷ (−5) = −9 −72 ÷ (−9) = 8

PracticeFor 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 =

Name Date

Algebra


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.

ExampleSimplify using the Distributive Property: 4(5x + 2)

Step 1 Multiply the number outside the parentheses by each number inside the parentheses.

4 (5x + 2) = (4 • 5x) __ (4 • 2)

Step 2 Place the operation symbol inside the parentheses between the two multiplication expressions.

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

Step 3 Simplify using the order of operations. (4 • 5x) + (4 • 2) = 20x + 8

PracticeUse 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)

Name Date

Algebra


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 ExponentsTo 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.

ExampleSolve. 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.

The exponent is 4 so you multiply 6 by itself 4 times; 64 = 6 × 6 × 6 × 6 = 1296.

Step 2 Count the number of times a number is multiplied by itself, that amount is your exponent.

7 is multiplied by itself 5 times; the exponent is 5.

Step 3 The number being multiplied is the base.

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 =

Name Date

Algebra


Negative Exponents

When you have a negative exponent, you treat the negative exponent as follows:

x−2 = 1 __ x2

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.

ExampleSimplify 5−3.

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

Step 2 Create a fraction with 1 as the numerator and the power as the denominator with a positive exponent.

5−3 = 1 __ 53

Step 3 Evaluate the power by multiplying the base by itself the number of times equal to the exponent.

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

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))0

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

00 ___ 000 = 00

_____________ 000000000000000 = 00 ____ 0000

2. 8−1 4. (−5)−3 6. a−2

3. 3−2 5. (−3)−4 7. 4x−3

exponent

base

Name Date

Algebra


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.

ExampleWrite 376,700 in scientific notation.

Step 1 Move the decimal point to the left or right to get a number between 1 and 10.

376,700 (5 decimal places): 3.767

Step 2 Multiply the number by 10 with an exponent.

3.767 × 10?

Step 3 The exponent is equal to the number of places the decimal point moved.

The decimal point moved 5 places: 3.767 × 105

Step 4 The exponent is positive if the decimal point is moved to the left; negative if moved to the right.

The decimal point moved to the left: 3.767 × 105

PracticeWrite 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.

0.0404 ( decimal places): 4.04

Multiply the number by 10 with an exponent.

4.04 × 10?

The exponent is equal to the number of places the decimal point moved.

4.04 × 10

The exponent is positive if the decimal point is moved to the left; negative if moved to the right.

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

Name Date

Algebra


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.

a x _ y

5 3 _ 2

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.

ExampleSolve. 7

Step 1 Identify the power the number is raised to; identify the root you will find.

7

Step 2 Raise the number to the identified power.

Raise 7 to the 4th power:

74 = 7 x 7 x 7 x 7 = 2401

Step 3 Take the identified root of the answer from Rule #2.

take the cube root of 2401

3 √_____

2401 = 13.4

Practice Solve.

1. 12

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

12

Raise the number to the identified power.

Raise 12 to the power:

12 =

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

Take the of the answer from

Rule #2:

2. 8 4. –4

3. 10 5. 4

cube 5

then take the square root

raise to the x power

take the y root of ax

4 __ 3

4 _ 3 raise to the 4th power

take the cube root

5 _ 2

5 _ 2 raise to the power

take the

3 _ 4 4 _ 3

2 _ 5 3 _ 9

Name Date

Algebra


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.

ExampleWrite 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 _____ x2 – x1

= 2 − (−2)

______ 5−3 = 4 _ 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.

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

Step 3 Place the values into the point-slope form.

y − y1 = m(x − x1) → y − (−2) = 2(x − 3)

PracticeWrite an equation to model the data in each table.

1. x —5 —1 1 7

y —9 —7 —6 —3

Choose two sets of ordered pairs. Find the slope.

slope = y2 − y1 _____ x2 − x1

= –7 – (–9)

_______ –1 – (–5)

=

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

(−5, −9) → = x-coordinate; = y-coordinate

Place the values into the point-slope form.

y − y1 = m(x − x1) → y − =

(x −

2. x 4 3 0 —2

y 8 6 0 —4

3. x —10 —4 —1 3

y 12 0 —6 —14

Name Date

Algebra


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.

ExampleDVDs 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

from the problem as a word-based mathematical sentence.

Total cost = 15 times the number of DVDs.

Step 2 Identify unknown values—assign a variable to each unknown.

Total cost is variable c. The number of DVDs bought is variable n.

Step 3 Replace each phrase with a mathematical expression. Connect the expressions with operation symbols and equal sign.

Total cost is 15 times number of DVDs

c = 15 × n

c = 15 × n

PracticeWrite 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.

is 12 the number of

DVDs plus .

Identify unknown values—assign a variable to each unknown.

is variable c.

is variable .

Replace each phrase with a mathematical expression. Connect the expressions with operation symbols and equal signs.

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.

Name Date

Algebra


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.

ExampleSolve. 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 number on each side of the equation.

You will “−24” on each side. x + 24 − 24 = 36 − 24 x + 0 = 12 or x = 12

PracticeSolve.

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 to each side of the equation.

172 = 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

Name Date

Algebra


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.

ExampleSolve. 3x = 36

Step 1 Identify the operation and number on the same side of the equation as the variable.

You are multiplying the variable 3 times.

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

The opposite of multiplication is division; so divide each side by 3.

3x ÷ 3 = 36 ÷ 3

x = 12 or

x = 12

PracticeSolve.

1. 24 = x _ 8

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 = x _ 8 ×

=

= x

2. 5x = 45 5. x _ 7 = 84

3. 16x = 32 6. x __ 10 = 240

4. 14x = 140

Name Date

Algebra


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.

ExampleSolve. 4x − 16 = 24

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

Step 2 Isolate the term with the variable using addition or subtraction.

4x − 16 + 16 = 24 + 16

4x − 0 = 40

4x = 40

Step 3 Isolate the variable using multiplication or division.

4x = 40

4x ÷ 4 = 40 ÷ 4

x = 10 or x = 10

PracticeSolve.

1. 7x + 7 = 70

Write the equation. 7x + 7 = 70

Isolate the term with the variable using addition or subtraction.

7x + 7 = 70

7x + 0 =

7x =

Isolate the variable using multiplication or division.

7x =

7x = 63

x = or x =

2. x _ 4 + 10 = 17 5. 24 = 5x − 96

3. x _ 7 − 3 = 21 6. x _ 2 − 17 = 3

4. 6x − 11 = 37 7. 7x − 9 = 68

Name Date

Algebra


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.

ExampleSolve. 2x + x + 12 = 78

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

Step 2 Combine like terms. 2x and x are like terms

2x + x + 12 = 78

3x + 12 = 78

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

PracticeSolve.

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

=

Isolate the variable. ÷ 7 = ÷ 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

Name Date

Algebra


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.

ExampleSolve 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 all numbers without a variable.

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

Step 3 Use multiplication or division to solve for the variable.

6x = 42 6x ÷ 6 = 42 ÷ 6 x = 7

PracticeSolve 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

− 4 = 0 + 16

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

5x − 4 = 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

Name Date

Algebra


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.

ExampleDecide 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 only one output.

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

PracticeDecide whether each table represents a function.

1. Hours 3 4 4 5 6 6

Envelopes stuffed 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 the outputs are 25

and .

2. Dollars 10 20 30 40 50

Tip 1.50 3.00 4.50 6.00 7.50

3. x 4 4 9 9 16

y 2 —2 3 —3 4

4. Month 1 2 3 4 5

Inches of Rain

3.5 4.1 5.4 3.5 1.25

Name Date

Algebra


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.

ExampleWrite 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 what is done to get from x to y. If it is not the same as in Step 1, then consider another pattern.

In the second ordered pair, (2, 7) you add 5 to get from the x–value (2) to 7.

Step 3 Write a function rule from the pattern. y = x + 5

PracticeWrite a function rule for each table.

1. 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

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

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

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

Write a function rule from the pattern. y =

2. x 2 4 6 8

y 0 2 4 6

Name Date

Algebra


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 = a __ x + b + c Equation with a variable in the denominator. y = — 5 __ x + 4 + 2

ExampleClassify the following function as linear, quadratic, exponential, or rational: a __ 4 + 3 = 6

Step 1 Does any variable have a 2 as an exponent?

No

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.

PracticeClassify each function as linear, quadratic, exponential, or rational.

1. y = 1 _ x − 1

Does the variable have a 2 as an exponent?

Is the variable an exponent?

Is the variable in the denominator?

Identify the function. It is a function.

2. y = x2 __ 2 + 4x − 3 5. y = 5 __ 6x × 1 _ 2

3. y = 4x +2 6. y = 3(0.5)x

4. y = 1 __ 2x + 3 7. y = 3x + 22 − 1

Name Date

Algebra


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)

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.

ExampleGraph the following point: (5, −3)

Step 1 Move across the x-axis the number of units of the x-coordinate.

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

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

Move down from the x-axis 3 units.

PracticeGraph the following points.

1. (−7, −4)

Move across the x-axis the number of units of the x-coordinate.

Move across the x-axis

units to the .

Move up or down from the x-axis the number of units of the y-coordinate.

Move from the

x-axis units.

2. (6, 1)

3. (−7, 0)

4. (0, −5)

Give the coordinates for each point.

5. A

6. B

7. C

8. D

Symbol of Coordinate x y

+ right up

− left down

D

C

B

A

1.

Name Date

Algebra


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).

ExampleList 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.

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

PracticeList 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 .

2. y = 5x − 4

3. y = x _ 2 + 1

4. y = −x + 3

5. y = 4x – 2

6. y = x _ 3 − 1

7. y = x + 1

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

y = 3 (1) − 2

y = 3 (2) − 2

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Algebra


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.

ExampleGraph the following equation y = x + 3



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.

PracticeGraph 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 = x _ 3 + 1

6. y = 2x + 3

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)

2 y = 3(2) − 1 5 (2, 5)

0 y = 3(0) − 1 —1 (0, −1)

y = 3(1) − 1 2 (1, 2)

y = 3(—2) − 1 —7 (—2, −7)

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Algebra


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.

ExampleTell whether the data shows a direct variation.

Time (hours) 1 2 3 4

Distance (miles) 40 80 120 160

Step 1 Examine the data. When one variable changes does the other variable change in the same direction?

Yes, as hours increase, the distance increases.

Step 2 Does the ratio of the two variables result in the same number?

Yes, the ratio of distance ______

time is 40 (e.g. 80

__ 2 = 40)

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, then the data shows a direct variation.

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

PracticeTell whether the data shows a direct variation.

1. 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?

The ratio of distance ______

time the same

number.

Does the graph go through the origin? The graph go through the origin.

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

The answer to all three questions

“yes.” The data show a direct variation.

2. Drop Height (cm) 10 20 30 40 50

Bounce (cm) 9 18 27 36 45

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Algebra


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.

ExampleDoes 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 increase while the other one decreases?

Yes, as the value for x increases, the value for 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 shows an inverse variation.

The answer to both is yes. The data shows an inverse variation.

PracticeDoes 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 = ; 6 × 6 = .

If the answer to 1 and 2 is “yes,” the data shows an inverse variation.

The answer to question #1 is , the

answer to question #2 is , the data

show an inverse variation.

2. x 2 4 6 8 10

y 4 8 12 16 20

4. x 18 12 9 3

y 4 6 8 24

3. x 4 12 20 30

y 15 5 3 2

Name Date

Algebra


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).

slope = vertical change

_____________ horizontal change

= y2 −y1 _____ x2 − x1

ExampleFind the slope of a line that passes through (6, 4) and (2, 1).

Step 1 Make one set of coordinates (x1, y1) and the other set, (x2, y2)

(x1, y1) (x2, y2)

(3, 5) (2, 1)

Step 2 Use the equation for slope, place the numbers into the formula.

slope = y2 −y1 _____ x2 − x1

= 1 − 4 ____ 2 − 6

Step 3 Solve. slope = 1 − 4 ____ 2 − 6 = 3 _ 4

The slope is 3 _ 4 .

PracticeFind 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)

(3, 5)

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

slope = y2 −y1 _____ x2 − x1

= − 5 ________ − 3

Solve. slope = − 5 ________ − 3 = ______ 0000 =

The slope is .

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)

Name Date

Algebra


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

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

ExampleFind the slope and the y-intercept of the line y = 4x − 2

Step 1 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 = 4x − 2

m = slope = 4

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 = 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

m = slope =

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

b = y–intercept =

2. y = 1 _ 3 x − 3

3. 2y = 2x + 2

4. y = 3 _ 4 x − 13

5. 3y = −2x + 9

y-intercept

slope

Name Date

Algebra


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.

Point–slope form: y − y1 = m (x − x1)

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)

ExampleWrite 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 x-coordinate and the y-coordinate.

The ordered pair is (2, 5)

The x-coordinate is 2; the y-coordinate is 5.

Step 3 Use the point–slope form to write the equation.

y − y1 = m(x − x1)

y − 5 = 3 (x − 2)

PracticeWrite 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 = − 1 _ 2 , (7, 1)

3. slope = 2, (−3, −3)

4. slope = 2 _ 3 , (4, −5)

5. slope = −3, (−1, 3)

y-coordinate

slopey-coordinate

Name Date

Algebra


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

1. Use the formula for slope (slope = vertical change

_____________ horizontal change

= y2 − y1 _____ x2 − x1

) to find the slope.

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

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

Step 1 Use the formula for slope ( y2 − y1 _____ x2 − x1


Slope = y2 − y1 _____ x2 − x1

= 5 − (−3)

______ 1 − (−3)

= 8 _ 4 = 2

Step 2 Use one set of ordered pairs for the x-coordinate and the y-coordinate.

Use the ordered pair (1, 5)

The x-coordinate is 1; the y-coordinate is 5.

Step 3 Use point–slope form to write the equation.

y − y1 = m (x − x1)

y − 5 = 2(x − 1)

PracticeUse the point–slope form to write an equation.

1. (−2, −2), (0, −4)

Use the formula for slope ( y2 − y1 _____ x2 − x1


Slope = y2 − y1 _____ x2 − x1

= −4 − (−2)

_______ 0 − (−2)

= −2 __ 2 = −1

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

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)

Name Date

Algebra


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.

ExampleAre the graphs of y = x + 4 and 6y – 3x = 6 parallel?

Step 1 Write all equations in slope-intercept form.

y = 1 _ 2 x + 4: is in slope-intercept form.

6y − 3x = 6 → y = 1 + x = 1 + x

Step 2 Identify the slope of each line. y = x + 4: slope = 1 _ 2

y = 1 + x: slope = 1 _ 2

Step 3 If the slopes are equal the lines are parallel.

The slopes are equal so the lines are 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 =– 1 _ 4 x + 5 and 12y + 3x = 24

3. 8x + 4y = 8 and Y = −2x + 4

4. y = 2x + 6 and −2x + 2y = 12

5. y = – 1 _ 4 x + 12 and 8x + 6y = 9

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Name Date

Algebra


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 = – 1 _ 2 x + 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

y = – 1 _ 2 x + 4 − 1 _ 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.

ExampleWrite 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 is the slope of the new line.

m = 3, the reciprocal is 1 _ 3 .

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

The slope of the first line is positive; make the new slope negative: − 1 _ 3

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

y = mx + b = − 1 _ 3 x + 2

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

1. y = − 1 _ 2 x + 2; new y-intercept: −3

Identify the slope of the known line. y = − 1 _ 2 x + 2; slope = m = − 1 _ 2

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

m = − 1 _ 2 ; the reciprocal of 1 _ 2 is

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

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

The slope of the first line is

; make the slope of the new

line :

y = mx + b = x +

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

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

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Name Date

Algebra


Using Reciprocals

Two numbers are reciprocals if their product is 1. For example, the reciprocal of 3 _ 4 is 4 _ 3 . 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.

ExampleWrite the reciprocal and negative reciprocal of 7 __ 8 .

Step 1 Start with the original number— identify the numerator and denominator.

7 _ 8 = numerator __________

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.

7 _ 8 → 8 _ 7

8 _ 7 is the reciprocal of 7 _ 8 .

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

The sign in front of 7 _ 8 is understood to be

positive. Change the sign in front of 8 _ 7 to a

negative. The negative reciprocal of 7 _ 8 is − 8 _ 7 .

Practice Write the reciprocal and negative reciprocal of each number.

1. − 2 _ 5

Start with the original number— identify the numerator and denominator.

2 _ 5 = numerator __________

denominator

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

− 2 _ 5 →

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

The sign in front of − 2 _ 5 is .

Change the sign in front of to

. The negative reciprocal of

− 2 _ 5 is .

2. 1 _ 7 4. 5 __ 13

3. 5 _ 3 5. 9 _ 8

Name Date

Algebra


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.

ExampleSolve. 1.5x + 2.1 = 17.1

Step 1 Count the number of decimal places in each number.

1.5 → one place 2.1 → one place 17.1 → one place

Step 2 Find the most number of decimal places.

The most number of decimal places is one.

Step 3 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 with one “ 0” is 10. (10)(1.5x) + (10)(2.1) = (10)(17.1) 15x + 21 = 171

Step 4 Solve for the variable. 15x + 21 −21 = 171 − 21 15x = 150 x = 10

Practice 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) + (16.25) = (22.65)

x + =

Solve for the variable. x = −

x =

x =

2. 2.5x − 2 = 15 3. 1.2x + 10.25 = 14.05

Name Date

Algebra


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 __ 5 + 2x

__ 3 = 5 or x + x _ 2 = 7 _ 8

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.

ExampleSolve. 2x __ 3 + 3x __ 4 = 4

Step 1 Place each fraction in front of each variable.

2 _ 3 x + 3 _ 4 x = 4

Step 2 Rewrite the fraction so each has the same denominator.

12 is the common denominator.

8 __ 12 x + 9 __ 12 x = 4

Step 3 Combine like terms. 8 __ 12 x + 9 __ 12 x = 4

17 __ 12 x = 4

Step 4 Isolate the variable by multiplying each side by the reciprocal of the fraction.

( 12 __ 17 ) ( 17 __ 12 )x = 4( 12 __ 17 )

x = 48 __ 17 = 2 14 __ 17

PracticeSolve.

1. x _ 2 + 3x __ 5 = 10

Place each fraction in front of each variable.

1 _ 2 x + 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.

x = 10

x =

2. x _ 2 − x _ 3 = 6 4. x _ 6 − 2 _ 3 = 1 _ 6

3. x _ 5 + x _ 2 = 7 5. 4 x _ 9 − x _ 3 = 12

Name Date

Algebra


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.

ExampleGraph the following equation y < 3x + 4.



Step 3 Substitute each value of x into the inequality. Solve the inequality for y.

Step 4 Plot each solution on a coordinate 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.

PracticeGraph the following equations.

1. y 4x − 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.

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

2. y 5x − 2 3. y −2x + 3

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)

x y 4x − 6 y (x, y)

2 y 4(2) − 6

0 y 4(0) − 6

y 4( ) − 6

y 4( ) − 6

Name Date

Algebra


Writing Inequalities from a Graph

There are four inequality symbols. The symbol of each type and its graph are summarized below.

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 1. Identify the point on the number line. Write x (the number). 2. Look at the direction the graph points to and if you have an open or closed circle.Point Direction of graph Symbol Point Direction of graph Symbol to the left to the right • to the left • to the right

ExampleWrite an inequality for the graph.

Step 1 Identify the point on the number line.

Write x (the number).

The point is located at: −1;

x −1.

Step 2 Look at the direction the graph points to and if you have an open or closed circle.

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 (the number).

The point is located at : .

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

The graph points to the , with a (n)

circle.

x

2. 3.

Inequality Symbol Meaning Example Graph

less than x 9

less than or equal to x 9

greater than x 9

greater than or equal to x 9

90

90

90

90

0-1

60

-5 0 0 3

Name Date

Algebra


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.

ExampleSolve. 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 number on each side of the inequality.

You will “+ 10” to each side.

x − 10 + 10 21 + 10

x − 0 31 or x 31

PracticeSolve.

1. 44 16 + x

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

is on the same side of the inequality

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

Name Date

Algebra


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.

ExampleSolve. 10x 240

Step 1 Identify the operation and number on the same side of the inequality symbol as the variable.

You are multiplying the variable times 10.

Step 2 Perform the opposite operation of that number on each side of the inequality symbol.

The opposite of multiplication is division; so divide each side by 10.

10x ÷ 10 240 ÷ 10

x 24

PracticeSolve.

1. − x _ 3 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.

− x _ 3 9

x ?

x

2. 3x 24 4. −4x −36 6. x _ 5 −10

3. 2x −16 5. x _ 2 14 7. − x _ 4 12

Name Date

Algebra


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.

ExampleSolve. 3x + 5 < 26

Step 1 Isolate the term with the variable using addition or subtraction.

3x + 5 26

3x + 5 − 5 26 − 5

3x + 0 21 or 3x 21

Step 2 Isolate the variable using multiplication or division.

3x 21

3x ÷ 3 21 ÷ 3

x 7

PracticeSolve.

1. −2x − 7 17

Write the equation. −2x − 7 17

Isolate the term with the variable using addition or subtraction.

−2x − 7 17

−2x = −2x 24

Isolate the variable using multiplication or division.

−2x 24

−2x ? 24

x ?

x

2. 3x + 2 − 10 5. x _ 5 + 10 −3

3. 4 − 2x 20 6. x _ 5 − 3 9

4. −5x − 7 18 7. − x _ 4 + 4 7

Name Date

Algebra


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 (b)

leg (a)

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.

ExampleFind 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 theorem. Square the numbers.

52 + b2 = 132

25 + b2 = 169

Step 3 If the unknown side is a leg, solve the equation for the unknown leg.

25 − 25 + b2 = 169 − 25

b2 = 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 + 2 =

64 + =

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

64 + =

= =

2. b = 8; c = 12 4. a = 20; b = 15

3. a = 3; b = 4 5. a = 200; c = 250

Name Date

Algebra


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.

√__

5 = 2.23606797 √__

1 _ 7 = 0.377964473

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.

ExampleIs the expression a rational or irrational? a. √

___ 81 b. √

___ 10

Step 1 Look at the number under the square root sign. Ask if the number is a perfect square.

a. √___

81 = 9 is a perfect square. So the 81 is rational.

b. √___

10 is not a perfect square.

Step 2 Find the square root. b. √___

10 = 3.16227766

Step 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.

b. 3.16227766 does not terminate and does not repeat. √

___ 10 is irrational.

Practice Is each expression rational or irrational? 1. a. √

__ 7 b. √

____ 100

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 .

2. √____

1.21 4. √__

1 _ 3

3. √___

25 5. √___

11

Name Date

Algebra


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 √_____

x + 5 √___

4x2

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.

ExampleSimplify. √

____

8x2 × √_____

18x5

Step 1 Place both numbers under the square root symbol under a single square root symbol, separated by a multiplication sign.

√___

8x2 × √____

18x5 = √_________

8x2 × 18x5

Step 2 Multiply. √_________

8x2 × 18x5 = √______________

(8 × 18)(x2 × x5) = √_____

144x7

Step 3 Simplify the radical by looking for a perfect square. Follow the rules for simplifying a radical expression by removing a perfect square.

√_____

144x7 144 is a perfect square: √____

144 = 12

√_____

144x6 × √__

x = 12x3 √__

x

Practice Simplify.

1. √___

3x × √___

6x

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.

√______

= √______

× √___

=

2. √___

5x3 × √___

8x2 4. √___

5x4 × √____

10x

3. √___

2x5 × √___

2x4 5. √___

2x6 × √___

8x2

Name Date

Algebra


Simplifying Radical Expressions by Removing Perfect Squares


√____

121 √__

2 √_____

x + 5 √___

4x2


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.


___ 50

Step 1 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 50 are: 1 and 50; 2 and 25; 5 and 10; 25 is a perfect square; so use the factors 2 and 25.

Step 2 Rewrite the radical expression as the product of two square roots of the factor found in # 1.

√___

50 = √___

25 × √__

2

Step 3 Find the square root of the perfect square. Place the result next to the square root symbol of the other factor.

√___

25 × √__

2 = 5 √__

2

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 . 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.

√___

27 = √_________

3 × = √__

3 × √___

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

√__

3 × √___

= √__

3 × =

2. √____

500 = 4. √___

48 =

3. √___

80 = 5. √___

75 =

Name Date

Algebra


Simplifying Radical Expressions with Variables


√____

121 √__

2 √_____

x + 5 √___

4x2


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.


_____

27a5

Step 1 Find two factors (including variables) of the expression under the square root symbol; one of the factors must be a perfect square.

Look for a number that is a perfect square; with the perfect square place the variable raised to an even power. The factors of 27a5 are 3a and 9a4.

9a4 × 3a = (9 × 3)(a4 × a) = 27a5

Step 2 Rewrite the radical expression as a product of two square roots of the factor found in #1.

√____

27a5 = √_______

9a4 × 3a = √___

9a4 × √___

3a

Step 3 Find the square root of the perfect square. Place the result next to the square root symbol of the other factor.

Find √__

9 and divide the exponent by 2.

√___

9a4 × √___

3a = 3a2 √___

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.

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 and 2x.

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

√____

32x7 = √__________

× 2x = √___________

× √___

2x

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

2. √____

50x2 4. √____

63x6

3. √____

48x5 5. √_____

180x9

Name Date

Algebra


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.

ExampleSolve. √

__ x − 5 = 1

Step 1 Isolate the radical on one side of the equation.

√__

x − 5 + 5 = 1 + 5

√__

x = 6

Step 2 Square both sides to remove the square root symbol.

√__

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

Name Date

Algebra


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.

9.5 = 19 __ 2 0.33 3 = 1 _ 3

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.

ExampleBetween what two integers is √

___ 10 ?

Step 1 Look at the number and find a perfect square that is less than the number.

The perfect square less than 10 is 9 ( √__

9 = 3).

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

The perfect square greater than 10 is 16. ( √

___ 16 = 4).

Step 3 The square root of the number is between the two perfect squares.

√___

10 is between 3 and 4.

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

( √___

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.

√___

22 is between and .

2. 5

3. 34

4. 29

5. 52

6. 130

Name Date

Algebra


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.

Cube root: 3 √__

Fourth Root: 4 √__

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.

ExampleFind the root:

3 √__

8

Step 1 Look at the number outside the symbol —determine the root to calculate.

3 √__

8

The “3” outside the symbol means you find the cube root.

Step 2 Determine if the number under the symbol is a perfect power. If it is a perfect power, find the root.

8 is a perfect cube (23 = 8)

So, 3 √__

8 = 2

Practice Find each root.

1. 3 √____

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 a perfect cube.

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

Then find a perfect power greater than the number.

The estimated root is between the two perfect roots.

The perfect cube less than 100 is .

( 3 = 64). The perfect cube greater than

100 is .

( 3 = 125)

3 √____

100 is between and .

2. 5 √_____

5000 4. 3 √____

168 6. 4 √____

625

3. 5 √___

32 5. 4 √____

250 7. 3 √____

216

Name Date

Algebra


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.)

ExampleMultiply. 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

PracticeMultiply.

1. −x4 (4x2 + x _ 2 + 10)

Multiply each term in the polynomial by the monomial.

(−x4) (4x2) =

(−x4) =

(−x4) =

−4x6 + (− x5 __ 2 ) + (−10x4) = −4x6 − x

5 __ 2 − 10x4

2. 2x3 (−x2 − 4x + 2)

3. x4 (3x4 + 7x2 − 10)

4. −3x2 (−2x3 − 5x2 + x _ 4 )

Name Date

Algebra


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.

ExampleMultiply. (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

Combine like terms. 20x2 + 10x − 12x − 6 = 20x2 − 2x − 6

PracticeMultiply.

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)

Name Date

Algebra


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.

ExampleSolve. (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 terms; use the same operation sign as the one between the two terms.

2(x × 3) = 2(3x) = 6x

Use the plus sign.

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) =

Use the sign.

Square the last term. −22 = 4

(5x − 2)2 =

2. (x + 4)2

3. (x − 8)2

4. (2x + 6)2

5. (4x − 4)2

6. (6x + 12)2

Name Date

Algebra


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).

ExampleAdd. (4x2 + 2x − 5) + (3x4 − 3x + 5x2)

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

Step 2 Line up like terms. 4x2 + 2x −5

3x4 + 5x2 − 3x

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)

Name Date

Algebra


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).

ExampleSubtract. (2x2 + 5x3 + 1) − (5 + 5x2 + 3x3)Step 1 Write polynomials in standard form. (5x3 + 2x2 + 1) − (3x3 + 5x2 + 5)Step 2 Line up like terms. 5x3 + 2x2 + 1

−(3x3 + 5x2 + 5)

Step 3 Change the sign of each term in the second polynomial.

5x3 + 2x2 +1

−3x3 − 5x2 − 5

Step 4 Add the numbers in front of each variable

5x3 + 2x2 + 1

+ (−3x3 − 5x2 − 5)

2x3 − 3x2 − 4

PracticeSubtract.

1. (x + 5 + 4x2) − (10 + 3x2 – 5x)

Write polynomials in standard form. (4x2 + x + 5) –

Line up like terms. 4x2 + x + 5

− Change the sign of each term in the

second polynomial. 4x2 + x + 5

Add the numbers in front of each

variable 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)

Name Date

Algebra


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 MonomialMultiply 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.)

ExampleMultiply. 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

PracticeMultiply.

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)

Name Date

Algebra


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.

ExampleFactor out the greatest common factor: 5x2 + 10x

Step 1 Find the greatest common factor of all the terms.

List the factors of 5x2 and 10x.

5x2: 1, 5, x2

10x: 1, 2, 5, 10, x

The greatest common factor is 5x.

Step 2 Determine the terms that when multiplied by the greatest common factor will result in each original term.

(5x) (x) = 5x2

(5x) (2) = 10x

Step 3 Rewrite the expression with the greatest common factor outside the parentheses and the terms you found in Step 2 inside the parentheses.

5x (x + 2)

Practice Factor out the greatest common factor.

1. 10x4 − 15x3

Find the greatest common factor of all the terms.

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 .

Determine the terms that when multiplied by the greatest common factor will result in each original term.

( )( ) = 10x4

( )( ) = −15x3

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 4. 5x2y5 + 15xy7

Name Date

Algebra


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.

ExampleFind 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).

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

Step 2 From the list, identify the greatest common factor; this is the coefficient- part of the greatest common factor.

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 lowest exponent.

The variable with the lowest exponent is x3.

The greatest common factor is 10x3.

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).

12: 1, 2, 3, 4, 6, 12

18: From the list, identify the greatest

common factor; this is the coefficient- part of the greatest common factor.

Of the factors listed, is the greatest

common factor.

List the variables with their exponents.

From the list, select the variable with the lowest exponent.

m3n2 = m3 × n2 m5n4 = ×

The variable with the lowest exponents are m3

and .The greatest common factor is m3

2. 6x4 and 8x6 4. 12x8 and 20x9

3. 9m2 and 3m5 5. 4x4y3 and 6xy2

Name Date

Algebra


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.

ExampleFactor out the greatest common factor. 20x5 + 10x6 − 15x4

Step 1 Find the greatest common factor for all of the terms.

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.Step 2 Determine the terms that when

multiplied by the greatest common factor will result in each original term.

(5x4) (4x) = 20x5; (5x4)(2x2) = 10x6

(5x4)(−3) = −15x4

Step 3 Rewrite the expression with the greatest common factor outside the parentheses and the terms found in Step 2 inside the parentheses.

5x4 (4x + 2x2 − 3)

Practice Factor out the greatest common factor.

1. 13x8 + 26x4 − 39x2

Find the greatest common factor for all of the terms.

List all the factors of 13x8, 26x4, and −39x2.

13x8:

26x4:

−39x2:

The greatest common factor is 13x2. Determine the terms that when

multiplied by the greatest common factor will result in each original term.

13x2 (x6) = 13x8 13x2 ( ) = 26x4

13x2 ( ) = −39x2

Rewrite the expression with the greatest common factor outside the parentheses and the terms found in Step 2 inside the parentheses.

13x2 (x6 )

2. 8x3 − 24x2 − 80x 4. x2 + 2x5 − 9x7

3. 3x4 + 30x3 + 48x2 5. 3x5 − 6x4y − 45x3y2

Name Date

Algebra


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.

ExampleFactor: 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.

FACTORS OF 18 SUM OF FACTORS

1 and 18 1 + 18 = 19

2 and 9 2 + 9 = 11

3 and 6 3 + 6 = 9

Step 2 Choose the pair of factors in the right column whose sum equals b.

Step 3 Create two expressions of “x +” the factors.

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.

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

FACTORS OF 12 SUM OF FACTORS

−1 and −12 −1 + −12 = −13

−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

Name Date

Algebra


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.

ExampleFactor. 6x2 + 23x + 7Step 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.

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.

6x2 23x 7

F O + I = ? L

1 × 6 1 × 71 × 1

++

1 × 67 × 6

==

1343

1 × 7

2 × 3 2 × 72 × 1

++

1 × 37 × 3

==

1723

1 × 7

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.

5x2 − 14x − 3 = ( x + )( x + )

=

2. 2x2 + 8x + 8 5. 2x2 − 8x + 6

3. 2x2 − 3x − 9 6. 2x2 − 7x − 4

4. 7x2 + 50x + 7

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

==

Value for b

Name Date

Algebra


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.

ExampleSolve. (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 two squared terms.

Result of the first squaring: x4

Result of the second squaring: 16

x4 − 16

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)

Name Date

Algebra


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

PracticeSolve the system of equations using graphing for the following.

1. y = 4x − 7; x + y = 8

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

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)

x y = 4x − 7 y (x, y)

2 y = 4(2) − 7

0 y = 4(0) − 7

x x + y = 8 y (x, y)

3 3 + y = 8

0 0 + y = 8

Step 1

Step 2

Step 4 The lines cross at .

y = x

+ 1

2x + y = 4

Step 3

Name Date

Algebra


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.

ExampleSolve 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.

y = 6 − x x − 2 = y

6 − x = x− 2

Step 3 Rearrange the resulting equation so x is on one side.

6 − x + x = x + x − 2

6 = 2x − 2

Step 4 Isolate x. 6 + 2 = 2x − 2 + 2

8 = 2x 8 ÷ 2 = 2x ÷ 2 4 = x

Step 5 Substitute the value of x into one of the original equations.

y = 6 − x y = 6 − 4 = 2

The solution is (4, 2).

PracticeSolve 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 =

Isolate x. 5 =

6 = 6 ÷ 2 = 2x ÷ 2 x = 3

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

y = x − 1 y = 3 − 1= 2

The solution is .

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

Name Date

Algebra


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.

ExampleSolve: 2x + 3y = 10 x − 3y = 2

Step 1 Line up the equation by like terms. 2x + 3y = 10

x − 3y = 2Step 2 Look for a variable to eliminate. You can eliminate 3y by adding the two

equations since 3y + (−3y) = 0Step 3 Eliminate the variable and solve for

the remaining variable. 2x + 3y = 10 + x − 3y = 2

3x + 0 = 12 3x = 12 x = 4Step 4 Solve for the eliminated variable

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

2x + 3y = 10 2(4) + 3y = 10 8 + 3y = 10 3y = 2 y = 2 _ 3

The solution is (4, 2 _ 3 ).

Practice Solve

1. 4x + 3y = 12; −4x + y = 4

Line up the equations by like terms. −4x + 3y = 12

−4x + y = 4 Look for a variable to eliminate. You can eliminate by adding the two

equations. Eliminate the variable and solve for

the remaining variable. 4x + 3y = 12

−4x + y = 4

0x + y =

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

4x + 3y = 12

4x + 3 = 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

Name Date

Algebra


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.

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

is a multiple of the same variable in the other equation.

2x + 3y = 10 −4x + 2y = 4

You can multiply the first equation by 2.

Step 2 Multiply all terms in one equation to eliminate one variable.

2(2x + 3y = 10) = 4x + 6y = 20

Step 3 Eliminate one variable and solve for the remaining variable.

4x + 6y = 20

+(−4x + 2y = 4)

0x + 8y = 24 y = 3Step 4 Solve for the eliminated variable by

substituting the solution in Step 3 into one of the equations.

2x + 3y = 10 2x + 3(3) = 10

2x + 9 = 10 2x =1 x = 1 _ 2

The solution is ( 1 _ 2 , 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.

−2x + 10y = 20 5x − 5y = 6

You can multiply the second equation by .

Multiply all terms in one equation to eliminate one variable.

(5x − 5y = 6) =

Eliminate one variable and solve for the remaining variable.

−2x + 10y = 20

x = Solve for the eliminated variable by

substituting the solution in Step 3 into one of the equations.

−2x + 10y =20 −2 + 10y = 20

+ 10y = 20 10y = 28 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

Name Date

Algebra


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.

ExampleSolve: 2x2 − 32 = 0

Step 1 Get the x2 term on one side of the equation.

2x2 − 32 + 32 = 0 + 32

2x2 = 32

Step 2 Isolate x2. 2x2 ÷ 2 = 32 ÷ 2

x2 = 16

Step 3 Find the square roots. Remember the solution of a square root is both a positive number and a negative number.

√__

x2 = √___

16

x = 4, x = −4

Practice Solve.

1. 3x2 − 25 = 50

Get the x2 term on one side of the equation.

3x2 − 25 + 25 = 50 + 25

3x2 =

Isolate x2. 3x2 ÷ 3 = 75 ÷ 3

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

Name Date

Algebra


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.

The quadratic formula is x = –b ±

√_______

b2 – 4ac __________ 2a

ExampleUse the quadratic formula to solve the following quadratic equation: x2 + 5x − 50 = 0

Step 1 Identify a, b, and c. a = 1 b = 5 c = −50

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

x = –b ±

√_______

b2 – 4ac __________ 2a

x = –5 ±

√___________

52–4(1)(–50) ______________

2(1)

Step 3 Solve. x = –5 ± √________

25 + 200 ___________ 2 = –5± √

____ 225 _______ 2

x = –5 ± 15 ______ 2

x = 10 __ 2 or x = −20

___ 2

x = 5 or x = −10

Practice Solve using the quadratic formula.

1. x2 + 3x – 4 = 0

Identify a, b, and c. a = b = c =

Plug the values for a, b, and c into the quadratic formula.

x = –b ±

√_______

b2 – 4ac __________ 2a

x = − ± √____________________

__________________________ 2

Solve. x = −3 ± √____________

000000000000 _______________ 2 = −3 ± √

______ __________ 2

x = −3 ± 0000 ________ 2

x = 0 _ 2 or x = 0 _ 2

x = or x =

2. x2 + 15x + 26 = 0 4. 2x2 − 10x + 12 = 0

3. x2 – 6x − 72 = 0 5. 3x2 − 12x − 15 = 0

2 − 4( )( )

+

Name Date

Algebra


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.

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

equation.3x2 − 5x − 1 = 0

a = 3 b = −5 c = −1Step 2 Plug the numbers for a, b, and c into

b2 − 4ac.b2 − 4ac = (−5)2 − 4 (3)(−1)

Step 3 Solve. (−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.

x2 + 3x + 7 = 0

a = 1 b = c = Plug the numbers for a, b, and c into

b2 − 4ac.b2 − 4ac = 2 − 4(1)

Solve. 2 − 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

Name Date

Algebra


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.

ExampleSolve. (x + 3)(2x − 1) = 0

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

(x+3) = 0 2x − 1 = 0

Step 2 Solve each factor/equation. x + 3 − 3 = 0 − 3 2x − 1 + 1 = 0 + 1

x = − 3 2x ÷ 2 = 1 ÷ 2

x = 1 _ 2

Practice Solve.

1. (x + 5)(x − 2) =0

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

(x + 5) = 0 = 0

Solve each factor/equation. x + 5 − 5 = 0 − 5 = 0

x = −5 x =

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 =

Name Date

Algebra


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.

ExampleSolve. 2x2 + 5x + 2 = 0

Step 1 Be sure the equation is in the form ax2 + bx + c = 0. Set up a FOIL table to help factor the equation.

2x2 +5x +2

F O + I = ? L

2 × 1 2 × 12 × 2

++

2 × 11 × 1

==

45

1 × 2

Step 2 Use the FOIL table to identify the numbers in each binomial. Write the factored form of the original equation.

The outer terms are 2 and 2; the inner terms are 1 and 1.

2x2 + 5x + 2 = (2x + 1)(x + 2) = 0Step 3 Set each binomial equal to 0 and solve

for the two values of x.(2x + 1)(x + 2) = 0

2x + 1 = 0 x + 2 = 0

x = − 1 _ 2 x = −2

Practice 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.

Use the FOIL table to identify the numbers in each binomial. Write the factored form of the original equation.

The outer terms are 2 and –2; the inner terms are 1 and –1.

2x2 – 5x + 2 = ( )( ) = 0 Set each binomial equal to 0 and solve

for the two values of x.(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

2x2 5x 2

F O + I = ? L

2 × 1 2 × 2 ×

++

× 1 × 1

==

−4−5

Name Date

Algebra


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 1 _ 2 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.

ExampleSolve. x2 − 8x = 5

Step 1 Identify the coefficient in front of the variable that is not squared.

The coefficient in front of the variable that is not squared is −8.

Step 2 Take 1 _ 2 of the coefficient in Step 1 and square that number.

Half of −8 is −4; (−4)2 = 16.

Step 3 Add the result to both sides of the equation.

x2 − 8x + 16 = 5 + 16

Step 4 Factor the expression on the left side; add the terms on the right side.

x2 − 8x + 16 = 5 + 16

(x − 4)2 = 21

Step 5 Take the square root of each side. √_______

(x − 4)2 = √___

21 → x − 4 = √___

21

Step 6 Isolate the variable. x = 4 ± √___

21

Practice Solve.

1. x2 + 2x = 5

Identify the coefficient in front of the variable that is not squared.

The coefficient in front of the variable that is

not squared is .

Take 1 _ 2 of the coefficient in Step 1 and square that number.

Half of is ; 2 = .

Add the result to both sides of the equation.

x2 + 2x + = 5 +

Factor the expression on the left side; add the terms on the right side.

x2 + 2x + = 5 +

2 =

Take the square root of each side. √____

000 = √___

00 → = √___

00

Isolate the variable. x = ± √__

2. x2 − 4x = −3 4. x2 + 2x = 5

3. x2 − 2x = 8 5. x2 + 4x = −1

Name Date

Algebra


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.

ExampleEvaluate. y = 2 × 0.75x, for x = 2

Step 1 For each value of x, plug the number into the variable exponent.

y = 2 × 0.75x → replace x with 2

y = 2 × 0.752

Step 2 Follow order of operations by simplifying the power.

y = 2 × 0.752 → (0.752 = 0.5625)

y = 2 × 0.5625

Step 3 Multiply or divide from left to right. y = 2 × 0.5625

= 1.125

Practice Evaluate

1. y = −3 × 3x, for x = 4

For each value of x, plug the number into the variable exponent.

y = −3 × 3x → replace x with

y = −3 × 3

Follow order of operations by simplifying the power.

y = −3 × 3 → 3 =

y = −3 ×

Multiply or divide from left to right. y = −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

Name Date

Algebra


Exponential Growth Functions

You can model exponential growth using the following formula:

y = C(1 + r)t

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.

ExampleA 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 $200.00r, 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 exponential growth.

y = 200(1 + 0.06)10

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?

} time period

(1 + r) is the growth factorinitial amount

r = growth rate

Name Date

Algebra


Exponential Decay Functions

You can model exponential decay using the following formula:

y = C(1 – r)t

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.

ExampleA 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.

C, the initial amount, is $15,000.00.

r, the decay rate, is 10% or 0.10.

t, the time period, is 5 years.

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

y = 15,000(1 – 0.10)5

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?

} time period

(1 − r) is the decay factor

r = decay rate

initial amount

Name Date

Algebra


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.

ExampleSimplify. (x5)3

Step 1 Multiply the two exponents. The product is the new exponent. (x5)3 → 5 × 3 = 15

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 → 6 × =

Do not change the base. x

Simplify if needed. The expression simplified.

2. (x5)4 5. (x−2)−4

3. (x2)−3 6. (x−4)−2

4. (x5)5 7. (x−6)−3

exponents

base

Name Date

Algebra


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.

ExampleSimplify: (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 power.

44 = 256

Step 3 Multiply the exponent to the right of the variable by the power.

(x3)4 = x3 × 4 = x12

Step 4 Simplify. 256x12

Practice Simplify.

1. (2x3)5

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

(2x3)5 = 2 (x3)

Simplify the number raised to the power.

2 =

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

(x3) = x

Simplify. x

2. (5x3)2

3. (4x4)3

4. (3x−4)3

5. (4x2)−3

Name Date

Algebra


Raising a Quotient to a Power

Suppose you raise the following quotient to the 4th power.

( x _ y )4 = x _ y × x _ y × x _ y × x _ y = x4 __

y4

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.

ExampleSimplify. ( x

3 __

y2 )2

Step 1 Apply the power to the numerator and the denominator.

( x3 __

y2 )4 = (

(x3)4

____ (y2)4 )

Step 2 Multiply any exponents. ( (x3)4

____ (y2)4 ) = ( x

3×4 ___

y2×4 )

Step 3 Simplify. ( x3×4

___ y2×4 ) = ( x

12 ___

y8 )

Practice Simplify.

1. ( x5 __

y3 )3

Apply the power to the numerator and the denominator.

( x5 __

y3 )3 = (

(x5) ______

(y3) )

Multiply any exponents. ( (x5)3

____ (y3)3 ) = ( x

5 × _______

y3 × )

Simplify. ( x5 × 3

____ y3 × 3 ) = ( x _____ y )

2. ( x4 __

y2 )5

3. ( x6 ___

y–2 )–2

4. ( 3 __ x4 )

2

5. ( x5 __ 4 )3

Name Date

Algebra


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 __

72 = 7 × 7 × 7 × 7 × 7

____________ 7 × 7 = 73

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.

ExampleSimplify.

Step 1 Subtract the exponent in the denominator from the exponent in the numerator.

x7 __

x3 = x7−3

Step 2 Keep the base the same. x7−3 = x4

Practice Simplify

1. x10 ___

x3

Subtract the exponent in the denominator from the exponent in the numerator.

x10 ___

x3 = x

Keep the base the same. x = x

2. x4 __

x8 =

3. x7 ___

x12 =

4. x2y5

___ x5y3 =

5. x4y6

____ x2y–3

=

Name Date

Algebra


Simplifying Rational Expressions

The following expressions are examples of rational expressions:

4 _ x 5 ___ x+3 x

___ y–2 x2+2

____ y2–1

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.

ExampleSimplify. 10x2

____ 15x4

Step 1 Factor the numerator and denominator.Use guess and test to find two factors that create each expression.

10x2 ____

15x4 = (5x2)(2)

________ (5x2)(3x2)

Step 2 Divide out common factors. A factor in the numerator cancels out the same factor in the denominator.

10x2 ____

15x4 = (5x2)(2)

________ (5x2)(3x2)

Step 3 Simplify. 2 ___ 3x2

Practice Simplify.

1. 18x3 ____ 3x+6

Factor the numerator and denominator.Use guess and test to find two factors that create each expression.

18x3 _____ 3x + 6 =

(6x3) __________

(x + 2)

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

( )(6x3)

____________ ( )(x + 2)

Simplify. ( )

2. 6x+15 _____ 18 5. 3x+12

_____ 2x+8

3. 5x4 ____

35x7 6. 20x+5x ______ 10+5x

4. 4x+12 _____

2x2

Name Date

Algebra


Multiplying Rational Expressions

When multiplying a fraction, you multiply the numerators and multiply the denominators. You then simplify the product as needed.

1 _ 2 × 3 _ 4 = 1 × 3 ____ 2 × 4 = 3 _ 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.

ExampleMultiply. 4 __

x2 × 5 __ x3

Step 1 Multiply the numerators. 4 __ x2 × 5 __

x3 = 4×5 ______

(x2)(x3) = 20

______ (x2)(x3)

Step 2 Multiply the denominators. (Multiplying like variables with exponents means adding the exponents.)

20 ______

(x2)(x3) = 20

__ x5

Step 3 Express the resulting rational expression in simplest form.

20 __

x5 is in the simplest form.

Practice Multiply.

1. x2 ___ x–1 × 5x

__ 4

Multiply the numerators. x2 ____ x − 1 × 5x

__ 4 = (x2)(5x)

_______ (x − 1)(4)

= ______

_______ (x − 1)(4)

Multiply the denominators. ______

_______ (x − 1)(4)

= 00000000 _______ 00000000

Express the resulting rational expression in simplest form.

00000000 _______ 00000000 in the simplest form.

2. 3x2 ___ 2 × 6

___ 5x4

3. x+4 ___ x × 2x2

____ 2x−3

4. 2 _ 3 × 4x−1 ____

x4

5. 3x ___ x−2 × x+1

___ 3

6. 6x2 ____ 2x+2 × 5−2x3

_____ 8x

Name Date

Algebra


Dividing Rational Expressions

When dividing fractions you flip the second fraction in the expression and then follow the rules for multiplying fractions.

1 _ 2 ÷ 2 _ 3 = 1 _ 2 × 3 _ 2 = 1 × 3 ____ 2 × 2 = 3 _ 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.

ExampleDivide. x − 3 _____

x2 ÷ 2 _____ x + 1

Step 1 Flip the second expression and change the division sign to a multiplication sign.

x − 3 ____

x2 ÷ 2 ____ x + 1 = x − 3 ____

x2 × x + 1 ____ 2

Step 2 Multiply the numerators.

Leave the numerator in factored form.

x − 3 ____

x2 × x + 1 ____ 2 =

(x − 3)(x + 1) __________

(x2)(2)

Step 3 Multiply the denominators. (x − 3)(x + 1)

__________ (x2)(2)

= (x − 3)(x + 1)

__________ 2x2

Step 4 Express the resulting rational expressions in simplest form.

(x − 3)(x + 1)

__________ 2x2 is in the simplest form.

Practice Divide

1. x–2 ___ x+2 ÷ 2 __

x2

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

x − 2 ____ x + 2 ÷ 2 __

x2 = x − 2 ____ 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.

2. 3x2 ___ 2 ÷ −5x4

____ 4 4. x – 1 ____ x – 3 ÷ 5x + 6

_____ x2

3. x + 2 ____ x + 3 ÷ x − 1

____ 2 5. 2x + 6 _____ 4x ÷ 8x2

____ x + 1

Name Date

Algebra


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.

ExampleFind the LCD for the following pair of rational expressions. Then rewrite each expression with the LCD. 1 __ 8x and 3 ___ 10x

Step 1 Factor denominators into prime factors.

8x = 2 × 2 × 2 x 10x = 2 × 5 x

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 the least common denominator.

2 × 2 × 2 × 5 × x = 40x

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

1 __ 8x : multiply by 5 _ 5 ; 3 ___ 10x : multiply by 4 _ 4 ´

1 ___ 8x × 5 _ 5 = 5 ___ 40x 3 ___ 10x × 4 _ 4 = 12 ___ 40x

Practice Find the LCD of the following pair of rational expressions. Then rewrite each expression with the LCD.

1. 4 ___ 3x,

5 ____

x2–4x

Factor each denominator into prime factors.

3x = 3 × x x2 − 4x = 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 × = ________

________ (3x)(x − 4)

5 _____

x2−4x × =

________ ________

(3x)(x − 4)

2. 2 __ 5x , 10 __ 6x 4. 5

____ 12x2 ,

8 ___

8x3

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

____ 4x2y3 ,

9 ____

3x4y

Name Date

Algebra


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.

ExampleAdd. 2x ____

x2–1 + 4 ___ x–1

Step 1 Find the least common denominator. (x2 − 1) = (x − 1)(x + 1)

x − 1 = (x − 1)

The LCD = (x + 1)(x − 1)

Step 2 Rewrite each expression as an equivalent expression with the LCD as the denominator.

2x ____

x2 – 1 = 2x

__________ (x – 1)(x + 1)

4 ____ x – 1 = 4 ____ x – 1 × x + 1 ____ x + 1 = 4x + 4

__________ (x + 1)(x – 1)

Step 3 Add the numerators. The LCD is the denominator of the answer.

2x __________

(x – 1)(x + 1) + 4x + 4

__________ (x + 1)(x – 1)

= 6x + 4 __________

(x + 1)(x – 1)

Step 4 Express the answer in simplest terms. 6x + 4 __________

(x + 1)(x – 1) is in simplest terms.

Practice Add.

1. 3 __ 4x + 6 ___

7x2

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 __ 4x = 3 __ 4x × 7x __ 7x =

6 ___

7x2 = 6 ___

7x2 × =

Add the numerators. The LCD is the denominator of the answer.

+ =

Express the answer in simplest terms. in simplest terms.

2. 1 ___ x+2 + 3 __

x2 4. 4 ___ x+3 + 2 ___ x–2

3. x ___ x+3 +

(–4) ____

x2–9 5. 4 ___

2x2 + 3y

__ 2x

Name Date

Algebra


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.

ExampleSubtract. 8 ___ x–2 − 3 __ x

Step 1 Find the least common denominator (LCD).

List the factors of each denominator.

x − 2 = x − 2 x = x

So the LCD is x(x − 2)

Step 2 Rewrite each expression as an equivalent expression with the LCD as the denominator.

8 ___ x–2 = 8

___ x–2 × x _ x = 8x ______

(x)(x–2)

3 _ x = 3 _ x × x–2 ___ x–2 = 3x–6

______ (x)(x–2)

Step 3 Subtract the numerators. The LCD is the denominator of the answer.

8x ______

(x)(x–2) − 3x–6

______ (x)(x–2)

= 8x–3x+6 _______

(x)(x–2) = 5x+6

______ (x)(x–2)

Step 4 Express the answer in simplest terms. The answer is in simplest terms.

Practice Subtract.

1. 7x ____

x2–1 − 6

___ x–1

Find the least common denominator (LCD).

x2 − 1 =

x − 1 = (x − 1)

The LCD is

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

7x _____

x2 − 1 already has the LCD

7x _____

x2 − 1 = ( 7x

__________ (x − 1)(x + 1)

).

6 ___ x–1 × x + 1

____ x + 1 =

Subtract the numerators. The LCD is the denominator of the answer.

7x __________

(x – 1)(x + 1) – =

7x − 6x + 6 __________

(x − 1)(x + 1) =

Express the answer in simplest terms. The answer in simplest terms.

2. 4 ____ x − 1 − 2 _ x 4. 5 − x

____ x − 4 _____ x2 + 2x

3. x ____ x − 3 −

(–8) ____

x2 – 9 5. x + 3

____ 3x − 2x + 2 _____ 4x

Name Date

Algebra


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.

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.

sine of A= length of the leg opposite A

___________________ length of the hypotenuse

= sin A = opposite

________ hypotenuse

cosine of A = length of the leg adjacent A


= cos A = adjacent

________ hypotenuse

tangent of A = length of the leg opposite A

___________________ length of the leg adjacent A

= tan A = opposite

______ adjacent

ExampleFind sin A, cos A, and tan A.

Step 1 Find the sin A sine of A = length of the leg opposite A


= 3 _ 5

Step 2 Find the cos A cosine of A = length of the leg adjacent A


= 4 _ 5

Step 3 Find the tan A tangent of A = length of the leg opposite A


= 3 _ 4

Practice

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

Find the sin A sine of A = length of the leg opposite A


=

Find the cos A cosine of A = length of the leg adjacent A


=

Find the tan A tangent of A = length of the leg opposite 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.

A

5

4

3

A

B Cleg opposite to A

leg adjacent to Ahypotenuse

A

13

12

5

B Triangle 1

X

10

8

6

YTriangle 2

Name Date

Algebra


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.

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

Step 1 Identify the location of the sides in relation to the given angles.

The missing side is opposite the angle.

You know the measure of the hypotenuse.

Step 2 Determine the trigonometric ratio to use.

The trigonometric ratio using the opposite side and the hypotenuse is sine.

Step 3 Plug the numbers into the formula for the selected trigonometric ratio.

sin A = length of the leg opposite A


= sin 45˚ = x __ 16

Step 4 Solve for the unknown. Use your calculator to find the value of the angle.

sin 45˚ = x __ 16 = sin 45˚(16) = x

sin 0.707(16) = x = 11.31Practice Find the length of the unknown side.

1.

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

The missing side is the angle.

You know the measure of the .

Determine the trigonometric ratio to use.

The trigonometric ratio that uses the

and the is

the

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

cosine of A = length of the leg adjacent to A

_____________________ length of the hypotenuse

=

cos =

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

cos = = cos = x

cos = 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

12

25˚x

16

x

45˚

Name Date

Algebra


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.

theoretical probability = number of favorable outcomes ________________________

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.

ExampleWhat 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.Step 2 Count the number of favorable

outcomes.A number cube has only one number “2.” The number of favorable outcomes is 1.

Step 3 Plug the outcomes into the formula for theoretical probability.

theoretical probability =

number of favorable outcomes ________________________

total number of possible outcomes = 1 _ 6

The probability of tossing a “2” is 1 _ 6 .Step 4 Express the probability as a fraction in

simplest terms. 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”?

Count the total number of outcomes. There are possible outcomes:

, ,

, ,

, ,

.

Count the number of favorable outcomes.

There are favorable outcomes,

and .

Plug the outcomes into the formula for theoretical probability.


Express the probability as a fraction in simplest terms.

The fraction 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”?

Name Date

Algebra


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.

ExampleA 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 occurs.

The number of times the event occurs (shots made) is 122.

Step 2 Identify the number of times the experiment is done.

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

Step 3 Plug the numbers into the formula for experimental probability.


number of times event occurs ________________________

number of times experiment done = 122 ___ 250

Step 4 Express the probability as a fraction in simplest terms or as a decimal.

122 ___ 250 = 61 ___ 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

done: .

Plug the numbers into the formula for experimental probability.

theoretical probability = =

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.

Name Date

Algebra


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.

ExampleFind 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 numbers are in the data set.

There are 5 numbers in the data set.

160 ___ 5 = 32

PracticeFind 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

Name Date

Algebra


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.

ExampleFind the median of this set: 10, 2, 14, 12, 17, 6, 15, 9, 19

Step 1 Place the numbers in order from least to greatest.

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

Step 2 Count the number of data points. There are nine data points.

Step 3 If there is an odd number of data, you find the middle value.

Nine is an odd number.

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

The middle value is 12.

PracticeFind 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 data points.

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

There is an 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

Name Date

Algebra


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.

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.

ExampleA softball team fields 10 players. How many different batting orders can there be?

Step 1 What is n (the number of things from which to chose)?

There are 10 players to place in the order.

Step 2 What is r (number of things taken at a time)?

All ten are taken at a time.

Step 3 Plug the numbers into the equation for combinations.

10P10 = 10! ______

(10–10)!

= 10! ___ 0!

10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 __________________________ 1 =

3,628,800 _______ 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 students running for class office.

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

Only can be elected.

Plug the numbers into the equation for combinations.

8P3 = 8! ______ 0000000

= 8! _____ 00000

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

40,320 _____ 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?

nPr = n! _____

(n–r)!

Name Date

Algebra


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.

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)

ExampleFive 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 from which to chose)?

There are 5 people from which to choose.

Step 2 What is r (number of things chosen at a time)?

There are 2 people chosen at a time.

Step 3 Plug the numbers into the equation for combinations.

5C2 = 5! ______

2!(5–2)!

= 5! ____

2!(3)!

= 5 × 4 × 3 × 2 × 1 _____________

(2 × 1)(3 × 2 × 1) = 120

___ 12 = 10

Practice

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

_______ 000 = 120

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?

nCr = n! ______

r!(n–r)!

Name Date

Algebra


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

Middle School 23,200 2,100

High School 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.

ExamplePlace the data from the table into a matrix.

1 2 3 410 20 30 402 4 6 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.

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

Practice 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.

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

The data in the table is in rows and

columns; create a ×

matrix.

2. Northeast North central South West

1,720 2,646 2,306 2,6922,889 3,963 2,889 2,808

29,300 2,900

23,200 2,100

15,400 1,200

1 2 3 4

10 20 30 40

2 4 6 8

Name Date

Algebra


Matrix Addition

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

+

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. +

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.

Practice Add.

1. +

Combine corresponding elements in each matrix to form one large matrix. Place a plus sign between each corresponding element.

Add the corresponding elements.

2. + =

3. + =

4. + =

–2 0

4 2

3 –1

4 7

corresponding elements

–4 2

–10 75 –9

9 –3

–4 2

−10 7

5 −9

9 −3

–4 + 5 2+ (–9)

(–10) + 9 7 + (–3)+ =

1 –7

–1 4

–4 + 5 2 + (–9)

(–10) + 9 7 + (–3) =

–5 83 –3

11 –1

5 –7

–5 8

3 –3

11 –1

5 –7+ =

=

2 –9 –4

3 –1 5

4 16 9

11 1 2

–4 7

–9 3

7 2

–4 20

2

17

5

–12

Name Date

Algebra


Matrix Subtraction

When subtracting matrices, you subtract the corresponding elements in each matrix.

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.

ExampleSubtract.

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.

Practice Subtract.

1.

Combine corresponding elements in each matrix to form one large matrix; place a minus sign between corresponding elements.

Subtract corresponding elements.

2. –

3. –

4. –

–2 0

4 2

3 –1

–4 7

corresponding elements

–

–2 5

0 –2–4 6

8 5

–2 5

0 −2

–4 6

8 5

–2–(–4) 5–6

0–8 –2–5– =

2 –1

–8 –7

–2–(–4) 5–6

0–8 –2–5 =

–

3 3

–4 –1

6 –2

8 –2–

3 3

–4 –1

6 –2

8 –2– =

=

9 –12 15

4 7 –5

2 4 –3

5 –1 –3

–3 5

0 –4

–5 9

10 3

6 12

–8 –4

–9 0

–2 –9

Name Date

Algebra


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

Step 1 Create an expanded matrix in which each element is multiplied by the scalar. 2 =

Step 2 Find the product of each element times each element.

=

Practice Solve.

1. 5

Create an expanded matrix in which each element is multiplied by the scalar.

5 =

Find the product of each element times each element

=

2. –3

3. 4

4. –6

–6 4

7 –3

–6 47 –3

–6 × 2 4 × 27 × 2 –3 × 2

–6 × 2 4 × 27 × 2 –3 × 2

–12 814 –6

11 –9 –4–5 6 3

11 –9 –4–5 6 3

11 × 5 –5 × 5

55 –25

–8 –4 10 2 –9

2 169 –2

–11 6

5 –128 –2

11 × 5 –5 × 5

Name Date

Algebra


3 6 5

5 3 8

5

2

2

× =3×5 + 6×2 + 5×2

5×5 + 3×2 + 8×2

3 5

2 –3

2

9×

3×2 + 5×9

2×2 + –3×9

3 5

2 –3

2

9× =

3×2 + 5×9

2×2 + –3×9= =

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.

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.

ExampleMultiply.

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.

Identify the elements to be multiplied.

Create a large matrix showing the multiplication of the elements.

Find the products of each element times each element. Add the products in each row.

2. 3.

6 + 45

4 + –27

51

−23

3 5

2 –3

2

9×

2 –4

3 7×

1 9

3 2

2 –4

3 7×

1 9

3 2

2 –4

3 7× =

1 9

3 2

=

6 3 8

9 2 4

7

4

2

×

2 5

4 6×

9 4

2 3

Name Date

Algebra


Answer Key

PAGE 1 Classifying Numbers 1. – √

__ 2 ,

0, 1.35, 5 2 _ 3 , 2, 5, 0, 0, 2, 2

2. – √___

10 , √__

3 3. −4.2, −3, −1, − 1 _ 3 , 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

+, 9, 17 subtract, positive 8, 8

5. −21 6. −4 7. −14 8. 8 9. 18 10. 0

PAGE 9 Multiplying Integers 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

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

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 13 Negative Exponents 1. exponent, base

1 __ 25

1 __ 25 ,

1 ____________ 2 × 2 × 2 × 2 × 2 , 1 __ 32

2. 1 _ 8

3. 1 _ 9

4. − 1 ___ 125

5. 1 __ 81

6. 1 __ a2

7. 4 1 __ x3 = 4 __

x3

Algebra


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. 2 _ 4 = 1 _ 2

−5, −9

(−9) = 1 _ 2 (x − (−5))

2. y – 8 = 2 (x − 4) 3. y − 12 = −2 (x − (−10)

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 Equations by Adding or Subtracting

1. −125 + 125, + 125, + 125, 0 2. x = 96 3. x = 71 4. x = 207 5. x = 0 6. x = 107 7. x = 211

PAGE 19 Solving One-Step Equations by Multiplying or Dividing

1. divided, division, 8, 192, x __ 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 Equations

1. 7x 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 with Variables on Both Sides

1. 5x 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 Coordinate Plane

1. 7, left down, 4, check graph 2. check graph 3. check graph 4. check graph 5. (2, 2) 6. (5, -4) 7. (−3, 7) 8. (−9, −5)

PAGE 27 Finding Solutions of Linear Equations

1. (−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. check graphs, possible answers

2. (−1, 4),(0, 3),(1, 2),(2, 1) 3. (−1, 1),(0, −1),(1, −3),(2, −5) 4. (−1, 1),(0, 5),(1, 9),(2, 13) 5. (−3, 0),(0, 1),(3, 2),(6, 3) 6. (−1, 1),(0, 3),(1, 5),(2, 7)

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

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)

Algebra


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

8 _ 6 , 3 _ 3 , 1, 1

2. 6 _ 7

3. 1

4. – 1 _ 3

5. 1

6. −2 __ 3

7. 3 _ 4

PAGE 32 Slope Intercept Form 1. −3 7

2. slope = 1 _ 3 , y–int. =−3

3. slope = 1, y-int. = 1

4. slope = 3 _ 4 , y-int. = −13

5. slope = – 2 _ 3 , y-int. = 3

PAGE 33 Point Slope Form I 1. 6 −3−1 y – (−1) = 6 (x − (−3))

2. y − 1 = – 1 _ 2 (x − 7)

3. y− (−3) = 2(x − (−3))

4. y − (−5) = 2 _ 3 (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 = 1 _ 2 (x − 2)

3. y − 4 = −1(x − (−6)) 4. y − 6 = 3(x − 2) 5. y − 2 = 1(x − 5)

6. y − 0 = 2 _ 3 (x −6)

PAGE 35 Parallel Lines 1. 1 _ 2 x − 1

3, 1 _ 2

are not, are not

2. m = – 1 _ 4 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 = – 1 _ 4 and – 4 _ 3 , the graphs are not parallel.

PAGE 36 Perpendicular Lines 1. 2 negative, positive, 2, 2, −3

2. y = − 4 _ 3 x + 4

3. y = − 1 _ 2 x+3

PAGE 37 Using Reciprocals

1. – 5 _ 2

negative, –5 __ 2 , positive, 5 _ 2

2. 7 _ 1 or 7 and − 7 _ 1 or −7

3. 3 _ 5 and − 3 _ 5

4. 13 __ 5 and − 13

__ 5

5. 8 _ 9 and − 8 _ 9

PAGE 38 Solving Equations That Contain Decimals

1. 2, 2 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. 3 _ 5

6 __ 10

6 __ 10 , 11 __ 10

10 __ 11 , 11 __ 10 , 10

__ 11 , 100 ___ 11 =9 1 __ 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 from a Graph

1. 6, 6 left, open, <, 6 2. x –5 3. x > 3

PAGE 42 Solving One-Step Inequalities by Adding or Subtracting

1. + 16 −16, −16 +, −16, 28, 0, 28 2. x < −2 3. x > 8 4. 53 > x 5. 12 < x 6. 37 > x 7. 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, ÷ −2, −12, <, −12 2. x −4 3. x > −8 4. x < −5 5. x < −65 6. x < 60 7. x −12

PAGE 45 The Pythagorean Theorem

1. a leg 15, c2, 225, c2

225, c2, 289, c2, 17, c 2. a = √

___ 80

3. c = 5 4. c = 25 5. b = 150

Algebra


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 Expressions by Multiplying Two Radicals

1. 6x 6x, (3×6)(x×x), 18x2

18x2, 9x2, 2, 3x √__

2

2. √____

40x5 = 2x2 √____

10x

3. √___

4x9 = 2x4 √__

x

4. √____

50x5 = 5x2 √___

2x

5. √____

16x8 = 4x4

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

1. 16x6

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 51 Estimating Square Roots

1. 4 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 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

1. −4x6, x5 __ 2 , 10, −10x4

2. −2x5 − 8x4 + 4x3

3. 3x8 + 7x6 − 10x4

4. 6x5 + 15x4 − 3x3 ___ 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

PAGE 57 Subtracting Polynomials

1. (3x2 − 5x + 10) (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 Polynomial

1. 12x5, 6x4, −3x3, 12x5 + 6x4 − 3x3

2. −21x5 + 14x2

3. 8x6 − 2x5 + 6x4

4. 3x7 + 6x5 + x3

5. −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 the Form x2 + bx + c

1. −2, −6, −8 2. (x + 3)(x + 1) 3. (x + 1)(x + 8) 4. (x + 5)(x + 3) 5. (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


3. (x − 3)(2x + 3) 4. (7x + 1)(x + 7) 5. (2x − 2)(x − 3) 6. (2x + 1)(x − 4)

PAGE 64 The Difference of Two Squares

1. x8

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 of Equations by Substitution

1. x − 1 + x, x − 1, + x, 2x − 1 + 1, 2x − 1, + 1, 2x (3, 2) 2. (3, 5) 3. (1, 6) 4. (2, 6)

5. ( 5 _ 2 , 5)

PAGE 67 Solving Systems of Equations by Elimination

1. 4x 4y = 16, 4 (4), 12, 0, 0 2. (–9, 7) 3. (3, 4) 4. (4, –2) 5. (3, –1)

PAGE 68 Solving Linear Systems by Multiplying

1. 2 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 Equations Using Square Roots

1. 75 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. 7 _ 2 , 3

4. – 5 _ 6 , −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 = 1 _ 3 , 7 _ 2

3. (2x+ 3)(x − 1) = 0; x = − 3 _ 2 , 1

4. (x − 3)(3x + 2) = 0; x = 3, − 2 _ 3

PAGE 74 Solving a Quadratic Equation by Completing the Square

1. 2 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 Functions

1. 4, 4 4, 4, 81, 81 81, −243 2. −108 3. 0.046875

4. 1 __ 27

5. 4 __ 25 = 4 __ 32 = 1 _ 8 = 0.125

6. 0.09375

7. 3.2 ___ 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

Algebra


Raising a Product to a Power

1. 5, 5 5, 32 5, 15 32, 15 2. 25x6

3. 64x12

4. 27x−12 = 27 ___

x12

5. 4−3(x−6) = 1 ____ 64x6

PAGE 80 Raising a Quotient to a Power

1. 3, 3, 3, 3, 15, 9

2. x20 ___

y10

3. x−12

___ y4 = 1 ____

x12y4

4. 9 __ x8

5. x15 ___ 64

PAGE 81 Dividing Powers with the Same Base

1. 10-3 10-3, 7 2. x−4

3. x−5

4. x−3y2

5. x2y9

PAGE 82 Simplifying Rational Expressions

1. 3, 3

2. 2x+5 ____ 6

3. 1 ___ 7x3

4. 2x+6 ____

x2

5. 3x + 12 ______ 2x + 8

6. 4+x ____ 2+x

PAGE 83 Multiply Rational Expressions

1. 5x3, 5x3, 5x3

5x3 _____ 4x − 4 , is

2. 18x2 ____

10x4

3. 2x3+8x2 ______

2x2−3x

4. 8x−2 ____

3x4

5. 3x2+3x _____ 3x−6 = x

2 + x _____ x − 2

6. 30x2−12x5 ________

16x2+16x = 15x − 6x4

_______ 8x + 8

PAGE 84 Dividing Rational Expressions

1. x2 __ 2 , x

2 __ 2

x3–2x2 _______

(x+2)(2) , x

3–2x2 _____ 2x+4

is

2. 12x2 _____

–10x4 = 6x ____

−5x2

3. 2x+4 ________

(x+3)(x–1)

4. x3–x2 _________

(x–3)(5x+6)

5. (2x + 6)(x + 1)

___________ 32x3

PAGE 85 Finding the LCD of a Rational Expression

1. once, once, once x × (x − 4), 3x (x − 4)

x−4 ___ x−4 , 4x−16, 3 _ 3 , 15

2. 12 ___ 30x , 50 ___ 30x

3. 48 ___ 60x , 105

___ 60x

4. 10x ____

24x3 , 24 ____

24x3

5. 18x2 _____

12x4y3 , 36y2

_____ 12x4y3

PAGE 86 Adding Rational Ex pressions

1. 7 × x × x , × 7 × x × x = 28x2

21x ____

28x2 , 4 _ 4 , 24 ____

28x2

21x ____

28x2 , 24 ____

28x2 , 21x + 24

______ 28x2

21x + 24 ______

28x2 , is

2. x2 + 3x + 6

________ (x2)(x + 2)

3. x2 − 3x − 4 __________

(x + 3)(x − 3)

4. 6x − 2 __________

(x + 3)(x − 2)

5. 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 __________

(x − 1)(x + 1) , 7x − 6x − 6

__________ (x − 1)(x + 1)

,

x − 6 __________

(x − 1)(x + 1)

is

2. 2x + 2 ______

x(x − 1)

3. x2 + 3x + 8 __________

(x − 3)(x + 3)

4. −x2 + 3x + 6 _________

x(x + 2)

5. −2x + 6 ______ 12x = −x + 3

_____ 6x

PAGE 88 Finding Trigonometric Ratios

1. 5 __ 13 , 12 __ 13 , 5 __ 12

2. 12 __ 13 , 5 __ 13 , 12 __ 5

3. 3 _ 5 , 4 _ 5 , 3 _ 4

4. 4 _ 5 , 3 _ 5 , 4 _ 3

PAGE 89 Using Trigonometric Ratios to Find a Missing Length

1. adjacent, hypotenuse adjacent, hypotenuse, cosine

25°, x __ 12

25°, x __ 12 , 12, 0.906, 12, 10.87

2. x = 1 3. x = 25.98

PAGE 90 Theoretical Probability 1. 7, Monday, Tuesday, Wednesday,

Thursday, Friday, Saturday and Sunday

2, Saturday and Sunday

2 _ 7

is

2. 2 __ 12 = 1 _ 6

3. 4 __ 12 = 1 _ 3

PAGE 91 Experimental Probability

1. 44, 250,

number of times event occurs ________________________

number of times experiment done

44 ___ 250

44 ___ 250 , 22 ___ 125 , 0.176

2. = 0.376 3. = 0.344 4. = 0.495

PAGE 92 Mean 1. 331, 6, 331

___ 6 , 55.2 2. 21 3. 43.6 4. 60 5. 40 6. 23.7 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


Permutations 1. 8 3 (8–3), 5!, 5 × 4 × 3 × 2 × 1,

120, 336 2. 56 3. 20 4. 870

PAGE 95 Combinations 1. 10, 3, 3!(10 − 3)!, 3!7!,

(3 × 2 × 1)(7 × 6 × 5 × 4 × 3 × 2 × 1), 30,240

2. 15 3. 35

PAGE 96 Matrices 1. 2, 6, 2, 6

25 50 75 100 125 150

200 180 160 140 120 100

2. 1,720 2,646 2,306 2,692

2,889 3,963 2,889 2,808

PAGE 97 Matrix Addition 1.

−5 + 11 8 + (−1)

3 + (5) −3 + (−7)

−5 + 11 8 + (−1)

3 + 5 −3 + (−7),

6 7

8 −10

2. 6 7 5

14 0 7

3. 3 9

−13 23

4. 7

5

PAGE 98 Matrix Subtraction 1.

3 − 6 3 − (−2)

−4 − 8 −1 − (−2)

3 − 6 3 − (−2)

−4 − 8 −1 − (−2),

−3 5

−12 1

2. 2 −4

−10 −7

3. 7 −16 18

−1 8 −2

4. 15 12

−6 5

PAGE 99 Scalar Multiplication 1. −9 × 5, −4 × 5, 6 × 5, 3 × 5

−9 × 5, −4 × 5, −45, −20 6 × 5, 3 × 5, 30, 15

2. −6 −48

−27 6

33 −18

3. 20 −48

32 −8

4. 48 24 −6

0 −12 54

PAGE 100 Matrix Multiplication 1.

2 × 1 + (−4) × 3) 2 × 9 + (−4) × 2

3 × 1 + 7 × 3 3 × 9 + 7 × 2

2 + (−12) 18 + (−8)

3 + 21 27 + 14,

−10 10

24 41

2. 70

79

3. 28 23

48 34

Algebra


Algebra (Curriculum Binders (Reproducibles))

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