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Holt McDougal Florida

Larson Algebra 1

Ready to Go On?Intervention and Enrichment

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ISBN 13: 978-0-547-24246-0ISBN 10: 0-547-24246-8

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Copyright © by Holt McDougal. iii Holt McDougal Algebra 1All rights reserved.

INTRODUCTIONUsing the Ready to Go On? Intervention and Enrichment ..................... v

Item Analysis ChartsChapters 1 and 2 Quiz ................................. vi

Chapters 3 and 4 Quiz .................................vii

Chapters 5 and 6 Quiz ................................viii

Chapters 7 and 8 Quiz ................................. ix

Chapters 9 and 10 Quiz ................................ x

Chapters 11 and 12 Quiz ............................. xi

Skills Checklist ..........................................xii

INTERVENTION WORKSHEETS

Chapters 1 and 2A. Expressions and Equations ..................... 1

B. Problem Solving ....................................... 4

C. Representations of Functions .................. 6

D. Operations ............................................. 10

E. Properties and Real Numbers ................ 15

Chapters 3 and 4A. Solving Equations .................................. 19

B. Proportion and Percent .......................... 23

C. Equations in Two Variables ..................... 27

D. Graphing Linear Equations .................... 29

E. Slope-Intercept Form ............................. 33

Chapters 5 and 6A. Writing Linear Equations ........................ 36

B. Parallel and Perpendicular ..................... 41

C. Linear Models ........................................ 44

D. Graphing Inequalities ............................. 49

E. Solving Inequalities ................................ 52

F. Absolute Value Equations ....................... 56

Chapters 7 and 8A. Solving Systems .................................... 58

B. Solving Systems Using Algebra ............. 63

C. Systems of Linear Inequalities ............... 67

D. Properties of Exponents ........................ 70

E. Scientifi c Notation .................................. 74

F. Exponential Functions ............................ 76

Chapters 9 and 10A. Polynomial Operations ........................... 80

B. Factoring Polynomials ............................ 84

C. Graphing Quadratic Functions ............... 88

D. Solving Quadratic Functions .................. 92

E. Comparing Models ................................. 96

Chapters 11 and 12A. Radical Functions ................................ 100

B. Radicals in Geometry .......................... 106

C. Rational Expressions ........................... 109

D. Solving Rational Equations .................. 115

Contents

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Copyright © by Holt McDougal. iv Holt McDougal Algebra 1All rights reserved.

Contents continued

QUIZZESChapters 1 and 2 Quiz .............................. 119

Chapters 3 and 4 Quiz .............................. 125



Chapters 9 and 10 Quiz ............................ 144

Chapters 11 and 12 Quiz .......................... 150

CUMULATIVE REVIEWChapter 1 Cumulative Review .................. 154

Chapter 2 Cumulative Review .................. 156








Chapter 10 Cumulative Review ................ 172

Chapter 11 Cumulative Review ................ 174

Chapter12 Cumulative Review ................. 176

ENRICHMENTChapter 1 Enrichment .............................. 178

Chapter 2 Enrichment .............................. 179








Chapter 10 Enrichment ............................ 187



Answer Key ............................................. 190

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Copyright © by Holt McDougal. v Holt McDougal Algebra 1All rights reserved.

Using the Ready to Go On? Intervention and Enrichment in Your Class

InterventionThe Ready to Go On? Intervention helps students to perform successfully by providing opportu-nities for you to address students’ weaknesses before the students are given summative assess-ments.

The Ready to Go On? Intervention is separated into six sections, which each cover 4 to 6 big ideas. Each big idea provides examples, vocabulary, helpful hints, and practice exercises. At the end of each big idea there is check to assess the student’s understanding.

Assessment and ReviewThe Ready to Go On? Intervention is correlated with the Ready to Go On? Quizzes. The Item Analysis Charts can be used to diagnose students’ weaknesses.

Use the Ready to Go On? Quizzes to assess the student’s profi ciency after you have provided Intervention, or use the Ready to Go On? Quizzes to assess the student’s retention on new concepts taught in the lessons.

The Ready to Go On? Cumulative Review can be used to prepare for the Ready to Go On? Quizzes, or to review concepts that have not been mastered.

EnrichmentFor those students who show profi ciency on the Ready to Go On? Quizzes provide them with the appropriate Enrichment worksheets. The worksheets extend the concepts taught in the les-sons.

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Copyright © by Holt McDougal. vi Holt McDougal Algebra 1All rights reserved.

Ready to Go On? Ch1–2 Quiz Item Analysis

Ready to Go On? Intervention Quiz Exercises

Items Correct

A. Expressions, Equations, and Inequalities

1. Evaluate Expressions 1, 2, 3 /3

2. Order of Operations 4, 5, 6, 7 /4

3. Write Expressions 8, 10, 11 /3

4. Write Equations and Inequalities 9 /1

B. Problem Solving

1. Check Possible Solutions 12, 13, 14, 15 /4

2. Read and Understand a Problem 16 /1

3. Make a Plan 17, 18 /2

C. Representations of Functions

1. Identify the Domain and Range of a Function 25, 26, 27 /3

2. Write a Function Rule19, 20, 21, 24b, 24c, 28a

/6

3. Make a Table for a Function 22, 23 /2

4. Graph a Function 24a, 28b /2

D. Operations

1. Find Opposites of Numbers 29, 30, 31 /3

2. Find Absolute Values of Numbers 29, 30, 31 /3

3. Add Numbers 32, 33, 34 /3

4. Subtract Numbers 35, 36, 37, 49 /4

5. Multiply Numbers and Find Multiplicative Inverses of Numbers

38, 39, 40, 47 /4

6. Divide Real Numbers 41, 42, 43 /3

7. Find Square Roots 44, 45, 46, 50, 51 /5

E. Properties and Real Numbers

1. Classify Numbers 52, 53 /2

2. Order Numbers 54, 55 /2

3. Identify Properties of Addition 57 /1

4. Identify Properties of Multiplication 58, 59 /2

5. Apply the Distributive Property 56, 60, 61, 62 /4

Name ——————————————————————— Date ————————————

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Copyright © by Holt McDougal. vii Holt McDougal Algebra 1All rights reserved.



Items Correct

A. Solving Equations in One Variable

1. Solve an Equation Using Addition or Subtraction 1, 2 /2

2. Solve an Equation Using Multiplication or Division 3, 4 /2

3. Solve a Two-Step Equation 5, 6, 7 /3

4. Solve Multi-Step Equations 8, 9, 10, 13 /4

5. Solve Equations with Variables on Both Sides 11, 12 /2

6. Identify the Number of Solutions to an Equation 14, 15 /2

B. Proportion and Percent Problems

1. Write a Ratio 16, 17 /2

2. Solve a Proportion 18, 19, 20 /3

3. Use the Cross Products Property 21, 22, 23, 24 /4

4. Find a Percent Using a Proportion 25, 26, 27 /3

5. Use the Percent Equation 28, 29 /2

C. Rewriting Equations in Two or More Variables

1. Solve a Literal Equation 31a, 32a, 33a, 34 /4

2. Use the Solution to a Literal Equation 31b, 32b, 33b /3

3. Rewrite an Equation 35, 36, 37, 38 /4

D. Graphing Linear Equations

1. Plot Points in a Coordinate Plane 39, 40, 41, 42 /4

2. Identify Solutions To Equations in Two Variables 43 /1

3. Graph an Equation Using a Table 44 /1

4. Graph Horizontal and Vertical Lines 45, 46, 47, 48 /4

E. Slope-Intercept Form and Direct Variation

1. Find the Intercepts of the Graph of an Equation 49, 50, 51 /3

2. Find the Slope of a Line 52, 53, 54 /3

3. Graph an Equation Using Slope-Intercept Form 55, 56, 57, 58 /4

4. Identify Direct Variation Equations59, 60, 61, 62, 63, 64

/6

5. Write and Use a Direct Variation Equation 65 /1

Name ——————————————————————— Date ————————————

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Copyright © by Holt McDougal. viii Holt McDougal Algebra 1All rights reserved.



Items Correct

A. Writing Linear Equations

1. Write an Equation in Slope-Intercept Form 1, 2, 3 /3

2. Write an Equation of a Line Given the Slope and a Point 4 /1

3. Write an Equation of a Line Given Two Points 5 /1

4. Write an Equation in Point-Slope Form 6 /1

5. Write an Equation in Standard Form 7, 8 /2

B. Parallel and Perpendicular Lines

1. Determine Whether Lines are Parallel or Perpendicular 9 /1

2. Write an Equation of a Parallel Line 10, 11 /2

3. Write an Equation of a Perpendicular Line 12, 13 /2

C. Linear Models

1. Describe the Correlation of Data 14, 15, 16, 17 /4

2. Make a Scatter Plot 19a, 20a /2

3. Draw a Line of Fit to Data 18, 19b, 20b /3

4. Interpolate Using an Equation 20c /1

5. Extrapolate Using an Equation 19c /1

D. Graphing Inequalities

1. Graph an Inequality 21, 22 /2

2. Write an Inequality Represented by the Graph 24, 25 /2

3. Write and Graph Compound Inequalities 26, 27 /2

4. Graph a Linear Inequality in Two Variables 23, 28, 29, 30, 31 /5

E. Solving Inequalities

1. Solve an Inequality Using Addition or Subtraction 32, 33 /2

2. Solve an Inequality Using Multiplication or Division 34, 35 /2

3. Solve a Multi-Step Inequality 36, 37, 38, 39 /4

4. Solve a Compound Inequality with “and” 40, 41 /2

5. Solve a Compound Inequality with “or” 42, 43 /2

F. Absolute Value Equations and Inequalities

1. Solve an Absolute Value Equation 46 /1

2. Solve an Absolute Value Equation in Multiple Steps 47, 48, 49, 50, 51 /5

3. Solve an Absolute Value Inequality 52, 53, 54, 55, 56 /5

Name ——————————————————————— Date ————————————

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Copyright © by Holt McDougal. ix Holt McDougal Algebra 1All rights reserved.



Items Correct

A. Solving Linear Systems by Graphing

1. Checking Solutions to Linear Systems 1 /1

2. Graphing Linear Systems 2 /1

3. Graph-and-Check Method 3 /1

4. Special Types of Linear Systems 4, 5 /2

B. Solving Linear Systems Using Algebra

1. Solve Linear Systems by Substitution 6, 7 /2

2. Solve Linear Systems by Adding or Subtracting Equations 8, 9, 10 /3

3. Solve Linear Systems by Multiplying11, 12, 13, 14, 15, 16, 17

/7

C. Solving Systems of Linear Inequalities

1. Graphing Systems of Two Linear Inequalities 18, 19 /2

2. Graphing Systems of Three Linear Inequalities 20 /1

3. Write a System of Linear Inequalities 21, 22, 23 /3

D. Properties of Exponents

1. Product of Powers Property 24, 25 /2

2. Power of Power Property 26 /1

3. Power of a Product Property 27 /1

4. Quotient of Powers Property 28, 33 /2

5. Power of a Quotient Property 29 /1

6. Zero and Negative Exponents 30, 31, 32, 34 /4

E. Scientifi c Notation

1. Writing Numbers in Scientifi c Notation 35, 36, 37, 38 /4

2. Computing with Scientifi c Notation 39, 40, 41, 42 /4

F. Exponential Functions

1. Graph an Exponential Growth Function 43, 44 /2

2. Use an Exponential Growth Model 45 /1

3. Graph an Exponential Decay Function 46, 47 /2

4. Use an Exponential Decay Model 48 /1

Name ——————————————————————— Date ————————————

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Copyright © by Holt McDougal. x Holt McDougal Algebra 1All rights reserved.



Items Correct

A. Adding, Subtracting, and Multiplying Polynomials

1. Identify and Classify Polynomials 1, 2, 3 /3

2. Add and Subtract Polynomials 4, 5, 6, 7 /4

3. Multiply Polynomials 8, 9, 10, 11 /4

4. Find Special Products of Polynomials 12, 13, 14 /3

B. Factoring Polynomials

1. Use the Zero Product Property 15, 16 /2

2. Solve an Equation by Factoring 17, 18, 19 /3

3. Factor x2 1 bx 1 c 20, 21, 22 /3

4. Factor ax2 1 bx 1 c 23, 24, 25, 26 /4

5. Factor Special Products 27, 28, 29 /3

6. Factor by Grouping 30 /1

C. Graphing Quadratic Functions

1. Graph y 5 ax2 31 /1

2. Graph y 5 x2 1 c 32 /1

3. Find the Axis of Symmetry and the Vertex 33, 34 /2

4. Graph y 5 ax2 1 bx 1 c 35, 36 /2

5. Find the Minimum or Maximum Value 37, 38 /2

D. Solving Quadratic Equations

1. Solve a Quadratic Equation by Graphing 39 /1

2. Solve a Quadratic Equation Using Square Roots 40, 41 /2

3. Solve a Quadratic Equation by Completing the Square 42, 43 /2

4. Solve a Quadratic Equation Using the Quadratic Formula 44, 45 /2

5. Use the Discriminant to Find the Number of Solutions 46, 47, 48, 49 /4

E. Comparing Linear, Exponential, and Quadratic Models

1. Choose Functions Using Sets of Ordered Pairs 50 /1

2. Identify Functions Using Differences or Ratios 51, 52 /2

3. Write an Equation for the Function 53 /1

Name ——————————————————————— Date ————————————

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Copyright © by Holt McDougal. xi Holt McDougal Algebra 1All rights reserved.



Items Correct

A. Radical Functions

1. Graph y 5 a Ï} x , y 5 Ï}

x + k, and y = Ï}

x 2 k 1, 2, 3 /3

2. Use Properties of Radicals and Rationalize the Denominator

4, 5, 6, 7 /4

3. Simplify Radical Expressions 8, 9, 10 /3

4. Solve a Radical Equation 11, 12, 13, 14 /4

5. Solve an Equation with an Extraneous Solution 15, 16 /2

B. Radicals in Geometry

1. Use the Pythagorean Theorem 17 /1

2. Use the Distance Formula 18, 19 /2

3. Use the Midpoint Formula 20, 21 /2

C. Operations on Rational Expressions

1. Divide a Polynomial by a Monomial 22 /1

2. Divide a Polynomial by a Binomial 23, 24, 25 /3

3. Find Excluded Values 26, 27 /2

4. Multiply and Divide Rational Expressions 28, 29, 30, 31, 32 /5

5. Find the LCD of Rational Expressions 33 /1

6. Add and Subtract Rational Expressions 34, 35, 36, 37 /4

D. Graphing and Solving Rational Equations

1. Use an Inverse Variation Equation 38, 39, 40 /3

2. Graph y 5 a } x , y 5

1 } x 1 k and y 5

1 }

x 2 h 41 /1

3. Solve a Rational Equation Using Cross Products 42, 43 /2

4. Solve a Rational Equation Using the LCD 44, 45 /2

Name ——————————————————————— Date ————————————

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Student’s Name __________________________________________________________

Individual Lesson Skills Checklist

Chapter Lesson Prescription Skill Mastered

Copyright © by Holt McDougal. xii Holt McDougal Algebra 1All rights reserved.

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Name ——————————————————————— Date ————————————

Copyright © by Holt McDougal. All rights reserved. Holt McDougal Algebra 1 1

A. Expressions, Equations, and Inequalities

A variable is a letter used to represent one or more numbers. An algebraic expression is made from numbers, variables, and algebraic operations. The following examples describe how expressions can be evaluated, combined, written, and used to write algebraic equations and inequalities.

1. Evaluate Expressions (Lesson 1.1)Evaluate an expression Substitute a number for the variable, perform the operation(s), and simplify the result if necessary.

Evaluate the expression.

a. 16n when n 5 4 b. 25

} k when k 5 5 c. h 2 8 when h 5 12.2

d. 4 }

3 1 h when h 5

1 }

3 e. x3 when x 5 4 f. a2 when a 5 1.2

Solution:

a. 16n 5 16 p 4 b. 25

} k 5

25 } 5 c. h 2 8 5 12.2 2 8

5 64 5 5 5 4.2

d. 4 }

3 1 h 5

4 }

3 1

1 }

3 e. x3 5 43 f. a2 5 1.22

5 5 }

3 5 4 p 4 p 4 5 (1.2)(1.2)

5 64 5 1.44


1. 5b when b 5 6 2. 42

} h when h 5 14

3. 14 2 b when b 5 11.3 4. v 1 7 }

6 when v 5

1 }

3

5. y4 when y 5 3 6. q2 when q 5 2.1

2. Order of Operations (Lesson 1.2)Order of operations Established rule for evaluating an expression involving more than one operation:

Step 1: Evaluate expressions inside grouping symbols.

Step 2: Evaluate powers.

Step 3: Multiply and divide from left to right.

Step 4: Add and subtract from left to right.

EXAMPLE

PRACTICE

When a number is followed directly by a variable, the operation ofmultiplication is always implied.

Vocabulary

Vocabulary

Ready to Go On? Chapters 1– 2 Intervention

Inte

rve

ntio

nC

h1

–2

A.

Ex

pre

ssio

ns a

nd

Eq

uatio

ns

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Copyright © by Holt McDougal. All rights reserved.Holt McDougal Algebra 12


a. 3 p 24 2 5 p 6 b. 4(32 1 5) c. 5[12 2 (4 1 5)]

Solution:

a. 3 p 24 2 5 p 6 5 3 p 16 2 5 p 6 Evaluate power.

5 48 2 30 Multiply.

5 18 Subtract.

b. 4(32 1 5) 5 4(9 1 5) Evaluate power.

5 4(14) Add within parentheses.

5 56 Multiply.

c. 5[12 2 (4 1 5)] 5 5(12 2 9) Add within parentheses.

5 5(3) Subtract within brackets.

5 15 Multiply.


7. 4(10 2 3) 2 5 p 2 8. 21 1 (32 2 4) 9. 2[42 4 (9 2 3)]

3. Write Expressions (Lesson 1.3)Translate verbal phrases into expressions.

a. The product of 8 and m increased by 5

b. The quotient of 8 and the difference of a number x and 2

c. The sum of 20 and the square of a number n

Solution:

a. 8m 1 5 b. 8 }

x 2 2 c. 20 1 n2

Translate the verbal phrases into expressions.

10. The quotient when the quantity of a number y increased by 4 is divided by 6

11. 4 less than twice the square of a number q

12. 8 more than the product of a number w and 6

4. Write Equations and Inequalities (Lesson 1.4)Open sentence A mathematical statement that contains two expressions and a symbol that compares them.

Equation An open sentence that contains the symbol 5.

Inequality An open sentence that contains one of the symbols <, ≤, >, or ≥.

EXAMPLE

Keep a glossary of terms that describe each of the four basic operations.

PRACTICE

The multiplicationthat could be written in two steps (3 p 16 evaluated fi rst, followed by 5 p 6)is combined as one step.

PRACTICE

Vocabulary

EXAMPLE

Ready to Go On? Chapters 1– 2 Intervention cont’d

Inte

rve

nti

on

Ch

1–2

A.

Ex

pre

ssio

ns a

nd

Eq

uati

on

s


Name ——————————————————————— Date ————————————


Write an equation or an inequality.

a. The difference of a number p and 12 is at most 15.

b. The product of 5 and a number m is 14.

c. A number x is at least 6 and less than 9.

Solution:

a. p 2 12 ≤ 15 b. 5m 5 14 c. 6 ≤ x < 9

Write an equation or inequality.

13. The quotient of 12 and a number q is at most 5.

14. The sum of twice a number h and 5 is the same as 23.

15. The difference of a number w and 4 is greater than 12 and no more than 20.

CheckEvaluate the expression.

1. h }

3 1

1 }

3 when h 5 5 2.

64 }

b2 when b 5 4 3. 12 2 5a

} 4 when a 5 4


4. (42 2 3) p (2 1 3) 1 1 5. 4[(22 2 3) 1 1] 6. [54 4 (6 2 3)2]2

}} 8 2 2

Translate the verbal phrases into expressions.

7. The product of twice the number y and 4 increased by 8

8. The difference of 6 times the square of a number x and 15

Write an equation or inequality.

9. The sum of the number b and 12 is twice the number b.

10. The product of a number q and 3 is no less than 10 and no more than 15.

Tell whether the statement is always, sometimes, or never true.

11. For a given whole number n, the expression n2 represents an even number.

12. For a given whole number n that is greater than 0, the expression 2n represents an even number.

EXAMPLE

PRACTICE

“No less than” (≥) and “no greater than” (≤) are opposites of “less than” (<) and “greater than” (>), respectively.


Inte

rve

ntio

nC

h1

–2

A.

Ex

pre

ssio

ns a

nd

Eq

uatio

ns


Name ——————————————————————— Date ————————————


B. Problem Solving (Lesson 1.5)One way to try solving a math problem is to use an organized strategy, or problem-solving plan. Read the problem to fi nd what information is given and what you need to fi nd out. Decide on the strategy you will use, and apply it to solve the problem. Finally, check that your solution makes sense.

1. Check Possible Solutions Solution of an equation or inequality A number that can be substituted for the variable in an equation or inequality to make a true statement.

Check whether the given number is a solution of the equation or inequality.

a. 2x 2 8 5 22; 3 b. x } 3 1 1 5 7; 6 c. x 2 5 ≤ 3; 2

Solution:

a. 2(3) 2 8 0 22 b. 6 }

3 1 1 0 7 c. 2 2 5 ≤? 3

6 2 8 0 22 2 1 1 0 7 23 ≤ 3 ✔

22 5 22 ✔ 3 Þ 7 ✘

3 is a solution. 6 is not a solution. 2 is a solution.

Check whether the given number is a solution of the equation or inequality.

1. 5 1 a < 10; 24 2. n 2 3 }

12 5 1; 4 3. r }

4 1 3 5 5; 22

4. 28p 2 6 ≤ 0; 21 5. 9d 2 3 5 60; 7 6. m 1 8 > 27; 214

2. Read and Understand a ProblemRead the problem below. Identify what you know and what you need to fi nd out. You do not need to solve the problem.

You run in a city where the short blocks on north-south 0.2 mi

0.03 mistreets are 0.03 miles long. The long blocks on east-west streets are 0.2 mile long. You will run 2 long blocks east, a number of short blocks south, 2 long blocks west, then back to your starting point. You want to run 1.1 miles. How many short blocks should you run?

Solution:

What do you know? Each short block is 0.03 miles long. Each long block is 0.2 miles long. You will run 4 long blocks total (2 east 1 2 west). You will run s short blocks total (south and north). You want to run a total of 2 miles.

What do you want to fi nd out? How many short blocks should you run so that the distance you run on short blocks and the distance you run on 4 long blocks makes a total of 1.1 miles?

Vocabulary

EXAMPLE

PRACTICE

EXAMPLE

There may be more than one method that can be used to solve a problem.


Inte

rve

nti

on

Ch

1–2

B.

Pro

ble

m S

olv

ing

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7. Read the problem below. Identify what you know and what you need to fi nd out. You do not need to solve the problem.

A bicycle park has a long trail and a short trail. The long

start/finish

2 km

5 km

trail is 5 km long. The short trail is 2 km long. You will ride 3 laps on the short trail and some number of laps on the long trail. You want to ride 21 km. How many laps should you ride on the long trail?

3. Make a PlanWrite a verbal model of the statement below.

How many short blocks should you run so that the distance you run on short blocks and the distance you run on 4 long blocks makes a total of 1.1 miles?

Solution:

Distance run on

short blocks 1

Distance run onlong blocks

5 Total

distance

(miles) (miles) (miles)

Length of a short block

p Number of short blocks

1 Length of a long block

p Number of long blocks

5 Total

distance

(miles/block) (block) (miles/block) (block) (miles)

8. Write a verbal model for the problem in Exercise 7.

CheckCheck whether the given number is a solution of the equation or inequality.

1. 6 1 j < 4; 21 2. n 1 5 } 2 5 6; 7 3. m }

3 2 8 5 25; 9

4. 22y 1 3 ≤ 0; 2 5. 4g 2 5 5 35; 10 6. b 1 12 > 0; 213

Read the problem below. Identify what you know and what you need to fi nd out. Then, write a verbal model of the problem. You do not need to solve the problem.

7. Jim’s grandmother exercises by walking the main

90 yd

start

20 yd

rectangular hall of a local shopping mall. She walks 90 yards down the length of the hall, turns right, and walks 20 yards across the width of the hall. Then, she turns right and walks up the length of the hall again. Finally, she turns right one more time, and walks 20 yards across the width of the hall and ends up at her starting point. Jim’s grandmother wants to walk 970 yards. She will walk the length of the hall 9 times. How many times will she walk across the width of the hall?

PRACTICE

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C. Representations of FunctionsFunctions can be represented by mapping diagrams, tables, verbal or algebraic function rules, and graphs. Each input value and its corresponding output value make up an ordered pair. An ordered pair can be written as (input, output) or plotted as a point on a coordinate grid.

1. Identify the Domain and Range of a Function (Lesson 1.6)

Function A pairing of input values to output values, where the value of each output depends on the value of the corresponding input, and each input corresponds to exactly one output.

Domain The set of input values for a function.

Range The set of output values for a function.

Identify the domain and range of the function.

a. Input Output

0 10

1 11

2 12

3 13

b. 3 21

6 22

9 23

12 24

c. Input Output

28 24

4 2

6 3

10 5

Solution:

a. Domain: 0, 1, 2, 3 b. Domain: 3, 6, 9, 12 c. Domain: 28, 4, 6, 10

Range: 10, 11, 12, 13 Range: 21, 22, 23, 24 Range: 24, 2, 3, 5

Identify the domain and range of each function.

1. Input Output

0 1

1 3

2 5

3 7

2. Input Output

25 1

210 2

215 3

220 4

3. 26 600

24 400

21 100

0 0

EXAMPLE

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Tell whether the relation is a function.

a. Input Output

26 2

27 3

28 4

29 5

b. 24 12

23 13

0 16

19

c. Input Output

35 0

40 5

40 10

45 15

Solution:

a. Yes b. No; 0 maps to c. No; 40 maps to two two outputs. outputs.


4. Input Output

214 4

28 4

22 4

4 4

5. Input Output

9 0

15 6

27 18

35 26

6. 21 0

0 21

1 3

2

2. Write a Function Rule (Lesson 1.6)Independent variable A function’s input variable.

Dependent variable A function’s output variable.

Write a rule for the function.

Input, x 0 1 2 3 4

Output, y 0 3 6 9 12

Solution:

Each value of y is 3 times the corresponding x value. The function rule is y 5 3x.


7. Input, x 24 22 0 2 4

Output, y 22 21 0 1 2

8. Input, x 5 10 15 20 25

Output, y 1 6 11 16 21

9. Input, x 3 5 9 12 16

Output, y 12 14 18 21 25

10. Input, x 26 22 3 7 8

Output, y 6 2 23 27 28

A function rule usually states the dependent variable y as a function of the independent variable x, such as y 5 x 1 3.

PRACTICE

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3. Make a Table for a Function (Lesson 1.6)Make a table for the function and identify the range of the function.

y 5 x 1 2.6

Domain: 2, 3, 4, 5, 6

Solution:

Input, x 2 3 4 5 6

Output, y 4.6 5.6 6.6 7.6 8.6

Range: 4.6, 5.6, 6.6, 7.6, 8.6

Make a table for the function and identify the range of the function.

11. y 5 2 }

3 x 12. y 5 x 2 1.1 13. y 5 22x 1 5

Domain: 3, 6, 9, 12, 15 Domain: 25, 24, 23, 22, 21 Domain: 1, 2, 4, 7, 9

14. y 5 x 1 1

} 2 15. y 5 x 1 14 16. y 5 25x


4. Graph a Function (Lesson 1.7)Graph the function y 5 x 2 2 with domain 4, 5, 6, 7, and 8.

Solution:

Step 1: Make an input-output table.

Input, x 4 5 6 7 8

Output, y 2 3 4 5 6

Step 2: List the ordered pairs (x, y).

(4, 2), (5, 3), (6, 4), (7, 5), (8, 6)

Step 3: Plot a point for each ordered pair (x, y).

1

2

3

4

5

6

7

8y

1 x2 3 4 5 6 7 8

Graph the function.

17. y 5 x 2 3 18. y 5 1 }

3 x 19. y 5 3x 2 3


PRACTICE

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20. y 5 1.5x 1 2 21. y 5 1 }

2 x 2 2 22. y 5

x 1 2 }

3


CheckIdentify the domain and range of each function.

1. Input Output

7 23

10 35

14 51

17 63

2. Input Output

210 90

28 54

25 15

24 6

3. 23 2

3 } 8

22 2 1 } 4

21 2 1 } 8

0 0


4. Input Output

5 8

8 5

8 4

10 3

5. 16 4

24

36 26

6

6. Input Output

25 22

2 5

7 10

11 14


7. Input, x 25 23 0 4 6

Output, y 225 215 0 20 30

8. Input, x 2 4 5 7 10

Output, y 24 28 210 214 220

9. Input, x 3 6 8 10 13

Output, y 24 21 1 3 6

10. Input, x 10 24 32 48 50

Output, y 15 36 48 72 75

Make a table for the function and identify the range of the function.

11. y 5 24x 1 10 12. y 5 3 }

4 x 2 1 13. y 5

2x 1 3 }

4


Graph the function.

14. y 5 x 2 6 15. y 5 5x 16. y 5 2x 1 4



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D. OperationsWhole numbers, integers, and rational numbers are part of the set of real numbers. The following examples describe different operations with real numbers.

1. Find Opposites of Real Numbers (Lesson 2.1)Opposite of a real number a 2a (read “the opposite of a”) is the same distance from 0 on a number line as a, but it is on the opposite side of 0.

For the given value of a, fi nd 2a.

a. a 5 3 b. a 5 7 3 } 5 c. a 5 25.4

Solution:

a. 2a 5 2(3) b. 2a 5 2 1 7 3 } 5 2 c. 2a 5 2(25.4)

2a 5 23 2a 5 27 3 } 5 2a 5 5.4

For the given value of the variable, fi nd the opposite.

1. x 5 26.2 2. u 5 809 3. m 5 0.25

4. w 5 45

} 8 5. k 5 2

6 } 11 6. c 5 28

1 } 7

2. Find Absolute Values of Real Numbers (Lesson 2.1)Absolute value of a real number a ZaZ (read “the absolute value of a”) is the distance between a and 0 on a number line. If a is greater than or equal to 0, ZaZ is a. If a is less than zero, ZaZ is the opposite of a.

For the given value of a, fi nd ZaZ.

a. a 5 8 b. a 5 2 4 } 9 c. a 5 11.5

Solution:

a. ZaZ 5 Z8Z 5 8 b. ZaZ 5 ) 2 4 } 9 ) 5 2 1 2

4 } 9 2 5

4 }

9

c. ZaZ 5 Z11.5Z 5 11.5

For the given value of the variable, fi nd the absolute value.

7. b 5 0.4 8. y 5 250 9. p 5 21.6

10. v 5 29 11. n 5 23

} 5 12. h 5 10

8 }

9

3. Add Real Numbers (Lesson 2.2)Sum The result of adding two or more real numbers.

To use a number line to fi nd the sum of a 1 b:

• Start at a.

• If b > 0, you will move to the right. If b < 0, you will move to the left.

• Find ZbZ and move that many units.

• The number you stop on is the sum.

EXAMPLE

Note that 2a is postive when a is negative.

PRACTICE

EXAMPLE

Vocabulary

Vocabulary

PRACTICE

Vocabulary

The absolute value of a number is always positive.


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Use a number line to fi nd the sum.

a. 4 1 (27) b. 22 1 3

Solution:

a.

�6 �5 �4 �3 �2 �1 0 1 2 3 4 5 6

Move 7 units to the left. End at �3. Start at 4. b.

�6 �5 �4 �3 �2 �1 0 1 2 3 4 5 6

Move 3 units to the right. Start at �2. End at 1.

4 1 (27) 5 23 22 1 3 5 1

Use a number line to fi nd the sum.

13. 21 1 (24) 14. 2 1 (28) 15. 25 1 9

16. 27 1 1 17. 0 1 (22) 18. 23 1 (26)

Use the rules of real number addition to fi nd the sum.

a. 219 1 (221) b. 2 1 } 2 1

3 }

2 c. 22.8 1 1.5

Solution:

If two numbers have the same sign, add their absolute values. The sum has the same sign as the numbers added.

a. 219 1 (221) 5 2(Z19Z 1 Z21Z) 5 2(19 1 21) 5 240

If two numbers have different signs, subtract the absolute value of the smaller number from the absolute value of the larger number. The sum has the same sign as the number with the larger absolute value.

b. 2 1 } 2 1

3 }

2 5 ) 3 }

2 ) 2 ) 2

1 } 2 ) 5

3 }

2 2

1 }

2 5 1

c. 22.8 1 1.5 5 2(Z22.8Z 2 Z1.5Z) 5 2(2.8 2 1.5) 5 21.3

Use the rules of real number addition to fi nd the sum.

19. 6.4 1 (20.3) 20. 8 1 (210) 21. 2100 1 (234)

22. 2 2 }

3 1 4

3 }

4 23. 216 1 5

1 }

4 24. 55.1 1 (247.7)

4. Subtract Real Numbers (Lesson 2.3)Difference The result of subtracting one real number from another real number.

Find the difference.

a. 5 2 18 b. 26 2 9 c. 28 2 (23)

Solution:

To subtract b from a, add a and the opposite of b.

a. 5 2 18 5 5 1 (218) b. 26 2 9 5 26 1 (29) c. 28 2 (23) 5 28 1 3

5 213 5 215 5 25

PRACTICE

EXAMPLE

PRACTICE

Vocabulary

You can fi nd the sum of three or more numbers together by fi rst adding two of the numbers and then adding the result to the third.

EXAMPLE

Use grouping symbols around negative numbers in your work to keep track of signs as you simplify expressions.

EXAMPLE


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25. 22.8 2 0.7 26. 1.9 2 21.1 27. 234 2 57

28. 3 3 } 5 2 (24) 29. 2

24 } 5 2 1 2

16 }

15 2 30. 73 2 (282)

5. Multiply Real Numbers (Lesson 2.4)Product The result of multiplying two or more real numbers.

Find the product.

a. 21.7(4) b. 2 4 } 5 (210) c. 2(23)(28)

Solution:

The product of two numbers with the same sign is positive and the product of two numbers with different signs is negative.

a. 21.7(4) 5 26.8 b. 2 4 } 5 (210) 5

40 } 5 c. 2(23)(28) 5 [2(23)](28)

5 8 5 26(28)

5 48

Find the product.

31. 10(23) 32. 2 11

} 4 (6) 33. 2 5 } 8 1 2

24 }

15 2 1 2

25 }

47 2

34. 212(4)(23) 35. 2.5(10.4)(27) 36. 22.4(29.1)

Multiplicative inverse of a real number a The reciprocal of a, or 1 } a . The product of a

and its multiplicative inverse is 1.

Find the multiplicative inverse of a.

a. a 5 9 b. a 5 24 c. a 5 2 1 } 8

Solution:

a. 1 } a 5

1 }

9 b.

1 } a 5

1 }

24 5 2

1 }

4 c.

1 } a 5

1 }

2 1 }

8 5 2

8 }

1 5 28

Find the multiplicative inverse of the number.

37. 26 38. 1 39. 3 } 4

40. 2 7 } 5 41. 9

1 }

2 42. 25

15 }

32

EXAMPLE

PRACTICE

To multiply three or more real numbers, fi rst multiply two of the numbers, then multiply the result with the third number.

Vocabulary

PRACTICE

Vocabulary

You can check your answer by multiplying the original number by its inverse and making sure the product is 1.

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6. Divide Real Numbers (Lesson 2.6)Quotient The result dividing a real number by another real number.

Find the quotient.

a. 35 4 (27) b. 226 4 (213) c. 2 }

3 4 1 2

12 } 18 2

Solution:

a. 35 4 (27) 5 35 p 1 2 1 } 7 2 b. 226 4 (213) 5 226 p 1 2

1 } 13 2

5 2 35

} 7 5 26

} 13

5 25 5 2

c. 2 }

3 4 1 2

12 } 18 2 5

2 }

3 p 1 2

18 } 12 2

5 2 36

} 36

5 21

Find the quotient.

43. 292 4 (24) 44. 22 1 }

4 4

5 }

8 45. 9 4

1 }

9

46. 1 4 1 2 5 } 2 2 47. 2

32 }

15 4 (28) 48. 26

2 }

3 4 10

4 }

9

7. Find Square Roots (Lesson 2.7)Square root of a If b2 5 a, then b is the square root of a. Every positive nonzero real number a has two square roots, 2 Ï

} a and Ï}

a .

Radicand The number or expression inside a radical symbol.


a. 6 Ï}

49 b. Ï}

1 c. 2 Ï}

144

Solution:

a. 67 b. 1 c. 212


49. 2 Ï}

400 50. 6 Ï}

9 51. Ï}

81

52. Ï}

0 53. Ï}

4 54. 6 Ï}

900

PRACTICE

EXAMPLE

The symbol 6 in front of a number refers to the number and its opposite. For example, “66” is the same as “6 and 26.”

PRACTICE

Vocabulary

Vocabulary

EXAMPLE

Division by 0 is undefi ned, because 0 does not have a multiplicative inverse.


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CheckFor the given value of the variable, fi nd the opposite, absolute value, and multiplicative inverse.

1. a 5 216 2. y 5 7 3 }

10 3. r 5 20.3


4. 51 2 (265) 5. 2 5 } 7 p 21

} 40

6. 6 5 }

9 4 1 21

2 }

3 2

7. 8 1 (215) 8. 22(235) 9. Ï}

64

10. 218 4 12

} 5 11. 6 Ï}

81 12. 22.3 2 4.9

13. 3 Ï

}

27 14. 3 Ï}

(227) 15. 3 Ï}

1000

16. Find A < B and A > B.

A 5 {0, 2, 4, 6} and B 5 {1, 2, 3, 4, 5}


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E. Properties and Real NumbersTaken together, the rational and irrational numbers make up the set of real numbers. The following examples illustrate some characteristics and properties of real numbers.

1. Classify Real Numbers (Lesson 2.1)Whole numbers A subset of the real numbers; A whole number is either 0 or one of the “counting numbers,” 1, 2, 3, . . .

Integers A subset of the real numbers; The integers are the set of whole numbers and their opposites, . . . 23, 22, 21, 0, 1, 2, 3, . . .

Rational numbers A subset of the real numbers; A rational number can be expressed as the ratio of two integers, and its decimal form terminates or repeats.

Irrational numbers A subset of the real numbers; An irrational number cannot be expressed as the ratio of two integers, and its decimal form neither terminates nor repeats.

Choose the word that best describes each: whole, integer, rational, or irrational.

a. 28 b. 3 } 4 c. 2 Ï

} 7

d. 20.1 e. Ï}

100 f. 4 1

} 3

Solution:

a. Integer; opposite of a b. Rational; ratio of two c. Irrational; cannot be whole number integers written as ratio of two integers nor as a terminating or repeating decimal

d. Rational; terminating e. Whole; Ï}

100 5 10 f. Rational; 13

} 3 5 4.333,

decimal a repeating decimal

Choose the word that best describes each: whole, integer, rational, or irrational.

1. 227 2. 22 } 7 3. 26

5 }

9

4. Ï}

64 5. 213.2 6. 0

Vocabulary

Use a Venn diagram to help remember which numbers are part of other numbers.

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Remember that the additive inverse of a, 2a, is not necessarily a negative number.

2. Order Real Numbers (Lesson 2.7)To order real numbers from least to greatest, graph them fi rst. Then read the numbers from left to right.

Order the numbers from least to greatest: 2 4 2 5 , 24.5, 25, 2 �

�

4 .

Solution:

45

�2 �3 �4 �5

�5 �4.5 � 4

�6 �1 0 1 2 3 4 5 6

� �

In order from least to greatest, the numbers are 25, 24.5, 2 Ï}

4 , and 2 4 } 5 .

Order each group of numbers from least to greatest.

7. 1.23, 1 2 }

3 ,

3 }

2 , Ï

}

3 8. 0.08, 21.9, 2 Ï}

0.04 , Ï}

2 9. 6.01, Ï}

6 , 6 1 }

6 , 6.1

3. Identify Properties of Addition (Lesson 2.2)Additive identity The number 0 is the additive identity. When 0 is added to a real number a, the sum equals a.

Additive inverse The opposite of a real number a is its additive inverse. The sum of a real number and its additive inverse always equals 0.

Identify the property of addition being illustrated.

a. (24 1 b) 1 3 5 24 1 (b 1 3) b. 7 1 x 5 x 1 7

c. 0 1 2 5 }

8 5 2

5 }

8 d. 17.3 1 (217.3) 5 0

Solution:

a. Associative; b. Commutative; c. Identity; d. Inverse; a changing the changing the adding 0 to a number plus grouping does order does not number does its opposite not change change the sum not change equals 0 the sum the number

Identify the property of addition being illustrated.

10. 2k 1 0 5 2k 11. 7 }

8 1 1 2

8 } 7 2 5 2

8 } 7 1

7 }

8

12. 0 1 (21 1 n) 5 (0 1 (21)) 1 n 13. 0 5 25.6 1 5.6

14. r 1 (s 1 t) 5 (r 1 s) 1 t 15. 3 1 5 1 7 5 7 1 5 1 3

EXAMPLE

To order numbers, it is sometimes helpful to write decimal approximations of rational and irrational numbers.

PRACTICE

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Remember that associative is related to grouping and commutative is related to order.

PRACTICE

4. Identify Properties of Multiplication (Lesson 2.4)Multiplicative identity The number 1 is the multiplicative identity. The product of a real number a and 1 equals a.

Identify the property of multiplication being illustrated.

a. 0 p w 5 0 b. 56 p (92 p 11) 5 (56 p 92) p 11

c. g p h 5 h p g d. 24,978 p 1 5 24,978

e. 277 p (21) 5 77 f. (r p s) p t 5 r p (s p t)

Solution:

a. Zero; a number b. Associative; changing c. Commutative; changing times 0 equals 0 the grouping does not the order does not change the product change the product

d. Identity; multiplying e. Negative one; the f. Associative; changing a number by 1 does product of a number the grouping does not not change the number and 21 is the opposite change the product of the number

Identify the property of multiplication being illustrated.

16. 6 p p 5 p p 6 17. 247 5 47 p (21) 18. (25a)b 5 25(ab)

19. 235 5 1 p (235) 20. 200 p 0 5 0 21. 21 p (29) 5 9

5. Apply the Distributive Property (Lesson 2.5)Distributive property The product of two factors, where one factor is a sum, is the sum of the product of the fi rst factor times the fi rst addend plus the product of the fi rst factor times the second addend. For example, a(b 1 c) 5 ab 1 ac.

Equivalent expressions Expressions that are equal in value, for any value of the variable.

Use the distributive property to write an equivalent expression.

a. 4(9 1 2) b. 8(b 2 3) c. 22n(n 1 6)

Solution:

a. 4(9 1 2) 5 (4)(9) 1 (4)(2) b. 8(b 2 3) 5 8[b 1 (23)] 5 36 1 8 5 8b 1 8(23) 5 44 5 8b 1 (224) 5 8b 2 24 c. 22n(n 1 6) 5 (22n)(n) 1 (22n)(6)

5 22n2 1 (212n) 5 22n2 2 12n


22. 27(4 1 v) 23. c(2c � 1) 24. 5m(�3 � m)

25. 9(t � 8) 26. 2 } 5 (15 � 20r) 27. �3p(�5 � 7p)

PRACTICE

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CheckChoose the word that best describes each number in the list. Then, write the numbers in order from least to greatest.

1. Ï}

8 , 4.1, 3, 8 }

3 2. 29.1, 29.02, 2

2 } 9 , 2 Ï

}

9

3. 0 }

1 , 20.01, 21.1, 2 Ï

}

1

Identify the property being illustrated.

4. (�8 � 3b) � 2 � �8 � (3b � 2) 5. 7q � �7q(�1)

6. � 3 } 7 t �

3 } 7 t � 0 7. 1(�3x2) � �3x2

8. 15(6 � s) � 90 � 15s 9. 12 � h � h � 12

10. 6y(2) � 2(6y) 11. 18(jk) � 18j(k)

12. 6c(1 � 3c) � 6c � 18c2 13. 28m(0) � 0

14. �3(2z � 1) � �6z � 3 15. �37w � 0 � �37w


16. 5(h � 2) 17. � 1 }

2 d(8 � 6d)

18. 8f(�4f � 10)


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A. Solving Equations in one VariableTo solve an equation in one variable, isolate the variable on one side of the equation. The following examples illustrate different ways to isolate the variable.

1. Solve an Equation Using Addition or Subtraction (Lesson 3.1)

Inverse operations Two operations that undo each other, such as addition and subtraction or multiplication and division.

Equivalent equations Equations that have the same solution(s).

Use addition or subtraction to solve the equation.

a. x 1 9 5 3 b. x 2 5 5 2 c. x 1 4.1 5 6

Solution:

a. x 1 9 5 3 Write original equation.

x 1 9 2 9 5 3 2 9 Use subtraction property of equality. Subtract 9 from each side.

x 5 26 Simplify.

The solution is 26.

b. x 2 5 5 2 Write original equation.

x 2 5 1 5 5 2 1 5 Use addition property of equality. Add 5 to each side.

x 5 7 Simplify.

The solution is 7.

c. x 1 4.1 5 6 Write original equation.

x 1 4.1 2 4.1 5 6 2 4.1 Subtract 4.1 from each side.

x 5 1.9 Simplify.

The solution is 1.9.

Use addition or subtraction to solve the equation.

1. x 1 5 5 4 2. c 2 3 5 8 3. t 1 6 5 10

2. Solve an Equation Using Multiplication or Division (Lesson 3.1)

Use multiplication or division to solve the equation.

a. x }

6 5 3 b. 27x 5 249 c. 2

3 } 8 x 5 5

Solution:

a. x } 6 5 3 Write original equation.

6 p x } 6 5 6 p 3 Multiply each side by 6.

x 5 18 Simplify.

The solution is 18.

Vocabulary

EXAMPLE

PRACTICE

Be sure to subtract (or add)the same number from each side, so that the new equation is equivalent to the original equation.

EXAMPLE

Ready to Go On? Chapters 3 – 4 Intervention

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b. 27x 5 249 Write original equation.

27x

} 27 5

249 }

27 Divide each side by 27.

x 5 7 Simplify.

The solution is 7.

c. 2 3 } 8 x 5 5 Write original equation.

2 8 } 3 1 2

3 } 8 x 2 5 2

8 } 3 (5) Multiply each side by the reciprocal � 8 } 3 .

x 5 2 40

} 3 Simplify.

The solution is 2 40

} 3 .


4. r }

10 5 2 5. 4q 5 32 6.

a }

9 5 23

3. Solve a Two-Step Equation (Lesson 3.2)Order of operations The rules for evaluating an expression involving more than one operation.

Solve the equation.

a. 7x 1 1 5 4 b. x } 5 2 10 5 20

Solution:

a. 7x 1 1 5 4 Write original equation.

7x 1 121 5 4 21 Subtract 1 from each side.

7x 5 3 Simplify.

7x

} 7 5 3 } 7 Divide each side by 7.

x 5 3 } 7 Simplify.

The solution is 3 } 7 .

b. x } 5 2 10 5 20 Write original equation.

x } 5 2 10 1 10 5 20 1 10 Add 10 to each side.

x } 5 5 30 Simplify.

5 p x } 5 5 5 p 30 Multiply each side by 5.

x 5 150 Simplify.

The solution is 150.

Solve the equation.

7. 3 1 4x 5 11 8. 7.5a 2 10 5 232.5 9. t }

8 1 6 5 3

PRACTICE

Vocabulary

EXAMPLE

PRACTICE

Recall that the product of a number and its reciprocal is 1.

Ready to Go On? Chapters 3 – 4 Intervention cont’d

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4. Solve Multi-Step Equations (Lesson 3.3)

Solve 4x � 3(x � 5) � �12.

Solution:

4x 1 3(x 2 5) 5 212 Write original equation.

4x 1 3x 2 15 5 212 Eliminate the parentheses by using the distributive property.

7x 2 15 5 212 Combine like terms.

7x 2 15 1 15 5 212 1 15 Add 15 to each side.

7x 5 3 Simplify.

7x

} 7 5 3 } 7 Divide each side by 7.

x 5 3 } 7 Simplify.

The solution is 3 } 7 .

Solve the equation.

10. 22(4a 1 5) 2 6a 5 10 11. 6 1 3(n 2 7) 5 12 12. 2 }

3 (6g 1 1) 1

1 }

3 5 219

5. Solve Equations with Variables on Both Sides (Lesson 3.4)

Solve x � 4 � 3x � 8.

Solution:

x 2 4 5 3x 1 8 Write original equation.

x 2 4 2 3x 5 3x 1 8 2 3x Subtract 3x from each side.

22x 2 4 5 8 Simplify each side.

22x 2 4 1 4 5 8 1 4 Add 4 to each side.

22x 5 12 Simplify.

22x

} 22

5 12

} 22

Divide each side by 22.

x 5 26 Simplify.

The solution is 26.

Solve the equation.

13. 8t 2 10 5 5 1 3t 14. 29 2 4h 5 22h 1 3 15. 12d 1 4 5 6 2 d

EXAMPLE

PRACTICE

EXAMPLE

Distributive Property :a(b 1 c) 5 ab 1 ac a(b 2 c) 5 ab 2 ac

You could also begin solving the equation by adding 4 to each side to obtainx 5 3x 1 12. You will get the same solution when you fi nish solving for x.

PRACTICE


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6. Identify the Number of Solutions to an Equation (Lesson 3.4)

Identity An equation that is true for all values of the variable.

Solve the equation, if possible.

a. 28x 5 24(2x 1 1) b. 25x 2 15 5 25(x 1 3)

Solution:

a. 28x 5 24(2x 1 1) Write original equation.

28x 5 28x 2 4 Distributive property

28x 1 8x 5 28x 2 4 1 8x Add 8x to each side.

0 5 24 ✘ Simplify.

The statement 0 5 24 is not true, so the equation has no solution.

b. 25x 2 15 5 25(x 1 3) Write original equation.

25x 2 15 5 25x 2 15 Distributive property

The statement 25x 215 5 25x 2 15 is true for all values of x. So the equation is an identity, and the solution is all real numbers.


16. 7(3s 2 3) 5 3(7s 2 7) 17. 26 1 3(n 2 9) 5 6n 1 27

18. 6a 1 1 5 2(3a 2 1)

CheckUse addition or subtraction to solve the equation.

1. w 1 3.6 5 8.9 2. p 2 7.2 5 25 3. n 1 12 5 23


4. 5n 5 26 5. g }

2.1 5 5 6. 1.2z 5 8.4

Solve the equation.

7. 9x 2 2 5 0 8. 24 1 b }

2.5 5 40

9. 24 5 n }

6 2 3 10. 9y 1 5(y 2 9) 5 39

11. 2c 1 4(8 2 c) 5 243 12. 1 }

4 m 2 1 3 }

4 m 1 2 2 5 8

13. 1 }

2 n 1 4 5 2

3 } 2 n 2 18 14. 4.3 1 2.3r 5 7.1 2 1.9r

15. z 2 26 5 5z 2 36


16. 4(b 2 5) 5 5(b 2 4) 17. 8p 2 12 5 24(22p 1 3)

18. 5(k 1 3) 2 k 5 3k 1 5

An equation can have one solution, no solution, or all real numbers as solutions.

PRACTICE

Vocabulary

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B. Proportion and Percent ProblemsThe comparison of two quantities by division is called a ratio. The following examples illustrate how to write and use ratios.

1. Write a Ratio (Lesson 3.5)Kim has a jar containing 45 pennies, 18 nickels, 30 dimes, and42 quarters. Write the specifi ed ratio in simplest form.

a. number of nickels to b. number of quarters c. number of pennies to number of pennies to number of dimes total number of coins

Solution:

a. nickels } pennies

5 18

} 45

b. quarters

} dimes

5 42

} 30

c. pennies

} total

5 45 }}

45 1 18 1 30 1 42

5 2 } 5 5

7 } 5 5

45 }

135 5

1 }

3

On his last report card, Jay earned 2 A’s, 3 B’s, and 1 C. Write the specifi ed ratio in simplest form.

1. the number of A’s to the number of B’s

2. the number of C’s to the number of A’s and B’s

3. the number of B’s to the total number of grades

A school band orders t-shirts. They order 12 smalls, 10 mediums, and 15 larges. Write the specifi ed ratio in simplest form.

4. the number of smalls to the total number of t-shirts

5. the number of mediums to the number of larges

6. the number of larges and smalls to the number of mediums

2. Solve a Proportion (Lesson 3.6)Proportion An equation showing that two ratios are equivalent.

Solve the proportion 2 ⎯ 3 5 x ⎯

15 .

Solution:

2 }

3 5

x }

15 Write original proportion.

15 p 2 } 3 5 15 p x }

15 Multiply each side by 15.

30

} 3 5 x Simplify.

10 5 x Divide.

Vocabulary

EXAMPLE

PRACTICE

EXAMPLE

The ratio of two quantities a and b can be written in three ways:a to b, a : b, or a }

b .


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Solve the proportion.

7. 7 }

42 5

t }

84 8.

5 }

6 5

k }

72 9.

a }

65 5

6 }

39

10. v }

3 5

85 }

51 11.

8 }

20 5

n }

15 12.

q }

54 5

8 }

36

3. Use the Cross Products Property (Lesson 3.6)Cross product The product of the numerator of one ratio in a proportion and the denominator of the other ratio in the proportion.

Solve the proportion 2 ⎯ 5 5 x ⎯

20 .

Solution:

2 } 5 5

x }

20 Write original proportion.

2 p 20 5 5 p x Cross products property

40 5 5x Simplify.

8 5 x Divide each side by 5.

Use the cross products property to solve the proportion.

13. y }

12 5

19 }

4 14.

15 }

35 5

p }

63 15.

13 }

36 5

65 }

w

16. 7 } 30

5 196

} m 2 5 17. 5 }

8 5

n }

n 1 9 18.

18 }

v 2 1 5

48 }

3v 2 5

4. Find a Percent Using a Proportion (Lesson 3.7)What percent of 50 is 12?

Solution:

a }

b 5

p }

100 Write proportion.

12 } 50

5 p }

100 Substitute 12 for a and 50 for b.

1200 5 50p Cross products property

24 5 p Divide each side by 50.

12 is 24% of 50.

Use a proportion to answer the question.

19. What is 34% of 60? 20. 5 is 4% of what number?

21. 16 is what percent of 20? 22. 9 is what percent of 50?


EXAMPLE

PRACTICE

EXAMPLE

Represent “a isp percent of b”using the proportion

a } b

5 p }

100 .

PRACTICE

Vocabulary

Cross Products Property:The cross products of a proportion are equal.

PRACTICE


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5. Use the Percent Equation (Lesson 3.7)Use the percent equation to answer the question.

a. What number is 13% of 42?

b. What percent of 15 is 13.8?

c. 84 is 60% of what number?

Solution:

a. a 5 p% p b Write percent equation.

5 13% p 42 Substitute 13 for p and 42 for b.

5 0.13 p 42 Write percent as decimal.

5 5.46 Multiply.

5.46 is 13% of 42.

b. a 5 p% p b Write percent equation.

13.8 5 p% p 15 Substitute 13.8 for a and 15 for b.

0.92 5 p% Divide each side by 15.

92% 5 p% Write percent as decimal.

13.8 is 92% of 15.

c. a 5 p% p b Write percent equation.

84 5 60% p b Substitute 84 for a and 60 for p.

84 5 0.6 p b Write percent as decimal.

140 5 b Divide each side by 0.6.

84 is 60% of 140.

Use the percent equation to answer the question.


27. 72 is what percent of 288? 28. 26 is what percent of 78?


EXAMPLE

PRACTICE

Represent “a is p percent of b” using the equation a 5 p% p b where a is a part of the base b and p% is the percent.


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CheckA shelter has 24 cats, 18 dogs, and 3 birds. Write each ratio in simplest form. 1. the number of dogs to the total number of animals

2. the number of dogs to the number of cats

3. the number of birds to the number of cats and dogs

Solve the proportion.

4. 48

} 92

5 8 } n 5.

84 }

d 5

70 }

25 6.

20 }

24 5

3a }

90

7. 76

} d 5

19 }

13 8.

2 } 7 5

46 }

s 2 2 9.

18 }

7x 1 3 5

3 }

x 1 2

Answer the question.


12. 51 is what percent of 60? 13. 18 is 3% of what number?

14. 78 is what percent of 120? 15. What is 51% of 9?

Identify the percent change as an increase or decrease. Then fi nd the percent of change.

16. Original: 15 17. Original: 90 New: 25 New: 40

Find the new amount.

18. The price of $15 is increased by 35%.

19. The price of $110 is decreased by 25%.


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A literal equation describes the relationship between two or more variables. The equation can be rewritten, or “solved,” to isolate any one of the variables. The following examples show how to solve and use literal equations.

1. Solve a Literal Equation (Lesson 3.8)Literal equation An equation, such as a formula, with two or more variables where the coeffi cients and constants have been replaced by letters.

Solve y � mx � b for m.

Solution:

y 5 mx 1 b Write original equation.

y 2 b 5 mx Subtract b from each side.

y 2 b

} x 5 m Assume x � 0. Divide each side by x.

Solve the literal equation for the specifi ed variable.

1. I 5 Prt for P 2. A 5 πr2 for r 3. F 1 V 5 E 1 2 for V

4. V 5 ,wh for w 5. V 5 1 }

3 πr2h for h 6. A 5

1 }

2 (b

1 1 b

2)h for b

1

2. Use the Solution to a Literal Equation (Lesson 3.8)Use the solution to the literal equation from the example in Part 1 to solve 14 � m p 6 � 4.

Solution:

y 2 b

} x 5 m Solution of literal equation.

14 2 (24)

} 6 5 m Substitute 14 for y, 6 for x, and �4 for b.

3 5 m Simplify.

Solve the given formula for the unknown variable. Then use the solu-tion to answer the question.

7. The density d of a substance is given by d 5 m

} V , where m is the mass in grams (g)

and V is volume in cubic centimeters (cm3). A scientist completely fi lls a beaker with 28 g of a substance that has density 0.4375 g/cm3. What is the beaker’s volume?

8. The strength s of a radio signal is given by s 5 1600

} d 2

, where d is the distance in

miles from the transmitter. If s is 100, how far are you from the transmitter?

Vocabulary

EXAMPLE

Remember that inverse operations apply to variables as well as to constants.

PRACTICE

EXAMPLE

After solving a literal equation, you can use unit analysis to check your work.

PRACTICE

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9. A roller coaster car goes down a hill and then makes a loop. The velocity v of the car at the top of the loop is v 5 8 Ï

}

h 2 2r , where h is the hill’s height and r is the loop’s radius. If v is 32 ft/s and r is 15 ft, how tall is the hill?

3. Rewrite an Equation (Lesson 3.8)Write 3x � 5y � 15 so that y is a function of x.

Solution:

3x 2 5y 5 15 Write original equation.

25y 5 15 2 3x Subtract 3x from each side.

y 5 23 1 3 } 5 x Divide each side by �5.

Write the equation so that y is a function of x.

10. 2x 2 y 5 10 11. 8 1 3y 5 24x 12. 9y 1 27 5 2x

13. 3 }

4 y 2 2x 5 12 14. 5x 1

2 } 5 y 5 30 15. 224 2 16y 5 8x

CheckSolve the literal equation for the specifi ed variable.

1. P 5 4, for , 2. V 5 πr2h for h

3. s 5 1 }

2 (a 1 b 1 c) for a 4. S 5 πr, 1 πr2 for ,

5. h 5 216t2 1 vt 1 c for v 6. S 5 2,w 1 2wh 1 2,h for w

Solve the given formula for the unknown variable. Then use the solution to answer the question.

7. For an electrical circuit, I 5 Ï}

P

} R gives the relationship between amperes of

current I, watts of power P, and ohms of resistance R. For a certain circuit, I is 5 amperes and P is 75 watts. What is the circuit’s resistance?

8. The sum s of the interior angles of an n-sided polygon is s 5 (n 2 2)180. If the sum of the interior angles of a polygon is 2340, how many sides does it have?

9. The formula e 5 c 2 Ï}

c } c gives the effi ciency e of a car’s engine. The variable c

is the engine’s compression ratio. If an engine has e 5 0.75, what is the compression ratio?

Write the equation so that y is a function of x.

10. 6y 2 3x 5 212 11. 3x 1 7y 5 14 12. 7 2 3y 5 21x

13. 24x 2 5y 5 9 14. 2 1 3y 5 28x 15. 24y 1 36 5 224x

EXAMPLE

PRACTICE

Remember to fi rst isolate the term containing the dependent variable. Then multiply or divide to isolate the variable.

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D. Graphing Linear EquationsThe x-axis and y-axis divide a coordinate plane into four equal parts called quadrants. The quadrants are labeled with roman numerals I, II, III, and IV, moving counter-clockwise from the upper right quadrant. Each point in a coordinate plane has a unique ordered pair (x, y) that describes the point’s location with respect to the origin (0, 0).

The solution of an equation is the set of all ordered pairs (x, y) that make the equation a true statement. The graph of an equation is a graph of all the ordered pairs that make up the solution of the equation.

1. Plot Points in a Coordinate Plane (Lesson 4.1)Plot each point and describe its location.

a. P(24, 21) b. Q(3, 24) c. R(22, 0) d. S(0, 23)

Solution:

a. Start at the origin. Move 4 units left, then 1 unit

1

2

3

4y

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22

23

24

P

R

S Q

down. Point P is in Quadrant III.

b. Start at the origin. Move 3 units right, then 4 units down. Point Q is in Quadrant IV.

c. Start at the origin. Move 2 units left. Point R is on the x-axis.

d. Start at the origin. Move 3 units down. Point P is on the y-axis.

Plot each point and describe its location.

1. A(3, 5) 2. B(24, 0) 3. C(21, 4)

4. D(0, 21) 5. E(22, 23) 6. F(1, 24)

2. Identify Solutions to Equations in Two Variables (Lesson 4.2)

Tell whether the ordered pair is a solution of the equation.

a. x 1 2y 5 8; (24, 6) b. 5x 2 2y 5 10; (2, 1)

Solution:

a. x 1 2y 5 8 Write original equation.

(24) 1 2(6) 0 8 Substitute �4 for x and 6 for y.

8 5 8 ✓ Simplify.

(24, 6) is a solution.

EXAMPLE

Another name forthe x-coordinate is abscissa. Another name for the y-coordinate is ordinate.

PRACTICE

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b. 5x 2 2y 5 10 Write original equation.

5(2) 2 2(1) 0 10 Substitute 2 for x and 1 for y.

8 Þ 10 ✗ Simplify.

(2, 1) is not a solution.


7. 22x 1 3y 5 4; 1 0, 4 }

3 2 8. 28 5 y; (25, 28)

9. 3x 2 4y 5 21; (23, 24) 10. x 5 22; (21, 22)

11. y 2 5x 5 23; (22, 213) 12. 24y 1 2x 5 0; 1 2 1 } 2 ,

1 }

4 2

3. Graph an Equation Using a Table (Lesson 4.2)Graph the equation �3x � y 5 1.

Solution:

Step 1: Solve the equation for y:

23x 1 y 5 1

y 5 3x 1 1

Step 2: Make a table by choosing a few values for x 2

4

6

8y

2 x22242428 4 6 822

24

26

28

and fi nding the values of y.

x 22 21 0 1 2

y 25 22 1 4 7

Step 3: Plot the points. Notice that the points appear to lie on a line.

Step 4: Connect the points by drawing a line through them. Use arrows to indicate that the graph goes on without end.

Graph the equation.

13. x 1 y 5 3 14. y 2 2x 5 21 15. 23x 1 2y 5 2

16. x 2 3y 5 3 17. 4y 2 3x 5 8 18. 2y 2 5x 5 0

4. Graph Horizontal and Vertical Lines (Lesson 4.2)Linear equation An equation that can be written in the form Ax 1 By 5 C, where A, B, and C are real numbers and A and B are not both equal to zero. The graph of a linear equation is a straight line. When A 5 0, the graph of the linear equation is a horizontal line. When B 5 0, the graph of the linear equation is a vertical line.

PRACTICE

EXAMPLE

PRACTICE

Vocabulary

You can choose any (x, y) pair from the graph and substitute it in the equation to make a true statement.


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Graph the equation.

a. y 5 23 b. x 5 1

Solution:

a. Notice that x can be any real number, but that y 1

y

1 x21222324 2 3 421

22

23

24

25

26

27

is always 23. The graph of the equation y 5 23 is a horizontal line 3 units below the x-axis.

b. Notice that x will always be 1, but that y can be

1

2

3

4y

1 x21222324 2 3 421

22

23

24

any real number. The graph of the equation x 5 1 is a vertical line 1 unit to the right of the y-axis.

Graph the equation.

19. x 5 25 20. y 5 1 21. 2y 5 23

22. 2x 2 1 5 0 23. x 2 3 5 0 24. y 1 2 5 0

CheckPlot each point and describe its location.

1. A(4, 27) 2. B(29, 22) 3. C(0, 7)

4. D(1, 3) 5. E(26, 0) 6. F(24, 8)


7. 25 5 y; (5, 25) 8. 2x 1 4y 5 4; (4, 5) 9. 28y 1 4x 5 0; 1 22, 2 1 }

4 2

10. y 2 2x 5 26; 1 1 } 2 , 25 2 11. x 5 29; (1, 29) 12. 3x 2 7y 5 24; (1, 1)

Graph the equation.

13. x 2 y 5 22 14. y 1 3x 5 24 15. 25x 1 3y 5 2

16. y 5 27 17. 23y 2 2x 5 9 18. x 5 8

19. 4y 2 6x 5 0 20. 3 5 2x 21. 2y 1 5 5 23

EXAMPLE

PRACTICE

All the solutions of y 5 23 are ordered pairs in the form (x, 23).


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For any two points, there is one and only one line that contains both points. This fact can help you graph a linear equation. Many times, it will be convenient to use the points where the line crosses the x-axis and y-axis. These points are the intercepts. Knowing how steep the line is, or the slope of the line, also can help you graph a linear equation. If the graph of a linear equation passes through the origin (0, 0), the relation-ship between x and y is called a direct variation.

1. Find the Intercepts of the Graph of an Equation (Lesson 4.3)

x-intercept The x-coordinate of the point where a graph intersects the x-axis.

y-intercept The y-coordinate of the point where a graph intersects the y-axis.

Find the x-intercept and the y-intercept of the graph of 3x � 4y � 12.

Solution:

To fi nd the x-intercept, substitute 0 for y and solve for x.


3x 1 4(0) 5 12 Substitute 0 for y.

x 5 12

} 3 5 4 Solve for x.

To fi nd the y-intercept, substitute 0 for x and solve for y.


3(0) 1 4y 5 12 Substitute 0 for x.

y 5 12

} 4 5 3 Solve for y.

The x-intercept is 4. The y-intercept is 3.

Find the x-intercept and the y-intercept of the graph of the equation.

1. x 1 y 5 26 2. 23y 1 8 5 212x 3. 4.5x 1 0.5y 5 9

4. 27y 5 14x 5. 215 1 10y 5 60x 6. 3 2 18x 5 26y

2. Find the Slope of a Line (Lesson 4.4)Slope Describes how quickly a line rises or falls as it moves from left to right. Slope is the ratio m of the vertical change between two points on the line to the horizontal change between the same two points.

For points (x1, y

1) and (x

2, y

2), m 5

y2 2 y

1 } x

2 2 x

1 .

Find the slope of the line that passes through the points.

a. (1, 5) and (4, 6) b. (25, 7) and (3, 21)

c. (22, 7) and (8, 7) d. (6, 28) and (6, 2)

Vocabulary

EXAMPLE

PRACTICE

Vocabulary

EXAMPLE

Remember that the x- and y-intercepts are numbers, NOT ordered pairs.


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Solution:

a. Let (x1, y

1) 5 (1, 5) and (x

2, y

2) 5 (4, 6).

m 5 y

2 2 y

1 } x

2 2 x

1 Write formula for slope.

5 6 2 5

} 4 2 1

5 1 }

3 Substitute and simplify.

b. Let (x1, y

1) = (25, 7) and (x

2, y

2) 5 (3, 21).

m 5 y

2 2 y

1 } x

2 2 x


5 21 2 7

} 3 2 (25)

5 28

} 8 5 21 Substitute and simplify.

c. Let (x1, y

1) 5 (22, 7) and (x

2, y

2) 5 (8, 7).

m 5 y

2 2 y

1 } x

2 2 x


5 7 2 7

} 8 2 (22)

5 0 }

10 5 0 Substitute and simplify.

The slope is 0. The line is horizontal.

d. Let (x1, y

1) 5 (6, 28) and (x

2, y

2) 5 (6, 2).

m 5 y

2 2 y

1 } x

2 2 x


5 2 2 (28)

} 6 2 6

5 10

} 0 Substitute. Division by 0 is undefi ned.

The slope is undefi ned. The line is vertical.


7. (6, 29) and (29, 6) 8. (4, 2) and (4, 0)

9. (211, 8) and (13, 5) 10. (21, 27) and (1, 27)

11. (2.5, 25) and (5.5, 29) 12. (23, 25) and (22, 0)

3. Graph an Equation Using Slope-Intercept Form (Lesson 4.5)

Slope-intercept form A linear equation in the form y 5 mx 1 b, where m is the slope and b is the y-intercept of the graph of the equation.

Graph the equation �x � 2y � 4.

Solution:

Step 1: Rewrite the equation in slope-intercept form.

1

2

3

4

5

6y

1 x21222324 2 3 421

22

(2, 3)

(0, 2)

y 5 1 }

2 x 1 2

Step 2: Identify the slope and the y-intercept.

m 5 1 }

2 and b 5 2.

Step 2: Plot the point that corresponds to the y-intercept, (0, 2).

Step 4: Use the slope to fi nd another point on the line. Draw a line through the two points.

PRACTICE

Vocabulary

EXAMPLE

If you can substitute the coordinates of the second point in the original equation and get a true statement, then your graph is correct.

Think of (x1, y1) as “the coordinates of the fi rst point” and (x2, y2) as “the coordinates of the second point.” Be sure to subtract the x- andy-coordinates in the same order.


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Graph the equation.

13. y 5 2 2 } 5 x 1 7 14. 23x 5 4y 1 8 15. 3x 2 3y 5 6

16. y 5 24 17. 214x 2 7y 5 21 18. 1.5y 2 6x 2 12 5 0

4. Identify Direct Variation Equations (Lesson 4.6)Direct variation An equation in the form y 5 ax, where a Þ 0, represents direct variation. The variable y varies directly with x.

Constant of variation The constant a in the direct variation equation y 5 ax.

Tell whether the equation represents direct variation. If so, identify the constant of variation.

a. 6x 2 4y 5 0 b. x 1 y 5 8

Solution:

Try to rewrite the equation in the form y 5 ax.

a. 6x 2 4y 5 0 Write original equation.

24y 5 26x Subtract �6x from each side.

y 5 3 }

2 x Simplify.

Because the equation 6x 2 4y 5 0 can be rewritten in the form y 5 ax,

it represents direct variation. The constant of variation is 3 }

2 .

b. x 1 y 5 8 Write original equation.

y 5 2x 1 8 Subtract x from each side.

Because the equation x 1 y 5 8 cannot be rewritten in the form y 5 ax, it does not represent direct variation.

Tell whether the equation represents direct variation. If so, identify the constant of variation.

19. y 5 2 7 }

8 x 20. x 1 4 5 16y 21. 9y 5 5x

22. x 5 247y 23. 23 1 x 1 7 5 2y 1 4 24. 13 5 26x

5. Write and Use a Direct Variation Equation (Lesson 4.6)The graph of a direct variation equation is shown.

a. Write the direct variation equation.

1

2

3

4

5y

1 x2122 2 3 4 5 621

22

23

(6, 5)

b. Find the value of y when x 5 36.

PRACTICE

EXAMPLE

PRACTICE

Vocabulary

EXAMPLE


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Solution:

a. Because y varies directly x, the equation has the form y 5 ax. Use the fact that y 5 5 when x 5 6 to fi nd a.

y 5 ax Write direct variation equation.

5 5 a(6) Substitute.

5 }

6 5 a Solve for a.

A direct variation equation that relates x and y is y 5 5 }

6 x.

b. When x 5 36, y 5 5 }

6 (36) 5 30.

Write the direct variation equation that passes through the given point. Then fi nd the value of y for the given x.

25. (3, 21); x 5 12 26. (24, 28); x 5 32 27. (26, 3); x 5 18

28. (9, 2); x 5 27 29. (25, 7); x 5 100 30. (22, 21); x 5 74

CheckFind the x-intercept and the y-intercept of the graph of the equation.

1. 221 1 14y 5 84x 2. 23 1 x 5 3y 3. 3.2x 1 0.8y 5 4


4. (8, 25) and (23, 4) 5. (1, 7) and (22, 7) 6. (29, 7) and (3, 25)

Graph the equation. Then fi nd the value of y when x 5 6.

7. y 5 x 1 1 8. y 5 22 9. 4x 2 6y 5 12

Does the equation represent direct variation? If so, fi nd the constant of variation.

10. y 5 2 4 } 5 x 11. x 1 3 5 9y 12. 4y 5 7x

Write the direct variation equation that passes through the given point. Then fi nd the value of y for the given x.

13. (2, 25); x 5 20 14. (23, 29); x 5 43 15. (24, 6); x 5 64

PRACTICE

Check the sign of the constant of variation in your equation. If the graph of y 5 ax passes through Quadrants I and III, the constant should be positive.If the graph ofy 5 ax passes through Quadrants II and IV, the constant should be negative.


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A. Writing Linear EquationsYou can describe a line with equations in three different forms. You can write these equations if you know the slope and y-intercept of the line, if you know the slope and a point on the line, or if you know two points on the line. The following examples illustrate these three different forms of the equation of a line and show how to fi nd them.

1. Write an Equation in Slope-Intercept Form (Lesson 5.1)Slope-intercept form The equation y 5 mx 1 b, for a line with slope m and y-intercept b.

Write an equation of the line with a slope of 1 } 3 and a y-intercept of 22.

Solution:

y 5 mx 1 b Write slope-intercept form. y

x1 2 3 4 5 6

1

23

24

212221

13

(0, 22)

25

26

27

y 5 1 }

3 x 2 2 Substitute 1 }

3 for m and 22 for b.

Write an equation of the line with the given slope and y-intercept.

1. Slope is 6; y-intercept is 24. 2. Slope is 21; y-intercept is 3.

3. Slope is 3 } 5 ; y-intercept is 25. 4. Slope is

2 } 5 ; y-intercept is 23.

5. Slope is 24; y-intercept is 5. 6. Slope is 2 1 } 3 ; y-intercept is 22.

2. Write an Equation of a Line Given the Slope and a Point (Lesson 5.2)

Write an equation of the line that passes through (4, 23) and has a slope of 22.

Solution:

1

2

3

4

5

6

7

8 y

1 x 21 22 23 24 2 3 4

(0, 5) 2

1Step 1: Identify the slope. The slope is 22.

Step 2: Find the y-intercept. Substitute the slope and the coordinates of the given point in y 5 mx 1 b. Solve for b.

y 5 mx 1 b Write slope-intercept form.

23 5 22(4) 1 b Substitute 22 for m,

4 for x, and 23 for y.

5 5 b Solve for b.

EXAMPLE

Make sure you don’t switch the x and y values when you substitute.

Vocabulary

EXAMPLE

PRACTICE

Ready to Go On? Chapters 5 –6 Intervention

Inte

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Step 3: Write an equation of the line.


y 5 22x 1 5 Substitute 2 for m and 5 for b.

Write an equation of the line that passes through the given point and has the given slope.

7. (26, 22); m 5 4 }

3 8. (21, 3); m 5 2

1 } 4

9. (3, 4); m 5 26 10. (5, 23); m 5 3 }

2

11. (23, 6); m 5 2 2 } 3 12. (21, 24); m 5 2

3. Write an Equation of a Line Given Two Points (Lesson 5.2)

Write an equation of the line shown.

1

2

3

4

5

6

x212223242526 1 221

22

y(25, 6)

(23, 2)

Solution:

Step 1: Calculate the slope using the formula.

m 5 y

2 2 y

1 } x

2 2 x

1

5 2 2 6 }

23 2 (25) 5

24 }

2 5 22

Step 2: Find the y-intercept. Use the point (25, 6).

y 5 mx 1 b Write slope-intercept

form.

6 5 22(25) 1 b Substitute 6 for y, 22 for m, and 25 for x.

6 2 10 5 b Solve for b.

24 5 b

Step 3: Write an equation of the line.



Write an equation of the line shown.

13.

2

3

4

5

y

21

6

7

1 x21 2 3 4 5 6 7(3, 21)

(6, 1)1

14. y

1 x2 3 4 5 6 7 8

22

23

24

25

26

27

1 (8, 0)(6, )1

2

2121

PRACTICE

EXAMPLE

You also could fi nd b by substituting the x and y values from the other known point,(23, 2).

PRACTICE

Ready to Go On? Chapters 5 –6 Intervention cont’d

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15.

x1212223 2 3 4 521

22

23

24

25

26

1

2

(3, 21)(21, 23)

y 16. y

x1 2 3 4

1

2

3

4

21

22

23

24

21222324

(1, 1)

(23, 22)

17.

3

4

5

6

7

8

x2223242526 1 2

1

2

y

(25, 3)

(22, 0)

18.

24

22

2

4

6

8y

2 x22242628 4 6 8

26

28

(5, 21)

(24, 2)

4. Write an Equation in Point-Slope Form (Lesson 5.3)Point-slope form The equation y 2 y

1 5 m(x 2 x

1), for the nonvertical line through a

given point (x1, y

1) with slope m.

Write an equation in point-slope form of the

1

2

3

4

5

6

7

8y

1 x21222324 2 3 4

(22, 1)2

1

line that passes through the point (22, 1) andhas a slope of 2.

Solution:

y 2 y1 5 m(x 2 x

1) Write point-slope form.

y 2 1 5 2(x 1 2) Substitute 1 for y1, 2 for

m, and 22 for x1.

Write an equation in point-slope form of the line that passes through the given point and has the given slope.

19. (3, 21); m 5 2 }

3 20. (4, 0); m 5 2

1 } 4

21. (23, 24); m 5 1 }

2 22. (1, 1); m 5

3 }

4

23. (25, 3); m 5 21 24. (24, 2); m 5 2 1 } 3

Notice that (x1, y1) is a point of the line, and that m is the slope of the line.

PRACTICE

Vocabulary

EXAMPLE


Inte

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5–6

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5. Write an Equation in Standard Form (Lesson 5.4)Standard form The equation Ax 1 By 5 C, where A, B, and C are real numbers and A and B are not both zero.

Write an equation in standard form of the y

1 x21222324 2 3 4

25

24

23

22

21

1

2

3

(2, 22)(4, 21)

line shown.

Solution:

Step 1: Calculate the slope.

m 5 y

2 2 y

1 } x

2 2 x

1 5

21 2 (22) }

4 2 2 5

1 }

2

Step 2: Write an equation in point-slope form. Use (2, 22).

y 2 y1 5 m(x 2 x

1) Write point-slope form.

y 2 (22) 5 1 }

2 (x 2 2) Substitute 22 for y

1, 1 }

2 for m, and 2 for x

1.

Step 3: Rewrite the equation in standard form.

y 1 2 5 1 }

2 x 2 1 Apply the distributive property.

2y 1 4 5 x 2 2 Multiply each term by 2.

2x 1 2y 5 26 Simplify. Collect variable terms on one side,

constants on the other.

Write an equation in standard form of the line shown.

25. y

x21222324252621

22

23

24

25

26

27

28

21

(21, 21)

(22, 25)

26.

x1212223 2 3 4 521

22

23

24

25

26

1

2

(1, 21)

(2, 24)

y

27.

1

2

3

4

x212223242526 1 221

22

23

24

y

(1, 3)

(23, 23)

28.

1

2

3

4

5

x212223242526 1 221

22

23

y(24, 5)

(2, 23)

PRACTICE

Vocabulary

EXAMPLE


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29.

1

2

3

4

5

6

x212223242526 1 221

22

y

(23, 3)

(22, 1)

30.

1

2

3

4

x2122232425 1 2 321

22

23

24

y(25, 4)

(3, 22)

CheckWrite an equation in slope-intercept form of the line with the given slope and y-intercept.

1. Slope is 4; y-intercept is 3. 2. Slope is 22; y-intercept is 1.

3. Slope is 5 }

2 ; y-intercept is 24. 4. Slope is 2

1 } 3 ; y-intercept is 25.

Write an equation in the given form of the line that passes through the given point and has the given slope.

5. (23, 24); m 5 1 } 5 6. (22, 7); m 5 2

3 }

4 7. (1, 5); m 5 24

slope-intercept form point-slope form point-slope form

Write equations in slope-intercept form and standard form of the line shown.

8. y

1 x2 3 4 5 6 7 8

23

24

25

26

27

28

(8, 22)

(4, 0)

21

22

9.

x21222324252627 121

y

(26, 5)

(23, 1) 1

2

3

4

5

6

7

The table shows the price per T-shirt for different quantities of T-shirts. Tell whether the cost of the T-shirts represents an arithmetic sequence. If it does, fi nd the cost of 12 T-shirts.

10. Number of T-shirts 1 2 3 4 5

Price of T-shirts ($) 8 16 24 32 40


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B. Parallel and Perpendicular LinesIf two non-vertical lines in the same plane have the same slope, then they are parallel. If their slopes are negative reciprocals, then they are perpendicular. The converse is also true. If two non-vertical lines in the same plane are parallel, then they have the same slope. If they are perpendicular, then their slopes are negative reciprocals.

1. Determine Whether Lines are Parallel or Perpendicular (Lesson 5.5)

Perpendicular lines Lines in a plane that intersect to form a right (90°) angle.

Determine which lines, if any, are parallel or perpendicular.

Line a: y 5 4x 2 1 Line b: 24x 1 y 5 3 Line c: 2x 1 8y 5 4

Solution:

Step 1: Write each equation in slope-intercept form. Find the slopes of the lines.

Line a: The equation is in slope-intercept form. The slope is 4.

Line b: 24x 1 y 5 3 y 5 4x 1 3

Line c: x 1 4y 5 4

y 5 2 1 } 4 x 1 1

Step 2: Compare the slopes. Line a and line b have slopes of 4,

so they are parallel. Line c has a slope of 2 1 } 4 . 4 1 2

1 } 4 2 5 21,

so it is perpendicular to lines a and b.

Determine which lines, if any, are parallel or perpendicular.

1. Line a: y 5 3 }

4 x 1 2 Line b: 4x 2 3y 5 23 Line c: 3x 2 4y 5 20

2. Line d: x 2 2y 5 4 Line e: 2x 1 y 5 0 Line f: x 1 2y 5 3

3. Line g: 5x 1 7y 5 7 Line h: y 5 7 } 5 x 1 3 Line j: 7x 2 5y 5 2

2. Write an Equation of a Parallel Line (Lesson 5.5)Write an equation of the line that passes through (1, 22) and is parallel to the line y 5 5x 1 2.

Solution:

Step 1: Identify the slope. The graph of the given equation has a slope of 5. So, the parallel line through (1, 22) will also have a slope of 5.

Step 2: Find the y-intercept. Use the slope and the given point.


22 5 5(1) 1 b Substitute 22 for y, 5 for m, and 1 for x.

27 5 b Solve for b.

EXAMPLE

Vocabulary

PRACTICE

EXAMPLE

The product of a non-zero slope m and its negative reciprocal is 21:

m 1 2 1 } m 2 5 21


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Step 3: Write an equation of the line in slope-intercept form.



Write an equation of the line that passes through the given point and is parallel to the given line.

4. (23, 21); y 5 4 }

3 x 1 1 5. (28, 5); y 5 2

1 } 4 x 2 2

6. (2, 3); y 5 26x 1 4 7. (2, 0); y 5 3 }

2 x 2 7

8. (26, 4); y 5 2 2 } 3 x 1 3 9. (25, 22); y 5 2x 2 9

3. Write an Equation of a Perpendicular Line (Lesson 5.5)Write an equation of the line that passes through (4, 3) and is perpendicular to the line y 5 2x 2 3.

Solution:

Step 1: Identify the slope. The graph of the given equation has a slope of 2. So, the slope of the perpendicular line through (4, 3) will be the

negative reciprocal of 2, which is 2 1 } 2 .

Step 2: Find the y-intercept. Use the slope and the given point.


3 5 2 1 } 2 (4) 1 b Substitute 3 for y, 2 1 } 2 for m, and 4 for x.

5 5 b Solve for b.

Step 3: Write an equation of the line in slope-intercept form.


y 5 2 1 }

2 x 1 5 Substitute 2 1 } 2 for m and 5 for b.

Write an equation of the line that passes through the given point and is perpendicular to the given line.

10. (23, 22); y 5 3 }

2 x 1 2 11. (26, 1); y 5 2

3 } 4 x 2 1

12. (2, 5); y 5 28x 1 3 13. (4, 0); y 5 1 }

3 x 2 4

14. (4, 6); y 5 2 2 } 3 x 1 3 15. (28, 22); y 5 2x 2 6

EXAMPLE

You can graph both lines to check your answer.

PRACTICE

PRACTICE


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CheckDetermine which lines, if any, are parallel or perpendicular.

1. Line a: y 5 2 3 }

2 x 1 4 Line b: 3x 1 2y 5 2 Line c: 2x 2 3y 5 3

2. Line d: x 1 3y 5 9 Line e: y 5 3x 2 2 Line f: 3x 1 y 5 2

3. Line g: x 1 4y 5 2 Line h: x 2 4y 5 0 Line j: y 5 1 }

4 x 1 1

Write an equation of the line that passes through the given point and is parallel to the given line.

4. (8, 1); y 5 3 }

8 x 5. (23, 3); y 5 2

2 }

3 x 2 5 6. (25, 22); y 5 2x 1 2

7. (26, 2); y 5 4 }

3 x 1 4 8. (28, 0); y 5 2

1 }

4 x 2 3 9. (3, 2); y 5 25x 1 1

Write an equation of the line that passes through the given point and is perpendicular to the given line.

10. (1, 7); y 5 1 }

3 x 2 2 11. (6, 4); y 5 2

2 } 3 x 1 6 12. (24, 23); y 5 2x 2 7

13. (26, 2); y 5 3 }

2 x 1 5 14. (3, 21); y 5 2

3 }

4 x 2 8 15. (8, 2); y 5 24x 1 1


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C. Linear Models Paired data graphed in a scatter plot may show a positive correlation, a negative correlation, or no correlation. If there is a positive or negative correlation, the data can be modeled by a line of fi t drawn close to the points on the scatter plot. The equation of this line will be in the form y 5 mx 1 b. Using linear regression, you can fi nd the line that best fi ts the data. This best-fi tting line or its equation can be used to approximate data points between or beyond known data points.

1. Describe the Correlation of Data (Lesson 5.6)Correlation The relationship between paired data; If the value of y tends to increase as the value of x increases, the correlation is positive. If the value of y tends to decrease as the value of x increases, the correlation is negative.

Scatter plot A graph that shows the relationship, if any, between paired data.

Describe the correlation, if any, of the data graphed in the scatter plot.

a. y

x1 2 3 4

1

2

3

4

21

22

23

24

21222324

b. y

x1 2 3 4

1

2

3

4

21

22

23

24

21222324

c. y

x2 3 4

1

2

3

4

21

22

23

24

21222324

Solution:

a. The value of y decreases as the value of x increases: negative correlation.

b. There is no apparent relationship between the value of y and x: no correlation.

c. The value of y increases as the value of x increases: positive correlation.

EXAMPLE

Vocabulary


Inte

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Describe the correlation, if any, of the data graphed in the scatter plot.

1. y

x21 3 4

1

2

3

4

21

22

23

24

21222324

2. y

x21 3 4

1

2

3

4

21

22

23

24

21222324

3. y

x21 3 4

1

2

3

4

21

22

23

24

21222324

2. Make a Scatter Plot (Lesson 5.6)Tracy is training for a swim race. The table shows her fastest time from each practice session for six days.

Practice day 1 2 3 4 5 6

Fastest time (min) 6.6 6.5 6.5 6.3 6.2 6.0

a. Make a scatter plot of the data.

b. Describe the correlation of the data.

Solution:

a. Treat the data as ordered pairs. Let x represent

Training day 1 2 3 4 5

Fast

est t

ime

(min

)

6.0 6.1 6.2 6.3 6.4 6.5 6.6

x

y

6 70

0

the training day and let y represent the fastest time each day. Plot the ordered pairs as points in a coordinate plane.

b. The scatter plot shows a negative correlation.The more Tracy trains, the less time she takes to fi nish the race.

Notice that a negative correlation is not always an “ undesirable” outcome.

EXAMPLE

PRACTICE


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Make a scatter plot of the data in the table. Describe the correlation,if any.

4. x 2 2 3 5 5 6y 21 0 2 3 6 6

5. x 23 22 21 21 0 2y 1 0 22 23 23 24

3. Draw a Line of Fit to Data (Lesson 5.6)Line of fi t A line on a scatter plot that appears to fi t the data closely.

The table shows the shoe size and height for nine customers of a men’s shoe store.

Shoe Size 9 9 10 10 1 }

2 11 11 11

1 }

2 12 12

Height (in.) 67 69 68 70 72 73 75 74 76

Write an equation that models the height of

Shoe size 8 9 10 11 14 13 12

Hei

ght (

in.)

66 68 70 72 74 76 78

x

y

0 0

a customer as a function of his shoe size.

Solution:

Step 1: Make a scatter plot of the data. Let x represent shoe size. Let y represent height.

Step 2: Decide whether the data can be modeledby a line. As shoe size increases, height tends to increase, so the scatter plot shows a positive correlation. You can fi t a line to the data.

Step 3: Draw a line of fi t. The line should be close to the data points,with about the same number of points above and below the line.

Step 4: Write an equation using two points on the line. Use (9, 68) and (12, 75).

Find the slope of the line. m 5 y

2 2 y

1 } x

2 2 x

1 5

75 2 68 }

12 2 9 5

7 }

3

Find the y-intercept of the line. Use the point (9, 68).


68 5 7 }

3 (9) 1 b Substitute 68 for y, 7 }

3 for m, and 9 for x.

47 5 b Solve for b.

The height in inches y of a customer can be modeled by the function

y 5 7 }

3 x 1 47, where x is the customer’s shoe size.

Write an equation that models y as a function of x.

6. x 5 4 3 2 1 0y 0 0 22 23 24 23

7. x 0 0 1 1 2 2y 25 23 23 21 0 2

Vocabulary

EXAMPLE

PRACTICE

A line of fi t and its equation model the trend in the data.

PRACTICE


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4. Interpolate Using an Equation (Lesson 5.7)Linear interpolation To use a line or its equation to estimate a value between two known values.

Use the data about shoe store customer’s sizes and heights to fi nd the equation of the best-fi tting line for the data. Then approximate the

height of a customer who wears shoe size 9 1 } 2 .

Solution:

8 9 10 11 12 13

666870727476Step 1: Enter the data into lists on a graphing

calculator. Make a scatter plot of the data. Let the x-values be shoe size andthe y-values be height.

Step 2: Perform linear regression using the paired data. The equation of the best-fi tting line is approximately y 5 2.6x 1 44.

Step 3: Find the value of y when x 5 9.5.

8 9 10 11 12 13

666870727476

y 5 2.6x 1 44 Equation of best-

fi tting line

y 5 2.6(9.5) 1 44 Substitute 9.5 for x.

y ø 68.7 Simplify.

A customer who wears a size 9 1 }

2 shoe is probably

about 68.7 inches tall.

Make a scatter plot of the data. Find the equation of the best-fi tting line. Approximate the value of y for x 5 4.

8. x 25 23 21 1 3 5y 4 3 1 0 22 23

9. x 0 1 3 6 7 8y 22 21 0 1 2 4

5. Extrapolate Using an Equation (Lesson 5.7)Linear extrapolation To use a line or its equation to estimate a value beyond the range of known values.

Use the data about shoe store customer’s sizes and heights to estimate the height of a customer who wears size 13.

Solution:

Use the equation of the best-fi tting line to fi nd the value of y when x 5 13.

y 5 2.6x 1 44

y 5 2.6(13) 1 44 5 77.8

A customer who wears a size 13 shoe is probably about 77.8 inches tall.

PRACTICE

Vocabulary

EXAMPLE

Vocabulary

EXAMPLE


Inte

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Make a scatter plot of the data. Find the equation of the best-fi tting line. Approximate the value of y for x 5 7.

10. x 24 23 0 0 2 5y 2 2 1 0 21 24

11. x 22 21 1 2 4 5y 25 24 21 0 0 3

CheckDescribe the correlation, if any, of the data graphed in the scatter plot.

1. y

x21 3 4

1

2

3

4

21

22

23

24

21222324

2. y

21 3 4

1

2

3

4

5

6

7

2121222324 x

3. y

1 x2 3 4 5 6 7 821

22

23

24

25

26

1

2

4. A bookstore is interested in the relationship between the number of rainy days in a month and the number of sales. The table shows data for six months.

Apr May Jun Jul Aug Sep

Rainy days 8 6 5 5 6 3

Sales 105 91 85 90 92 75


b. Describe the correlation.

c. Draw a line of fi t. Write the equation of the line.

d. Perform linear regression to fi nd the equation of the best-fi tting line.

e. Estimate the number of sales during a month with 4 rainy days. Estimate the number of sales during a month with 10 rainy days.

PRACTICE


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Copyright © by Holt McDougal. All rights reserved. 49Holt McDougal Algebra 1

D. Graphing InequalitiesAn inequality is a statement that compares unequal quantities. A compound inequality is the intersection or union of two inequalities. The graph of an inequality is the set of points that represents all solutions of the inequality. The graph of an inequality in one variable is a line, line segment, or ray graphed on a number line. The graph of an inequality in two variables is a half-plane and boundary line graphed on a coordinate plane.

1. Graph an Inequality (Lesson 6.1)The least expensive book at the City Bookstore costs $2. Graph an inequality that describes prices of books at City Bookstore.

Solution:

Let P represent the price of a book at City Bookstore. The value of P must be greater than or equal to 2. So, an inequality is P ≥ 2.

0 1 2 3 4 5 6

Graph an inequality that describes the situation.

1. The shortest player on a basketball team is 74 inches tall.

2. A diver’s depth is at least 30 feet below sea level.

3. A pitcher holds a maximum of 8 cups of water.

4. A computer has 162 megabytes of open storage space.

2. Write an Inequality Represented by a Graph (Lesson 6.1)

Write an inequality represented by the graph.

a. �1 0 1 2 3 4 5

b. 20 21 22 23 24 25 26

Solution: a. The open circle means that 3 is NOT

a solution of the inequality. Because the arrow points to the left, allnumbers less than 3 are solutions.

The graph represents theinequality x < 3.

b. The closed circle means that 24 is a solution of the inequality. Because the arrow points to the right, all numbers greater than 24 also are solutions.

The graph represents the inequality x ≥ 24.


5. �1 0 1 2 3�2�3�4

6. �9 �8 �7 �6 �5�12 �11 �10

7. 0 1 2 3 4 5 6 7

8. �1 0 1�2�3�4�5�6

EXAMPLE

PRACTICE

EXAMPLE

To graph an inequality in one variable, use an open circle for < or > and a closed circle for ≤ or ≥.

PRACTICE


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Copyright © by Holt McDougal. All rights reserved.50 Holt McDougal Algebra 1

3. Write and Graph Compound Inequalities (Lesson 6.4)Compound inequality Two inequalities joined by and or or. The graph of a compound inequality with and contains only the points that the graphs of the separate inequalities have in common. The graph of a compound inequality with or contains all the points on the graphs of the separate inequalities.

Translate the verbal phrase into a compound inequality. Then graph the inequality. a. All real numbers that are less than or

equal to 7 and greater than 1. b. All real numbers that are greater than

or equal to 4 or less than 24.

Solution:

a. 1 < x ≤ 7 b. x ≥ 4 or x < 24

0 1 2 3 4 5 6 7 8

0 2 4 6�2�4�6

Translate the verbal phrase into a compound inequality. Then graph the inequality.

9. All real numbers that are greater than 23 and less than 0.

10. All real numbers that are less than or equal to 2 or greater than or equal to 9.

11. All real numbers that are less than 6 and greater than or equal to 21.

12. All real numbers that are greater than or equal to 60 or less than 55.

4. Graph a Linear Inequality in Two Variables (Lesson 6.7)Linear inequality in two variables A statement that can be written in one of the following forms: y > mx 1 b; y ≥ mx 1 b; y < mx 1 b; or y ≤ mx 1 b. The solution of a linear inequality in two variables is the set of ordered pairs (x, y) that makes the inequality a true statement.

Graph the inequality y � 2x 1 1.

Solution:

Step 1: Graph the boundary line, y 5 2x 1 1. y

x1 2 3 4

1

2

3

4

21

22

23

24

21222324

The inequality is <, so use a dashed line.

Step 2: Test a point NOT on the boundaryline of the inequality. Test (0, 0) in y < 2x 1 1.

0 < 2(0) 1 1

0 < 1 ✓

Step 3: Shade the half-plane that contains (0, 0),because (0, 0) is a solution of the inequality.

Vocabulary

Be sure that the point you test is not on the boundary line. In this example, you could not use(0, 1) because it is on the boundary line y 5 2x 1 1.

EXAMPLE

Vocabulary

EXAMPLE

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EXAMPLE Graph the inequality 3x 1 2y � 4.

Solution:

Step 1: Graph the equation 3x 1 2y 5 4. y

x2 3 4

1

2

3

4

23

24

2122232421

22

1

The inequality is ≥, so use a solid line.

Step 2: Test (1, 0) in 3x 1 2y ≥ 4.

3(1) 1 2(0) ≥ 4

3 ≥ 4 ✗

Step 3: Shade the half-plane that does NOTcontain (1, 0), because (1, 0) is NOT a solution of the inequality.

Graph the inequality.

13. y > 2x 1 3 14. 2x 2 y ≤ 2 15. y < 5x

16. 22x 1 3y ≤ 26 17. 6x 1 4y ≥ 28 18. x 1 y < 10

CheckGraph an inequality that describes the situation.

1. Jay reads a maximum of 20 pages. 2. Mandy hikes more than 7.5 miles.


3. �1 0 1 2 3 4 65

4. �1 0 1�2�3�4�5�6

Translate the verbal phrase into a compound inequality. Then graph the inequality.

5. All real numbers that are less than 210 and greater than or equal to 213.

6. All real numbers that are greater than 6 or less than or equal to 0.

Graph the inequality.

7. y ≥ 23x 2 4 8. 22x < y 9. x 2 2y ≥ 4

PRACTICE

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E. Solving InequalitiesIf two inequalities have the same solution, then they are equivalent. Solving inequalities is similar, but not identical, to solving equations. Whenever you multiply or divide both sides of the inequality by the same negative number, you must reverse the direction of the inequality symbol.

1. Solve an Inequality Using Addition or Subtraction (Lesson 6.1)

Solve the inequality. Graph the solution.

a. x 2 3 < 2 b. b 1 1 ≥ 28

Solution:

a. x 2 3 < 2 Write original inequality.

x 2 3 1 3 < 2 1 3 Add 3 to each side.

x < 5 Simplify.

The solutions are all real numbers 0 1 2 3 4 5 6 7less than 5.

b. b 1 1 ≥ 28 Write original inequality.

b 1 1 2 1 ≥ 28 2 1 Subtract 1 from each side.

b ≥ 29 Simplify.

The solutions are all real numbers greater �9 �8 �7 �6 �5 �4�11 �10than or equal to 29.


1. p 1 1.2 > 4.5 2. m 2 2 5 }

8 ≤ 4

3 }

8 3. 26 < 27 1 k

4. 6 } 5 1 t ≥ 25 5. r 2 7.3 < 23.9 6. 14 1 m ≤ 54

2. Solve an Inequality Using Multiplication or Division (Lesson 6.2)


a. x }

2 ≥ 3 b. 25x > 15

Solution:

a. x }

2 ≥ 3 Write original inequality.

2 p x } 2 ≥ 3 p 2 Multiply each side by 2.

x ≥ 6 Simplify.

The solutions are all real numbers greater 1 2 3 4 5 6 7 8than or equal to 6.

EXAMPLE

PRACTICE

EXAMPLE

Check by substituting a number less than 5 for x in the original inequality. Try x 5 4:

x 2 3 < 24 2 3 < 2 1 < 2 ✔

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b. 25x > 15 Write original inequality.

25x

} 25 <

15 }

25 Divide each side by 25 and reverse inequality symbol.

x < 23 Simplify.

The solutions are all real numbers less �2 �1 0�3�4�5�6�7 than 23.


7. 16x < 4 8. 2 d } 3 ≤ 29 9. 7c ≥ 242

10. 26z > 54 11. w

} 5 < 2.3 12. n }

8 ≤ 25

3. Solve a Multi-Step Inequality (Lesson 6.3)Solve the inequality. Graph the solution.

a. 2c 1 1 ≤ 6c 2 9 b. 23(m 1 2) < 215

Solution:

a. 2c 1 1 ≤ 6c 2 9 Write original inequality.

1 ≤ 4c 2 9 Subtract 2c from both sides.

10 ≤ 4c Add 9 to both sides.

2.5 ≤ c Divide both sides by 4.

The solutions are all real numbers greater 0 1 2 3than or equal to 2.5.

b. 23(m + 2) < 215 Write original inequality.

23m 2 6 < 215 Distributive property

23m < 29 Add 6 to both sides.

m > 3 Divide both sides by 23 and reverse inequality symbol.

The solutions are all real numbers greater 0 1 2 3 4 5 6 7than 3.


13. 3x 2 2 > 6 1 x 14. 26t 1 10

} 3 ≤

2 }

3 t

15. 1.8q 1 3.4 < 2.1q 2 2.6 16. 1 }

2 (3v 1 1) ≤

7 }

2

17. 22n 2 4 ≤ 3(n 1 2) 18. 2 3 } 4 (8 1 12a) > 3a

EXAMPLE

PRACTICE

Here’s an alternate way to solve this example. Subtract 6c and 1 from both sides:

2c + 1 ≤ 6c – 9

–4c + 1 ≤ –9

–4c ≤ –10

c ≥ 2.5

c ≥ 2.5 and 2.5 ≤ c are equivalent inequalities.

PRACTICE

Multiplying or dividing each side of an inequality by a negative number and reversing the direction of the inequality produces an equivalent inequality.


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EXAMPLE

4. Solve a Compound Inequality With and (Lesson 6.4)Solve 3 � x 2 2 � 8. Graph the solution.

Solution:

3 ≤ x 2 2 and x 2 2 ≤ 8 Write as two separate inequalities.

3 1 2 ≤ x 2 2 + 2 and x 2 2 1 2 ≤ 8 1 2 Add 2 to each side.

5 ≤ x and x ≤ 10 Simplify.

The compound inequality can be written 4 5 6 7 8 9 10 115 ≤ x ≤ 10. The solutions are all real numbers

greater than or equal to 5 and less than equal to 10.

Solve 27 � 22x 1 1 � 5. Graph the solution.

Solution:

27 < 22x 1 1 < 5 Write original inequality.

27 2 1 < 22x 1 1 2 1 < 5 2 1 Subtract 1 from each expression.

28 < 22x < 4 Simplify.

28

} 22

> 22x

} 22

> 4 }

22 Divide each expression by 22. Reverse both

inequality symbols.

4 > x > 22 Simplify.

22 < x < 4 Rewrite in the form a < x < b.

The solutions are all real numbers greater �1 0 1 2 3 4�2 5than 22 and less than 4.


19. 5 < 4a 2 7 < 13 20. 24 ≤ 2x

} 3 ≤ 8 21. 11 ≤ 23p 1 2 ≤ 20

22. 27 < 6n + 5 ≤ 21 23. 5 < 2 5w

} 3 < 15 24. 27 < 22j 2 3 < 21

5. Solve a Compound Inequality With or (Lesson 6.4)Solve 4x 1 2 � 5 or 3x 2 8 � 1. Graph the solution.

Solution:

4x 1 2 < 5 or 3x 2 8 ≥ 1 Write as two separate inequalities.

4x 1 2 2 2 < 5 2 2 or 3x 2 8 1 8 ≥ 1 1 8 Addition or subtraction

property of inequality

4x < 3 or 3x ≥ 9 Simplify.

4x

} 4 < 3 }

4 or

3x }

3 ≥

9 }

3 Multiplication or division

property of inequality

x < 3 }

4 or x ≥ 3 Simplify.

The solutions are all real numbers less than 3 }

4 0 1 2 3

or greater than or equal to 3.

EXAMPLE

PRACTICE

You can also solve a compoundinequality with and by applying the same operation to each expression in the inequality.

If you get overlappinginequalities, such as x > 1 or x < 3, the solution is all real numbers.

You can solve a compound inequality with and by separating the inequality.

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25. 22a 1 3 ≥ 15 or 1 2 a < 4 26. 2n 2 5 ≥ 3 or 6 2 n > 9

27. 4y + 20 < 0 or y 1 3 > 2 28. k 1 7 ≤ 3 or 22k 1 5 < 2

29. b 2 2 ≥ 0 or b 1 4 ≤ 5 30. 2 j } 5 1 2 < 12 or j 1 40 < 25

CheckSolve the inequality. Graph the solution.

1. y 1 5.3 > 2.9 2. b 2 1 3 }

4 ≤ 5

1 }

4 3. 26 < 24 1 n

4. 4 }

3 1 k ≥ 22 5. d 2 1.9 < 25.7 6. 12 1 a ≤ 63

7. 15x < 30 8. 2 x } 7 ≤ 24 9. 8c ≥ 232

10. 23z > 72 11. y }

9 < 4.6 12. b }

12 ≤ 240

13. 3x 2 6 > 4 1 x 14. 25s 1 1 ≤ 1 }

2 s 15. 22.5p 1 0.9 < 4.3p 2 0.8

16. 2 }

3 (6r 1 9) ≤ 26 17. 211m 1 1 ≥ 2(m 2 6) 18. 2

2 } 5 (5 2 10z) > 3z

19. 6 < 3g 2 3 < 15 20. 26 ≤ 2x

} 7 ≤ 4 21. 11 ≤ 24a 1 7 ≤ 35

22. 24 < 6n + 2 ≤ 0 23. 8 ≤ 2 4t

} 7 < 24 24. 29 < 2j 2 19 < 25

25. x 1 4 > 14 or x 2 2 ≤ 4 26. 2x 1 6 ≤ 210 or x 1 10 > 9

27. 22x 2 8 < 2 or x 2 2 < 9 28. 1 }

2 x 2 7 < 6 or 32 2 x ≤ 0

29. x 2 3 > 13 or x 1 5 < 9 30. x 2 7 ≥ 23 or 23x 1 6 ≥ 23

PRACTICE


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F. Absolute Value Equations and Inequalities

The rules for solving absolute value equations and inequalities are different from those for solving linear equations and inequalities. You must isolate the absolute value expression on one side of the equation or inequality before you can solve it.

1. Solve an Absolute Value Equation (Lesson 6.5)Absolute value equation An equation that includes an absolute value expression.

Solve �x � 5 6.

Solution:

uxu 5 6 means “the distance between x and 0 on a number line is 6 units.” Only two numbers are 6 units from 0: 6 and 26.

uxu 5 6 0 2 4 6�2�4�6

x 5 6 or x 5 26

Solve the equation.

1. uxu 5 12 2. u yu 5 2.4 3. ubu 5 2 }

3

4. umu 5 44 5. uru 5 6 5 }

8 6. upu 5 0.9

2. Solve an Absolute Value Equation in Multiple Steps (Lesson 6.5)

Solve 4�3x 2 1� 2 6 5 10.

Solution:

Step 1: Rewrite the equation in the form ) ax 1 b ) 5 c.

4u3x 2 1u 2 6 5 10 Write original equation.

4u3x 2 1u 5 16 Add 6 to each side.

u3x 2 1u 5 4 Divide each side by 4.

Step 2: Solve the absolute value equation.

u3x 2 1u 5 4 Write absolute value equation.

3x 2 1 5 4 or 3x 2 1 5 24 Rewrite as two equations.

3x 5 5 or 3x 5 23 Add 1 to each side.

x 5 5 }

3 or x 5 21 Divide each side by 3.

Solve the equation.

7. 5ux 1 3u 1 4 5 19 8. 3u2c 2 4u 1 5 5 17 9. 1 }

2 u4p 2 7u 2 2 5 8

10. 2u3t 2 8u 2 5 5 29 11. 3u22z 1 5u 2 2 5 7 12. 3 }

2 ) 6 } 5 a 2 3 ) 2 2 5 1

EXAMPLE

Vocabulary

u3x 2 1u 5 4 and u3x u 5 5 are NOT equivalentstatements.

PRACTICE

EXAMPLE

PRACTICE

You cannot add 1 to both sides ofu3x 2 1u 5 4.


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3. Solve an Absolute Value Inequality (Lesson 6.6)Solve the inequality. Graph the solution.

a. u x 1 11u ≥ 4 b. u2x 2 6u 1 1 < 9

Solution:

a. u x 1 11u ≥ 4 Write original inequality.

x 1 11 ≤ 24 or x 1 11 ≥ 4 The inequality symbol is ≥. Rewrite as compound

inequality in form ax 1 b ≤ 2c or ax 1 b ≥ c.

x ≤ 215 or x ≥ 27 Subtract 11 from each side.

�8 �6 �4 �2�16 �14 �12 �10

b. u2x 2 6u 1 1 < 9 Write original inequality.

u2x 2 6u < 8 Subtract 1 from each side.

28 < 2x 2 6 < 8 The inequality symbol is <. Rewrite as compound

inequality in form 2c < ax 1 b < c.

22 < 2x < 14 Add 6 to each expression.

21 < x < 7 Divide each expression by 2.

0 2 4 6 8 10�2�4

Solve the equation. Graph the solution.

13. ux 2 7u < 2 14. 5u2a 2 4u ≥ 15 15. un 1 5u 2 8 > 3

16. 26 ) 3t 1 5 }

6 ) 2 4 ≤ 25 17.

2 }

3 u4q 1 1u 1 3 < 9 18. 6u25f 1 2u 1 4 ≤ 16

CheckSolve the equation.

1. uau 5 14 2. ucu 5 9.7 3. uxu 5 4 }

9

4. unu 5 38 5. usu 5 1 6 } 7 6. |q| 5 0.3

7. 2ux 1 4u 1 1 5 7 8. 6u3u 2 2u 1 5 5 41 9. 1 }

4 u2q 2 3u 2 6 5 4

10. 2u8t 1 7u 2 2 5 25 11. 4u24y 1 3u 2 9 5 15 12. 3 }

4 ) 2 } 5 r 2 6 ) 2 23 5 1

Solve the equation. Graph the solution.

13. u x 2 35u < 68 14. 4u7a 2 2u ≥ 20 15. ut 1 3u 2 7 > 53

16. 24u6t 1 4.5u 2 3 ≤ 25.4 17. 1 }

6 u3b 1 2u 1 4 < 8 18. 5u22k 1 1u 1 3 ≤ 18

EXAMPLE

The solution of an absolute value inequality with < or ≤ is values between a number and its opposite.

The solution of an absolute value inequality with > or ≥ is values beyond a number and its opposite.

PRACTICE


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A. Solving Linear Systems by GraphingA linear system can have zero, one, or infi nitely many solutions. One way to fi nd this solution is to graph each equation in the system. The point where the lines intersect is the solution to the system. If the graph of the equations are parallel lines, then no point of intersection exists, and the system has no solution. If each equation represents the same line, the system has infi nitely many solutions.

1. Checking Solutions to Linear Systems (Lesson 7.1)System of linear equations Two or more linear equations in the same variables; also called a linear system.

Tell whether the ordered pair is a solution of the linear system.

a. (2, 1); b. (0, 5);

2x 2 y 5 3 Equation 1 4x 2 2y 5 210 Equation 1

2x 1 3y 5 1 Equation 2 3x 1 y 5 15 Equation 2

Solution:

a. Substitute 2 for x and 1 for y in b. Substitute 0 for x and 5 for y in each equation. each equation.

2x 2 y 5 32(2) 2 (1) 0 3 4 2 1 0 3 3 5 3 ✔

2x 1 3y 5 122 1 3(1) 0 1 22 1 3 0 1 1 5 1 ✔

4x 2 2y 5 2104(0) 2 2(5) 0 210 0 2 10 0 210 210 5 210 ✔

3x 1 y 5 153(0) 1 5 0 15 0 1 5 0 15 5 5 15 ✗

The ordered pair (2, 1) is a solution The ordered pair (0, 5) is NOT the of each equation, so it is the solution solution of 3x 1 y 5 15, so it is NOT of the system. the solution of the system.


1. (4, 22); 2. (23, 3); 3. (21, 0); 3x 1 3y 5 6 x 1 2y 5 3 6x 1 4y 5 26 22x 1 4y 5 216 22x 2 3y 5 23 5x 2 2y 5 2

4. (3, 6); 5. (2, 28); 6. (1, 5); 2x 2 3y 5 12 10x 1 5y 5 220 4x 2 2y 5 26 2x 1 5y 5 27 23x 2 y 5 22 23x 1 4y 5 17

Vocabulary

EXAMPLE

For a linear system with one solution, the solution is an ordered pair that satisfi es each equation in the system.

PRACTICE

Ready to Go On? Chapters 7–8 Intervention

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2. Graphing Linear Systems (Lesson 7.1)

Use the graph to solve the system. Then check your solution algebraically.

x 2 y 5 23 Equation 1 x 1 y 5 1 Equation 2

Solution:

The lines seem to intersect at (21, 2). Check by

y

21 3 4

1

2

3

4

5

6

7

2121222324 x

x 1 y 5 1

x 2 y 5 23substituting 21 for x and 2 for y in each equation.

x 2 y 5 2321 2 2 0 23

23 5 23 ✔

x 1 y 5 121 1 2 0 1

1 5 1 ✔

The ordered pair (21, 2) is the solution of each equation. So, (21, 2) is the solution of the system.


7.

1

2

3

4

1 x21 2 3 4 5 6 7

5

6

7

8y

2x 2 4y 5 24

x 1 2y 5 10

8.

1

2

1 x23 22 21 2 3 4 5

3

4

5

64x 2 5y 5 23

23x 1 y 5 26

y

21

22

9. y

21 3 4

1

2

3

4

5

6

7

2121222324 x

x 2 3y 5 211

25x 1 2y 5 16

3. Graph-and-Check Method (Lesson 7.1)Consistent independent system A linear system with exactly one solution.

Solve the linear system: 3x 2 y 5 2 Equation 1

x 1 2y 5 10 Equation 2

EXAMPLE

PRACTICE

Vocabulary

EXAMPLE

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Solution:

Step 1: Graph both equations.

Step 2: Estimate the point of intersection. The two lines appear to intersect at (2, 4).

Step 3: Check whether (2, 4) is a solution by substituting 2 for x and 4 for y in each of the original equations.

Equation 13x 2 y 5 2

3(2) 2 4 0 2 2 5 2 ✔

Equation 2 x 1 2y 5 10

2 1 2(4) 0 10 10 5 10 ✔

Because (2, 4) is a solution of each equation, it is a solution of the linear system.

Solve the linear system by graphing. Check your solution.

10. 3x 1 y 5 5 11. 4x 2 2y 5 210 12. 22x 1 y 5 29 5x 2 2y 5 12 2x 1 y 5 1 x 2 3y 5 12

13. 2x 2 y 5 22 14. 6x 2 4y 5 4 15. 2x 1 2y 5 6 23x 1 3y 5 9 x 2 y 5 0 2x 1 2y 5 29

4. Special Types of Linear Systems (Lesson 7.5)Inconsistent system A linear system with no solution; the graphs of the equations are parallel.

Consistent dependent system A linear system with infi nitely many solutions; the graphs of the equations are the same line.

Show that the linear system has no solution. 2x 2 3y 5 23 Equation 1

2x 2 3y 5 15 Equation 2

Solution:

You can use either of two methods to solve the problem.

Method 1: Graphing

Graph the linear system.

The lines have the same slope, but different y-intercepts. So, the equations represent two parallel lines, which do not intersect. The system has no solution.

PRACTICE

Vocabulary

EXAMPLE

y

21 3 4

1

2

3

4

5

6

7

2121222324 x

y 5 2 x 1 512

y 5 3x 2 2

x1212223 2 3 4 521

22

23

24

25

26

1

2y

2x 2 3y 5 15

2x 2 3y 5 23Lines that do not intersect are said to be inconsistent. So, a linear system with no solutions is an inconsistent system.


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Method 2: Elimination

Subtract the equations.

2x 2 3y 5 23 2x 2 3y 5 23

2 (2x 2 3y 5 15) 22x 1 3y 5 215

0 5 218 ✗ This is a false statement.

Subtracting the equations leads to a false statement, so the system has no solution.

Show that the linear system has infi nitely many solutions.

y 5 3 }

4 x 22 Equation 1


Solution:

You can use either of two methods to solve the problem.

Method 1: Graphing

Graph the linear system.

Both equations represent the same line. So, all points on the line are solutions. The system has infi nitely many solutions.

Method 2: Substitution

Substitute 3 }

4 x 22 for y in Equation 2 and solve

for x.

3x 2 4y 5 8 Write Equation 2.

3x 2 4 1 3 } 4 x 22 2 5 8 Substitute

3 } 4 x � 2 for y.

3x 2 3x 1 8 5 8 Simplify.

8 5 8 Simplify.

Substitution leads to a statement that is always true. The system has infi nitely many solutions.

Tell whether the linear system has no solution or infi nitely many solutions. Explain.

16. x 2 4y 5 220 17. y 5 2 3 } 4 x 2 6 18. 3x 2 2y 5 28

x 2 4y 5 8 3x 1 4y 5 224 23x 1 2y 5 26

19. 3x 1 1 }

3 y 5 1 20. y 5 25x 1 3 21. 14y 2 6x 5 228

9x 1 y 5 26 10x 1 2y 5 6 3x 2 7y 5 14

Lines that intersect are said to be consistent, and equations that are equivalent are said to be dependent. So, a linear system in which all the equations represent the same line is a consistent, dependent system.

PRACTICE

x1212223 2 3 4 521

22

23

24

25

26

1

2y

3x 2 4y 5 8

y 5 x 2 234

EXAMPLE


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Check


1. (23, 2); 2. (27, 25); 3. (9, 210); x 2 2y 5 27 x 1 y 5 212 2x 1 y 5 8 3x 2 2y 5 213 2x 2 4y 5 234 23x 1 4y 5 267


4.

x 1 21 22 23 2 3 4 5 21

22

23

24

25

26

1

2 y

5x 1 4y 5 6

x 1 3y 5 21

5.

1

2

3

x2122232425 1 2 321

22

23

25

24

y2x 2 3y 5 0

3x 1 y 5 211

6.

1

2

1 x23 22 21 2 3 4 5

3

4

5

6y

21

22

y 5 22x 1 9

y 5 x 213

13

Solve the linear system. If the system has one solution, check the solution. If the system has no solution or infi nitely many solutions, explain.

7. x 1 4y 5 2 8. 3x 1 y 5 24 9. 8x 2 6y 5 22 3x 2 5y 5 6 y 5 23x 1 1 x 1 3y 5 24

10. x 2 4y 5 23 11. 2x 1 y 5 1 12. 2y 5 2x 1 8 22x 1 6y 5 2 5x 1 3y 5 0 2x 1 4y 5 16


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B. Solving Linear Systems Using AlgebraA system of linear equations is two or more linear equations with the same variables. The solution of a system of linear equations is an ordered pair that satisfi es each equation in the system. The following examples describe methods which can be used to fi nd the solution of a system of two linear equations.

1. Solve a Linear System by Substitution (Lesson 7.2)Substitution Replacing a variable in an equation by an expression or a value.

Solve the linear system:

2x 2 3 5 y Equation 1


Solution:

Step 1: Solve for a variable in one of the equations.

y 5 2x 2 3 Equation 1 is already solved for y.

Step 2: Substitute the expressions from Step 1 into the other equation and solve.


3x 2 2(2x 2 3) 5 4 Substitute 2x � 3 for y into Equation 2.

3x 2 4x 1 6 5 4 Simplify.

2x 5 22 Combine like terms.

x 5 2 Multiply each side by �1.

Step 3: Use the value found in Step 2 to fi nd the value of the other variable.

y 5 2x 2 3 5 2(2) 2 3 5 1 Substitute x � 2 into Equation 1.

The solution is (2, 1).

Use the substitution method to solve the linear system.

1. y 5 2x 1 5 2. x 5 y 1 3 3. 2x 1 4y 5 0 3x 1 y 5 10 x 1 2y 5 26 3x 1 7 5 y

4. x 1 2y 5 25 5. 3x 2 2y 5 4 6. 3x 1 y 5 22 4x 2 3y 5 2 x 1 3y 5 5 x 1 3y 5 2

2. Solve a Linear System by Adding or Subtracting (Lesson 7.3)

Elimination Performing operations on and combining two or more equations in a system in such a way that one of the variables is no longer present.

Vocabulary

EXAMPLE

Don’t stop after solving for one variable. The system is not solved until a solution is found for both variables.

PRACTICE

Vocabulary


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2y 2 2x 5 24 Equation 2

Solution:

Step 1: Add the equations to eliminate one variable and solve.

2x 1 3y 5 9 Add Equation 1 and Equation 2.

1 22x 1 2y 5 24

5y 5 5

y 5 1 Solve for y.



2x 1 3(1) 5 9 Substitute y � 1 into Equation 1.

2x 5 6 Simplify.

x 5 3 Divide each side by 2.




9 5 2x 1 3y Equation 2

Solution:

Step 1: Subtract the equations to eliminate one variable and solve.

4x 1 3y 5 3 Subtract Equation 1 and Equation 2.

2 2x 1 3y 5 9

2x 5 26

x 5 23 Solve for x.



4(23) 1 3y 5 3 Substitute x � �3 into Equation 1.

3y 5 15 Simplify.

y 5 5 Divide each side by 3.


EXAMPLE

Check solutions by substituting them into the original equations to see if the resulting statements are true.

When subtracting equations, make sure each term of the second equation is subtracted from each like term in the fi rst equation.

EXAMPLE


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Use the method of adding or subtracting equations to solve the linear system.

7. 2x 5 3y 8. 4x 1 3y 5 8 9. 5x 2 2y 5 20 x 1 3y 5 9 2y 1 4x 5 28 3x 1 4 5 22y

10. 3y 2 4x 5 3 11. 3x 2 7 5 5y 12. 22x 5 y 2 3 2x 1 3y 5 215 3x 1 4y 5 211 7x 1 y 5 3

3. Solve a Linear System by Multiplying First (Lesson 7.4)




Solution:

Step 1: Modify the original system by multiplying both equations by a constant. Use the least common multiple of the coeffi cients for one of the variables to determine the constants.

22x 1 4y 5 2 (3 3) → 26x 1 12y 5 6 The least common

25x 1 6y 5 23 (3 22) → 10x 2 12y 5 6 multiple of 4 and 6 is 12,

so multiply Equation 1 by

3 to get 12y and Equation

2 by �2 to get �12y.

Step 2: Add the modifi ed equations from Step 1 and solve.

26x 1 12y 5 6 Add the modifi ed equations.

1 10x 2 12y 5 6

4x 5 12

x 5 3 Solve for x.



22(3) 1 4y 5 2 Substitute x � 3 into Equation 1.

4y 5 8 Simplify.

y 5 2 Divide each side by 4.


Multiply one or both equations by constants to solve the linear system.

13. 2x 1 12 5 2y 14. 5x 1 2y 5 14 15. 5y 2 4x 5 21 3x 1 y 5 210 2x 2 3y 5 22 6x 2 7y 5 3

16. 3y 2 2x 5 210 17. 6y 2 5x 5 8 18. 3x 2 2y 5 23 4x 1 7y 5 26 7x 2 8y 5 212 9x 2 4y 5 3

EXAMPLE

PRACTICE

PRACTICE


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Check

Use the substitution method to fi nd the solution to the set of equations.

1. x 1 3y 5 10 2. 5x 5 y 2 1 3. 5y 2 2x 5 24 2x 2 y 5 6 3x 1 9 5 2y x 2 3y 5 3

Use the elimination method to fi nd the solution to the set of equations.

4. 4x 5 3y 2 6 5. 2x 1 5y 5 3 6. 4y 2 5x 5 3 x 1 3y 5 29 2x 1 7y 5 5 3x 5 4y 2 5

7. 2x 2 3y 5 0 8. 4y 2 3x 5 21 9. 3x 1 2y 5 2 3x 2 4y 5 1 5x 2 2y 5 23 6x 1 5y 5 8

For each linear system, name the method which is most appropriate for solving the system: substitution, adding equations, subtracting equations, or multiplying fi rst. Then solve the system.

10. 2y 5 5x 1 3 11. 4x 5 3y 2 4 12. 5x 1 6y 5 24 23x 1 2y 5 5 3x 2 y 5 2 4x 1 3y 5 25

13. You have 50 tickets to ride the Ferris wheel and the roller coaster. If you ride 12 times total, using 3 tickets for each Ferris wheel ride and 5 tickets for each roller coaster ride, how many times did you go on each ride?


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C. Solving Systems of Linear Inequalities One way to solve a system of inequalities is to graph each inequality. The graph of the system is the intersection of these graphs.

1. Graphing Systems of Two Linear Inequalities (Lesson 7.6)System of linear inequalities Two or more linear inequalities in the same variables; also called a system of inequalities.

Graph the system of inequalities.

y ≤ x 1 3 Inequality 1

y > 22x 2 1 Inequality 2

Solution:

Graph both inequalities in the same coordinate plane. The graph of the system is the intersection of the two half-planes, which is shown as the darkest gray.


1. y > 4x 1 1 2. y ≥ x 2 4 3. x 2 2y < 24 y < 2x 2 2 y ≤ 23x 1 4 y ≥ 22

4. y ≤ 2x 2 2 5. x > 23 6. y > 1 2x 1 3y < 1 x ≥ 21 x ≤ 4

2. Graphing Systems of Three Linear Inequalities (Lesson 7.6)


y ≤ 23x 1 2 Inequality 1

x > 21 Inequality 2

x ≤ 2 Inequality 3

Solution:

Graph all three inequalities in the same coordinate plane. The graph of the system is the shaded region shown.


7. y ≥ 22x 2 1 8. y ≥ x 9. y < x 1 4 y ≥ 2x 2 1 x < 4 4x 1 y < 4 y > 4 y > 21 y ≥ 24

10. y < 2 11. 2x 1 3y ≤ 22 12. x < 23 y > 2x y ≥ 22x 2 8 x < 0 y ≤ 2x 1 3 y > 3x 1 6 y ≥ 0

Vocabulary

EXAMPLERecall that inequalities with < or > symbols are graphed with dashed lines, while inequalities with ≤ or ≥ symbols are graphed with solid lines.

PRACTICE

EXAMPLE

1 x 2324 22 21 2 3 4

y

21

22

23

24

1

2

3

4

Choose a point in the shaded region and substitute it in each inequality. The solution checks if each substitution results in a true statement.

1 x 23 24 22 21 2 3 4

y

21

22

23

24

1

2

3

4

PRACTICE


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3. Write a System of Linear Inequalities (Lesson 7.6)

Write a system of inequalities for the shaded region.

Solution:

Inequality 1 One boundary line for the shaded region is y = 5. Because the shaded region is below the solid line, the inequality is y ≤ 5.

Inequality 2 Another boundary line for the shaded region has a slope of 1 and a y-intercept of 3. So, its equation is y = x 1 3. Because the shaded region is below the dashed line, the inequality is y < x 1 3.

The system of inequalities for the shaded region is:

y ≤ 5 Inequality 1

y < x 1 3 Inequality 2


13.

1 x 23 24 22 21 2 3 4

y

21

22

23

24

1

2

3

4 14.

1 x 23 24 22 21 2 3 4

y

21

22

23

24

1

2

3

4

15.

1 x 2324 22 21 2 3 4

y

21

22

23

24

1

2

3

4

EXAMPLE

Find the equation of the lines from the slope and y-intercept, from two points, or from a point and the slope.

PRACTICE

1 x 2324 22 21 2 3 4

y

21

22

1

2

3

4 5

6


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Check


1. x ≤ 25 2. x 1 2y > 21 3. x 1 y ≥ 2 2 y ≥ 21 y > 23 x 1 y > 26 4x 1 y < 2 4

4. y ≥ 22 5. x > 1 6. y < 25x 1 3 y ≤ 1 x > 2 y ≤ 24x 2 2 x ≥ 0 x > 3 4x 2 3y < 23


7.

1 x2324 22 21 2 3 4

y

21

22

23

24

25

26

1

2 8.

x23 24 22 21 1 2 3 4

y

21

22

23

24

1

23

4

9.

x

y

2

46

8

2628 24 22

24

26

28

224 6 82


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D. Properties of ExponentsA power represents repeated multiplication. The following examples illustrate several properties that you can use to simplify expressions with powers.

1. Product of Powers Property (Lesson 8.1)Product of Powers Property To multiply powers having the same base, add the exponents.

Simplify the expression.

a. 42 p 48 b. 3 p 37 p 36 c. (212)(212)4 d. x2 p x10 p x5

Solution:

a. 42 p 48 b. 3 p 37 p 36 c. (212)(212)4 d. x2 p x10 p x5

5 42 + 8 5 31 p 37 p 36 5 (212)1 p (212)4 5 x2 + 10 + 5

5 410 5 31 + 7 + 6 5 (212)1 + 4 5 x17

5 314 5 (212)5


1. 87 p 82 2. 53 p 5 p 59 3. (26)7(26)4(26)

4. d10 p d7 5. r5 p r7 p r8 6. k2 p k3 p k

2. Power of a Power Property (Lesson 8.1)Power of a Power Property To fi nd a power of a power, multiply exponents.


a. (32)4 b. [(24)9]2 c. (x3)3 d. [(y 23)5]4

Solution:

a. (32)4 5 32 p 4 b. [(24)9]2 c. (x3)3 5 x3 p 3 d. [(y 23)5]4

5 38 5 (24)9 p 2 5 x9 5 (y 2 3)5 p 4

5 (24)18 5 (y 2 3)20


7. (103)6 8. [(26)2]8 9. (t5)3

10. (b4)4 11. [(p + 5)7]2 12. [(h 2 1)9]5

Vocabulary

EXAMPLE

If a is a real number and m and n are positive integers,am p an � am�n.

PRACTICE

Vocabulary

EXAMPLE

If a is a real number and m and n are positive integers, (am)n � amn.

PRACTICE


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3. Power of a Product Property (Lesson 8.1)Power of a Product Property To fi nd a power of a product, fi nd the power of each factor and multiply.


a. (9 p 7)3 b. (5xy)2 c. (23z)4 d. 2(3z)4

Solution:

a. (9 p 7)3 b. (5xy)2 c. (23z)4 d. 2(3z)4

5 93 p 73 5 (5 p x p y)2 5 (23 p z)4 5 2(3 p z)4

5 52 p x2 p y2 5 (23)4 p z4 5 2(34 p z4) 5 25x2y2 5 81z4 5 281z4


13. (5 p 4)6 14. (3gh)3 15. (6cd)2

16. (22p)4 17. 2(5t)3 18. 2(28a)2

4. Quotient of Powers Property (Lesson 8.2)Quotient of Powers Property To divide powers having the same base, subtract exponents.


a. 97

} 93 b.

(26)10

} (26)8 c.

39 p 35

} 34 d.

1 }

x5 p x12

Solution:

a. 97

} 93 b.

(26)10

} (26)8 c.

39 p 35

} 34 5

314

} 34 d.

1 }

x5 p x12 5 x12

} x5

5 97 2 3 5 (26)10 2 8 5 314 2 4 5 x12 2 5

5 94 5 (26)2 5 310 5 x7


19. 1823

} 1817 20.

235

} 23 21.

(225)3

} (225)

22. 46 p 49

} 43 23.

1 }

n8 p n12 24. w6 p 1 } w4

If a and b are real numbers and m is a positive integer,(ab)m � ambm.

If a is a nonzero real number and m and n are positive integers such that m > n, a

m } an � am�n, a � 0.

Vocabulary

EXAMPLE

PRACTICE

Vocabulary

EXAMPLE

PRACTICE

Evaluate the numerical power when simplifying powers with both numerical and variable bases.


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5. Power of a Quotient Property (Lesson 8.2)Power of a Quotient Property To fi nd a power of a quotient, fi nd the power of the numerator and the power of the denominator and divide.


a. 1 a } b 2 7 b. 1 2

2 } x 2 3

Solution:

a. 1 a } b 2 7 5

a7

} b7 b. 1 2 2 } x 2 3 5 1 22

} x 2 3 5 (22)3

} x3 5 2

8 }

x3


25. 1 2 } 5 2 3 26. 1 2 5 } 8 2 2 27. 1 1 } p 2 8

28. 1 r } s 2 6 29. 1 2

u } v 2 9 30. 1 2

2 }

b 2 5

6. Zero and Negative Exponents (Lesson 8.3)Zero power a to the zero power is 1.Negative exponents a2n is the reciprocal of an, an is the reciprocal of a2n.


a. 523 b. (212)0 c. 1 1 } 2 2 25

d. 024

Solution:

a. 523 5 1 }

53 b. (212)0 5 1 c. 1 1 } 2 2 25

5 1 }

1 1 } 2 2 5

d. 024 5 1 }

04

5 1 }

125 5

1 }

1 }

32

Division by 0

is undefi ned.

5 32


31. 1122 32. 1 5 } 8 2 0 33. 1 1 }

4 2 23

34. (23)24 35. 1 }

622 36. (225)0

Vocabulary

EXAMPLE

PRACTICE

Vocabulary

EXAMPLE

PRACTICE

If a and b are real numbers and m is a positive integer,

1 a } b

2 m � a

m }

bm , b � 0.

If a is a real number (a � 0) and n is an integer,a0 � 1,a�n � 1 } an , and

an � 1 } a�n .


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7. Fractional Exponents (Lesson 8.3)


a. 16 1 } 2 b. 125

2 1 } 3

Solution:

a. 16 1 } 2 5 Ï

} 16 b. 125

2 1 } 3 5

1 }

125 1 } 3

5 4 5 1 }

3 Ï}

125

5 1 } 5


37. 9 1 }

2 38. 27 1 }

3 39. 362 1 }

2

Check


1. 74 p 7 p 75 2. (28)3(28)9(28) 3. y12 p y8 p y9

4. (65)3 5. [(24)3]7 6. (s6)2

7. [(a + 11)8]4 8. (8 p 2)13 9. (25jk)7

10. 39 p 36

} 32 11.

1 }

v8 p v17 12. 1 2 3 } 4 2 3

13. 1 4 } h 2 2 14. 1 2

p } q 2 15

15. 1 2 } 7 2 0

16. (22)25 17. 1 }

722 18. (213)0

19. 64 1 } 2 20. 64

1 } 3 21. 49

2 1 } 2

PRACTICE

EXAMPLE


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E. Scientifi c NotationAny real number can be written in standard form or in scientifi c notation. A number written in scientifi c notation is in the form c 3 10n, where c is greater than or equal to 1 and less than 10, and n is an integer. For example, the number two million can be written in standard form as 2,000,000, or in scientifi c notation as 2 3 106. The number fi ve thousandths can be written in standard form as 0.005, or in scientifi c notation as 5 3 1023.

1. Writing Numbers in Scientifi c Notation (Lesson 8.4)

Write each number in scientifi c notation.

a. 0.00000317 b. 8,356,000,000

Solution:

a. Move the decimal point 6 places to b. Move the decimal point 9 places to the right. The exponent is 26. the left. The exponent is 9.

0.00000317 5 3.17 3 1026 8,356,000,000 5 8.356 3 109

Write each number in standard form.

a. 1.982 3 107 b. 8.04 3 1025

Solution:

a. The exponent is 7. Move the b. The exponent is 25. Move the decimal point 7 places to decimal point 5 places to the right. the left.

1.982 3 107 5 19,820,000 8.04 3 1025 5 0.0000804

Write the number in scientifi c notation.

1. 42,635,000,000,000 2. 0.0000000000582 3. 73,009

Write the number in standard form.

4. 7.81 3 1012 5. 5.403 3 1028 6. 6.034 3 102

2. Computing with Scientifi c Notation (Lesson 8.4)

Evaluate the expression. Write the answer in scientifi c notation.

a. (4.3 3 105)(2.8 3 109) b. 4.5 3 107

} 7.5 3 1024 c. (9.2 3 1026)4

For a positive number written in scientifi c notation, c � 10n:if n < 0, the number is between 0 and 1,if n ≥ 0, the number is greater than 1.

EXAMPLE

EXAMPLE

PRACTICE

EXAMPLE


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Solution:

a. (4.3 3 105)(2.8 3 109)

5 (4.3 p 2.8) 3 (105 p 109) Commutative property and

associative property

5 12.04 3 1014 Product of powers property

5 (1.204 3 101) 3 1014 Write 12.04 in scientifi c notation.

5 1.204 3 (101 3 1014) Associative property


b. 4.5 3 107

} 7.5 3 1024 5

4.5 } 7.5 3

107

} 1024 Product rule for fractions

5 0.6 3 1011 Quotient of powers property

5 (6.0 3 1021) 3 1011 Write 0.6 in scientifi c notation.



c. (9.2 3 1026)4 5 (9.2)4 3 (1026)4 Power of a product property

5 7163.9296 3 10224 Power of a power property

5 (7.1639296 3 103) 3 10224 Write in scientifi c notation.




7. (3.6 3 106)(8.14 3 105) 8. (4.9 3 1015)3 9. (6.02 3 1023)(3.1 3 102)

10. 5.6 3 107

} 7 3 1025 11.

1.5 3 1022

} 7.5 3 1012 12. (2.6 3 1024)25

Check

If the number is in standard form, write it in scientifi c notation. If the number is in scientifi c notation, write it in standard form.

1. 5.026 3 1028 2. 0.00002707 3. 618,090,300,000,000

4. 0.001001001001001 5. 3.11 3 1014 6. 9.45824 3 1023


7. (7.2 3 1026)(5.3 3 1010) 8. (1.6 3 1012)2 9. (3.14 3 1025)(8.2 3 103)

10. 2.1 3 1015

} 2.4 3 108 11.

2.56 3 10212

} 6.4 3 109 12. (8.8 3 106)3

PRACTICE


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F. Exponential FunctionsA function in the form y 5 abx, where a ≠ 0, b > 0, and b ≠ 1, is an exponential function. If a is greater than 0 and b is greater than 1, the function represents exponential growth. If a is greater than 0 and b is between 0 and 1, the function represents exponential decay. The graph of an exponential function is a curved line.

1. Graph an Exponential Growth Function (Lesson 8.5)Exponential growth Each time the value of x increases by 1, the value of y is multiplied by a constant amount b, where b > 1.

Graph the function y � 3x. Identify its domain and range.

Solution:

Step 1: Choose several values for x and fi nd the values of y. The domain is all real numbers.

x 22 21 0 1 2

y 1 }

9

1 }

3 1 3 9

Step 2: Plot the points.

Step 3: Draw a smooth curve through the points. The range is all positive real numbers.

Graph the function. Identify its domain and range.

1. y 5 5x 2. y 5 6x 3. y 5 (3.1)x

4. y 5 (2.7)x 5. y 5 1 3 } 2 2

x

6. y 5 1 6 } 5 2

x

2. Use an Exponential Growth Model (Lesson 8.5)Exponential growth model The equation y 5 a(1 1 r)t is a model of exponential growth where a is the initial value, r is the growth rate, 1 1 r is the growth factor, and t is the time period.

In 1999, Tom’s Cafe paid $12,500 for electricity. The cafe’s electricity costs since then have increased at a rate of 2.4% per year.

a. Write a function that models the cafe’s electricity costs over time.

b. How much did the cafe pay for electricity in 2004? Round your answer to the nearest whole dollar.

Vocabulary

EXAMPLE

PRACTICE

Vocabulary

EXAMPLE

1 x2324 22 21 2 3 4

y

22

24

2

46

8

10

12

y � 3x


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Solution:

a. Let E be the cost of electricity, and let t be the time (in years) since 1999. The initial value a is $12,500, and the growth rate r is 0.024.

E 5 a(1 1 r)t Write exponential growth model.

5 12,500(1 1 0.024)t Substitute 12,500 for a and 0.024 for r.

512,500(1.024)t Simplify.

b. 2004 is 5 years after 1999. Substitute 5 for t.

E 5 12,500(1.024)5 Substitute 5 for t.

ø 14,074 Use a calculator.

In 2004, Tom’s Cafe paid about $14,074 for electricity.

Solve.

7. The population of Hapsville was 245,000 in 1995. The population grows exponentially at 1.8% per year.

a. Write a function that models the population of Hapsville over time.

b. What was the population of Hapsville in 2005? Round your answer to the nearest whole number.

8. A newly-discovered plant grows in a lab at a rate of 4.3% per month. On March 1, the plant was 3.75 inches tall.

a. Write a function that models the plant’s height over time.

b. How tall is the plant by December 1? Round your answer to the nearest hundredth of an inch.

3. Graph an Exponential Decay Function (Lesson 8.6)Exponential decay Each time the value of x increases by 1, the value of y is multiplied by a constant amount b, where 0 < b < 1.

Graph the function y � (0.625)x. Identify its domain and range.

Solution:

Step 1: Choose several values for x and fi nd the values of y. The domain is all real numbers.

x 23 22 21 0 1

y 4.096 2.56 1.6 1 0.625


Step 3: Draw a smooth curve through the points. The range is all positive real numbers.

1 x2324 22 21 2 3 4

y

21

22

1

23

4

5

6

y � (0.625)x

PRACTICE

Vocabulary

EXAMPLE


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9. y 5 1 2 } 5 2 x 10. y 5 (0.125)x 11. y 5 1 5 } 6 2 x

12. y 5 (0.9)x 13. y 5 1 10 }

11 2 x 14. y 5 (0.3)x

4. Use an Exponential Decay Model (Lesson 8.6)Exponential decay model The equation y 5 a(1 2 r)t is a model of exponential decay where a is the initial value, r is the decay rate, 1 2 r is the decay factor, and t is the time period.

In 1999, 355 customers ate at Tom’s Café every day, on average. Since then, the average number of customers who eat there each day has decreased at a rate of 4% per year.

a. Write a function that models the average daily number of customers who eat at the cafe over time.

b. How many customers ate at Tom’s café each day, on average, in 2003? Round your answer to the nearest whole number.

Solution:

a. Let C be the average number of customers who eat at Tom’s Café each day, and let t be the time (in years) since 1999. The initial value a is 355, and the decay rate r is 0.04.

W 5 a(1 2 r)t Write exponential decay model.

5 355(1 2 0.04)t Substitute 355 for a and 0.04 for r.

5 355(0.96)t Simplify.

b. 2003 is 4 years after 1999. Substitute 4 for t.

W 5 355(0.96)4 Substitute 4 for t.

ø 302 Use a calculator.

In 2003, about 302 customers ate at the café each day.

Solve.

15. An essay contest starts with 95 entries. In each round of judging, the judges eliminate 25% of the essays.

a. Write a function that models the number of essays in the contest over time.

b. How many essays are still in the contest after 5 rounds of judging? Round your answer to the nearest whole number.

16. An oven is heated to 475°F and then turned off. The temperature of the oven decreases 8% per minute.

a. Write a function that models the oven’s temperature over time.

b. What is the temperature of the oven after 15 minutes? Round your answer to the nearest whole degree.

PRACTICE

Vocabulary

EXAMPLE

PRACTICE


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1. y 5 10x 2. y 5 1 1 } 7 2 x 3. y 5 1 5 } 3 2 x

4. y 5 (0.27)x 5. y 5 (4.1)x 6. y 5 (0.72)x

Solve.

7. In 1960, a museum bought a statue worth $5000. The statue’s value has increased at a rate of 5.8% per year.

a. Write a function that models the value of the statue over time.

b. What was the value of the statue in 2005? Round your answer to the nearest whole dollar.

8. A chef makes an ice sculpture with 45 pounds of ice. As the sculpture melts, the weight of the remaining ice decreases by 2.5% per minute.

a. Write a function that models the weight of the remaining ice over time.

b. What is the weight of the remaining ice after 120 minutes? Round your answer to the nearest whole pound.


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A. Adding, Subtracting, and Multiplying Polynomials

Real numbers can be represented by polynomials. Polynomials can be added, subtracted, and multiplied. The result of any of these operations on polynomials is another polynomial.

1. Identify and Classify Polynomials (Lesson 9.1)Monomial A number, a variable, or the product of a number and one or more variables with whole number exponents.

Degree of a monomial The greatest sum of the exponents of the variables in a monomial.

Polynomial A monomial or a sum of monomials, each called a term.

Degree of a polynomial The greatest sum of the exponents of the variables in a monomial.

Binomial A polynomial with two terms.

Trinomial A polynomial with three terms.

Tell whether the expression is a polynomial. If it is a polynomial, fi nd its degree and classify it by the number of its terms. Otherwise, tell why it is not a polynomial.

Expression Is it a polynomial?Classify by degree and

number of terms

a. 7 Yes 0 degree monomial

b. 3x3 1 2x Yes 3rd degree binomial

c. 4d x 2 d 2 1 3 No; variable exponent

d. 2x24 1 3x2 No; negative exponent

e. 6m3n 2 5mn2 2 1 Yes 4th degree trinomial

Tell whether the expression is a polynomial. If it is a polynomial, fi nd its degree and classify it by the number of its terms. Otherwise, tell why it is not a polynomial.

1. 5x 2. 4xa 2 3x 1 2 3. 2z4 1 5z2

4. 2ab4 1 4a2b 2 1 5. 3y23 2 y2 2 4 6. 4m3 2 5m2 1 m 2 7

2. Add and Subtract Polynomials (Lesson 9.1)Find the sum.

a. (4x3 2 3x2 1 2) 1 (x2 1 2x 2 3) b. (4x2 1 3x 2 4) 1 (x2 2 2x 1 1)

Solution:

a. Vertical format: Align the terms in vertical columns.

4x3 2 3x2 1 2

1 x2 1 2x 2 3

4x3 2 2x2 1 2x 2 1

Vocabulary

EXAMPLE

PRACTICE

EXAMPLE

An exponent of a variable in a polynomial must be a whole number.

When a power of the variable appears in one polynomial but not the other, leave a space in that column, or write the term with a coeffi cient of 0 as a placeholder.

Ready to Go On? Chapters 9 –10 Intervention

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b. Horizontal format: Group like terms and simplify.

(4x2 1 3x 2 4) 1 (x2 2 2x 1 1) 5 (4x2 1 x2) 1 [3x 1 (22x)] 1 (24 1 1)

5 5x2 1 x 2 3


a. (3x2 1 2) 2 (2x2 1 3x 2 1) b. (2x2 2 4x 1 4) 2 (x2 2 x 1 1)

Solution:

a. 3x2 1 2

(2) 2x2 1 3x 2 1

3x2 1 2

1 x2 2 3x 1 1

4x2 2 3x 1 3

b. (2x2 2 4x 1 4) 2 (x2 2 x 1 1) 5 2x2 2 4x 1 4 2 x2 1 x 2 1

5 (2x2 2 x2) 1 (24x 1 x) 1 (4 2 1)

5 x2 2 3x 1 3

Find the sum or difference.

7. (2x2 1 3x 2 1) 1 (x3 2 4x 1 3) 8. (x2 1 2) 2 (4x2 1 5x 2 3)

9. (3x4 2 3x2 1 5) 1 (2x3 2 x2 1 2x 2 3) 10. (5x3 1 2x2 2 x 1 1) 2 (x3 2 x2 2 2)

11. (4x2 1 x 2 6) 2 (2x2 2 3x 1 2) 12. (4x2 1 3x 2 4) 1 (x2 2 2x 1 1)

3. Multiply Polynomials (Lesson 9.2)Find the product.

a. 3x2 (2x3 1 x2 2 3) b. (2a2 2 a 2 6)(5a 1 2) c. (y2 2 3y 1 4)(2y 2 5)

Solution:

a. 3x2 (2x3 1 x2 2 3) Write product.

5 3x2 (2x3) 1 3x2 (x2) 2 3x2(3) Distributive property

5 6x5 1 3x4 2 9x2 Product of powers property

b. Step 1: Multiply by 2. 2a2 2 a 2 6

3 5a 1 2

4a2 2 2a 2 12

Step 2: Multiply by 5a. 2a2 2 a 2 6

3 5a 1 2

4a2 2 2a 2 12

10a3 2 5a2 2 30a

EXAMPLE

PRACTICE

EXAMPLE

To subtract a polynomial, add its opposite. Write the subtraction as addition, and multiply each term in the polynomial by 21.

Align like terms vertically to help you add correctly.


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Step 3: Add products. 2a2 2 a 2 6

3 5a 1 2

4a2 2 2a 2 12

10a3 2 5a2 2 30a

10a3 2 a2 2 32a 2 12

c. (y2 2 3y 1 4)(2y 2 5) Write product.

5 y2(2y 2 5) 2 3y(2y 2 5) 1 4(2y 2 5) Distributive property

5 2y3 2 5y2 2 6y2 1 15y 1 8y 2 20 Distributive property

5 2y3 2 11y2 1 23y 2 20 Combine like terms.

Find the product.

13. 4b3(2b2 1 b 2 3) 14. 2x4(2x3 2 5x2 2 x 1 6)

15. (y2 2 3y 1 5)(2y 2 1) 16. (2z 2 3)(z2 1 4z 2 1)

17. (a2 2 5a 2 2)(2a 2 3) 18. (3x 1 2)(2x2 2 3x 1 4)

4. Find Special Products of Polynomials (Lesson 9.3)Find the product.

a. (2x 1 5)2 b. (3y 2 2)2 c. (2x 1 y)(2x 2 y)

Solution:

a. (2x 1 5)2 5 (2x)2 1 2(2x)(5) 1 (5)2 Square of a binomial pattern

5 4x2 1 20x 1 25 Simplify.

b. (3y 2 2)2 5 (3y)2 2 2(3y)(2) 1 (2)2 Square of a binomial pattern

5 9y2 2 12y 1 4 Simplify.

c. (2x 1 y)(2x 2 y) 5 (2x)2 2 y2 Sum and difference pattern

5 4x2 2 y2 Simplify.

Find the product.

19. (2m 1 5)2 20. (z 2 7)(z 1 7) 21. (3x 2 4)2

22. (2x 2 4)(2x 1 4) 23. (2s 1 t)2 24. (5x 1 2)(5x 2 2)

PRACTICE

EXAMPLE

PRACTICE

(a 1 b)2 5 a2 1 2ab 1 b2

(a 2 b)2 5 a2 2 2ab 1 b2

(a 1 b)(a 2 b)5 a2 2 b2


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CheckTell whether the expression is a polynomial. If it is a polynomial, fi nd its degree and classify it by the number of its terms. Otherwise, tell why it is not a polynomial.

1. 26x22 2. 3x 1 1 3. x3 1 2x2 2 5x 1 4

4. x3 2 2x2 1 xn 2 7 5. 3x3 y2 1 2x2 y 1 xy 6. 4x2 2 3x 1 1


7. (x2 1 6x 2 3) 1 (2x3 2 2x 1 5) 8. (5x2 1 2x 2 1) 2 (3x2 2 4x 1 2)

9. (x2 1 3) 2 (23x2 2 2x 1 6) 10. (2x2 1 x 2 2) 1 (x2 2 5x 2 4)

11. 3z2(2z3 2 4z 1 5) 12. (5x 1 1)(3x2 2 x 2 7)

13. 4b3(b3 2 2b2 2 1) 14. (3y 2 4)(3y 1 4)

15. (k 2 9)2 16. (2x 2 3)(x2 1 5x 2 2)

17. (2p 1 3r)(2p 2 3r) 18. (5x 1 1)(2x 2 3)

19. (3y 1 5)2


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B. Factoring PolynomialsPolynomial equations can often be solved by factoring. To factor a polynomial, write it as a product of other polynomials. Then use the zero-product property to fi nd the roots of the equation. The following examples describe methods that can be used to factor polynomial equations and fi nd their solutions.

1. Use the Zero-Product Property (Lesson 9.4)Greatest common factor (GCF) of a polynomial A monomial with an integer coeffi cient that divides evenly into each term of the polynomial.

Zero-Product Property Let a and b be real numbers. If ab 5 0, then a 5 0 or b 5 0.

Roots The solutions of an equation in which one side is zero and the other side is a product of polynomial factors.

Solve (x 2 3)(x 1 1) 5 0.

(x 2 3)(x 1 1) 5 0 Write original equation.

x 2 3 5 0 or x 1 1 5 0 Zero-product property

x 5 3 or x 5 22 Solve for x.

Use the zero-product property to solve the equation.

1. (x 2 2)(x 1 6) 5 0 2. (x 2 1)(x 2 8) 5 0 3. (x 1 3)(x 1 4) 5 0

2. Solve an Equation by Factoring (Lesson 9.4)Solve the equation.

a. 3x2 2 6x 5 0 b. 5a2 5 20a

Solution:

a. 3x2 2 6x 5 0 Write original equation.

3x(x 2 2) 5 0 Factor left side.

3x 5 0 or x 2 2 5 0 Zero-product property


b. 15a2 5 20a Write original equation.

15a2 2 20a 5 0 Subtract 20a from each side.

5a(3a 2 4) 5 0 Factor left side.

5a 5 0 or 3a 2 4 5 0 Zero-product property

a 5 0 or a 5 4 }

3 Solve for a.

Solve the equation by factoring.

4. 4b2 2 16b 5 0 5. x2 1 7x 5 0 6. 8m2 5 26m

7. 5z2 5 7z 8. 6a2 2 3a 5 0 9. 10x2 5 2x

Vocabulary

EXAMPLE

PRACTICE

EXAMPLE

Write the equation so that one side is 0 in order to use the zero-product property.

PRACTICE


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3. Factor x2 � bx � c (Lesson 9.5)Factor the polynomial.

a. x2 1 8x 1 12 b. t2 2 7t 1 10 c. z2 2 5z 2 14

Solution:

a. Factors of 12 Sum of factors

12, 1 12 1 1 5 13 ✗

6, 2 6 1 2 5 8 Correct sum

4, 3 4 1 3 5 7 ✗

The factors 6 and 2 have a sum of 8, so they are the correct values of p and q.

x2 1 8x 112 5 (x 1 6)(x 1 2)

b. Factors of 10 Sum of factors

210, 21 210 1 (21) 5 211 ✗

25, 22 25 1 (22) 5 27 Correct sum

t2 27t 1 10 5 (x 2 5)(x 2 2)

c. Factors of 214 Sum of factors

214, 1 214 1 1 5 213 ✗

14, 21 14 1 (21) 5 13 ✗

27, 2 27 1 2 5 25 Correct sum

7, 22 7 1 (22) 5 5 ✗

z2 2 5z 2 14 5 (x 2 7)(x 1 2)

Factor the polynomial.

10. z2 1 7z 1 12 11. n2 2 8n 1 7 12. m2 2 2m 2 24

13. y2 2 5y 1 6 14. t2 1 2t 2 15 15. x2 1 6x 1 5

4. Factor ax2 1 bx 1 c (Lesson 9.6)Factor the polynomial.

a. 3x2 2 5x 1 2 b. 2t2 1 t 2 3 c. 25m2 1 9m 1 2

Solution:

a. Factors of 3

Factors of 2

Possible factorization

Middle term when multiplied

1, 3 21, 22 (x 2 1)(3x 2 2) 22x 2 3x 5 25x Correct

1, 3 22, 21 (x 2 2)(3x 2 1) 2x 2 6x 5 27x ✗

3x2 2 5x 1 2 5 (x 2 1)(3x 2 2)

EXAMPLE

PRACTICE

EXAMPLE

To fi nd two positive factors of 12 whose sum is 8, make an organized list.

Both p and q must be negative, since b is negative and c is positive.

Since c is negative, p and q must have different signs.

When a is positive, consider the signs of b and c. Since b is negative and c is positive, both factors must be negative.

x 2 1 bx 1 c 5 (x 1 p)(x 1 q)if p 1 q 5 b and

pq 5 c.


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b. Factors of 2

Factors of 23



1, 2 1, 23 (t 1 1)(2t 2 3) 23t 1 2t 5 2t ✗

1, 2 21, 3 (t 2 1)(2t 1 3) 3t 2 2t 5 t Correct

1, 2 3, 21 (t 1 3)(2t 2 1) 2t 1 6t 5 5t ✗

1, 2 23, 1 (t 2 3)(2t 1 1) t 2 6t 5 25t ✗

2t2 1 t 2 3 5 (t 2 1)(2t 1 3)

c. 25m2 1 9m 1 2 5 2(5m2 2 9m 2 2)

Factors of 5

Factors of 22



1, 5 1, 22 (m 1 1)(5m 2 2) 22m 1 5m 5 3m ✗

1, 5 2, 21 (m 1 2)(5m 2 1) 2m 1 10m 5 9m ✗

1, 5 21, 2 (m 2 1)(5m 1 2) 2m 2 5m 5 23m ✗

1, 5 22, 1 (m 2 2)(5m 1 1) m 2 10m 5 29m Correct

25m2 1 9m 1 2 5 2(m 2 2)(5m 1 1)


16. 4x2 2 8x 2 5 17. 2y2 2 11y 1 9 18. 24z2 1 4z 1 15

19. 3m2 2 10m 1 8 20. 22x2 1 17x 2 21 21. 3t2 1 5t 2 12

5. Factor Special Products (Lesson 9.7)Factor the polynomial.

a. 4x2 2 25 5 0 b. 4m2 2 12m 1 9 c. 9z2 1 30z 1 25

Solution:

a. 4x2 2 25 5 (2x)2 2 52 Write as a2 2 b2.

5 (2x 1 5)(2x 2 5) Difference of two squares pattern

b. 4m2 2 12m 1 9 5 (2m)2 2 2(2m p 3) 1 32 Write as a2 2 2ab 1 b2.

5 (2m 2 3)2 Perfect square trinomial pattern

c. 9z2 1 30z 1 25 5 (3z)2 1 2(3z p 5) 1 52 Write as a2 1 2ab 1 b2.

5 (3z 1 5)2 Perfect square trinomial pattern


22. 9x2 2 1 23. 4s2 1 4s 1 1 24. 4m2 2 12mn 1 9n2

25. 16t2 2 9 26. 25z2 2 20z 1 4 27. x2 1 8xy 1 16y2

PRACTICE

EXAMPLE

PRACTICE

Since b is positive and c is negative, the factors of c must have different signs.

When a is negative, fi rst factor 21 from each term of the trinomial. Then factor as in the previous examples.

Remember to include the 21 that you factored out earlier.

Difference of two squares: a2 2 b2 5 (a 1 b)(a 2 b)

Perfect square trinomial:a2 1 2ab 1 b2

5 (a 1 b)2

a2 2 2ab 1 b2

5 (a 2 b)2


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6. Factor by Grouping (Lesson 9.8)Factor by grouping In a polynomial of four terms, factor a common monomial from pairs of terms, then look for a common binomial factor.


a. m3 1 4m2 2 3m 2 12 b. s2 1 2st 2 2s 2 4t

Solution:

a. m3 1 4m2 2 3m 2 12 5 (m3 1 4m2) 1 (23m 2 12) Group terms.

5 m2(m 1 4) 1 (23)(m 1 4) Factor each group.

5 (m2 2 3)(m 1 4) Distributive property

b. s2 1 2st 2 2s 2 4t 5 (s2 1 2st) 1 (22s 2 4t) Group terms.

5 s(s 1 2t) 1 (22)(s 1 2t) Factor each group.

5 (s 2 2)(s 1 2t) Distributive property


28. 2t3 1 3t2 2 10t 2 15 29. 2m2 2 10m 1 mn 2 5n

30. 3x3 2 21x2 1 x 2 7 31. x2 2 4x 1 2xy 2 8y

CheckFactor the polynomial.

1. 4y3 2 20y 2. 10m4 1 15m2 3. t2 2 7t 2 18

4. x2 1 6x 2 27 5. 4s2 2 27s 2 7 6. 10y2 1 9y 1 2

7. 16x4 2 y2 8. 25s2 2 4t2 9. m2 2 6mn 1 9n2

10. 4x2 1 24xy 1 36y2 11. x2 1 4xy 2 3x 2 12y 12. t2 1 3st 2 2t 2 6s

Factor and use the zero-product property to solve the equation.

13. 2x2 1 6x 5 0 14. r2 1 r 2 20 5 0 15. y2 2 6y 5 16

16. 3x2 5 5 2 14x 17. 4m2 2 8m 1 3 5 0 18. z2 5 64

19. 9n2 2 4 5 0 20. 9x2 1 30x 1 25 5 0 21. 4z2 1 1 5 4z

PRACTICE

EXAMPLE

Vocabulary

Check your factorization by multiplying the factors. Or graph the original polynomial and the factors on a graphing calculator. The two graphs should coincide.


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C. Graphing Quadratic FunctionsA quadratic function is a nonlinear function that can be written in the standard form y 5 ax2 1 bx 1 c where a Þ 0. The following examples describe how to graph quadratic functions, fi rst where b 5 0 and c 5 0, then where b 5 0, and fi nally where all three coeffi cients are nonzero. The examples also describe how to fi nd certain characteristics of graphs of quadratic functions.

1. Graph y 5 ax 2 (Lesson 10.1)Parent quadratic function The most basic quadratic

21

1

2

3

4

5

6

7y

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vertex (0, 0)

axis ofsymmetry

y 5 x2

function, y 5 x2 as shown.

Parabola The U-shaped graph of a quadratic function.

Vertex The lowest or highest point on a parabola.

Axis of symmetry The line that passes through aparabola’s vertex and divides the parabola into twosymmetric parts.

Graph the function. Compare the graph with the graph of y 5 x2.

a. y 5 2x2 b. y 5 2 1 } 5 x2

Solution:

a. Step 1: Make a table of values for y 5 2x 2.

1

2

3

4

5

6

7

8y

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y 5 x2y 5 2x2

x 22 21 0 1 2

y 8 2 0 2 8

Step 2: Plot the points from the table.

Step 3: Draw a smooth curve through the points.

Step 4: Compare the graphs of y 5 2x2 and y 5 x2. Both graphs open up and have the same vertex, (0, 0), and axis of symmetry, x 5 0. The graph of y 5 2x2 is narrower than the graph of y 5 x2 because the graph of y 5 2x2 is a vertical stretch (by a factor of 2) of the graph of y 5 x2.

b. Step 1: Make a table of values for y 5 2 1 } 5 x2.

1

2

3

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y 5 x2

y 5 2 x215

21

22

23

24

x 25 22 0 2 5

y 25 20.8 0 20.8 25



Vocabulary

EXAMPLE

Plot additional points if you are having trouble seeing the shape of the parabola with just fi ve points.

Choose values of x that result in whole numbers for y, whenever possible.


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Step 4: Compare the graphs of y 5 2 1 } 5 x2 and y 5 x2. Both graphs open up and

have the same vertex, (0, 0), and axis of symmetry, x 5 0. However, the

graph of y 5 2 1 } 5 x2 is wider than the graph of y 5 x2 and it opens down.

This is because the graph of y 5 2 1 } 5 x2 is a vertical shrink (by a factor

of 1 } 5 ) with a refl ection in the x-axis of the graph of y 5 x2.


1. y 5 4x2 2. y 5 2 1 } 3 x2 3. y 5

1 }

2 x2

2. Graph y 5 x2 1 c (Lesson 10.2)Graph y 5 x2 1 3. Compare the graph with the graph of y 5 x2.

Step 1: Make a table of values for y 5 x2 1 3.

y 5 x2 1 3

1

2

3

4

5

6

7

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y 5 x2

x 22 21 0 1 2

y 7 4 3 4 7



Step 4: Compare the graphs of y 5 x2 1 3 and y 5 x2. Both graphs open up and have the same axis of symmetry, x 5 0. However, the vertex of y 5 x2 1 3, (0, 3), is different than the vertex of the graph of y 5 x2, (0, 0), because the graph of y 5 x2 1 3 is a vertical translation (of 3 units up) of the graph of y 5 x2.


4. y 5 x2 1 1 5. y 5 x2 2 2 6. y 5 x2 1 4

3. Find the Axis of Symmetry and the Vertex (Lesson 10.2)Consider the function y 5 2x2 2 8x 1 1.

a. Find the axis of symmetry of the graph of the function.

b. Find the vertex of the graph of the function.

Solution:

a. For the function y 5 2x2 2 8x 1 1, a 5 2 and b 5 28.

x 5 2 b } 2a 5 2

(28) }

2(2) 5 2 Substitute 2 for a and 28 for b. Then simplify.

b. The x-coordinate of the vertex is 2. To fi nd the y-coordinate, substitute 2 for xin the function and fi nd y.

y 5 2(2)2 2 8(2) 1 1 5 27 Substitute 2 for x. Then simplify.

The vertex is (2, 27).

PRACTICE

EXAMPLE

PRACTICE

EXAMPLE

For the quadratic function y 5 ax 2 1 bx 1 c, the axis of symmetry is x 5 2

b } 2a , and the

y-coordinate of the vertex is

a 1 2 b

} 2a 2 2 1b 1 2

b } 2a 2 1 c

Draw the x-axis and y-axis so that you have room to graph the points in your table.

A parabola opens up when the coeffi cient of x 2 is positive and opens down when the coeffi cient of x 2 is negative.


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Find the axis of symmetry and the vertex of the graph of the function.

7. y 5 x2 1 2x 1 9 8. 2x2 2 16x 1 23

9. 23x2 1 12x 2 8 10. 2x2 2 4x 2 7

11. 4x2 2 24x 1 39 12. 22x2 1 6x 1 5

4. Graph y 5 ax2 1 bx 1 c (Lesson 10.2)Graph y 5 x2 2 4x 1 3.

Step 1: Determine whether the parabola opens up or down. Because a > 0, the parabola opens up.

Step 2: Find and draw the axis of symmetry: x 5 2 b } 2a 5 2

24 }

2(1) 5 2.

Step 3: Find and plot the vertex. The

x 5 1 axisof symmetry

1

2

3

4

5

6y

1 2 x2122 3 4 5 621

22

(4, 3)(0, 3)

(1, 0) (3, 0)

vertex(2, 21)

x-coordinate of the vertex is 2. To fi nd the y-coordinate,substitute 2 for x in the functionand simplify. y 5 (2)2 2 4(2) 13 5 21. So, the vertex is (2, 21).

Step 4: Plot two points. Choose two x-values less than the x-coordinate of the vertex. Then fi nd the corresponding y-values.

x 1 0

y 0 3

Step 5: Refl ect the points plotted in Step 4 in the axis of symmetry.

Step 6: Draw a parabola through the plotted points.

Graph the function.

13. y 5 2x2 2 8x 1 3

14. y 5 23x2 1 6x 2 4

15. y 5 2x2 2 10x 2 20

5. Find the Minimum or Maximum Value (Lesson 10.2)Minimum value The y-coordinate of the vertex of the function y 5 ax2 1 bx 1 c when a > 0.

Maximum value The y-coordinate of the vertex of the function y 5 ax2 1 bx 1 c when a < 0.

EXAMPLE

Choose the location of the x-axis and y-axis so that you can easily graph the position in your table.

PRACTICE

Vocabulary

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Tell whether the function f (x) 5 2x 2 2 4x 1 5 has a minimum value or a maximum value. Then fi nd the minimum or maximum value.

Because a 5 2 and 2 > 0, the parabola opens up and the function has a minimum value. To fi nd the minimum value, fi nd the vertex.

x 5 2 b } 2a 5 2

24 }

2(2) 5 1 The x-coordinate is 2 b } 2a .

ƒ(1) 5 2(1)2 2 4(1) 1 5 5 3 Substitute 1 for x. Then simplify.

The minimum value of the function is ƒ(1) 5 3.

Tell whether the function has a minimum value or a maximum value. Then fi nd the minimum or maximum value.

16. ƒ(x) 5 2x2 1 4x 2 1 17. ƒ(x) 5 2x2 2 6

18. ƒ(x) 5 3x2 1 4 19. ƒ(x) 5 2 1 } 2 x2 1 2x 1 3

20. ƒ(x) 5 25x2 1 10x 2 7 21. ƒ(x) 5 4x2 1 8x

CheckGraph the function. Label the vertex and the axis of symmetry.

1. y 5 23x2 2. y 5 1 }

2 x2 2 2 3. y 5 2x2 2 4x 1 5

4. y 5 x2 2 4x 5. y 5 22x2 2 8x 2 5 6. y 5 2x2 2 4x 2 7

Tell whether the function has a minimum value or a maximum value. Then fi nd the minimum or maximum value.

7. ƒ(x) 5 x2 1 7 8. ƒ(x) 5 23x2 2 6

9. ƒ(x) 5 2x2 2 x 10. ƒ(x) 5 1 }

2 x2 2 4x 1 1

11. ƒ(x) 5 22x2 1 6x 1 1 12. ƒ(x) 5 24x2 2 8x 1 3

EXAMPLE


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D. Solving Quadratic EquationsA quadratic equation is an equation that can be written in the standard form ax2 1 bx 1 c 5 0 where a Þ 0. In Chapter 9 you used factoring to solve a quadratic equation. The following examples describe several additional methods you can use for solving quadratic equations.

1. Solve a Quadratic Equation by Graphing (Lesson 10.3)Solutions, or roots, of a quadratic equation The real numbers that make the equation true.

Solve the quadratic equation by graphing.

a. x2 1 3x 5 4 b. 2x2 2 2x 5 1 c. x2 1 5 5 x

Solution:

a. Step 1: Write the equation in standard form. 1

2y

x21222324252621

22

23

24

25

26

21

x2 1 3x 5 4 Write original equation.

x2 1 3x 2 4 5 0 Subtract 4 from each side.

Step 2: Graph the function y 5 x2 1 3x 2 4. The x-intercepts are 24 and 1.

The solutions of the equation x2 1 3x 5 4 are 24 and 1.

Check:

You can check 24 and 1 in the original equation.

x2 1 3x 5 4 x2 1 3x 5 4 Write original equation.

(24)2 1 3(24) � 4 12 1 3(1) � 4 Substitute for x.

4 5 4 ✔ 4 5 4 ✔ Simplify. Each solution checks.

b. Step 1: Write the equation in standard form. y

x21222324252621

22

23

24

25

26

27

28

21 2x2 2 2x 5 1 Write original equation.

2x2 2 2x 2 1 5 0 Subtract 1 from each side.

Step 2: Graph the function y 5 2x2 2 2x 2 1.The x-intercept is 21.

The solution of the equation 2x2 2 2x 5 1 is 21.

c. Step 1: Write the equation in standard form.

1

2

3

4

5

6

7

8y

1 x21222324 2 3 4

x2 1 5 5 x Write original equation.

x2 2 x 1 5 5 0 Subtract x from each side.

Step 2: Graph the function y 5 x2 2 x 1 5.The graph has no x-intercepts.

The equation x2 1 5 5 x has no solution.

Vocabulary

EXAMPLE

The x-intercepts of the related function y 5 ax 2 1 bx 1 c are the solutions, or roots, ofax 2 1 bx 1 c 5 0.

Although the graph has a y-intercept, only x-intercepts are solutions.


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Solve the quadratic equation by graphing.

1. 2x2 1 2 5 x 2. x2 1 5x 5 28 3. x2 1 4x 1 4 5 0

4. 6x 2 9 5 x2 5. 2x2 13x 5 4 6. x2 1 x 5 6

2. Solve a Quadratic Equation Using Square Roots (Lesson 10.4)

Solve the equation.

a. 22x2 5 218 b. c2 1 5 5 5 c. t2 1 6 5 2

Solution:

a. 22x2 5 218 Write original equation.

x2 5 9 Divide each side by 22.

x 5 6 Ï}

9 5 63 Take square roots of each side. Simplify.

b. c2 1 5 5 5 Write original equation.

c2 5 0 Subtract 5 from each side.

c 5 0 The square root of 0 is 0.

c. t2 1 6 5 2 Write original equation.

t2 5 24 Subtract 6 from each side.

There is no solution.

Use square roots to solve the equation.

7. y2 1 12 5 12 8. 23t2 5 227 9. m2 1 10 5 4

10. x2 2 2 5 211 11. x2 2 5 5 25 12. 2z2 5 32

3. Solve a Quadratic Equation by Completing the Square (Lesson 10.5)

Completing the square The process of adding a constant c to an expression of the form x2 1 bx so that the expression x2 1 bx 1 c is a perfect square trinomial.

Solve x 2 2 10x 5 11 by completing the square.

x2 2 10x 5 11 Write original equation.

x2 2 10x 1 (25)2 5 11 1 (25)2 Add 1 210 } 2 2

2, or (25)2 to each side.

(x 2 5)2 5 11 1 (25)2 Write left side as the square of a binomial.

(x 2 5)2 5 36 Simplify the right side.

x 2 5 = 66 Take square roots of each side.

x 5 5 6 6 Add 5 to each side.

The solutions of the equation are 5 1 6 5 11 and 5 2 6 5 21.

Solve the quadratic equation by completing the square.

13. x2 1 4x 5 12 14. x2 1 2x 5 3 15. x2 2 2x 5 35

16. x2 1 12x 5 232 17. x2 2 6x 5 27 18. x2 1 16x 5 239

EXAMPLE

Negative real numbers do not have real square roots.

PRACTICE

Vocabulary

EXAMPLE

x 2 1 bx 1 1 b } 2 2

2 5

1 x 1 b } 2 2

2

PRACTICE

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4. Solve a Quadratic Equation Using the Quadratic Formula (Lesson 10.6)

The Quadratic Formula The solutions of the quadratic equation.

ax2 1 bx 1 c 5 0 are x 5 2b 6 Ï

}

b2 2 4ac }} 2a when a Þ 0 and b2 2 4ac ≥ 0.

Solve x2 2 5 5 2x.

x2 2 5 5 2x Write original equation.

x2 2 2x 2 5 5 0 Write in standard form.

x 5 2b 6 Ï

}

b2 2 4ac }}

2a Quadratic formula

x 5 2(22) 6 Ï

}}

(22)2 2 4(1)(25) }}}

2(1) Substitute values in the quadratic formula: a 5 1,

b 5 22, and c 5 25.

x 5 2 6 Ï

}

24 }

2 5 1 6 Ï

}

6 Simplify.

The solutions are 1 1 Ï}

6 < 3.45 and 1 2 Ï}

6 < 21.45.

Use the quadratic formula to solve the equation. Round your solutions to the nearest hundredth, if necessary.

19. x2 1 4x 5 7 20. 3x2 5 2x 1 5 21. 2x2 1 3x 5 1

Tell what method(s) you would use to solve the quadratic equation. Explain your choice(s).

a. x2 5 8x b. 6x2 2 22 5 0 c. 3x2 2 4x 2 7 5 0

Solution:

a. The equation can be solved by factoring because the expression x2 2 8x can be factored easily. Also, the equation can be solved by completing the square because the equation is of the form ax2 1 bx 1 c 5 0 where a 5 1 and b is an even number.

b. The quadratic equation can be solved using square roots because the equation can be written in the form x2 5 d.

c. The quadratic equation cannot be factored easily, and completing the square will result in many fractions. So, the equation can be solved using the quadratic formula.

Tell what method(s) you would use to solve the quadratic equation. Explain your choice(s).

22. 2x2 2 x 5 6 23. x2 2 6x 5 0 24. 4x2 1 2x 2 9 5 0

25. 23x2 5 254 26. x2 2 3x 5 6 27. x2 2 4x 5 3

Vocabulary

EXAMPLE

PRACTICE

EXAMPLE

Check your answer by graphing the related function on a graphing calculator. The x-intercepts should match your solutions.

The quadratic formula can be used for any quadratic equation.

PRACTICE


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5. Use the Discriminant to Find the Number of Solutions (Lesson 10.7)

Discriminant In the quadratic formula, the expression b2 2 4ac is the discriminant of the associated equation ax2 1 bx 1 c 5 0.

Use the discriminant to determine the number of solutions of the equation.

a. 9x2 1 6x 1 1 5 0 b. x2 2 2x 1 3 5 0 c. 2x2 2 3x 2 7 5 0

Solution:

Equation Discriminant Number of solutions

ax2 1 bx 1 c 5 0 b2 2 4ac

a. 9x2 1 6x 1 1 5 0 62 2 4(9)(1) 5 0 One solution

b. x2 2 2x 1 3 (22)2 2 4(1)(3) 5 28 No solution

c. 2x2 2 3x 2 7 5 0 (23)2 2 4(2)(27) 5 65 Two solutions

Tell whether the equation has two solutions, one solution, or no solution.

28. 3x2 2 5x 2 1 5 0 29. 5x2 1 2x 1 3 5 0 30. 2x2 2 7x 1 2 5 0

31. 16x2 2 8x 1 1 5 0 32. 4x2 2 x 1 5 5 0 33. 9x2 1 30x 1 25 5 0

CheckSolve the quadratic equation by graphing.

1. x2 1 16 5 8x 2. x2 1 4x 5 25 3. x2 5 x 1 12

Use square roots to solve the equation.

4. t2 1 5 5 9 5. m2 1 8 5 0 6. y2 1 7 5 7

Solve the quadratic equation by completing the square.

7. x2 2 8x 5 20 8. x2 1 12x 5 13 9. x2 2 14x 5 224

Use the quadratic formula to solve the equation. Round your solutions to the nearest hundredth, if necessary.

10. 3x2 1 1 5 5x 11. 2x2 1 2x 5 5 12. x2 2 2x 2 7 5 0

Tell whether the equation has two solutions, one solution, or no solution.

13. 2x2 2 3x 1 2 5 0 14. 13x2 1 x 2 5 5 0 15. 4x2 2 4x 1 1 5 0

b2 2 4ac > 0

Two solutions

b2 2 4ac 5 0

One solution

b2 2 4ac < 0

No solution


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The linear, exponential, and quadratic functions you have studied can be used to model data. The following examples describe methods for determining which type of function best models a set of ordered pairs.

1. Choose Functions Using Sets of Ordered Pairs (Lesson 10.8)

Use a graph to tell whether the ordered pairs represent a linear function, an exponential function, or a quadratic function.

a. (24, 10), (22, 4), (0, 2), (2, 4), (4, 10) b. (24, 6), (22, 4), (0, 2), (2, 0), (4, 22)

c. 1 22, 1 }

27 2 , 1 21,

1 }

9 2 , 1 0,

1 }

3 2 , (1, 1), (2, 3)

Solution:

a.

2

4

6

8

10

12y

2 x22242628 4 6 822

24

b.

1

2

3

4

y

1 x21222324 2 3 421

22

5

6

Quadratic function Linear function

c.

1

2

y

x21222324 1 2

21

3

Exponential function

Use a graph to tell whether the ordered pairs represent a linear function, an exponential function, or a quadratic function.

1. (22, 21), (21, 0), (0, 1), (1, 2), (2, 3)

2. 1 22, 3 }

8 2 , 1 21,

3 }

4 2 , 1 0,

3 }

2 2 , (1, 3), (2, 6)

3. (22, 3), 1 21, 3 }

2 2 , (0, 1), 1 1,

3 }

2 2 , (2, 3)

Draw a smooth curve through the points you plot.

Linear function: y 5 mx 1 b

Exponential function: y 5 ab x

Quadratic function: y 5 ax 2 1 bx 1 c

EXAMPLE

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2. Identify Functions Using Differences or Ratios (Lesson 10.8)

Use differences or ratios to tell whether the table of values represents a linear function, an exponential function, or a quadratic function.

a. x 21 0 1 2 3

y 0.0625 0.25 1 4 16

b. x 21 0 1 2 3

y 1 3 5 7 9

Ratios: 0.25

} 0.0625

5 4 4 4 4 Differences: 2 2 2 2

Exponential function Linear functionc.

x 21 0 1 2 3

y 4 2 4 10 20 First differences: 22 2 6 10 Second differences: 4 4 4 The table of values represents a quadratic function.


4. x 21 0 1 2 3

y 23 0 3 6 9

5. x 21 0 1 2 3

y 7 4 3 4 7

6. x 21 0 1 2 3

y 0.5 2 8 32 128

3. Write an Equation for the Function (Lesson 10.8)Tell whether the table of values represents a linear function, an exponential function, or a quadratic function. Then write an equation for the function.

x 22 21 0 1 2

y 16 4 0 4 16

Step 1: Determine which type of function the table of values represents.

x 22 21 0 1 2

y 16 4 0 4 16

First differences: 212 24 4 12 Second differences: 8 8 8

The table of values represents a quadratic function because the second differences are equal.

PRACTICE

EXAMPLE

The table of values represents:

• an exponential function if the ratios of successive y-values are all equal

• a linear function if the differences of successive y-values are all equal

• a quadratic function if the second differences are all equal

EXAMPLE

Check your function by plotting the ordered pairs from the table on the same grid as a graph of the function. The graph should pass through the plotted points.


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Step 2: Write an equation for the quadratic function. The equation has the form y 5 ax2. Find the value of a by using the coordinates of a point that lies on the graph, such as (1, 4).

y 5 ax2 Write equation for quadratic function.

4 5 a(1)2 Substitute 1 for x and 4 for y.

4 5 a Solve for a.

The equation is y 5 4x2.

Tell whether the table of values represents a linear function, an exponential function, or a quadratic function. Then write an equation for the function.

7. x 21 0 1 2 3

y 0.2 0 0.2 0.8 1.8

8. x 22 21 0 1 2

y 28 25 22 1 4

9. x 21 0 1 2 3

y 1 3 5 7 9

10. x 22 21 0 1 2

y 12 3 0 3 12

CheckUse a graph to tell whether the ordered pairs represent a linear function, an exponential function, or a quadratic function.

1. (24, 9), (22, 3), (0, 1), (2, 3), (4, 9)

2. (24, 210), (22, 24), (0, 2), (2, 8), (4, 14)

3. (24, 0.0625), (22, 0.25), (0, 1), (2, 4), (4, 16)

PRACTICE


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4. x 21 0 1 2 3

y 23 2 7 12 17

5. x 21 0 1 2 3

y 1 } 9

1 }

3 1 3 9

6. x 21 0 1 2 3

y 6 3 4 9 18

Tell whether the table of values represents a linear function, an exponential function, or a quadratic function. Then write an equation for the function.

7. x 21 0 1 2 3

y 23 21 1 3 5

8. x 22 21 0 1 2

y 16 4 0 4 16

9. x 21 0 1 2 3

y 0.1 0 0.1 0.4 0.9

10. x 22 21 0 1 2

y 26 22 2 6 10


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A. Radical FunctionsA radical expression is an expression that contains a radical, such as a square root, cube root, or other root. A radical function contains a radical expression with the independent variable in the radicand. If the radical is a square root, then the function is called a square root function. In the following examples, you will graph radical functions, simplify radical expressions, and solve radical equations.

1. Graph y � a � � x , y � a � � x � k , and y � a � � x � h (Lesson 11.1)

Parent square root function The most basic square root function, y 5 Ï} x .

y

x1 2 3 4 6 7 85

y 5 x�

(1, 1)1

2

3

4

5

6

7

8

Graph the function and identify its domain and range. Compare the graph with the graph of y � � � x .

a. y 5 4 Ï} x b. y 5 20.7 Ï}

x

c. y 5 Ï} x 1 4 d. y 5 Ï

} x 2 5

Solution:

Step 1: Make a table.

a. x 0 1 2 3 4

y 0 4 5.7 6.9 8

The domain is x ≥ 0.

b. x 0 1 2 3 4

y 0 20.7 21.0 21.2 21.4


c. x 0 1 2 3 4

y 4 5 5.4 5.7 6


d. x 5 6 7 8 9

y 0 1 1.4 1.7 2


Vocabulary

EXAMPLE

To fi nd the domain, fi nd the values of x for which the radicand is nonnegative.


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a. y

x1 2 3 4 6 7 85

1

2

3

4

5

6

7

8

y 5 x�

y 5 4 x�

b. y

x1 2 3 4 6 7 85

1

2

3

4

y 5 x�

y 5 20.7 x�

21

22

23

24

c. y

x1 2 3 4 6 7 85

1

2

3

4

5

6

7

8

y 5 x�

y 5 x 1 4�

d. y

x1 2 3 4 6 7 85

y 5 x�

y 5 x 2 5�1

2

3

4

5

6

7

8


Step 4: Compare the graph with the graph of y 5 Ï} x .

a. The range is y ≥ 0. The graph of y 5 4 Ï} x is a vertical stretch (by a factor of 4)

of the graph of y 5 Ï} x .

b. The range is y ≤ 0. The graph of y 5 20.7 Ï} x is a vertical shrink (by a factor of

0.7) with a refl ection in the x-axis of the graph of y 5 Ï} x .

c. The range is y ≥ 4. The graph of y 5 Ï} x 1 4 is a vertical translation

(of 4 units up) of the graph of y 5 Ï} x .

d. The range is y ≥ 0. The graph of y 5 Ï}

x 2 5 is a horizontal translation (of 5 units to the right) of the graph of y 5 Ï}

x .

Graph the function and identify its domain and range. Compare the graph with the graph of y � � � x .

1. y 5 5 Ï} x 2. y 5 20.8 Ï}

x 3. y 5 Ï} x 1 1 4. y 5 Ï

}

x 2 3

2. Use Properties of Radicals (Lesson 11.2)Radical expression in simplest form A radical expression with no perfect square factors (other than 1) in the radicand, no fractions in the radicand, and no radicals in the denominator of a fraction.

Product Property of Radicals Ï}

ab 5 Ï} a p Ï

}

b , where a ≥ 0 and b ≥ 0.

Quotient Property of Radicals Ï}

a }

b 5

Ï} a }

Ï}

b , where a ≥ 0 and b > 0.

You can determine the range by looking at either the table or the graph of the function.

PRACTICE

Vocabulary

Plot additional points if you are having trouble seeing the shape of the curve with just fi ve points.


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Rationalizing the denominator the process of eliminating a radical from an expression’s denominator

Use the product property of radicals.

a. Ï}

75 5 Ï}

25 p 3 Factor using perfect square factor.

5 Ï}

25 p Ï}

3 Product property of radicals

5 5 Ï}

3 Simplify.

b. Ï}

4y5 5 Ï}

4 p y4 p y Factor using perfect square factors.

5 Ï}

4 p Ï}

y4 p Ï}

y Product property of radicals

5 2y2 Ï}

y Simplify.

Use the quotient property of radicals.

a. Ï}

5 }

49 5

Ï}

5 }

Ï}

49 Quotient property of radicals

5 Ï

} 5 } 7 Simplify.

b. Ï}

3 }

z2 5 Ï

}

3 }

Ï}

z2 Quotient property of radicals

5 Ï

}

3 } z Simplify.

Rationalize the denominator.

a. 3 }

Ï}

5 5

3 }

Ï}

5 p Ï

} 5 }

Ï}

5 Multiply by

Ï}

5 }

Ï}

5 .

5 3 Ï

} 5 }

Ï}


5 3 Ï

} 5 } 5 Simplify.

b. Ï

} 5 }

Ï}

2m 5

Ï}

5 }

Ï}

2m p Ï

}

2m }

Ï}

2m Multiply by

Ï}

2m }

Ï}

2m .

5 Ï}

10m }

Ï}

4m2 Product property of radicals

5 Ï}

10m }

Ï}

4 p Ï}

m2 Product property of radicals

5 Ï}

10m }

2m Simplify.

Simplify the radical expression.

5. Ï}

48 6. Ï}

11

} 25

7. Ï

}

6x }

Ï}

5 8. Ï

}

15

} x4

9. Ï}

12x

} z2 10.

Ï}

10 }

Ï}

3a 11. Ï

}

63 12. Ï}

16y3

13. Ï}

28yz2 14. 4 }

Ï}

3x 15. Ï

}

20

} 9x2 16. Ï

}

25

} 4b3

EXAMPLE

PRACTICE

Multiply both the numerator and the denominator of the radical expression by the denominator of the radical expression.

EXAMPLE

EXAMPLE

Remember that variables with even exponents are perfect squares.


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3. Add, Subtract and Multiply Radical Expressions (Lesson 11.2)


a. 6 Ï}

6 1 Ï}

7 2 2 Ï}

6 5 6 Ï}

6 2 2 Ï}

6 1 Ï}

7 Commutative property

5 (6 2 2) Ï}

6 1 Ï}

7 Distributive property

5 4 Ï}

6 1 Ï}

7 Simplify.

b. 7 Ï}

2 1 Ï}

18 5 7 Ï}

2 1 Ï}

9 p 2 Factor using perfect square factor.

5 7 Ï}

2 1 Ï}

9 p Ï}


5 7 Ï}

2 1 3 Ï}

2 Simplify.

5 (7 1 3) Ï}


5 10 Ï}

2 Simplify.


a. Ï}

2 1 Ï}

32 2 5 2 5 Ï}

2 p Ï}

32 2 5 Ï}


5 Ï}

64 2 5 Ï}


5 8 2 5 Ï}

2 Simplify.

b. 1 Ï}

5 2 Ï}

3 2 1 2 Ï}

5 1 Ï}

3 2

5 Ï}

5 p 2 Ï}

5 1 Ï}

5 p Ï}

3 1 1 2 Ï}

3 2 1 2 Ï}

5 2 1 1 2 Ï}

3 2 1 Ï}

3 2 Multiply.

5 2 1 Ï}

5 2 2 1 Ï

}

5 p 3 2 2 Ï}

3 p 5 2 1 Ï}

3 2 2 Product property of

radicals

5 10 1 Ï}

15 2 2 Ï}

15 2 3 Simplify.

5 7 2 Ï}

15 Simplify.


17. 5 Ï}

6 2 Ï}

3 1 2 Ï}

6 18. 9 Ï}

2 1 2 Ï}

32

19. Ï}

3 1 5 1 Ï}

12 2 20. 8 Ï}

3 2 Ï}

75

21. 1 Ï}

6 2 Ï}

10 2 1 Ï}

6 1 2 Ï}

10 2 22. 1 Ï}

3 1 2 Ï}

2 2 1 2 Ï}

3 2 Ï}

2 2

4. Solve a Radical Equation (Lesson 11.3)Radical equation An equation that contains a radical expression with a variable in the radicand.

Solve the radical equation.

a. 5 Ï} x 2 10 5 0 Write original equation.

5 Ï} x 5 10 Add 10 to each side.

Ï} x 5 2 Divide each side by 5.

1 Ï} x 2 2 5 22 Square each side.

PRACTICE

Vocabulary

EXAMPLE

Use the FOIL method to multiply two binomials: First, Outer, Inner, Last.

Squaring both sides of an equation:if a 5 b then a2 5 b2.

EXAMPLE

You can combine expressions only if they have the same radicand.

EXAMPLE


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x 5 4 Simplify.

b. 5 Ï}

x 2 3 1 6 5 21 Write original equation.

5 Ï}

x 2 3 5 15 Subtract 6 from each side.

Ï}

x 2 3 5 3 Divide each side by 5.

1 Ï}

x 2 3 2 2 5 32 Square each side.

x 2 3 5 9 Simplify.

x 5 12 Add 3 to each side.

c. Ï}

3x 2 7 5 Ï}

x 1 11 Write original equation.

1 Ï}

3x 2 7 2 2 5 1 Ï

}

x 1 11 2 2 Square each side.

3x 2 7 5 x 1 11 Simplify.

2x 2 7 5 11 Subtract x from each side.

2x 5 18 Add 7 to each side.

x 5 9 Simplify.

Solve the radical equation.

23. 3 Ï} x 2 12 5 0 24. 3 Ï

}

x 2 8 1 11 5 32 25. 6 Ï} x 2 4 5 0

26. Ï}

6x 2 15 5 Ï}

2x 2 3 27. 2 Ï}

x 1 9 2 16 5 26 28. Ï}

5x 2 3 5 Ï}

2x 1 15

5. Solve an Equation with an Extraneous Solution (Lesson 11.3)

Extraneous solution A solution that results from squaring both sides of an equation that is not a solution to the original equation.

Solve � � 3 � 2x � x.

Solution:

Ï}

3 2 2x 5 x Write original equation.

1 Ï}

3 2 2x 2 2 5 x2 Square each side.

3 2 2x 5 x2 Simplify.

0 5 x2 1 2x 2 3 Write in standard form.

0 5 (x 1 3)(x 2 1) Factor.



Check:

Check x 5 23 and x 5 1 in the original equation.

x 5 23: Ï}

3 2 2(23) 0 2 3 x 5 1: Ï}

3 2 2(1) 0 1

3 5 23 ✗ 1 5 1 ✓

PRACTICE

Vocabulary

EXAMPLE

The radicand must be isolated on one side of the equation before you square each side. Always check the solution by substituting it in the original equation.

When a radical equation contains two radical expressions, be sure that each side of the equation has only one radical expression before squaring each side.


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Because 23 does not check in the original equation, it is an extraneous solution. The only solution of the equation is 1.

Solve the radical equation. Check for extraneous solutions.

29. x 5 Ï}

12 2 x 30. x 5 Ï}

5x 1 6 31. x 5 Ï}

7x 2 12

CheckGraph the function and identify its domain and range. Compare the graph with the graph of y � � � x .

1. y 5 2 Ï} x 2. y 5 23 Ï}

x 3. y 5 Ï} x 1 4 4. y 5 Ï

}

x 2 2


5. Ï}

99 6. Ï}

13

} 64

7. Ï}

17y

} 4z4

8. Ï}

12x5 9. Ï}

7x

} 9y2 10. Ï

}

45x3 y2

11. Ï}

44x8 12. Ï}

32y

} 25z2 13. Ï

}

3 1 Ï}

27 2 3 2

14. Ï}

8 2 Ï}

72 15. Ï}

7 2 2 Ï}

7 1 6 Ï}

7

16. 1 3 Ï}

7 2 Ï}

3 2 1 Ï}

7 1 Ï}

3 2 17. Ï}

48 2 Ï}

12

18. 1 2 Ï}

5 1 Ï}

6 2 1 Ï}

5 1 2 Ï}

6 2

Solve the radical equation. Check for extraneous solutions.

19. 6 Ï} x 2 30 5 0 20. 2 Ï

}

x 2 2 1 15 5 35

21. 10 Ï} x 2 5 5 0 22. Ï

}

2x 2 5 5 Ï}

4x 2 15

23. 4 Ï}

x 1 7 2 3 5 13 24. Ï}

6x 2 17 5 Ï}

3x 1 7

25. x 5 Ï}

4x 1 5 26. x 5 Ï}

3x 2 2

27. x 5 Ï}

2x 1 8

PRACTICE


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B. Radicals in GeometryAn important application of radical equations is found in geometry. A formula for fi nding the distance between two points can be derived from the Pythagorean theorem. The Pythagorean theorem, the distance formula and the midpoint formula are described in the following examples.

1. Use the Pythagorean Theorem (Lesson 11.4)Pythagorean theorem If a triangle is a right triangle, then the sum of the squares of the lengths of the legs equals the square of the length of the hypotenuse. (a2 1 b2 5 c2)

b

ac

A right triangle has one leg that is 3 inches longer than the other leg. Thelength of the hypotenuse is � � 17 inches. Find the unknown lengths.

x

x 1 3

17�

Solution:

Sketch a right triangle and label the sides with their lengths. Let x be the length of the shorter leg.

a2 1 b2 5 c2 Pythagorean theorem

x2 1 (x 1 3)2 5 ( Ï}

17 )2 Substitute.

x2 1 x2 1 6x 1 9 5 17 Simplify.

2x2 1 6x 2 8 5 0 Write in standard form.

2(x 1 4)(x 2 1) 5 0 Factor.



Because length is nonnegative, the solution x 5 24 does not make sense. The legs have lengths of 1 inch and 1 1 3 5 4 inches.

Find the unknown lengths.

1. A right triangle has one leg that is 5 inches longer than the other leg. The length of the hypotenuse is Ï

}

53 inches.

2. A right triangle has one leg that is 4 feet shorter than the other leg. The length of the hypotenuse is Ï

}

58 inches.

3. A right triangle has one leg that is twice as long as the other leg. The length of the hypotenuse is 5 meters.

Vocabulary

EXAMPLE

Don’t stop after solving for x. The problem asks for the unknown lengths of two sides.

PRACTICE


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2. Use the Distance Formula (Lesson 11.5)Distance formula The distance d between any two points (x

1, y

1) and (x

2, y

2)

is d 5 Ï}}

(x2

2 x1)2 1 (y

22 y

1)2 .

Find the distance between (2, 4) and (23, 1).

Solution:

Let (x1, y

1) 5 (2, 4) and (x

2, y

2) 5 (23, 1).

d 5 Ï}}

(x2

2 x1)2 1 (y

22 y

1)2 Distance formula

5 Ï}}

(23 2 2)2 1 (1 2 4)2 Substitute.

5 Ï}}

(25)2 1 (23)2 5 Ï}

34 Simplify.

The distance between the points is Ï}

34 units.

Find the distance between the points.

4. (5, 2), (1, 24) 5. (3, 23), (2, 6) 6. (24, 23), (2, 26)

7. (22, 23), (1, 26) 8. (1, 2), (4, 6) 9. (25, 23), (22, 21)

The distance between (22, 23) and (a, 3) is 10 units. Find the value of a.

Solution:

Use the distance formula with d 5 10. Let (x1, y

1) 5 (22, 23) and (x

2, y

2) 5 (a, 3).

Then solve for a.

d 5 Ï}}

(x2

2 x1)2 1 (y

22 y

1)2 Distance formula

10 5 Ï}}}

(a 2 (22))2 1 (3 2 (23))2 Substitute.

10 5 Ï}}

a2 1 4a 1 4 1 36 Multiply.

10 5 Ï}}

a2 1 4a 1 40 Simplify.

100 5 a2 1 4a 1 40 Square each side.

0 5 a2 1 4a 2 60 Write in standard form.

0 5 (a 1 10)(a 2 6) Factor.

a 1 10 5 0 or a 2 6 5 0 Zero-product property

a 5 210 or a 5 6 Solve for a.

The value of a is 210 or 6.

Find the value of a.

10. The distance between (2, a) and (26, 3) is 2 Ï}

17 units.

11. The distance between (6, 2) and (2, a) is Ï}

41 units.

12. The distance between (a, 5) and (22, 27) is 13 units.

The radical Ï}

34 is in its simplest form.

PRACTICE

EXAMPLE

Be sure to include the negative signs of the coordinates.

You can check the solutions by plotting (22, 23), (210, 3), and(6, 3) on the same coordinate grid.

PRACTICE

Vocabulary

EXAMPLE


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3. Use the Midpoint Formula (Lesson 11.5)Midpoint formula The midpoint M of the line segment with endpoints A(x

1, y

1) and

B(x2, y

2) is M 1 x1

1 x2 }

2 ,

y1 1 y

2 }

2 2 .

Find the midpoint of the line segment with endpoints (24, 1) and (2, 23).

Solution:

Let (x1, y

1) 5 (24, 1) and (x

2, y

2) 5 (2, 23).

1 x1 1 x

2 }

2 ,

y1 1 y

2 }

2 2 5 1 24 1 2

} 2 , 1 1 (23)

} 2 2 Substitute.

5 (21, 21) Simplify.

Find the midpoint of the line segment with the given endpoints.

13. (24, 3), (2, 7) 14. (8, 1), (2, 5) 15. (2, 1), (23, 24)

16. (24, 23), (4, 21) 17. (6, 23), (22, 1) 18. (3, 2), (0, 5)

CheckFind the unknown lengths.

1. A right triangle has one leg that is 1 inch longer than the other leg. The length of the hypotenuse is Ï

}

61 inches.

2. A right triangle has one leg that is three times as long as the other leg. The length of the hypotenuse is 4 Ï

}

10 feet.

Find the distance between the points.

3. (26, 25), (23, 21) 4. (1, 3), (4, 22) 5. (2, 5), (22, 3)

Find the value of b.

6. The distance between (0, 23) and (b, 2) is Ï}

29 units.

7. The distance between (24, b) and (3, 2) is Ï}

58 units.

Find the midpoint of the line segment with the given endpoints.

8. (1, 24), (25, 4) 9. (5, 2), (23, 24) 10. (4, 3), (1, 6)

Vocabulary

EXAMPLE

Each coordinate of the midpoint is an average of the coordinates of the endpoints.

PRACTICE


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C. Operations on Rational ExpressionsIn Chapter 9 you learned how to add, subtract and multiply polynomials. You will now learn how to divide polynomials and determine excluded values. The following examples also show how to apply the operations of addition, subtraction, multiplication, and division to rational expressions.

1. Divide a Polynomial by a Monomial (Lesson 12.3)Divide 3y 3 1 9y 2 2 6y by 3y .

Solution:

Method 1: Write the division as a fraction.

(3y3 1 9y2 2 6y) 4 3y 5 3y3 1 9y2 2 6y

}} 3y

Write as fraction.

5 3y3

} 3y

1 9y2

} 3y

2 6y

} 3y

Divide each term by 3y.

5 y2 1 3y 2 2 Simplify.

Method 2: Use long division.

(3y3 1 9y2 2 6y) 4 3y 5 y2 1 3y 2 2

Divide

1. (8x3 2 12x2 1 4x) 4 4x 2. (15y4 1 5y3 2 20y2) 4 5y

3. (10y3 1 6y2 2 20y) 4 2y 4. (12z4 2 3z3 2 15z2) 4 3z

2. Divide a Polynomial by a Binomial (Lesson 12.3)Divide x 2 1 3x 2 10 by x 2 2.

Solution:

Step 1: Divide the fi rst term of x2 1 3x 2 10 by the fi rst term of x 2 2.

x 2 2 qww x2 1 3x 2 10 Think: x2 4 x 5 ?

x2 2 2x Multiply x and x 2 2.

5x Subtract x2 2 2x from x2 1 3x.

x

Step 2: Bring down 210. Then divide the fi rst term of 5x 2 10 by the fi rst term of

EXAMPLE

PRACTICE

Check your answer by multiplying:3y(y2 1 3y 2 2).

3y qww 3y3 1 9y2 2 6y y2 1 3y 2 2

Think:

3y3 4 3y 5 ?

Think:

9y2 4 3y 5 ? Think:

26y 4 3y 5 ?

EXAMPLE

Be sure to subtract x2 2 2x from x2 1 3x to obtain 5x.


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x 2 2.

x 2 2 qww x2 1 3x 2 10

x2 2 2x

5x 2 10 Think: 5x 4 x 5 ?

x 1 5

5x 2 10 Multiply 5 and x 2 2.

0 Subtract 5x 2 10 from 5x 2 10.

(x2 1 3x 2 10) 4 (x 2 2) 5 x 1 5

Divide 2x2 1 7x 2 1 by 2x 2 1.

Solution:

2x 2 1 qww 2x2 1 7x 2 1

2x2 2 x Multiply x and 2x 2 1.

8x 2 1 Subtract 2x2 2 x. Bring down 21.

8x 2 4 Multiply 4 and 2x 2 1.

3 Subtract 8x 2 4 from 8x 2 1.

x 1 4

(2x2 1 7x 2 1) 4 (2x 2 1) 5 x 1 4 1 3 }

2x 2 1

Divide

5. (y2 1 y 2 12) 4 (y 1 4) 6. (b2 2 2b 2 8) 4 (b 1 2)

7. (3a2 1 8a 1 4) 4 (3a 2 1) 8. (2x2 2 5x 2 14) 4 (x 2 4)

3. Find the Excluded Values of Rational Expressions (Lesson 12.4)

Rational expression An expression that can be written as a ratio of two polynomials where the denominator is not 0.

Excluded value a number that makes a rational expression undefi ned.

Find the excluded values, if any, of the expression.

a. y 2 5

} 4y

b. 3 }

3b 2 15 c.

2x }

x2 2 4 d.

3m 2 1 }}

5m2 1 2m 1 4

Solution:

a. The expression y 2 5

} 4y

is undefi ned when 4y 5 0, or y 5 0.

The excluded value is 0.

b. The expression 3 }

3b 2 15 is undefi ned when 3b 2 15 5 0, or b 5 5.

The excluded value is 5.

c. The expression 2x }

x2 2 4 is undefi ned when x2 2 4 5 0, or (x 2 2) (x 1 2) 5 0.

The solutions of the equation are 22 and 2. The excluded values are 22 and 2.

EXAMPLE

PRACTICE

When you obtain a nonzero remainder, apply the rule:Dividend 4 Divisor 5 Quotient 1 Remainder }

Divisor .

To check your answer, multiply the quotient by the divisor, then add the remainder to the product.

Vocabulary

EXAMPLE

A rational expression is undefi ned if its denominator is 0.


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d. The expression 3m 2 1

}} 5m2 1 2m 1 4

is undefi ned when 5m2 1 2m 1 4 5 0.

The discriminant is b2 2 4ac 5 22 2 4(5)(4) < 0. So, the quadratic equation has no real roots.

There are no excluded values.


9. 2b 1 3

} 5b

10. 22 }

5y 1 15 11.

x }

x2 2 25 12. 7x }

8 2 2x

13. z 2 5 }

z2 2 3z 2 4 14.

m2 2 1 }

2m 15. 4a }

a2 1 9 16. n 1 3 }

n2 2 6n 1 9

4. Multiply and Divide Rational Expressions (Lesson 12.5)Rational expression An expression that can be written as a ratio of two polynomials where the denominator is not 0.

Excluded value A number that makes a rational expression undefi ned.

Find the product x 2 2 4x 1 4 }} x 2 1 3x

p 2x 2 1 2x }} 3x 2 2 3x 2 6 .

Solution:

x2 2 4x 1 4 } x2 1 3x

p 2x2 1 2x }

3x2 2 3x 2 6

5 (x2 2 4x 1 4)(2x2 1 2x)

}} (x2 1 3x)(3x2 2 3x 2 6)

Multiply numerators and denominators.

5 2x(x 2 2)(x 2 2)(x 1 1)

}} 3x(x 1 3)(x 1 1)(x 2 2)

Factor and divide out common factors.

5 2(x 2 2)

} 3(x 1 3)

Simplify.

Find the quotient x2 2 5x 1 4 }} x 2 1 4x

4 x 2 1 x 2 2 } 5x 1 20

.

Solution:

x2 2 5x 1 4 } x2 1 4x

4 x2 1 x 2 2

} 5x 1 20

5 x2 2 5x 1 4 } x2 1 4x

p 5x 1 20 }

x2 1 x 2 2 Multiply by multiplicative inverse.

5 (x2 2 5x 1 4)(5x 1 20)

}} (x2 1 4x)(x2 1 x 2 2)

Multiply numerators and denominators.

5 5(x 2 4)(x 2 1)(x 1 4)

}} x(x 1 4)(x 1 2)(x 2 1)

Factor and divide out common factors.

5 5(x 2 4)

} x(x 1 2)

Simplify.

PRACTICE

Vocabulary

EXAMPLE

Use a graphing calculator to check your simplifi cation. Graphy1 5 x

2 2 4x 1 4 } x2 1 3x

p

2x2 1 2x } 3x2 2 3x 2 6

and

y2 5 2(x 2 2) }

3(x 1 3) . If the

graphs coincide, your simplifi cation is correct.

EXAMPLE

To divide by a rational expression, multiply by its multiplicative inverse.

If the discriminant of the quadratic equation is nonnegative, then you must solve it to fi nd the excluded values.


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Find the product or quotient.

17. x2 1 x 2 6

} 12x2 2 6x

p 4x2 2 2x }

x2 2 x 2 12 18.

z2 2 25 }

2z 1 4 p 4z2 1 8z

} z2 2 3z 2 10

19. x2 2 6x

} 4x 1 12

4 x2 2 36

} x2 1 2x 2 3

20. w2 2 w 2 12

} 2w2 1 4w

4 w2 2 3w 2 4

} w3 1 2w2

5. Find the LCD of Rational Expressions (Lesson 12.6)Least common denominator (LCD) The LCD of two or more rational expressions is the product of the factors of the denominators of the rational expressions with each common factor used only once.

Find the LCD of the rational expressions.

a. 3 }

6z2 , 2 2 z

} 9z

b. a }

a2 2 4 ,

7 }

a2 2 4a 1 4 c.

2 }

x2 1 4 ,

x 2 3 }

3x 2 1

Solution:

a. Find the least common multiple (LCM) of 6z2 and 9z.

6z2 5 2 p 3 p z p z 9z 5 3 p 3 p z

The common factors are circled.

LCM 5 2 p z p 3 p 3 p z 5 18z2

The LCD of 3 }

6z2 and 2 2 z

} 9z

is 18z2.

b. Find the least common multiple (LCM) of a2 2 4 and a2 2 4a 1 4.

a2 2 4 5 (a 2 2) p (a 1 2)

a2 2 4a 1 4 5 (a 2 2) p (a 2 2)

LCM 5 (a 2 2) p (a 1 2) p (a 2 2) 5 (a 2 2)2 (a 1 2)

The LCD of a }

a2 2 4 and

7 }

a2 2 4a 1 4 is (a 2 2)2 (a 1 2).

c. Find the least common multiple (LCM) of x2 1 4 and 3x 2 1. Because x2 1 4 and 3x 2 1 cannot be factored, they don’t have any factors in

common. The least common multiple is their product, (x2 1 4)(3x 2 1).

The LCD of 2 }

x2 1 4 and

x 2 3 }

3x 2 1 and is (x2 1 4)(3x 2 1).


21. x 1 5

} 12x

, 4 }

18x3 22. 3b }

b2 1 2 ,

2 }

b 2 2 23.

3 2 y }

y2 1 2y 2 3 ,

6 }

5y2 2 5

Vocabulary

EXAMPLE

PRACTICE

Be sure to use the common factors only once when fi nding the LCD.

You can use the discriminant of a quadratic expression to check for factorability. If it is negative, the expression cannot be factored.

PRACTICE


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6. Add and Subtract Rational Expressions (Lesson 12.6)Find the sum or difference.

a. 2 }

12y 1

7 }

16y2 b. x 1 1 }

x2 1 x 2 6 2

x 2 1 }

x2 2 4

Solution:

a. 2 }

12y 1

7 }

16y2 5 2 p 4y

} 12y p 4y

1 7 p 3

} 16y2 p 3

Rewrite fraction using LCD, 48y2.

5 8y

} 48y2

1 21

} 48y2 Simplify numerators and

denominators.

5 8y 1 21

} 48y2 Add fractions.

b. x 1 1 }

x2 1 x 2 6 2

x 2 1 }

x2 2 4

5 x 1 1 }}

(x 2 2)(x 1 3) 2

x 2 1 }}

(x 2 2)(x 1 2) Factor denominators.

5 (x 1 1)(x 1 2)

}} (x 2 2)(x 1 3)(x 1 2)

2 (x 2 1)(x 1 3)

}} (x 1 2)(x 2 2)(x 1 3)

Rewrite fraction using LCD,

(x 2 2)(x 1 3)(x 1 2).

5 (x 1 1)(x 1 2) 2 (x 2 1)(x 1 3)

}}} (x 2 2)(x 1 3)(x 1 2)

Subtract fractions.

5 x2 1 3x 1 2 2 (x2 1 2x 2 3)

}}} (x 2 2)(x 1 3)(x 1 2)

Find products in numerator.

5 x 1 5 }}

(x 2 2)(x 1 3)(x 1 2) Simplify.


24. 3 }

4b2 1 4 }

10b 25.

5x }

x 2 3 2

9 }

4x

26. z 1 1 }

z2 2 2z 2 15 2

z 2 1 }

z2 1 z 2 6

To subtract x 2 1 2x 2 3 in the numerator, add the opposite of every term.

PRACTICE

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CheckDivide

1. (3a3 1 18a2 2 9a) 4 3a 2. (16x4 1 24x3 1 8x2) 4 4x

3. (20m3 2 5m2 2 15m) 4 5m 4. (14y4 2 6y3 1 18y) 4 2y

5. (n2 1 4n 2 12) 4 (n 2 2) 6. (2x2 2 5x 2 3) 4 (x 2 3)

7. (4b2 1 4b 2 7) 4 (2b 1 3) 8. (4y2 2 19y 1 1) 4 (y 2 5)


9. z 1 2

} 7z 10. m2

} m2 1 16

11. 4a 1 3

} a2 2 1

12. 3 }

3x2 1 x 1 6

Find the product or quotient.

13. a2 2 4

} a3 2 16a

4 a2 2 2a

} a3 2 4a2 14.

y2 2 9 }

y2 1 y p

y2 2 1 }

y2 1 5y 1 6

15. 4b 1 12

} 2b2 2 2b 2 4

p b3 2 4b2 2 5b

}} 22b 2 6

16. 3x2 2 12x 1 12

}} x3 2 4x2 4

6x2 2 6x 2 12 }}

x2 2 4x


17. 2y 2 1

} 15y3 ,

1 }

25y2 18. 4m 2 1

} 2m2 2 8

, 3 }}

m2 1 3m 2 10 19.

x 1 4 }

x2 1 3x 1 3 ,

7 }

3x


20. 5 }

12x 2

3 }

20x3 21. m }

m 1 2 1

7 }

3m2 22. b 1 3

} b2 2 16

2 b 1 2 }

b2 1 2b 2 8


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The inverse variation equation is a type of rational function. A rational function has a rule given by a fraction whose numerator and denominator are polynomials and whose denominator is not 0. The following examples describe how to use an inverse variation equation, and how to graph and solve rational equations.

1. Use an Inverse Variation Equation (Lesson 12.1)Inverse variation equation The variables x and y show inverse variation if y 5

a } x and

a Þ 0. y is said to vary inversely with x.

Constant of variation The number a in the inverse variation equation.

The variables x and y vary inversely, and y 5 4 when x 5 25.

a. Write an inverse variation equation that relates x and y. Because y varies inversely with x, the equation has the form y 5

a } x . Use the fact that x 5 25 and

y 5 4 to fi nd the value of a.

y 5 a } x Write inverse variation equation.

4 5 a }

25 Substitute 25 for x and 4 for y.

220 5 a Multiply each side by 25.

An equation that relates x and y is y 5 220

} x .


When x 5 8, y 5 220

} 8 5

25 }

2 .

Given that y varies inversely with x, use the specifi ed values to write an inverse variation equation that relates x and y. Then fi nd the value of y when x 5 4.

1. x 5 6, y 5 22 2. x 5 1, y 5 8 3. x 5 2, y 5 23

4. x 5 7, y 5 2 5. x 5 23, y 5 12 6. x 5 29, y 5 3

2. Graph y 5 a } x , y 5 1 } x 1 k, and y 5 1 } x 2 h

(Lesson 12.2)

Parent rational function y 5 1 } x The domain and y

x1 2 3 4

1

2

3

4

2121222324

22

23

24

y 5 1x

range are all nonzero real numbers. The horizontal asymptote is the x-axis. The vertical asymptote is the y-axis.

Vocabulary

EXAMPLE

PRACTICE

You can also write the inverse variation equation as xy 5 a. Therefore, a set of ordered pairs (x, y) shows inverse variation if all the products xy are constant.

Vocabulary

Ready to Go On? Chapters 11-12 Intervention cont’d

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Graph the function and identify its domain and range. Compare the graph with the graph of y 5 1 } x .

a. y 5 25

} x

The graph of y 5 25

} x is a vertical stretch and a refl ection in the x-axis

of the graph of y 5 1 } x . The domain

and range are all nonzero real numbers.

y

x1 2 3 4

1

2

3

4

2121222324

22

23

24

y 5 25x

y 5 1x

c. y 5 1 } x 1 2

The graph of y 5 1 } x 1 2 is a

vertical translation (of 2 units up) of

the graph of y 5 1 } x . The domain is all

real numbers except 0. The range is all real numbers except 2.

1

2

3

4

5

6y

1 x21222324 2 3 421

22y 5 1x

y 5 1 2 1x

b. y 5 1 }

4x

The graph of y 5 1 }

4x is a vertical

shrink of the graph of y 5 1 } x . The

domain and range are all nonzero real numbers.

y

x1 2 3 4

1

2

3

4

2121222324

22

23

24

y 5 1x14xy 5

d. y 5 1 }

x 2 3


x 2 3 is a horizontal

translation (of 3 units to the right) of

the graph of y 5 1 } x . The domain is all

real numbers except 3. The range is all real numbers except 0.

y 5 1x1

2

3

4y

1 x2 3 4 5 6 721

22

23

24

21

y 5 1x 2 3

EXAMPLE

To see the graph’s behavior at extreme values, plot points with x values close to 0 and far from 0.

The graph of an inverse variation equation lies in Quadrants I and III if a > 0, and in Quadrants II and IV if a < 0.


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Graph the function and identify its domain and range. Compare the

graph with the graph of y 5 1 } x .

7. y 5 3 } x 8. y 5

21 }

2x 9. y 5

1 } x 2 3 10. y 5

1 }

x 1 2

3. Solve a Rational Equation Using Cross Products (Lesson 12.7)

Rational equation An equation that contains one or more rational expressions.

Use the cross products property to solve x }

5 5

3 }

x 2 2 .

Solution:

x } 5 5

3 }

x 2 2 Write original equation.

x2 2 2x 5 15 Cross products property

x2 2 2x 2 15 5 0 Subtract 15 from each side.

(x 2 5)(x 1 3) 5 0 Factor polynomial.



The solutions are 23 and 5.

Solve the equation. Check your solution.

11. 1 }

m 1 4 5

m } 5 12.

z }

6 5

2 }

z 1 1 13.

x }

4 5

24 }

x 1 10

4. Solve a Rational Equation Using the LCD (Lesson 12.7)Solve the equation. Check your solution.

a. x }

x 1 4 1

1 }

9 5

6 }

x 1 4 b.

8 }

x 2 3 1 1 5

4 }

x2 2 x 2 6

Solution:

a.

x }

x 1 4 1

1 }

9 5

6 }

x 1 4 Write equation.

x }

x 1 4 p 9(x 1 4) 1

1 }

9 p 9(x 1 4) 5

6 }

x 1 4 p 9(x 1 4) Multiply by LCD, 9(x 1 4).

x p 9(x 1 4)

} (x 1 4)

1 9(x 1 4)

} 9 5

6 p 9(x 1 4) }

(x 1 4) Multiply and divide out

common factors.

9x 1 x 1 4 5 54 Simplify.

10x 1 4 5 54 Combine like terms.

10x 5 50 Subtract 4 from each side.

x 5 5 Divide each side by 10.

The solution is 5.

EXAMPLE

PRACTICE

Cross Products Property:

If a } b

5 c } d

where

b Þ 0 and d Þ 0,then ad 5 bc.

Vocabulary

Check your solutions by substituting them into the original equation.

EXAMPLE

The excluded value for this equation is 24. Since our solution of 5 is not excluded, we can check it in the original equation.

PRACTICE


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b. Write each denominator in factored form. The LCD is (x 2 3)(x 1 2).

8 }

x 2 3 1 1 5

4 }}

(x 2 3)(x 1 2)

8 }

x 2 3 p (x 2 3)(x 1 2) 1 1 p (x 2 3)(x 1 2) 5

4 }}

(x 2 3)(x 1 2) p (x 2 3)(x 1 2)

8(x 2 3)(x 1 2)

}} x 2 3

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

}} (x 2 3)(x 1 2)

8(x 1 2) 1 (x2 2 x 2 6) 5 4

x2 1 7x 1 10 5 4

x2 1 7x 1 6 5 0

(x 1 6)(x 1 1) 5 0

x 1 6 5 0 or x 1 1 5 0

x 5 26 or x 5 21

The solutions are 26 and 21.


14. 3z } z 1 4

1 1 }

6 5

7 }

z + 4 15. 2 }

a 1 2 1

1 }

a 2 1 5

a 2 2 }

a 2 1

16. 3 } x 2 5

1 1 5 6 }

x2 2 2x 2 15 17.

2y }

y 2 3 1

5 }

y 1 3 5

8y 2 13 }

y2 2 9

CheckGiven that y varies inversely with x, use the specifi ed values to write an inverse variation equation that relates x and y. Then fi nd the value of y when x 5 2.

1. x 5 1, y 5 8 2. x 5 5, y 5 22 3. x 5 24, y 5 3 4. x 5 12, y 5 2

Graph the function and identify its domain and range. Compare the graph with the graph of y 5 1 } x .

5. y 5 4 } x 6. y 5

21 }

3x 7. y 5

1 } x 2 2 8. y 5

1 }

x 1 3


9. a } 8 5

3 }

a 1 2 10.

4 } x 2 7 5

x }

2

11. 23 } y 5 y 2 8

} 4 12.

2a }

a 1 6 1

1 }

4 5

6 }

a 1 6

13. 4 } x 2 5 1

12 }

x 1 1 5

x } x 2 5 14.

2z }

z 2 4 5 1 2

1 }

z2 2 2z 2 8

15. 2b }

b 2 1 2

b }

b 1 3 5

5b 1 8 }

b2 1 2b 2 3

PRACTICE

Multiply all terms of the equation by the LCD.

Always check the solution by substituting the values into the original equation.


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Expressions, Equations, and Inequalities (Lessons 1.1–1.4)For Exercises 1–3, evaluate the expression for the given value of the variable.

1. x2 1 5 when x 5 3 2. 5 }

6 y when y 5 18

3. 20 2 a3 when a 5 2

For Exercises 4–6, evaluate the expression.

4. (5 1 2)2 5. 3[(4 1 6) 4 5]

6. 6 p (27) 2 8 1 (52 2 4)

7. The diagram shows a box and a gift box.

14 cm15 cm

15 cm

15 cm

14 cm

14 cm

gift box box

a. What is the volume of the gift box?

b. When you put the box in the gift box, is there any empty space left? If so, what is the volume of the empty space?

8. Which is equivalent to the sum of four and the square of b?

A. 4b2 B. 4 1 b2

C. (4 1 b)2 D. 42 1 b2

9. Write an inequality: The quotient of 16 and b is at most 4.

A. 16 } b ≥ 4 B. 16 }

b ≤ 4

C. b } 16

≥ 4 D. b } 16

≤ 4

10. At the hardware store, Maya buys n boxes of nails for $1.99 per box and b boxes of bolts for $2.49 per box. Which expression represents Maya’s purchase?

A. 1.99n 1 2.49b B. (1.99 1 n)(2.49 1 b)

C. (1.99 1 2.49)(n p b) D. 1.99n p 2.49b

Answers

1.

2.

3.

4.

5.

6.

7a.

7b.

8.

9.

10.

Ready to Go On? Chapters 1–2 Quiz

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Answers

11.

12.

13.

14.

15.

16.

17.

18a.

18b.

18c.

19.

11. Barry’s salary is $9.45 per hour. Which expression represents Barry’s earnings for h hours?

A. 9.45h B. h 2 9.45

C. 9.45 1 h D. h 4 9.45

Problem Solving (Lesson 1.5)For Exercises 12–15, state whether the given number is a solution of the equation or inequality.

12. 2x 1 4 5 10; 4 13. t 2 5 < 8; 11

14. 6c 5 42; 7 15. h } 5 5 25; 5

For Exercises 16 and 17, read the following problem and answer the questions.

You make 100 refrigerator magnets to sell at a local crafts fair. Each magnet costs $.60 to make. In addition, you spend $50 for a booth and set up. Assuming that you sell half of your magnets, how much should you charge for each magnet to make a profi t of $100?

16. What do you know and what do you need to fi nd out?

17. Write and solve an equation to answer the question.

18. A restaurant charges $7 for a regular meal and $5 for a child’s meal. Both prices include tax.

a. Write an expression for the cost of a given number of regular and child’s meals.

b. A family has $35 to spend at the restaurant. Write an inequality to represent the maximum number of regular and child’s meals they can buy.

c. Does the family have enough money to buy two regular meals and three child’s meals? Explain.

Representations of Functions (Lessons 1.6–1.7) 19. The table shows how many newspapers Malcolm delivers and how

much money he collects from his customers.

Newspapers, n 12 15 18 21

Money, m $24 $30 $36 $42

Write a rule to describe how much Malcolm would collect for delivering n newspapers.

Ready to Go On? Chapters 1–2 Quiz cont’d

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Answers

20.

21.

22. See table.

23. See table.

20. The table shows amounts in U.S. dollars and the equivalent amounts of Mexican pesos on a certain day.

U.S. dollars, x 20 30 50 60

Mexican pesos, y 80 120 200 240

Write a rule to describe the relationship between dollars and pesos on that day.

21. Which is a rule for the function shown?

y

x

5

4

3

2

1

67

8

6 7 851 2 3 4

(3, 0)

(5, 2)

(7, 4) (8, 5)

A. y 5 23x B. y 5 3x

C. y 5 x 2 3 D. y 5 3 2 x

22. Make a table for the function shown.

26

y

x22

26

28

24

22

2

4

6

8

6 82428 42

(22, 25)

(0, 21)

(2, 3)

(4, 7)

x

y

23. Make a table for the function y 5 23x 1 4, with domain {22, 21, 1, 2}.

x

y


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Answers

24a. See graph.

24b.

24c.

25.

26.

27.

24. Jess puts 2 pennies on the fi rst square of a checkerboard. He stacks 4 pennies in the second square. In the third square, he stacks 6 pennies. Jess stacks 8 pennies in the fourth square.

a. Graph the function for the fi rst four squares.

1

2

3

4

5

6

7

8

4 5 6 7 8321

p

n

b. Write a rule to represent the number of pennies in the nth square.

c. What is the combined number of pennies that Jess puts in the 16th, 17th, 18th, and 19th squares?

25. What is the range of the function y 5 2x 1 2 with domain {3, 6, 8, 9}?

A. {23, 26, 28, 29} B. {21, 24, 26, 27}

C. {3, 6, 8, 9} D. {5, 8, 10, 11}

For Exercises 26 and 27, write the domain and range of the function on the coordinate plane.

26. y

x

5

4

3

2

1

67

8

6 7 851 2 3 4

(1, 2) (3, 3)

(2, 1)

(4, 4) (5, 4)

27. y

x

10

8

6

4

2

1214

16

12 14 16102 4 6 8

(2, 12) (6, 10)

(8, 2) (10, 2)

(12, 6)


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Answers

28a.

28b. See graph.

29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

41.

42.

43.

44.

45.

46.

47a.

47b.

48.

49.

28. Each week, Nina saves $2.50 from the money she earns mowing lawns.

a. Write a rule to represent Nina’s total savings after w weeks.

b. Graph the function for the fi rst 4 weeks.

s

w

10

8

6

4

2

1214

16

6 7 851 2 3 4

Operations (Lessons 2.1–2.4)

For Exercises 29–31, given the value of a, fi nd 2a and | a |.

29. a 5 23 30. a 5 0.2 31. a 5 2 Ï}

5

For Exercises 32–46, evaluate or simplify the expression.

32. 2 7 } 8 1

2 }

3 33. 8 1 (26.2) 34. 25 1 (212)

35. 22.4 2 9.3 36. 9 2 (23) 37. 2 1 } 3 2 1 2

1 } 2 2

38. a(2b)(21) 39. 4 p 7t 40. 3 } 5 (210) (22)

41. 3 } 5 4 (29) 42.

1 }

6 4 6

1 } 4 43. 3 4 1 2

1 } 9 2

44. Ï}

121 45. 2 Ï}

144 46. 6 Ï}

64

47. Given a 5 22 1 } 5 ,

a. What number multiplied by a equals 1?

b. Explain how you found your answer.

48. Petra rides her bicycle 4.5 miles on Monday, 1.6 miles on Wednesday, and 3.2 miles on Friday. What is the mean distance Petra rides?

A. 3.1 miles B. 1.6 miles C. 3.2 miles D. 9.3 miles

49. A weather report predicts that the temperature will drop 3°F per hour. If the temperature starts at 12°F, what is the temperature after 6 hours?

A. 218°F B. 26°F C. 3°F D. 9°F


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Answers

50.

51a.

51b.

52.

53.

54.

55.

56.

57.

58.

59.

60.

61.

62.

50. Evaluate the expression 2 Ï}

36 1 Ï}

144 .

51. The lengths of two legs of a right triangle are 5 feet and 6 feet. By the Pythagorean Theorem, the length of the third side of the triangle

is Ï}

52 1 62 .

a. Approximate the length of the third side to the nearest whole foot.

b. Explain how you found your answer.

Properties of Real Numbers (Lessons 2.5–2.7) 52. Can a whole number also be considered a rational number? Explain.

53. Which is the most specifi c classifi cation of the number 2 5 }

8 ?

A. real B. integer C. rational D. irrational

For Exercises 54 and 55, write the numbers in order from least to greatest:

54. |27|, Ï}

50 , 26.9, 2 21

} 4 55. 2 Ï}

3 , 3 }

2 , 2|23.2|, 24

56. Which property is illustrated by the equation 25(3x 1 2) 5 215x 2 10?

A. inverse B. identity C. commutative D. distributive

57. Write a statement of the identity property of addition in terms of a

real number 1 }

3 m.

58. Which equation illustrates the commutative property of multiplication?

A. 5(x 1 2) 5 5x 1 10 B. 5(2x) 5 10x

C. 5(2x) 5 (2x)5 D. 5(2x) p 1 }

5(2x) 5 1

59. Jake writes the following:

22 1 2 1 } 2 x 2

5 1 22 p 2 1 } 2 2 x Step 1

5 x Step 2

List a property of real numbers to justify each of Jake’s steps.

For Exercises 60–62, use the distributive property to simplify the expression.

60. 2x(4x2 2 5x) 61. (4n 2 6) (23n) 62. x }

2 (4 1 2x 2 4y)


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Solving Equations in One Variable (Lessons 3.1–3.4)For Exercises 1–12, solve the equation.

1. a 2 6 5 13 2. 13 5 5 1 h

3. z }

26 5 22 4. 28b 5 72

5. 5x 1 3 5 28 6. m

} 3 2 9 5 25

7. 3 }

2 w 1 4 5 19 8. 9a 2 32.3 2 a 5 25.9

9. 5x 2 3(x 1 6) 5 28 10. 8(d 2 4) 2 4(d 2 3) 5 33

11. 5g 1 2 5 23g 2 4 12. 7(23v 1 1) 5 14v

13. The amount of simple interest I on a loan is I 5 Prt, where P is the amount borrowed, r is the interest rate written as a decimal, and t is time in years to pay off the loan.

Jack borrowed $500 at 6% and paid $90 in simple interest. How long did Jack take to pay back the loan?

A. 3 years B. 5.5 years C. 6 years D. 15 years

14. Solve the equation 5x 2 10 5 5(x 1 2), if possible.

A. x 5 2 B. x 5 5

C. all real numbers D. no solution

15. Solve the equation 4(2 2 2y) 1 6 5 2(7 2 4y), if possible.

A. y 5 4 B. y 5 8

C. all real numbers D. no solution

Proportions and Percents (Lessons 3.5–3.7) 16. A basketball team is on a 5-game road trip and plays 3 games in

Texas and 2 games in California.

a. Find the ratio of games in Texas to games in California.

b. Find the ratio of games in California to games on the road trip.

17. Melissa has 35 pins in her collection and her brother Roger has 40 in his collection. Find the ratio of the number of pins in Roger’s collection to the total number of pins.

A. 40

} 35

B. 40

} 75 C. 35

} 40

D. 35

} 75

For Exercises 18–23, solve the proportion.

18. w

} 7 5 8 }

35 19.

4 }

9 5

8 }

b 20.

3 }

8 5

x }

32

21. 2 56

} 48 5 7 }

6h 22.

3.2 }

12 5

4v } 7.5 23.

8 } 5 5

11.2 }

2y 2 1

Answers

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16a.

16b.

17.

18.

19.

20.

21.

22.

23.


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Answers

24a.

24b.

25.

26.

27.

28.

29.

30.

31a.

31b.

32a.

32b.

33a.

33b.

24. A recipe for biscuits calls for an 8 to 3 ratio of fl our to milk. You use 6 cups of fl our to make 24 biscuits.

a. How many cups of milk should you use to make 24 biscuits?

b. How many cups of milk should you use to make 36 biscuits?

25. What percent of 25 is 19?

A. 19% B. 25% C. 38% D. 76%

26. What percent of 380 is 76?

A. 20% B. 24% C. 25% D. 30%

27. The total height of the Statue of Liberty, plus its foundation and pedestal, is 93 m from the ground to the tip of the torch. The statue by itself is 46 m high. The statue by itself makes up what percent of the total height? Round to the nearest tenth of a percent.

28. Nina has 20 hours of free time during a week. She spends 7 hours of her free time exercising. What percent of her free time does Nina spend exercising?

A. 20% B. 35% C. 65% D. 70%

29. What number is 70% of 350?

A. 50 B. 105 C. 245 D. 500

30. Chad pays $22.50 for a pair of shoes. If they were on sale for 25% off, what was the original price of the shoes?

Rewriting Equations in Two or More Variables (Lesson 3.8)

For Exercises 31 and 32, (a) solve the literal equation for x and (b) use the solution to solve the specifi c equation.

31. ax 2 b 5 cx; 12x 2 5 5 4x

32. a 1 bx 5 cx 2 d; 2 1 3x 5 4x 2 5

33. The circumference C of a circle is given by the formula C 5 2πr, where π is a constant and r is the radius of the circle.

a. Solve the equation C 5 2πr for π.

b. Given the circle shown, calculate π. Round to the nearest hundredth.

7 ft

11 ft


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Answers

34.

35.

36.

37.

38.

39. See grid.

40. See grid.

41. See grid.

42. See grid.

43.

44. See table and grid.

34. Solve the equation s 5 2t 2 q } p for p.

A. p 5 2 s 2 2t

} q B. p 5 2 q } s 2 2t

C. p 5 2q(s 2 2t) D. p 5 2t 2 q } s

For Exercises 35–38, write the equation so that y is a function of x. Assume x � 0 and y � 0.

35. 4x 1 8y 5 3 36. 2x 2 y 5 21

37. 12xy 5 60 38. 18x 1 27

} 3y

5 23

Graphing Linear Equations (Lessons 4.1–4.2)

For Exercises 39–42, plot and label each point on the grid below.

39. (23, 1) y

2

4

6

8

28 26 2422

24

26

28

2 4 6 8 x22

40. (2, 0)

41. (4, 24)

42. (0, 22)

43. Which ordered pair is a solution of 2x 5 4 2 3y?

A. (4, 4) B. (2, 22) C. (0, 2) D. (21, 2)

44. For the equation 4x 1 3y 5 39, complete the table of values and graph the equation.

x 0 3 6 9

y

y

x

12

10

8

6

4

2

14

16

18

2100

3 4 5 6 7 8 9


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Answers

45. See grid.

46. See grid.

47. See grid.

48. See grid.

49.

50.

51.

52.

53.

54.

For Exercises 45–48, graph and label the equation on the grid below.

45. x 5 2 46. y 5 24

47. x 5 25 48. y 5 8

1

2

3

4

5

6

7

8

9

21

22

23

24

25

26

27

28

29

21 1 2 3 4 5 6 7 8 92223242526272829

y

x

Slope-Intercept Form and Direct Variation (Lessons 4.3–4.6) 49. Find the x-intercept and the y-intercept of the graph of 3x 2 8y 5 48.

50. Find the x-intercept and the y-intercept of the graph of 23x 1 5y 5 30.

51. Find the x-intercept and the y-intercept of the graph of 2x 1 30 5 5y.

52. What is the slope of a line that passes through (0, 6) and (23, 24)?

A. 0 B. 3 }

10 C.

10 }

3 D. undefi ned


A. 0 B. 2 5 } 4 C. 2

4 } 5 D. undefi ned


A. 0 B. 4 }

9 C.

9 }

4 D. undefi ned


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Answers

55a.

55b. See grid.

56a.

56b. See grid.

57a.

57b. See grid.

58.

59.

60.

61.

62.

63.

For Exercises 55–57, (a) write the equation in slope-intercept form and (b) graph and label the equation on the grid.

55. x 2 y 5 1 56. y 5 2 1 } 3 x 1 2

57. 5(x 1 1) 5 y 2 10

1

2

3

4

5

6

7

8

9

21

22

23

24

25

26

27

28

29

21 1 2 3 4 5 6 7 8 92223242526272829

y

x

58. What are the slope and y-intercept of the graph of 4x 2 9y 1 18 5 0?

A. slope: 24; y-intercept: 29 B. slope: 2 4 } 9 ; y-intercept: 218

C. slope: 4 }

9 ; y-intercept: 2 D. slope: 4; y-intercept:18

For Exercises 59–62, state whether the equation is a direct variation. If so, identify the constant of variation.

59. y 5 4 60. 3x 2 4y 5 0

61. 2x 1 2y 5 4 62. 2 1 4y 5 24x 1 2

63. What is the constant of variation for 2x 5 3y?

A. 23 B. 2 1 } 3 C.

1 }

3 D. 3


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64. The graph of a direct variation equation is shown.

y

x

2

22

24

26

28

4

6

8

22 212324 1 2 3 4

(3, 5)

(23, 25)

a. Write the direct variation equation.


65. Each month, Laura’s cell phone charges c vary directly with the number of minutes m that she talks. Last month, she talked 450 minutes and was charged $13.50.

a. Write a direct variation equation that relates c and m.

b. How much would Laura pay if she talked 300 minutes?

Answers

64a.

64b.

65a.

65b.


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s


Writing Linear Equations (Lessons 5.1–5.4) 1. Write an equation in slope-intercept form of the line with a slope

of 23 and a y-intercept of 2.

2. Which of the following is an equation of the line with a slope of 22 and a y-intercept of 23?

A. y 5 22x 1 3 B. y 5 2x 2 3

C. y 5 22x 2 3 D. y 5 2x 1 3

3. Which of the following is an equation in slope-intercept form of the line shown?

y

x

2

22

24

26

4

6

8

10

24 222628 2 4 6 8

(3, 23)

(26, 9)

A. y 5 x 2 4 }

3 B. y 5 x 2

2 }

3

C. y 5 2 4 } 3 x 1 1 D. y 5 2

2 } 3 x 1 1

4. Write an equation in slope-intercept form of the line that passes through the point (21, 4) and has a slope of 22.

5. Which of the following is an equation of the line that passes through (23, 4) and (1, 0)?

A. y 5 2x 1 1 B. x 1 y 5 4

C. y 5 x 1 7 D. x 2 y 5 1

6. Write an equation in point-slope form of the line that passes through the points (5, 29) and (26, 4).

7. Write an equation in standard form of the line that passes through(7, 28) and has a slope of 23.

8. a. What is the slope of the

(2q, 0)(0, p)

p

q

2p

2q

line in terms of p and q?

b. Write the equation of theline in standard form.

Answers

1.

2.

3.

4.

5.

6.

7.

8a.

8b.


Name Date


Qu

izze

s


Parallel and Perpendicular Lines (Lesson 5.5) 9. Determine which lines, if any, are parallel or perpendicular.

Line a: 2x 1 4y 5 5 Line b: x 5 2(y 2 5) Line c: 5x 2 10y 5 3

A. Lines a and b are perpendicular.

B. Lines a and c are parallel.

C. Lines b and c are parallel.

D. Lines a and c are perpendicular.

10. Write an equation of the line that passes through (1, 23) and is parallel to the line 3x 2 4y 5 5.

11. Which of the following is an equation of the line that passes through (5, 21) and is parallel to the line 4x 2 7y 5 3?

A. y 5 2 7 } 4 x 1

31 }

4 B. y 5

4 } 7 x 2

27 } 7

C. y 5 2 4 } 7 x 1

13 } 7 D. y 5

7 }

4 x 2

39 }

4

12. Write an equation of the line that passes through (4, 0) and is perpendicular to the line 2x 2 5y 5 10.

13. Which of the following is an equation of the line that passes through (21, 2) and is perpendicular to the line x 2 2y 5 5?

A. y 5 2 1 } 2 x 1

3 }

2 B. y 5

1 }

2 x 1

5 }

2

C. y 5 22x D. y 5 2x 1 4

Linear Models (Lessons 5.6–5.7)For Exercises 14–17, describe the correlation of the data graphed in the scatter plot.

14.

Year

Popu

latio

n

x

y

00

15.

Hours of Exercise

Bod

y M

ass

Inde

x

x

y

00

Answers

9.

10.

11.

12.

13.

14.

15.


Name Date


Qu

izze

s


16.

Weight

Shoe

Siz

e

x

y

00

17.

Time on Internet

Test

Sco

res

x

y

00

18. Which of the following equations best models the data in the table?

x 1 2 4 5 6 8 9

y 3 5 8 10 11 16 19

A. y 5 1 }

2 x 1 2 B. y 5 2x 1

1 }

2

C. y 5 x D. y 5 x 1 1

19. The table shows the year and average length of a critic’s top ten movies from 1930, 1950, 1970, and 1990.

Year t(in years since 1930) 0 20 40 60

Average Length m(in minutes over 100) 31 38 54 72

a. Make a scatter plot

4030Years since 1930

Aver

age

Leng

th (i

n m

inut

es o

ver 1

00)

2010 6050 70 80 90

2010

30

50

70

40

60

8090

t

m

00

of the data.

b. Draw and write an equation of a line of fit.

c. Predict the approximate average length (100 1 m) of the critic’s top ten movies in the year 2010. Explain how you made your prediction.

Answers

16.

17.

18.

19a. See graph.

19b. See graph.

19c.


Name Date


Qu

izze

s


20. The table shows the year and population of a small town from 1980, 1985, 1990, and 1995.

Year t(in years since 1980) 0 5 10 15

Population p 1050 998 935 864


10Years (in years since 1980)

Popu

latio

n

5 15 20

800

700

900

1000

1100

t

p

00

b. Draw and write an equation of a line of fit.

c. Approximate the population in the year 1992. Explain how you made your prediction.

Graphing Inequalities (Lessons 6.1, 6.7) 21. Graph the inequality x ≥ 0.

�3�4�5 �2 �1 0 3 4 51 2

22. Graph the inequality x < 22.

�3�4�5 �2 �1 0 3 4 51 2

Answers

20a. See graph.

20b. See graph.

20c.

21. See graph.

22. See graph.

22a.

See graph.


Name Date


Qu

izze

s


23. A school basketball team sells cups for $4 each and T-shirts for $12 each to raise money for their travel expenses. They need to raise a minimum of $360.

a. Write and graph an inequality to model the number of T-shirts y in terms of the number of cups x that the team needs to sell.

4030Cups Sold

T-sh

irts

Sol

d

2010 6050 70 80 90

2010

30

50

70

40

60

8090

x

y

00

b. If the team sells 18 T-shirts and 20 cups, will they raise enough money? Use your graph to explain your answer.

c. What is the y-intercept of the graph? What does it represent?

For Exercises 24 and 25, choose an inequality represented by the graph.

24.

�3�4�5 �2 �1 0 3

2.5

4 51 2

A. x ≤ 2.5 B. x ≤ 22.5 C. x ≥ 2.5 D. x ≥ 22.5

25. �3�4�5 �2 �1 0 3 4 51 2

A. x ≥ 21 B. x > 21 C. x < 21 D. x ≤ 21

For Exercises 26 and 27, translate the verbal phrase into an inequality. Then graph the inequality.

26. All real numbers that are less than or equal to 4 and greater than 23.

�3�4�5 �2 �1 0 3 4 51 2

27. All real numbers that are greater than or equal to 1 or less than 0.

�3�4�5 �2 �1 0 3 4 51 2

Answers

23b.

23c.

24.

25.

26.

See graph.

27.

See graph.


Name Date


Qu

izze

s


For Exercises 28–31, graph the inequality on the grid.

28. y ≥ 3x 2 2 29. x < 24

y

2

4

6

8

28 26 2422

24

26

28

2 4 6 8 x22

y

2

4

6

8

28 26 2422

24

26

28

2 4 6 8 x22

30. x 1 3y < 0 31. x 2 y ≤ 21

y

2

4

6

8

28 26 2422

24

26

28

2 4 6 8 x22

y

2

4

6

8

28 26 2422

24

26

28

2 4 6 8 x22

Solving Inequalities (Lessons 6.1–6.4)For Exercises 32–43, solve the inequality, if possible.

32. x 1 12 > 5 33. a 2 3 }

2 ≤ 28

34. 24x < 12 35. 2 p } 3 ≤ 2

36. 9 ≤ 2(ƒ 1 10) 37. 23(x 1 2) < 5 2 3x

38. 3(a 1 1) 1 5a > 4(3 1 2a) 39. 5q 2 2(3 2 q) ≥ 4 1 3q

40. 8 > 4b ≥ 220 41. 11 < 5x 2 4 < 26

42. 2(c 1 4) > 23 or 4 2 1 }

2 c ≥ 7 43. 27(n 1 2) ≥ 42 or 4n 2 1 > 15

44. At 9.00 A.M., you begin a hike on a six-mile trail at a park. Let t represent the hours past 9 A.M. If your walking speed is 2 miles per hour, then which inequality describes the possible values of t?

A. 9 < t < 12 B. | t | ≤ 12

C. 0 ≤ t ≤ 3 D. t > 0 or t < 3

Answers

28. See graph.

29. See graph.

30. See graph.

31. See graph.

32.

33.

34.

35.

36.

37.

38.

39.

40.

41.

42.

43.

44.


Name Date


Qu

izze

s


45. A teacher stacks 25 exams in order by score, with the highest score at the top. The fi rst 5 exams in the stack earn an A. The fi rst 10 exams, except those in the fi rst 5, earn a B.

a. Kelly earns a B. Write a compound inequality to describe the position p of Kelly’s exam in the stack.

b. Jason’s exam is 14th in the stack. Does Jason earn an A, B, or neither? Explain.

Absolute Value Equations and Inequalities (Lessons 6.5–6.6)For Exercises 46–53, solve the absolute value equation or inequality, if possible.

46. )x) 5 13 47. )b 1 3) 5 5

48. )2m 2 4) 5 11 49. )5 2 4y) 1 6 5 4

50. 23)d 1 3) 1 4 5 25 51. 3)4m 1 1) 2 3 5 24

52. 3|z| < 6 53. 9|g 2 1| ≥ 27

54. Graph the solution of |p 1 1| > 2.

�3�4�5 �2 �1 0 3 4 51 2

55. Which of the following is the solution of |x 1 58| < 100?

A. x < 2158 or x > 42 B. 2158 < x < 42

C. 242 < x < 158 D. x < 242 or x > 158

56. Which of the following is the graph of the solution to )v 2 2) < 1?

A. �3�4�5 �2 �1 0 3 4 51 2 v

B. �3�4�5 �2 �1 0 3 4 51 2 v

C. �3�4�5 �2 �1 0 3 4 51 2 v

D. �3�4�5 �2 �1 0 3 4 51 2 v

Answers

45a.

45b.

46.

47.

48.

49.

50.

51.

52.

53.

54. See graph.

55.

56.


Name Date


Qu

izze

s


Solving Linear Systems by Graphing (Lessons 7.1, 7.5) 1. The graph represents a system of linear equations. What is the

solution of the linear system?

y

x

1

21

22

23

24

2

3

4

22 21 1 2 3 4 5 6

A. (0, 0) B. (0, 1) C. (5, 22) D. 1 5 } 3 ,

5 }

2 2

2. Mari and James are friends. The difference between 9 times James’ age and 4 times Mari’s age is 72. Twelve years ago, James was half as old as Mari was.

a. Write a linear system to describe the relationship between Mari’s age and James’ age.

b. Graph the system on the grid.

j

m

10

210

220

230

240

20

30

40

220210230240 10 20 30 40

c. Five years from now, how old will Mari and James be?

3. Solve the linear system by graphing.

x 2 y 5 22

5x 1 3y 5 218 y

2

4

6

8

28 26 2422

24

26

28

2 4 6 8 x22

Answers

1.

2a.

2b. See graph.

2c.

3. See graph.


Name ——————————————————————— Date ————————————


LE

SS

ON

00.0

0

Mo

re C

op

y i

f n

ee

de

d


Qu

izze

s


4. The graph represents a system of linear equations. Which of the following best describes the number of solutions of the system?

y

x

2

22

24

26

28

4

6

8

24 222628 2 4 6 8

A. none B. one C. two D. infi nite

5. Describe the graph of a linear system of two equations with infi nitely many solutions.

Solving Linear Systems Using Algebra (Lessons 7.2–7.4)For Exercises 6–14, solve the linear system.

6. y 5 3 2 x 7. x 1 3y 5 5 8. 4x 1 3y 5 211 24(x 1 y) 5 212 2x 2 2y 5 26 24x 1 y 5 3

9. 2p 1 3q 5 10 10. x 2 3y 5 4 11. 5 1 4m 5 3n 5p 2 3q 5 27 2x 5 3y 1 5 6m 2 9n 5 1

12. 4a 2 5b 5 8 13. 5x 2 3y 5 3 14. 5y 2 3x 5 24

22a 1 b 5 1 3x 1 11 5 5y x 1 25 5 5 }

3 y

15. Which ordered pair is a solution of the linear system 14x 1 7y 5 7 and 24x 1 15y 5 3?

A. (2, 23) B. (3, 22) C. (2, 3) D. (3, 2)

16. Gina rents a car for a two-day trip. The fi rst day, she drives h hours, averaging 60 miles per hour. The second day, she drives one-half hour less and averages 51.2 miles per hour. Gina averages 56 miles per hour for the entire trip.

a. Write a linear system to model total distance d in terms of h hours.

b. How many total miles and hours did Gina drive?

17. A company makes rectangular tablecloths in two sizes. The perimeter of the smaller tablecloth is 196 inches. The larger tablecloth is 80% longer and 25% wider than the small tablecloth, with a perimeter of 300 inches. What is the area of the larger tablecloth?

A. 2000 square inches B. 2400 square inches

C. 5400 square inches D. 14,700 square inches

Answers

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16a.

16b.

17.


Name ——————————————————————— Date ————————————


Qu

izze

s


Answers

18. See graph.

19.

20. See graph.

21.

Solving Systems of Linear Inequalities (Lesson 7.6) 18. Graph the system of linear inequalities.

3y 2 x < 15 y 1 2x ≥ 21 y

2

4

6

8

28 26 2422

24

26

28

2 4 6 8 x22

19. Which ordered pair is a solution of the system of linear inequalitiesx 2 y > 6 and 3x 1 y < 10?

A. (0, 0) B. (3, 24) C. (4, 22) D. (6, 21)

20. Graph the system of linear inequalities x ≤ 5, y < 4, and x 2 2y ≥ 8.

y

x

1

2

21

22

23

24

3

41 2 3 6 7 85

4

21. Which system of inequalities is represented by the shaded region?

y

x

1

21

22

23

24

2

3

4

222324 1 2 3 421

A. x 2 y < 4 B. x 2 y > 4 x 1 y ≥ 22 x 1 y ≤ 22

C. x 2 y ≥ 4 D. x 2 y ≤ 4 x 1 y < 22 x 1 y > 22


Name ——————————————————————— Date ————————————


Qu

izze

s


Answers

22.

23a.

23b.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

22. Write a system of inequalities for the shaded region.

y

x22

24

26

28

4

6

8

24 222628 6 82

2

23. Yuri wants to use a special delivery service to send a package. The package’s width w can be no more than 3 times its height h. The sum all three measurements (length, width, and height) must be less than 9 feet. The diagram shows the measurements of the package.

w

h0.75 h

a. Write a system of linear inequalities to model w in terms of the package’s height h.

b. If the package is 1.5 feet high and 2.75 feet wide, can Yuri use the special delivery service? Explain.

Properties of Exponents (Lessons 8.1–8.3)For Exercises 24–32, simplify the expression. Write your answer using only positive exponents.

24. (23)(23)4 25. a3 p a25 p a23 26. (x3)5 p x3

27. (23pq22)5 28. 4wz3

} 5w2 29. 1 a 2

} 4 2 2 p 1 2 }

a3 2 5

30. (2x)0 1 03 31. 64 1 }

3 32. 64 2

1 } 2

33. Which expression is equivalent to 124?

A. 1 122 }

12 2 2 B. 1 124

} 12

2 2 C. 1 124 }

123 2 4 D. 1 124

} 122 2

4

34. Which expression is equivalent to (232 p 20 p 922)21?

A. 29 B. 2 1 } 9 C.

1 }

9 D. 9


Name ——————————————————————— Date ————————————


Qu

izze

s


Answers

35.

36.

37.

38.

39.

40.

41.

42.

43. See graph.

Scientifi c Notation (Lesson 8.4) 35. Which number represents 10,100,000,000 written in scientifi c notation?

A. 1.0 3 1010 B. 1.01 3 1010 C. 10.1 3 109 D. 101.0 3108

36. Which expression represents 0.00710 written in scientifi c notation?

A. 7.1 3 1023 B. 7.1 3 1024 C. 7.1 3 1022 D. 7.1 3 1025

37. Which number represents 1.678 3 1027 written in standard form?

A. 0.0001678 B. 0.00001678

C. 0.000001678 D. 0.0000001678

38. Which number is equivalent to (1.2 3 1024)2?

A. 1.44 3 1016 B. 1.44 3 1028 C. 1.44 3 1027 D. 1.44 3 108

For Exercises 39–42, evaluate the expression. Write your answer in scientifi c notation.

39. (7.2 3 103)(1.4 3 104) 40. (1.3 3 1023)(2.6 3 105)

41. (2.1 3 1027)2 42. 8.1 3 1023

} 2.7 3 105

Exponential Functions (Lesson 8.5) 43. Graph the function y 5

1 } 5 p 4x. Identify its domain and range.

y

x

1

2

3

4

5

6

7

8

222324 1 2 3 421


Name ——————————————————————— Date ————————————


Qu

izze

s


Answers

44.

45a.

45b.

46. See graph.

47.

48a.

48b.

44. The graph of which function is shown?

y

x

1

2

3

4

5

6

7

8

222324 1 2 3 421

(2, 1)

(4, 4)

A. y 5 1 }

2 p 4x B. y 5

1 }

4 p 2x

C. y 5 4 p 1 1 } 2 2

x D. y 5 2 p 1 1 }

4 2

x

45. The population of a country is growing by 1.5% per year. A census taken in 1999 showed a population of 9,800,000.

a. Write an exponential growth function to model the country’s population over time.

b. Assume the country’s population growth remains constant. What will the population be in 2009?

46. Graph the function y 5 (0.7)x. Identify its domain and range.

y

x

1

2

3

4

5

6

7

8

222324 1 2 3 421

47. Which of the following is an exponential decay function?

A. y 5 23 p x2 B. y 5 3 p x22

C. y 5 1 }

2 p 3x D. y 5 3 p 1 1 }

2 2

x

48. Jana buys a new car for $16,000. The value of the car decreases by 14.2% per year.

a. Write a function to model the value of Jana’s car.

b. What is the value of the car after Jana has owned it for 8 years? Round your answer to the nearest whole dollar.


Name ——————————————————————— Date ————————————


Qu

izze

s


Adding, Subtracting, and Multiplying Polynomials (Lessons 9.1–9.3)For Exercises 1–3, tell whether the expression is a polynomial. If it is a polynomial, fi nd its degree and classify it by the number of its terms.

1. 8a2x 1 3bx2 1 4c4

A. Yes; 2nd degree trinomial B. Yes; 3rd degree trinomial

C. Yes; 4th degree trinomial D. No

2. 4x2y 1 2x24

A. Yes; 3rd degree binomial B. Yes; 4th degree binomial

C. Yes; 3rd degree monomial D. No

3. 28m3 1 2mn5 2 6

4. Find the sum (y5 1 5y3 2 3y2) 1 (9y5 2 8y2 1 14).

5. Find the sum (4x2 2 3x 1 8) 1 (x3 2 2x 1 x2).

A. 5x3 2 5x 1 8 B. x3 1 5x2 2 5x 1 8

C. x3 1 5x2 2 x 1 8 D. x3 1 4x2 2 5x 1 8

6. Find the difference (4x2 2 2x 1 6) 2 (5x2 1 9).

7. Find the difference (3n2 2 n 1 8) 2 (n2 2 2n 2 5).

A. 2n2 1 n 1 13 B. 2n2 2 3n 1 13

C. 2n2 2 3n 1 3 D. 2n2 1 n 1 3

8. Find the product 24a3(5a3 1 a2 2 7).

9. Find the product (2p 2 4q)(5p 1 q).

10. Find the product (d 1 2)(d2 2 3d 1 1).

11. A sprinkler in the middle of a rectangular lawn sprays water in a circular pattern.

(4x 2 2)wateredlawn

unwateredlawn

sprinkler(3x 1 8)

x x

a. Write polynomials to model the area of the entire lawn and the area of the lawn that is watered.

b. Write a polynomial to model the area of the lawn that is not watered. Approximate π as 3.14.

Answers

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11a.

11b.


Name ——————————————————————— Date ————————————


Qu

izze

s



12. Find the product (w 2 6)2.

A. w 2 1 36 B. w 2 2 36

C. w 2 212w 1 36 D. w2 1 12w 2 36

13. Find the product (3x 1 y)2.

14. Find the product (9g 2 7)(9g 1 7).

A. 81g2 1 49 B. 81g2 2 49

C. 81g2 2 12g 1 49 D. 81g2 2 12g 2 49

Factoring Polynomials (Lessons 9.4–9.7) 15. Solve (2x 2 1)(x 1 3) 5 0.

16. Which of the following is a solution of (3a 1 12)(a 2 9) 5 0?

A. a 5 4 B. a 5 212 C. a 5 29 D. a 5 24

17. Solve 4x2 2 8x 5 0.

18. Solve 6n2 5 23n.

19. Which of the following is a solution of 3b2 2 15b 5 0?

A. b 5 3 B. b 5 23 C. b 5 25 D. b 5 5

20. Factor d2 2 2d 2 24.

21. Factor k2 1 k 2 56.

22. Which of the following is a factor of y 2 2 2y 2 15?

A. y 2 5 B. y 1 5 C. y 2 2 D. y 2 3

23. Factor 5y 2 2 3y 2 2.

24. Factor 23z2 2 z 1 4.

25. Which of the following is a factor of 210c2 1 43c 2 28?

A. 2c 1 7 B. 2c 2 14 C. 210c 1 2 D. 5c 2 4

26. A fi reworks expert is launching fi reworks from a platform. One of the fi reworks shot up in the air does not explode, and falls to the ground. The equation h 5 216t2 1 76t 1 20 models the height of the unexploded fi rework above the ground.

a. Factor 216t2 1 76t 1 20.

b. Explain how you can use these factors to fi nd the time it takes the unexploded fi rework to hit the ground.

27. Factor r 2 2 25.

28. Factor 4z2 2 49.

A. (4z 2 7)(z 1 7) B. 4(z 2 7)(z 1 7)

C. (2z 2 7)(2z 2 7) D. (2z 2 7)(2z 1 7)

Answers

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26a.

26b.

27.

28.


Name ——————————————————————— Date ————————————


Qu

izze

s



29. Factor 9x2 2 24x 1 16.

A. (3x 2 4)2 B. (3x 1 4)2

C. (3x 2 4)(3x 1 4) D. 9(x 2 4)2

30. Factor 2x3 1 6x2 1 3x 1 9.

Graphing Quadratic Functions (Lessons 10.1–10.2) 31. Graph y 5 2

1 }

2 x 2.

y

1

2

3

4

24 23 2221

22

23

24

1 2 3 4 x21

32. Graph y 5 2x 2 2 3.

y

2

4

6

8

28 26 2422

24

26

28

2 4 6 8 x22

33. Find the axis of symmetry of the graph of the function y 5 4x2 1 3x 2 1.

A. x 5 1 }

4 B. x 5 2

3 } 8 C. y 5

1 }

4 D. y 5 2

3 } 8

34. Which equation has a graph with vertex (6, 230)?

A. y 5 x2 2 12x 1 6 B. y 5 x2 1 6x 2 30

C. y 5 x2 1 6x 2 12 D. y 5 x2 230x 1 6

Answers

29.

30.

31. See graph.

32. See graph.

33.

34.


Name ——————————————————————— Date ————————————


Qu

izze

s



35. Graph y 5 2x2 1 3x 2 3.

y

2

4

6

8

28 26 2422

24

26

28

2 4 6 8 x22

36. Write an equation of the graph shown.

y

x

2

22

24

26

28

4

6

8

24 222628 2 4 6 8

37. Tell whether the function ƒ(x) 5 22x2 1 4x 2 10 has a minimum value or a maximum value. Then fi nd the minimum or maximum value.

A. minimum; 21 B. minimum; 28

C. maximum; 1 D. maximum; 28

38. The diagram shows a parabola formed by a telephone wire between two utility poles. The equation y 5 0.0008x2 2 0.08x 1 20 models the parabola. The value of y is the height in feet.

y

x

a. At its lowest point, what is the height of the wire above the ground?

b. What is the distance between the telephone poles?

Answers

35. See graph.

36.

37.

38a.

38b.


Name ——————————————————————— Date ————————————


Qu

izzes



Solving Quadratic Equations (Lessons 10.3–10.7) 39. Graph y 5 4x2 2 2x 2 12 on the grid below. Use the graph to solve

the equation 4x2 2 2x 2 12 5 0. Label the coordinates of any solutions. If the equation has no solution, write no solution.

2

4

6

8

10

12

14

16

18

22

24

26

28

210

212

214

216

218

22 2 4 6 8 10 12 14 16 18242628210212214216218

y

x

For Exercises 40�45, solve the equation. If the equation has no solution, write no solution.

40. 9p2 5 49 41. d 2 1 11 5 47

42. a2 1 4a 5 25 43. h2 2 20h 5 221

44. 218y 2 1 51y 5 8 45. x2 2 6x 2 14 5 0

46. What is the value of the discriminant of 3x2 2 5x 1 3 5 0?

A. 11 B. 25 C. 0 D. 211

47. What is the value of the discriminant of 9x 2 2 12x 1 4 5 0?

A. 0 B. 2144 C. 144 D. 236

48. How many solutions does 3b2 2 5b 5 24 have?

A. none B. one C. two D. three

49. How many solutions does (3c)2 5 24c 2 16 have?

A. none B. one C. two D. three

Answers

39. See graph.

40.

41.

42.

43.

44.

45.

46.

47.

48.

49.


Name ——————————————————————— Date ————————————


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Comparing Linear, Exponential, and Quadratic Models (Lesson 10.8) 50. What type of function does the ordered pairs ( 22, 24), (21, 22),

(0, 21), 1 1, 2 1 } 2 2 represent?

A. linear B. exponential

C. quadratic D. none of these

51. The table below represents a function.

x 22 21 0 1 2

y 19 9 3 1 3

Is the function linear, exponential, or quadratic? Explain.


x 1 2 3 4 5

y 40 60 90 135 202.5

What type of function does the table represent?

A. linear B. exponential

C. quadratic D. none of these


x 1 2 3 4 5

y 8.5 13 18.5 25 32.5

Write an equation for the function.

Answers

50.

51.

52.

53.


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Radical Functions (Lessons 11.1–11.3) 1. Graph y � Ï

}

x 2 3 � 2.

y

2

4

6

8

28 26 2422

24

26

28

2 4 6 8 x22

2. Which of the following is an equation of the graph shown?

y

x

2

22

24

26

28

4

6

8

24 222628 2 4 6 8

A. y � Ï} x � 3 B. y � � Ï}

x

C. y � �3 Ï} x D. y � � Ï}

x � 3

3. What are the domain and range of y 5 2 Ï}

x 1 5 1 1?

A. Domain: x ≥ 25; Range: y ≥ 1 B. Domain: x ≥ 0; Range: y ≥ 0

C. Domain: x ≥ 5; Range: y ≥ 21 D. Domain: x ≥ 1; Range: y ≥ 25

For Exercises 4–9, simplify the expression.

4. Ï}

147 5. Î}

12

} 121

6. Ï}

3x3 p Ï}

24x

7. Ï

}

3 }

Ï}

5c 8. 3 Ï

}

12 1 Ï}

75 9. (1 2 Ï}

7 )(3 1 2 Ï}

7 )

10. Simplify (8 1 Ï}

3 )2.

A. 16 1 2 Ï}

3 B. 73 1 8 Ï}

3 C. 64 1 16 Ï}

3 D. 67 1 16 Ï}

3

11. Which of the following is the solution of 3 Ï} x 2 12 5 0?

A. x 5 4 B. x 5 �16 C. x 5 16 D. x 5 �4

Answers

1. See graph.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.


Name Date


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For Exercises 12–15, solve the equation.

12. 3 Ï}

25 2 x 2 8 5 1 13. Ï}

23 1 2x 5 Ï}

11 2 x

14. Ï}

3 2 x 2 Ï}

5x 2 3 5 0 15. Ï}

3x 1 4 5 x

16. Which of the following is the solution of x 5 Ï}

5x 2 6 ?

A. x 5 2 only B. x 5 3 only

C. x 5 2 or x 5 3 D. no solution

Radicals in Geometry (Lessons 11.4–11.5) 17. A right triangle has one leg that is 4 inches shorter than the other

leg. The length of the hypotenuse is 2 Ï}

10 inches. Find the unknown lengths.

18. What is the distance between (6, 5) and (21, 8)?

A. Ï}

58 B. Ï}

82 C. 1 }

2 Ï}

194 D. 10

19. The distance between (3, 5) and (21, c) is 2 Ï}

5 . Find the value of c.

20. What is the midpoint of the line segment with endpoints (6, 5) and (21, 8)?

A. (5, 13) B. 1 5 } 2 ,

13 }

2 2 C. 1 7 }

2 ,

3 }

2 2 D. (3, 7)

21. Ann and Gina work at the offi ces shown on the map. They agree to meet for lunch at the place closest to the midway point between their offi ces. Where do they meet?

y

x

1

2

3

4

5

6

7

8

9

1 2 3 4 5 6 7 8 9

Ann’s Office

Gina’s Office

Snack Shack

Restaurant 99Cafe Rio

Jo’s Diner

A. Jo’s Diner B. Cafe Rio

C. Restaurant 99 D. Snack Shack

Answers

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.


Name Date


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s


Qu

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Operations on Rational Expressions (Lessons 12.3–12.6) 22. Divide 9x4 1 6x3 2 15x2 by 3x2.

23. Divide x2 2 15x 2 34 by x 1 2.

24. What is the remainder when you divide 3x2 2 2x 1 1 by x 1 1?

A. 24 B. 3x 2 5 C. 6 D. 0

25. Divide 2x2 2 5x 1 9 by 2x 2 3.

26. Which one of the following is an excluded value of the

expression 3v }

v2 1 4 ?

A. 0 B. 22

C. 2 D. no excluded values

27. Which one of the following is an excluded value of the

expression 3d 2 6

}} d2 1 7d 1 10

?

A. 5 B. 22

C. 2 D. no excluded value

28. Simplify the expression 3z2 2 24z

} z2 2 6z 2 16

.

A. 3z 2 8 } z 1 2

B. 3z2

} z 1 2

C. 3z }

z 1 2 D.

3z }

z 2 8

29. Simplify the expression 144r3 2 400r

} 3r 1 5

.

A. 16r(3r 1 5) B. 16r(3r 2 5)

C. 16r }

3r 1 5 D.

16r }

3r 2 5

For Exercises 30–32, fi nd the product or quotient.

30. 2w2 2 w 2 15

}} 3w2 1 7w 1 2

p 3w2 1 10w 1 3 }}

2w 1 5

31. (3x2 2 11x 1 6) 4 (x 2 4)

32. (25a2 2 15) 4 (5a 2 3)

33. Which one of the following is the LCD of 4x }

x2 2 9 and

3x 2 2 }

x2 2 x 2 6 ?

A. x 2 3 B. (x 1 2)(x 1 3)(x 2 3)

C. (x 1 2)(x 1 3)(x 2 3)2 D. (x2 2 9)(x2 2 x 2 6)

Answers

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.


Name Date


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s


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For Exercises 34–37, fi nd the sum or difference.

34. v 2 8

} v2 2 64

1 v 1 2 }}

v2 1 10v 1 16 35.

5x }

x 2 3 2

2x }

x 1 4

36. x }

x2 1 2x 1 1 2

3x 2 1 }

x2 2 2x 2 3 37.

3 } 7c 1

8c }

c 2 4

Graphing and Solving Rational Expressions (Lessons 12.1–12.2, 12.7) 38. Suppose that y varies inversely with x, and y 5 6 when x 5 3. What

is the constant of variation?

A. 2 B. 6 C. 12 D. 18

39. Suppose that y varies inversely with x, and y 5 3 when x 5 24. What is the value of y when x 5 3?

A. 24 B. 212 C. 3 D. 23

40. The table shows the average speed and fi nish time for different racers on the same model-car race track.

Racer A B C D

Average Speed (ft/sec) 3 4 5 6

Time (sec) 12 9 7.2 6

a. Is the relationship between speed and time an inverse variation? If yes, write an equation for time t in terms of speed s. If no, explain.

b. If a racer fi nishes in 8.4 seconds, what was the racer’s average speed? Round your answer to the nearest tenth.

41. Graph y 5 4 }

x 1 2 1 3.

y

2

4

6

8

28 26 2422

24

26

28

2 4 6 8 x22

For Exercises 42–45, solve the equation.

42. 10 } x 2 7 5

6 } x 1 3 43.

7 }

m 2 3 5

m }

4

44. 14c 2 21

} 6c 1 5

5 2c 1 3

} c 1 4 45. 5 }

n 2 1 1 3 5 n 2

4n 1 9 }

n 2 1

Answers

34.

35.

36.

37.

38.

39.

40a.

40b.

41. See graph.

42.

43.

44.

45.


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CHAPTER

1

Evaluate the expression. (Lesson 1.1)

1. 12x when x 5 6 2. 5.1 2 w when w 5 3.8

3. t }

8 when t 5 56 4. m2 when m 5 4

5. 2n when n 5 5 6. 1 }

2 k when k 5

3 }

4


7. 27 2 11 1 9 8. 8 1 4 p 3

9. 7 1 82 4 4 10. 14 4 (3 1 22)

Translate the verbal phrase into an expression. (Lesson 1.3)

11. The product of 9 and a number x

12. 15 more than a number y

13. The difference of a number z squared and 6

14. 11 less than three times a number w

Write an equation or an inequality. (Lesson 1.4)

15. The product of 7 and a number x is 42.

16. The sum of a number y and 17 is at most 36.

17. The product of 5 and the sum of a number z and 3 is less than 45.

Check whether the given number is a solution of the equation or inequality. (Lesson 1.4)

18. 13 1 m 5 19; 6 19. 3p 2 8 5 12; 7

20. n 2 1.2 ≤ 3.7; 5.1 21. r 2 1 8 > 21; 4

State the formula that is needed to solve the problem. Then solve the problem. (Lesson 1.5)

22. What is the interest earned on $500 invested for 7 years in an account that earns simple interest at a rate of 4% per year?

23. A bus travels at an average speed of 65 miles per hour. How many miles does the bus travel in 4.5 hours?

Ready to Go On? Cumulative ReviewFor use after Chapter 1

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CHAPTER

1

Identify the domain and range of the function. (Lesson 1.6)

24. Input 2 3 4 5 6

Output 3 8 13 18 23

25. Input 1 4 7 10 13

Output 1 2 3 4 5

Make a table for the function. Identify the range of the function. (Lesson 1.6)

26. y 5 x 2 7 27. y 5 4x 2 6

Domain: 11, 13, 15, 17, 20 Domain: 2, 3, 5, 7

28. A contractor buys screws for $1.55 per box and nails for $1.05 per box. Write an equation for the total cost. Then fi nd the total cost of 3 boxes of screws and 5 boxes of nails. (Lesson 1.6)

Graph the function. (Lesson 1.7)

29. y 5 5x 2 4 30. y 5 2 }

3 x 1

1 }

2

Domain: 1, 2, 3, 4, 5 Domain: 0, 3, 6, 9, 12

Write a rule for the function represented by the graph. Identify the domain and the range of the function. (Lesson 1.7)

31.

x

y

O 1 2 3

1

4

2

3

4

5

6

7

8

9 32.

x

y

O 1 2 3

1

4

2

3

4

5

6

7

8

9

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Evaluate the expression when x 5 3 and y 5 25. (Lessons 1.1, 2.3)

1. 7x 2 2 2. y 2 x 3. 2x 2 y 4. x2 2 (3 1 y)


5. 8 2 9 4 3 1 22 6. 17 2 13 } 32 2 1

7. 42 2 [(52 1 33) 4 4]

Write an algebraic expression, equation, or inequality. (Lesson 1.3, 1.4)

8. The difference of three times a number x and 4

9. 11 less than the product of 8 and a number y

10. The sum of a number x and 15 is 7.

11. 7 times a number y is less than the number y squared.

Check whether 3 is a solution of the equation or inequality. (Lesson 1.5)

12. 12 1 4x 5 19 13. x2 2 7 ≤ 12 14. 12x 2 5 > 3x2 1 5

15. Boarding Fees A dog kennel’s boarding fees are $11.50 per night for a small dog and $13.75 per night for a large dog. First write an algebraic expression for the total income from boarding fees. Then fi nd the total income if 7 small dogs and 5 large dogs were boarded for the night.

Write a rule for the function. Identify the domain and the range. (Lesson 1.6)

16. Input, x 5 6 7 8 9

Output, y 1 2 3 4 5

17. Input, x 0 2 4 6 8

Output, y 0 1 2 3 4

18. Graph the function y 5 3x 1 1 with domain 0, 1, 2, 3, and 4. (Lesson 1.7)

Order the numbers in the list from least to greatest. (Lesson 2.1, 2.7)

19. 2 1 } 3 , 0, 1.5, 20.6,

1 }

2 20. 3.2, 2 Ï

}

8 , 23.5, Ï}

9 , 2 Ï}

4

CHAPTER

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For the given value of x, fi nd 2x and ⏐x⏐. (Lesson 2.1)

21. x 5 221 22. x 5 20.75 23. x 5 2 } 7

Find the sum, difference, product, or quotient. (Lessons 2.2, 2.3, 2.4, 2.6)

24. –8 1 (27) 25. 1.3 1 (22.7) 26. 11 2 15

27. 23.8 2 7.3 28. 0.4 2 (20.8) 29. 2 7 }

16 2 1 2

3 }

8 2

30. 26(11) 31. 12 1 2 1 }

2 2 32. (235) 1 2

1 } 5 2

33. 242 4 (27) 34. 218 4 2 }

3 35.

9 }

28 4 1 2

3 }

4 2

Identify the property being illustrated. (Lesson 2.2, 2.4)

36. 5.7 1 (25.7) 5 0 37. [7 1 (29)] 1 (23) 5 7 1 [(29) 1 (23)]

38. 27 p 1 5 27 39. 28 p 9 5 9 p (28)

Simplify the expression. (Lesson 2.5, 2.6)

40. 26(x 1 9) 41. (y 2 11)(23y)

42. 9(z 2 6)

} 2 43. 8x 1 3(5x 2 4)

44. 1 } 4 (12y 2 2) 1 9y 45.

221z 1 7 } 7

Find the mean of the numbers. (Lesson 2.6)

46. 15, 27, 211 47. 12, 28, 9, 25

Rewrite the conditional statement in if-then form. Then tell whether the statement is true or false. If the statement is false, give a counterexample. (Lesson 2.7)

48. All irrational numbers are real numbers.

49. All integers are whole numbers.


50. 2 Ï}

144 51. 6 Ï}

36

52. Ï}

16 53. 2 Ï}

361


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Evaluate the expression. (Lessons 1.1, 1.2, 2.1)

1. 23 2 3 p 52 2. (17 2 8) 2 (5 2 7) 3. 7[(9 1 6) 4 3]

4. m2 2 8 when m 5 7 5. 21 2 n when n 5 13 6. 2p2 2 5 when p 5 3

7. 2q when q 5 23 8. ⏐r⏐ when r 5 212 9. 23s 2 2 when s 5 26

Write an algebraic expression, an equation, or an inequality. (Lessons 1.3, 1.4)

10. The difference of 3 times a number x and 11.

11. The quotient of a number y squared and 6 is greater than or equal to 20.

12. The sum of 17 and twice a number z is 32.

13. Newsstand A newsstand buys newspapers for $.75 each and magazines for $2.99 each. Write an algebraic expression for the total cost of the newsstand’s stock. Then fi nd the total cost of 30 newspapers and 20 magazines. (Lesson 1.5)

Write a rule for the function. (Lesson 1.6)

14. Input, x 15 17 21 22

Output, y 6 8 12 13

15. Input, x 1 2 3 4

Output, y 23 26 29 212

Tell whether the number is a real number, a rational number, an irrational number, an integer, or a whole number. (Lessons 2.1, 2.7)

16. Ï}

28 17. 2 2 } 5 18. 25 19. 1.2

Find the sum, difference, product, or quotient. (Lessons 2.2, 2.3, 2.4, 2.6)

20. 213 2 8 21. 2.8 1 (25.3) 22. 12.2 2 (26.7)

23. 2 3 } 10 1

2 } 5 24. 26(4) 25. 20.5(218)

26. 2 1 } 7 (221)(22) 27. 24 4 (28) 28. 226 4 1 2

13 } 17 2

Simplify the expression. (Lessons 2.5, 2.6)

29. 28(m 1 2) 30. 4(22x 1 1) 2 3x

31. 15x 2 40

} 25 32.

26x 1 21 }

23

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Solve the equation. Check your solution. (Lessons 3.1, 3.2, 3.3, 3.4)

33. x } 7 5 29 34. 2y 2 7 5 5

35. 47 5 24z 2 2z 2 7 36. 7 }

8 (a 2 8) 5 14

37. 3.1b 2 7 5 2.8b 1 8 38. 28(2c 2 5) 5 25(3c 1 7)

Solve the proportion. Check your solution. (Lessons 3.5, 3.6)

39. 7 }

8 5

21 }

22p 40.

7m 2 6 }

6 5

16 }

12 41.

4 } 7 5

23w 1 2 }

22w

42. Maps A map has a scale of 1 in. : 5 mi. The distance on the map between two cities is 11.5 inches. Estimate the actual distance between the cities. (Lesson 3.6)

Use the percent equation to answer the question. (Lesson 3.7)

43. What percent of 65 is 13? 44. 162 is what percent of 450?

45. What number is 12% of 140? 46. What percent of 144 is 151.2?

47. Survey In a school cell phone survey of 375 students, 210 of the students surveyed said they own a cell phone. What percent of the students surveyed own a cell phone? (Lesson 3.7)

Write the equation in function form. (Lesson 3.8)

48. 6x 1 2y 5 24 49. 5x 2 y 5 7 50. 7x 2 8 5 4y 2 3x

51. Suitcase The volume V of the suitcase is given by the formula V 5 lwh where l is the length, w is the width, and h is the height. (Lesson 3.8)

a. Solve the formula for l. b. Use the rewritten formula to fi nd the length of the suitcase shown.

V = 7560 in.3

30 in.

12 in.

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3 Ready to Go On? Cumulative Review cont’dFor use after Chapter 3

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1. Packaging A computer monitor is packaged in a cube-shaped box with an inside edge length of 2.5 feet. What is the volume of the box? (Lesson 1.1)

Evaluate the expression. (Lessons 1.1, 1.2, 2.2–2.6)

2. 32 p 23 2 17 3. 2 } 3 (9)(5 1 2)

4. x }

4 when x 5 224 5. 7(2y 2 1) when y 5 3

6. ⏐m⏐when m 5 28 7. 23x 2 (x 2 2) when x 5 24


8. The quotient of 5 times a number x and 7

9. The difference of a number y cubed and 4 is less than the product of 8 and the number y.

10. 6 times the sum of a number z and 3 is 27.

Check whether the given number is a solution of the equation or inequality. (Lessons 1.3, 1.4, 2.1–2.3)

11. 9 2 2x 5 23; 26 12. 21 ≥ 3y 2 5; 4 13. 1 }

2 z 1 6 < 23z 1 12; 8

Find the sum, difference, product, or quotient. (Lessons 2.2–2.6)

14. 22.7 1 (28.6) 15. 23 2 } 5 2 1 21

1 }

10 2 16. 28 p 9

17. 23(213)(24) 18. 18 4 1 2 3 } 5 2 19. 2

3 }

4 4 1 2

11 }

12 2

20. Checkbook You record the following deposits and withdrawals in your checkbook: $75, 2$36, $51, 2$17, 2$24. Find the mean of the account activity. (Lesson 2.6)

Order the numbers from least to greatest. (Lesson 2.7)

21. 2 Ï}

36 , 0.5, 26.2, 2, Ï}

2 22. Ï}

3 , 2 1 } 3 , 2.1, 2 Ï

}

6 , 23.5

Solve the equation. (Lessons 3.1–3.4)

23. a 2 17 5 9 24. k }

23 5 11 25.

2 }

3 x 2 7 5 17

26. 7q 1 18 2 11q 5 26 27. 32 5 2 } 5 (15p 2 10) 28. 7(2m 2 3) 5 12m 2 15

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Solve the proportion. (Lesson 3.6)

29. 7 }

8 5

42 }

x 30. 4 }

3y 5

16 }

60 31.

z }

z 1 4 5

9 }

21

32. Scale Drawing A zoo is designing a new polar bear exhibit using the scale 1 in. :4 ft. The drawing has a length of 25 inches and a width of 15 inches. What will be the actual length and width of the polar bear exhibit? (Lesson 3.6)


33. What number is 15% of 28? 34. What percent of 62 is 24.8?

35. 88.2 is what percent of 252? 36. What number is 225% of 180?

37. Favorite Vegetable In a school favorite vegetable survey of 375 students, 165 of the students surveyed said that corn is their favorite vegetable. What percent of the students surveyed said corn is their favorite vegetable? (Lesson 3.7)

Graph the equation. (Lesson 4.2, 4.3, 4.5)

38. 3x 1 y 5 5 39. y 5 6 40. 7x 2 14y 5 28

Find the slope of the line that passes through the points. (Lesson 4.4)

41. (2, 7) and (11, 15) 42. (23, 27) and (27, 5) 43. (21, 22) and (29, 6)

Identify the slope and y-intercept of the line with the given equation. (Lesson 4.5)

44. y 5 25x 1 8 45. x 1 2y 5 10 46. 4x 2 3y 5 24

Tell whether the equation represents direct variation. If so, identify the constant of variation. (Lesson 4.6)

47. 5x 2 7y 5 0 48. x 2 1 }

2 y 5 3 49.

1 }

2 x 1

1 }

3 y 5 0

50. Cereal The amount of sugar s (in grams) in a particular cereal varies directly with the amount of cereal C in the serving. The box recommends a serving size of 25 grams, which contains 5 grams of sugar. (Lesson 4.6)

a. Write a direct variation equation that relates C and s.

b. What would be the amount of sugar in a 35-gram serving of cereal?

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Evaluate the expression. (Lessons 1.2, 2.2–2.6)

1. 21 4 (5 2 2) 2. 1 } 2 [(5 1 6)2 2 85]

3. 2x }

36 2 x2 when x 5 23 4. 4y 1 8 2 3y2 when y 5 4

5. ⏐z⏐when z 5 23 6. 7 2 (m 2 p) when m 5 28 and p 5 29

7. Building Materials A contractor is putting wood baseboards around a room that is 31 feet long by 17 feet wide. The baseboards come in 8-foot long pieces. How many pieces of baseboards should the contractor buy to go around the room? (Lesson 1.5)


8. 211 1 18 9. 21.1 2 7.3 10. 3 } 10

2 1 2 1 } 5 2

11. 15(24) 12. 2 1 } 5 (20)(23) 13. 221 4 1 2

3 } 7 2

Simplify the expression. (Lessons 2.5, 2.6)

14. –3(x 1 4) 15. 5(b 2 9) 2 7b 16. 224a 2 8

} 4

Solve the equation. Check your solution. (Lessons 3.1–3.4)

17. m 2 8 5 217 18. 212p 5 60

19. 5q 1 11 5 26 20. 3r 2 17 1 7r 5 83

21. –5.1x 1 4.1 5 6.7x 2 1.8 22. 2 }

3 (6k 2 9) 5 3k 1 8

23. Discount Cards A bookstore sells frequent buyer discount cards for $12 each. The cost of each book with the discount card is $7. The cost of a book without the card is $9. After how many book purchases does a cardholder and a non-cardholder spend the same amount of money, if you include the cost of the card? (Lesson 3.2)

Solve the proportion. (Lesson 3.5)

24. w

} 6 5

11 }

15 25. 40 }

x 5

15 }

21 26.

4 }

18 5

y 1 3 }

54 27.

3 } 7 5

z }

z 1 16

28. Retirement Savings A company’s 401K-retirement program allows an employee to invest up to 15% of their gross earnings in a retirement account. If an employee invests the full 15% and earns $2500 a month, how much is the employee investing each month? (Lesson 3.7)

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Solve the literal equation. (Lesson 3.8)

29. ax 1 by 5 1 for y 30. y 2 1 x } 2 2 3z 5 0 for x


31. (7, 2) and (1, 21) 32. (0, 4) and (22, 22) 33. (5, 22) and (23, 21)

Identify the slope and y-intercept of the line with the given equation. (Lesson 4.3–4.5)

34. y 5 22x 1 7 35. 3x 1 1 }

2 y 5 5 36. 5x 2 4y 5 0

Tell whether the equation represents direct variation. If so, identify the constant of variation. (Lesson 4.6)

37. x 1 7y 5 0 38. 2x 1 3y 5 1 39. 3x 1 5 5 4y 1 5

Graph the equation. (Lesson 4.2–4.5)

40. y 5 23 41. y 5 2 }

3 x 2 3 42. 6x 1 3y 5 3

Evaluate the function for the given value. (Lesson 4.7)

43. f (x) 5 3x 2 5 when x 5 7 44. g(x) 5 2 } 7 x 1 8 when x 5 14

Write an equation in slope-intercept form of the line with the given characteristics. (Lesson 5.1–5.4)

45. slope: 8; y-intercept: 23 46. slope: 1 }

2 ; passes through (2, 8)

47. passes through (22, 5) and (3, 25) 48. horizontal line; passes through (4, 3)

Make a scatter plot of the data. Draw a line of fi t. Write an equation of the line. (Lesson 5.6, 5.7)

49. x 0 2 3 4 5

y 6 8 12 14 15

50. x 1 2 3 4 5

y 3 3 4 4 5

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Evaluate the expression. (Lessons 1.1, 1.2)

1. 11 1 8 p 32 2 8 2. 2 }

3 [(5 1 2)2 2 13] 3. 5k2 2 7k when k 5 3

Check whether the given number is a solution of the equation or inequality. (Lesson 1.4)

4. 7x 2 11 5 17; 4 5. 9y 1 3 < 13; 2 6. 5z 2 9 ≥ 7z 2 11; 22


7. 5 1 (213) 8. 217 1 (221) 9. 15 2 (212)

10. 27(211) 11. 2 1 }

2 (15)(28) 12.

72 }

26


13. Ï}

169 14. 2 Ï}

9 15. 6 Ï}

256 16. 6 Ï}

0.04 4

Solve the equation, if possible. (Lessons 3.1–3.4)

17. x 1 7 5 19 18. 23 2 y 5 14 19. 7p 2 11 5 10

20. q }

4 1 8 5 13 21. 15j 1 7j 5 11

22. 14m 1 7 2 8m 5 217 23. 23.1t 1 4.3 5 21.9t 1 3.7

24. 3(4k 2 2) 5 16k 2 32 25. 1 }

3 (9g 2 6) 5

3 } 5 (5g 1 15)

26. Reading A student can read 4 pages of a book in 5 minutes. How many pages can the student read in 60 minutes? (Lesson 3.6)


27. What percent of 38 is 15.2? 28. What number is 35% of 150?

29. At your school, 74% of students bought their lunch at the cafeteria. There are 1250 students in the school. How many students bought their lunch?

Solve the literal equation. (Lesson 3.8)

30. Solve ax 1 by 5 c for a. 31. Solve ax 2 y }

b 5 c for y.

Plot the point in a coordinate plane. Describe the location of the point. (Lesson 4.1)

32. A(3, 6) 33. B(22, 24) 34. C(0, 23)

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35. (4, 8) and (5, 13) 36. (23, 6) and (2, 7) 37. (28, 22) and (25, 22)

Graph the equation. (Lessons 4.2, 4.3, 4.5)

38. y 2 2x 5 23 39. x 5 24 40. y 5 3 } 5 x 1 3


41. f (x) 5 2 1 }

2 x 1 3 when x 5 22 42. g(x) 5 8x 2 7 when x 5 0.5

43. Spillways The amount of water w (in thousands of gallons) fl owing over a dam’s spillway varies directly with time t (in hours). In 4 hours, 6 thousand gallons of water fl ow over the spillway. Write a direct variation equation that relates w and t. How many thousands of gallons of water spill out of the dam in 12 hours?

Write an equation in slope-intercept form of the line that passes through the given point and has the given slope m. (Lessons 5.1, 5.2)

44. (21, 3), m 5 2 45. (4, 23), m 5 3 }

4 46. (2, 27), m 5 25

47. Write an equation in standard form of the line that has a slope of 27 and passes through the point (3, 27). (Lesson 5.4)

48. Make a scatter plot of the data. Describe the correlation of the data. (Lesson 5.6)

x 0 2 4 6 8

y 7 16 25 38 50

Solve the inequality, if possible. Graph your solution. (Lessons 6.1–6.5)

49. x 2 13 < 28 50. 5x 1 6 ≥ 16

51. 7x 2 3 > 2(6x 2 4) 52. 25 < 7 2 2x < 5

53. 4x 1 1 ≤ 9 or 2x 2 5 ≥ 9 54. ⏐2x 2 5⏐ 1 2 < 8

55. Graph the inequality y ≤ 3x 2 4. (Lesson 6.7)

56. Summer Work You work two part-time jobs during the summer. You earn $5 per hour washing cars, and $8 per hour as a swimming instructor. You want to earn at least $300 over the summer. Write an inequality that describes your goal in terms of hours spent washing cars and hours spent as a swim instructor. Graph the inequality. (Lesson 6.7)

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Evaluate the function. (Lessons 1.1, 1.2)

1. 25 2 42 2. 3 p 2 1 8

3. 4[28 2 (8 2 3)2] 4. 12m 1 13 when m 5 5

5. n3 when n 5 2 }

3 6.

9g 2 6 }

3g when g 5 4

Make a table for the function. Identify the range of the function. (Lesson 1.6)

7. y 5 2x 2 5; domain: 0, 3, 5, 7, 10 8. y 5 1 }

2 x 1 3; domain: 0, 2, 4, 6, 8


9. 29 1 Ï}

121 10. 2 2 } 6 2

5 }

12

11. 22.8 2 (27.3) 12. (22.5)(12)

13. 42 4 (26) 14. 4 }

9 4 1 2

6 } 7 2

Solve the proportion. Check your solution. (Lessons 3.5, 3.6)

15. 40

} 56

5 5 } y 16. m }

7 5

5m }

35 17.

38 }

72 5

2q 1 1 }

4q

18. Scale Drawings You are making a scale drawing of a sailboat using the scale 1 inch : 3 feet. The length of the sailboat at its longest point is 36 feet and the height is 54 feet. What should the length and height of the sailboat be in your drawing? (Lesson 3.6)


19. What percent of 72 is 28.8? 20. What number is 110% of 55?

21. 20 is what percent of 125? 22. What percent of 15 is 39?


23. (7, 2) and (11, 14) 24. (23, 6) and (5, 22)

Graph the equation. (Lessons 4.2, 4.3, 4.5)

25. x 5 y 26. y 5 5 }

3 x 2 1


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Write an equation in slope-intercept form of the line with the given characteristics. (Lessons 5.1–5.3, 5.5)

27. slope: 2 5 } 2 ; y-intercept: 23 28. passes through (2, 21); m 5 2

29. passes through (7, 2) and (12, 12) 30. parallel to y 5 3x 1 7; passes through (3, 0)

31. Write an equation in standard form of the line that passes through (1, 1) and (9, 7). (Lesson 5.4)


32. y 2 13 < 7 33. 23x ≤ 15

34. 8z 2 3 > 5z 1 12 35. 26 ≤ 5x 1 4 < 19

36. ⏐x 1 5⏐1 11 < 7 37. 3⏐2y 1 3⏐> 12

38. Summer budgeting You have saved $350 to spend on summer activities. The pool costs $8 per day and a movie costs $6 per movie. Write an inequality that describes the possible combinations of pool visits and movies. Then graph the inequality and give two possible combinations of activities. (Lesson 6.7)

Solve the linear system. (Lessons 7.1–7.4)

39. 7x 1 4y 5 2 40. 3x 1 y 5 6 41. 3x 2 4y 5 24

7x 2 2y 5 20 27x 1 6y 5 11 2x 2 15y 5 22

Tell whether the linear system has one solution, no solution, or infi nitely many solutions. (Lesson 7.5)

42. 9x 1 13y 5 10 43. 2x 1 5y 5 15 44. 24y 5 27x 2 4

9x 2 13y 5 8 10y 5 24x 1 30 7x 5 4 1 4y

Graph the system of linear inequalities. (Lessons 6.7, 7.6)

45. y ≥ 22x 1 1 46. x ≤ 3

y < 5x 2 2 y < 4

y ≥ 2x 2 3


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1. 4 p 7 2 11 2. 1 }

3 (15 1 32)

3. 63 4 9 p 2 1 7 4. 2 }

3 w when w 5

6 } 7

5. 12h

} h 2 4

when h 5 8 6. 6(t2 2 8) when t 5 4

7. Perimeter A rectangular table is 7 feet long and has an area of 42 square feet. What is the perimeter of the table? (Lesson 1.5)


8. 3.2 1 (28.4) 9. 2 2 } 3 2

1 }

8

10. 212(24) 11. 2 2 } 3 1 2

6 } 7 2 1 2

14 } 11 2

12. 2 } 7 4 (214) 13. 2

3 } 8 4

9 }

16

Simplify the expression. (Lessons 2.2–2.6)

14. 26(x 2 2)

} 23

15. 9(22x 1 7) 1 11x 16. (23z)(2z 1 3) 1 4z


17. 14 2 3x 5 23 18. 22 5 7y 2 3y 1 14

19. 5 }

3 (15w 2 9) 5 35 20. 27(3m 2 2) 5 24(2m 11)

21. Scale Drawing A map has a scale of 1 cm : 13 km. The distance bewteen two cities on the map is 5.2 centimeters. What is the actual distance between the cities? (Lessons 3.5, 3.6)


22. y 5 x 23 23. 5x 2 3y 5 30 24. 24x 5 8y 1 16


25. f (x) 5 2 2 } 3 x when x 5 29 26. g (x) 5

1 }

2 x 2 7 when x 5 212


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Write an equation in point-slope form of the line that passes through the given points. (Lessons 4.4, 5.3)

27. (6, 3) and (22, 1) 28. (23, 23) and (1, 26) 29. (0, 1) and (1, 0)

30. Write an equation in standard form of the line that passes through (3, 22) and is (a) parallel to and (b) perpendicular to y 5 22x 1 3. (Lesson 5.4, 5.5)


31. x 2 11 < 23 32. y }

23 ≥ 2 4

33. 3x 2 13 ≤ 5 34. 23 ≤ 7 1 2x < 1

35. 4x > 28 or 3x 2 1 ≤ 210 36. 2⏐3x 22⏐ 2 1 < 7

Tell whether the ordered pair is a solution of 26x 1 7y < 20. (Lesson 6.7)

37. (3, 5) 38. (22, 2) 39. (24, 26)


40. x 5 3y 1 7 41. 3x 2 7y 5 27 42. 3x 2 4y 5 7

2x 1 4y 5 25 5x 1 7y 5 211 5x 1 8y 5 247

Graph the system of linear inequalities. (Lesson 7.6)

43. y > x 1 5 44. x ≥ 5 45. y < 0

y ≤ 22x 1 1 y < 3 } 5 x 2 2 x ≥ 3

y ≥ 23x 1 2

Simplify the expression. Write your answer using only positive exponents. (Lessons 8.1–8.3)

46. 58 p 511

} 513 47. m6 p m p m7 48.

5x2 }

2xy3 p (2xy)2

} z

Evaluate the expression. Write your answer in scientifi c notation. (Lesson 8.4)

49. (2.1 3 106)(3.2 3 10215) 50. 4.2 3 108

} 1.2 31022

51. Graph the function y 5 1.4x. Identify the domain and the range of the function. (Lesson 8.5)

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Ready to Go On? Cumulative Review For use after Chapter 9


1. The product of 8 and a number x is equal to 32.

2. Three times the sum of a number y and 8

3. The difference of 10 and the square of a number z is less than 18.

4. Graph the function y 5 5x 2 3 with domain 1, 2, 3, 4, and 5. (Lesson 1.7)

Simplify the expression when x 5 22 and y 5 5. (Lessons 2.2–2.6)

5. ⏐x⏐ 2 (2y) 6. 23y 1 x2

7. 5 4 1 } x 8. 3(y 2 2) 2 2y

9. ⏐x 1 y⏐ 10. xy 2 4x

Solve the proportion. (Lessons 3.5, 3.6)

11. d }

21 5

18 }

63 12. 9 }

m 1 7 5

45 } 55 13.

y }

15 5

y 1 12 } 35

14. Survey In a school survey of 2550 students, 1428 of the students surveyed said they regularly attend the school’s football games. What percent of the students surveyed regularly attend the school’s football games? (Lesson 3.7)


15. y 5 4 } 3 x 2 1 16. 5x 2 3y 5 30 17. 6x 2 y 5 3

Tell whether the equation represents direct variation. If so, identify the constant of variation. (Lesson 4.5, 4.6)

18. 6x 2 3y 5 0 19. 2x 1 2y 5 4 20. 1 }

2 x 2 2y 5 0

Write an equation in slope-intercept form of the line with the given characteristics. (Lessons 4.4, 5.1–5.3)

21. slope: 5; y-intercept: 22 22. passes through (2, 7); slope: 1

23. passes through (2, 3) and (6, 5) 24. passes through (1, 4); parallel to y 5 7x 2 3

Solve the inequality, if possible. Graph your solution. (Lessons 6.1–6.4, 6.6)

25. 24x > 28 26. 9x 2 5 < 13 27. ⏐2x 2 1⏐ > 14

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Graph the inequality. (Lesson 6.7)

28. y ≤ 2x 2 8 29. y > 7


30. y 5 5x 2 2 31. 9x 1 2y 5 5 32. 3x 1 7y 5 43

7x 2 3y 5 210 9x 1 4y 5 19 2x 2 10y 5 274

Tell whether the linear system has one solution, no solution, or infi nitely

many solutions. (Lesson 7.5)

33. –2x 1 y 5 3 34. x 5 8y 1 7 35. 3x 2 7y 5 10

y 5 2x 1 5 3x 2 2y 5 12 6x 5 14y 1 20


36. (23)2(23)(23)8 37. 2(11x3)2

38. 1 m22n }

2mn3 2 2 39. (4a2b)3 p 2a–3

40. Write 19,321,000 in scientifi c notation. (Lesson 8.4)

41. Write 3.21 3 1024 in standard form. (Lesson 8.4)

42. Graph the function y 5 1 1 } 2 2 x. Identify the domain and the range of the function.

(Lesson 8.5)

Find the sum or difference. (Lesson 9.1)

43. (7h3 1 4h2 2 3h) 1 (9h3 2 2h2 2 4h) 44. (11k2 2 3k 1 5) 2 (23k2 1 7k 1 2)

Find the product. (Lessons 9.2, 9.3)

45. (w2 1 5w 2 6)(w 1 2) 46. (3y 2 7)(2y 1 3) 47. (5x 2 9)2

Factor the polynomial. (Lessons 9.5–9.8)

48. k2 2 15k + 36 49. 3m3 1 12m2 2 36m 50. 22a2 1 72

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1. 15 1 23 p 4 2. (32 2 18) 2 (17 2 11) 4 3

3. 3 }

2 [33 2 (4 1 1)2] 4. m4 when m 5

1 } 2

5. 9(r 2 7) when r 5 12 6. t2

} 9 1 2t when t 5 6


7. 2 2 } 5 1 1 2

1 } 10 2 8. 1.6 2 (22.7)

9. 22.2(1.5) 10. 3 } 5 1 2

4 } 9 2 1 2

5 } 12 2

11. 42 4 (27) 12. 2 5 } 11 4 255


13. 232 5 9x 2 5 14. y }

y 1 7 5 2 15.

2 }

3 (9z 2 21) 5 22z 1 10

16. Wood Stain A label on a can of wood stain states that one pint can stain 24 square feet. How many pints of stain are needed to stain a piece of furniture that has 108 square feet to stain? (Lessons 3.5, 3.6)

17. Survey In a classroom survey of 30 students, 18 of the students surveyed said that they eat spinach. What percent of the students surveyed eat spinach? (Lesson 3.7)


18. y 5 7x 1 2 19. 22x 1 3y 5 9 20. 9x 2 5y 5 45

Graph the equation. (Lessons 4.5–4.7)

21. y 5 2 1 } 2 x 1 6 22. 4x 2 2y 5 10

Write an equation in slope-intercept form of the line with the given characteristics. (Lessons 4.4, 5.1–5.3)

23. slope 5 2 1 } 2 : y-intercept 6 24. passes through (1, 4), m 5 23

25. passes through (3, 6) and (7, 14) 26. passes through (23, 22),

perpendicular to y 5 3x 1 2

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27. x 1 17 < 4 28. x } 28

> 22

29. 2(3x 2 1) ≤ 7x 1 9 30. 2 < 5x 2 3 ≤ 17

31. ⏐4x 2 5⏐ < 11 32. 1 } 2 ⏐x 1 3⏐ 1 4 < 3


33. y 5 0.5x 1 4 34. 5x 1 7y 5 213 35. 2x 1 3y 5 22

1.2x 2 2y 5 27.2 5x 2 11y 5 59 10x 2 7y 5 222

Graph the system of linear inequalities. (Lesson 7.6)

36. y < 2x 1 4 37. y ≤ 2x 1 5

y > 2 }

3 x 1 1 y ≥ 5x 1 1


38. 1 }

623 39. [(23)2]3 40. ( p6)4 41. 1 4x22y }

3xy23 2 2

Evaluate the expression. Write your answer in scientifi c notation. (Lesson 8.4)

42. (3.9 3 104)2 43. 5.7 3 109 }

1.2 3 104 44. (1.8 3 106)(9.2 3 10–9)


45. (3x 2 2)(x 1 11) 46. (7y 2 3)2 47. (5z 1 4)(z2 1 7z 2 6)


48. 21x2 5 12x 49. 6x3 5 24x 50. x2 2 56 5 2x

51. The area of a triangle with a base b of (5x 2 4) inches and a perpendicular height h of 2x inches is 672 square inches. Find the value of x. Then give the dimensions of the triangle. (Lesson 10.6)

Tell whether the equation has two solutions, one solution, or no solutions. (Lesson 10.7)

52. 2x2 1 3x 5 5 53. 29x2 1 12x 5 4 54. 11x2 2 6x 1 5 5 0

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1. 6[14 2 (11 2 9)3] 2. 45 } t3

when t 5 3 3. 3m 1 14 ÷ m when m 5 7

4. A bus travels 260 miles in 4 hours. Find the unit rate. (Lesson 1.3)


5. 23 1 7 2 (28) 6. 16 2 (27 2 4) 7. 212(9)

8. 2 2 } 3 (15)(24) 9. 254 4 6 10. 2

2 } 9 4 1 2

5 } 6 2


11. 1 }

2 (8x 2 7) 5

2 } 3 (15x 2 12) 12. x }

214 5 23 13. 21x 5 17x 2 12

14. Reading A student can read 15 pages of a book in 1 hour. How long will it take to read 88 pages of the book? (Lessons 3.5, 3.6)

Plot the point in a coordinate plane. Describe the location of the point. (Lesson 4.1)

15. A(3, 5) 16. B(22, 21) 17. C(0, 3) 18. D(4, 23)


19. (28, 26) and (2, 14) 20. (4, 9) and (7, 9) 21. (2, 24) and (21, 0)

Write an equation in standard form of the line with the given characteristics. (Lesson 5.4)

22. passes through (22, 2) and (3, 6) 23. slope: 28; passes through (24, 13)

24. Car Rentals Your family is renting a car for a day. The car costs $55 per day plus an additional $.25 per mile driven. Write an equation that gives the total cost C as a function of the number of miles driven m. (Lesson 5.3)


25. 25x > 7x 1 24 26. 27 ≤ 4x 1 1 < 17 27. ⏐3x 1 5⏐ 1 4 < 2


28. x 5 3y 2 8 29. 3x 2 7y 5 8 30. 7x 1 4y 5 219

7x 2 5y 5 48 9x 1 7y 5 232 5x 2 3y 5 4

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Simplify the expression. Write your answer using only positive expo-nents. (Lessons 8.1–8.3)

31. (126)4 32. 1 } r3

p r17 33. (9k7)2 p (3k2)23

Graph the function and identify its domain and range. (Lessons 8.5, 8.6)

34. y 5 (2.1)x 35. y 5 4 1 1 } 4 2

x


36. (5x 2 3)(7x 1 4) 37. (4x 2 13y)2 38. (4x2 2 x 1 7)(2x 2 3)


39. x2 2 81 40. 26y2 1 5y 1 21 41. 14z3 2 15 2 6z 1 35z2

Solve the equation. Round your solution to the nearest hundredth if necessary. (Lessons 10.4, 10.6)

42. 8x2 5 288 43. x2 1 12x 5 15 44. 7x2 2 6 5 2x

45. Tell whether the table of values represents a linear function, an exponential function, or a quadratic function. Then write an equation for the function. (Lesson 10.8)

x 22 21 0 1 2

y 26 210 212 212 210

46. Graph the function y 5 Ï}

x 1 2 and identify its domain and range. Compare the graph with the graph of y 5 Ï

}

x . (Lesson 11.1)

Simplify the expression. (Lesson 11.2)

47. Ï}

12x5 p Ï}

6x 48. Ï}

15 } 2y2

49. (6 1 Ï}

6 )(6 2 2 Ï}

6 )

Solve the equation. Check for extraneous solutions. (Lesson 11.3)

50. Ï}

20 2 x 5 x 51. Ï}

2x 2 14 5 Ï}

x 1 17 52. 3 Ï}

x 2 6 1 8 5 32

Find the distance between the given points. Then fi nd the midpoint of the line segment whose endpoints are the given points. (Lesson 11.5)

53. (11, 5), (21, 23) 54. (28, 22), (22, 23)


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1. The product of 8 and the quantity of 13 and a number y

2. The difference of 11 and the square of a number m is no more than twice the number.

Simplify the expression when k 5 24 and m 5 9. (Lessons 2.222.7)

3. 22k 1 4m 4. 2 1 } 2 km

5. m 1 3

} k 6.

7k 2 2m }

km

7. m(k2 2 6) 8. ⏐k⏐2 Ï}

m


9. 211x 1 8 5 63 10. 5x } 7x 1 6

5 5 11. 4 } 7 (14x 2 21) 5 7x 1 11


12. 342 is what percent of 95? 13. 97.2 is 27% of what number?


14. (9, 5) and (7, 8) 15. (29, 29) and (27, 25)

Graph the equation. (Lessons 4.6, 4.7)

16. y 5 x 1 7 17. y 5 27 18. 3x 1 4y 5 12

Write an equation in slope-intercept form of the line with the given characteristics. (Lessons 5.1, 5.2)

19. passes through: (3, 2); slope: 22 20. passes through (21, 25) and (3, 3)

Solve the inequality. Graph your solution. (Lesson 6.1-6.6)

21. 20.7 ≥ x 2 1.2 22. 7 2 2x < 19 23. 26 ≤ 5x 2 2 ≤ 12

24. Taxi Service A taxi service charges $5 per ride plus $1.50 per mile. Suppose you only have $20 to spend. Find the possible number of miles you can travel. (Lesson 6.6)

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Solve the linear system. (Lessons 7.2-7.4)

25. y 5 0.2x 1 7 26. 5x 2 2y 5 13 27. 2x 2 3y 5 218

0.5x 2 0.1 y 5 4.1 23x 2 2y 5 227 5x 1 4y 5 1


28. (23)4 29. t5 }

s4 p 3ts2

} t22 30. 1 q

2s21

} 5q 2 3

31. Write 0.00268 in scientifi c notation. (Lesson 8.4)

32. Write 7.19 3 104 in standard form. (Lesson 8.4)


33. (14 2 11b)(14 1 11b) 34. (x 2 8)(4x 1 7) 35. 7x(x 1 3)2


36. 2x2 1 13x 2 7 37. 3y3 2 75y 38. 5z3 2 2z 1 10z2 2 4

Solve the equation. (Lessons 10.3-10.6)

39. 7x2 5 63 40. x2 1 28 5 11x 41. 9(q 2 2)2 5 27

42. 2m2 1 8m 5 16 43. 3r2 2 9r 5 5 44. 6t 1 7 5 15 2 11t2

45. Area The area of a circle is 1962.5 square centimeters. Find the radius of the circle. Use 3.14 for π. (Lesson 11.3)

Let a and b represent the lengths of the legs of a right triangle, and let c represent the length of the hypotenuse. Find the unknown length. (Lesson 11.4)

46. b 5 5, c 5 5 Ï}

10 47. a 5 5.5, b 5 30 48. a 5 7.5, c 5 56.5

Given that y varies directly with x, use the specifi ed values to write an inverse variation equation that relates x and y. Then fi nd y when x 5 2. (Lesson 12.1)

49. x 5 5, y 5 8 50. x 5 3 }

4 , y 5 12

Divide. (Lesson 12.3)

51. (2r2 1 19r 1 29) 4 (r 1 8) 52. (6t2 2 19t 1 13) 4 (2t 2 5)


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Number PuzzlesGiven the following scenarios, fill in the blank with one of these numbers.

6.5 500 �4 1 �75 ��

17 �5 18 11 __ 9

5 __ 6 �

� 15 96 ___

4 4.5 � �

� 11 0 2 __

3 300 �12

1. When you square my value and subtract it from 23, the result is the square root of 36. I am an irrational number. What is my number?

2. If you cube my value and divide it by the quotient of 150 and 6, the result is my value. I am not a natural number, but I am a rational number. What is my number?

3. When you take my value and multiply it by �8, the result is an integer greater than �220. If you take the result and divide it by the sum of �10 and 2, the result is my value. I am a rational number. What is my number?

4. If you take my absolute value and subtract the square of 16, the result is a double-digit number whose prime factorization is 11 � 2 � 2. I am a natural number. What is my number?

5. When you add 5 to my value and subtract 1 1 __ 2 , the result is twice

the square root of 25. I am a terminating decimal. What is my number?

6. If you take 3 to the value of my power, the result is a non-terminating, non-repeating decimal. If you take 3 and raise it to the power of the absolute value of my value, the result is 81. I am an integer. What is my number?

7. When you divide my value by the absolute value of �6, the result is a fraction that when simplified has a numerator that is the first natural number and a denominator that is a perfect square. If you take the square root of the reciprocal of this fraction, the result is 3. I am a real number. What is my number?

En

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Holt McDougal Algebra 1178 Copyright © by Holt McDougal. All rights reserved.

Ready to Go On? EnrichmentCHAPTER

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Profit-Loss-Revenue

An important concept in business is the ability to make a profit. Profit is equal to the amount of sales minus the cost of production. If the sales are greater than the cost, the business makes a profit. If the sales are less than the cost, the business is losing money.

Use the information below to answer each question.

A manufacturer of compact-disc players sells them to a retailer for $45 each. It costs the manufacturer $200 plus $25 each to produce the compact-disc player.

1. Write a function, s, to represent the total amount of sales of

compact-disc players, n.

2. Write a function, c, to represent the total cost of producing the


3. Graph the functions s and c on the same coordinate grid.

x

y

2 6 10 14 18

50045040035030025020015010050

4 86 12 16 20

Am

ou

nt,

$

Number of Compact-Disc Players

4. For what dollar amount is the sales and the cost equal?

5. For what value of n is the sales and the cost equal?

6. Write an inequality that represents the value(s) of n for which the cost is more than the sales.

7. Write an inequality that represents the value(s) of n for which the manufacturer makes a profit.

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Holt McDougal Algebra 1 179Copyright © by Holt McDougal. All rights reserved.


2


Diophantine Equations

Diophantus (about 200-284) is known to some as the ‘father of algebra’. He studied primarily the solutions of algebraic equations and the theory of numbers.

One type of equation he studied has the form ax � by � c where a, b, and c are all integers and the solutions to the equation (x, y) are also integers. These types of equations are now known as Diophantine Equations. They can be quite difficult to solve and many times the only way to solve them is by guessing and checking.

Solve each Diophantine Equation. Find at least one pair of positive integers for x and y that make the equation true.

1. 3x � 4y � 12 2. �2x � 3y � �9

a. Solve the equation for y. a. Solve the equation for y.

b. What number must x be divisible by? b. What number must x be divisible by?

Why? Why?

c. Find at least one solution. c. Find at least one solution.

3. x � 2y � 10 4. �4x � y � 15

5. 8x � 19y � 100 6. 3x � 7y � 35

7. �5x � 11y � 30 8. 3x � 4y � 32

9. 7x � y � 14 10. �3x � 5y � 9

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3


Triangle Inequalities

The Triangle Inequality Theorem states that “For any triangle, the sum of the lengths of any two sides is greater than the length of the third side”. This inequality defines the existence of a triangle. There is a theorem in geometry that determines whether a given triangle is a right triangle, obtuse triangle, or acute triangle.

A right triangle has exactly one 90 degree angle.

An obtuse triangle has exactly one angle greater than 90 degrees.

An acute triangle has no angle with a measure greater than or equal to 90 degrees.

In a triangle with sides a, b, and c with c being the longest side:

If c 2 � a 2 � b 2 , the triangle is obtuse.

If c 2 � a 2 � b 2 , the triangle is a right triangle.

If c 2 � a 2 � b 2 , the triangle is acute.

Determine whether the triangles, with these given side lengths, are acute, right, or obtuse.

1. 2, 3, 4 2. 3, 4, 5

3. 6, 6, 7 4. 7, 20, 24

5.

12

13 5

6.

7

7 7

7.

8

11 6

8.

5

6 4

9. The longest side of an acute triangle measures 12 inches. One of the shorter sides is 7 inches. Express the length of the third side as an inequality.

10. The two shorter sides of an obtuse triangle measure 5 cm and 10 cm. Express the length of the third side as an inequality.

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4


Onto FunctionsA relation is a function if for every number in the domain, there is one and only one number in the range. Another way to describe a function is to determine if it is onto. A function is onto if and only if all the numbers of the range are paired with all of the numbers in the domain.

For example, the function in Table 1 is described as onto because each number in the range has a number from the domain assigned to it.

The function in the Table 2 is not onto because the number 7 in the range does not have a number in the domain assigned to it.

Determine if each function can be described as onto.

1. 2.

3. 4.

Domain

Range 6 7 8 9 10

1 2 3 4 5

Domain

Range 6 7 8 9 11

1 2 3 4 5

Domain

Range 0 2 4 6 8

2 4 6 8 10 Domain

Range 1 2

1 2 4 8

0 3 6 9 12

Domain

Range 6 7 8 9 10

5 10 15 20 25 Domain

Range 3 4 5 6 7

1 3 5 7 9

Table 1

Table 2

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5


Intercept–Intercept FormLook at the graph of the equation 1 __ 2 x � 1 __ 3 y � 1 and

1 ___ �4 x � 1 __ 5 y � 1 shown on the right. The y-intercept for the

first equation is 3, so you know that the point (0, 3) lies on the line. Another point containing 0 also lies on the line; it is the point (2, 0). The x-coordinate of the point at which the line crosses the x-axis is called the x-intercept.

You can see that for the second equation, the y-intercept is 5.

What is the x-intercept of the second equation?

An equation in the form 1 __ a x � 1 __ b y � 1 is in intercept-intercept form. In this

form, a is the x-intercept and b is the y-intercept.

Look back at the equation 1 ___ �4 x � 1 __ 5 y � 1. Since it is in

intercept-intercept form, you know the following relationships.

1 __ a � 1 ___ �4 1 __ b � 1 __ 5 How would you determine a and b?

a � �4 b � 5

You can use the intercept-intercept for to determine the slope of the line. Recall that slope equals rise over run. You can use the two intercepts to count the rise and run on the graph. Applying the definition of slope, you can calculate the slope using the intercepts a and b.

m � y 2 � y 1

_______ x 2 � x 1 � 0�b _____ a � 0 � � b __ a

What is the slope of the line described by the equation 1 __ 2 x � 1 __ 3 y � 1?

m �

For each equation, determine the x-intercept, the y-intercept and the slope of the line.

1. x � 1 __ 6 y � 1 2. x � y � 1 3. x � y � 1

4. 1 __ 2 x � 1 __ 6 y � 1 5. x � y � 2 6. x � 2y � 2

3

5

1

–3

–5

x

y

1 3 5–1–3–5

x + y = 112

13x + y = 11

–415

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6


Holt McDougal Algebra 1184

Using a Matrix to Represent a System of EquationsA matrix is a rectangular array of numbers enclosed in a single set of brackets. If each equation in a system of equations is written in standard form, you can represent the system with a matrix equation. The matrix equation is made up of three matrices; one for the coefficients on the variables x and y, one for the variables x and y, and one for the constants.

For example, the system of equations { 3x � y � 6

x � y � �2 is represented by the matrix

equation [ 3 �1 1 1

] [ x y ] � [ 6 �2

] .Determine which system of equations represents the correct matrix equation.

1. { x � y � 8

x � y � 2 a. [ 3 �1

6 2 ] [ x y ] � [ 4

�8 ]

2. { 3x � y � 4

6x � 2y � �8 b. [ 1 �1

2 3 ] [ x y ] � [ 2

9 ]

3. { x � y � 2

2x � 3y � 9 c. [ �5 8

10 3 ] [ x y ] � [ 21

15 ]

4. { �5x � 8y � 21

10x � 3y � 15 d. [ 1 1

1 �1 ] [ x y ] � [ 8

2 ]

Create a matrix equation for each system of equations.

5. { x � 5y � 0

2x � 3y � 7 6. { 4x � 3y � 19

3x � 4y � 8

7. { 5x � 3y � 12

4x � 5y � 17

8. { x � y � 7

x � y � 9 9. { 12x � 9y � 114

12x � 7y � 82

10. { 2x � 3y � �4

x � 3y � 7

11. { 1 __ 2 x � y � 12

x � 1 __ 5 y � 10 12. { 2 __ 3 x � 1 __ 3 y � �9

1 __ 4 x � 3 __ 4 y � 16

13. { 1.2x � 1.6y � 2.4

�0.8x � 0.2y � �1.2


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7 Ready to Go On? Enrichment

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Describing Geometric Regions with a System of Inequalities

Describe the shaded region of the graph by writing a system of inequalities consisting of three different linear inequalities. To write this system, follow these steps:

1. Determine the slope and y-intercept of each line.

2. Write inequalities in slope-intercept form.

3. If the line is solid use either � or �. If the line is dashed, use either � or �.

4. If the shaded region is “above” the line use the symbol �, and if the shaded region is “below” the line, use the symbol �.

For example, the system of inequalities that describes the region below is { x � �2y � �4

y � �3 ___ 2

x � 1

.

x

y5

–5

–5 5

Write a system of inequalities that describes each region.

1.

x

y5

–5

–5 5

2.

x

y5

–5

–5 5

3.

x

y5

–5

–5 5

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LAH_A1_11_FL_RTGO_185.indd 185LAH_A1_11_FL_RTGO_185.indd 185 3/1/09 6:52:13 PM3/1/09 6:52:13 PM

Fourth Degree TrinomialsSometimes it is possible to write a trinomial of the fourth degree, a 4 � a 2 b 2 � b 4 , as a difference of two squares and then factor.

Example: Factor 4 a 4 � 21 a 2 b 2 � 9 b 4 .

Step I Find the square roots of the first and last terms.

��

4 a 4 � 2 a 2 ��

9 b 4 � 3 b 2

Step II Find twice the product of the square roots from the terms in Step 1.

2(2 a 2 )(3 b 2 ) � 12 a 2 b 2

Step III Split the middle term of the trinomial into two parts. One part is either the answer from the Step II or its opposite. The other part should be the opposite of a perfect square.

�21 a 2 b 2 � �12 a 2 b 2 � 9 a 2 b 2

Step IV Rewrite the trinomial as the difference of two squares and then factor.

4 a 4 � 21 a 2 b 2 � 9 b 4 � (4 a 4 � 12 a 2 b 2 � 9 b 4 ) � 9 a 2 b 2

� � 2 a 2 � 3 b 2 � 2 � 9 a 2 b 2

� [(2 a 2 � 3 b 2 ) � 3ab] [(2 a 2 � 3 b 2 ) � 3ab]

� (2 a 2 � 3ab � 3 b 2 )(2 a 2 � 3ab � 3 b 2 )

Factor each trinomial.

1. 16 d 4 � 7 d 2 � 1

2. p 4 � p 2 � 1

3. 4 x 4 � 13 x 2 � 1

4. 4 x 4 � 9 x 2 y 2 � 16 y 4

5. 9r 4 � 26r 2 s 2 � 25 s 4

6. 4 a 4 � 5 a 2 c 2 � 25 c 4

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9


Graphing Circles by Completing the SquareCompleting the square can be used to graph circles. The general equation for a circle with its center at the origin is x 2 � y 2 � r 2 , where r is the radius of the circle. The general equation of a circle with its center translated from the origin is (x � h ) 2 � (y � k ) 2 � r 2 . An equation representing a circle can be transformed into the sum of two squares.

Example: x 2 � 14x � y 2 � 6y � 49 � 0( x 2 � 14x � ) � ( y 2 � 6y � ) � �49( x 2 � 14x � 49) � ( y 2 � 6y � 9) � �49 � 49 � 9(x � 7 ) 2 � (y � 3 ) 2 � 9(x � 7 ) 2 � (y � 3 ) 2 � 3 2

The center of the circle is (7, �3) and the radius is 3.

The circle is shown at the right.

Complete the square on the following equations. Identify the center and radius of the circle and then graph.

1. x 2 � 8x � y 2 � 2y � 13 � 0 2. x 2 � 6x � y 2 � 4y � 12 � 0

y

x

–4

–2

–6

2

2 4 6 8

–2 –4

y

x 2

–2

–4

4

2 4

Center: Center:

Radius: Radius:

3. x 2 � y 2 � 10y � 75 � 0 4. x 2 � 8x � y 2 � 84 � 0

–8 –16

y

x 8

–8

–16

16

8 16

–8 –16

y

x 8

–8

–16

16

8 16

Center: Center:

Radius: Radius:

y

x

–4

–2

–6

2

2 4 6 8 10

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10


Geometric Series

A series is the sum of a sequence of terms. In other words, a series is a list of numbers with addition operations between them. Series can be finite, meaning that there is an end, or infinite. A geometric series is a sequence of numbers such that the quotient of any two successive numbers form a common ratio.

A geometric series can be written as S n � � k�0

n

a r k where r � 0 is the common ratio

and a is a scale factor. The Greek letter sigma, �, represents the sum of each term in the sequence. For example:

S 4 � � k�0

4

2(�2 ) k

� 2(�2 ) 0 � 2(�2 ) 1 � 2(�2 ) 2 � 2(�2 ) 3 � 2(�2 ) 4 � 2 � (�4) � 8 � (�16) � 32 � 22

Find the value of each geometric series.

1. S 3 � � k�0

3

3(�3 ) k 2. S 4 � � k�0

4

2 � 1 __ 2 � k

3. S 4 � � k�0

4

4 � � 1 __ 2 � k 4. S 3 � � k�0

3

1 __ 2 (4 ) k

5. S 3 � � k�0

3

�1 � � 1 __ 3 � k 6. S 4 � � k�0

4

9 � 2 __ 3

� k

7. S 5 � � k�0

5

4 (�1 ) k 8. S 4 � � k�0

4

1 __ 3

(�3 ) k

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11


Pascal’s TrianglePascal’s Triangle is a geometric arrangement of numbers. These numbers represent the binomial coefficients. That is, they represent the coefficients of the terms of the expansion of (x � y ) n . The first seven rows of Pascal’s Triangle look like this.

Row 0 1

Row 1 1 1

Row 2 1 2 1

Row 3 1 3 3 1

Row 4 1 4 6 4 1

Row 5 1 5 10 10 5 1

Row 6 1 6 15 20 15 6 1

Notice that each number is the sum of the two numbers above it.

For example what two numbers were added to get 10 in the 5th row?

What are the numbers for the 7th row?

As an example, find (x � y ) 3 .

Look at row 3, what are the coefficients of the expansion?

The first term of the expansion starts with the highest power of x, namely x 3 , and the lowest power of y, namely y 0 � 1. The power of x increases by 1 for each successive term and the power of y increases by 1 for each successive term.

(x � y ) 3 � 1 � x 3 � y 0 � 3 � x 2 � y 1 � 3 � x 1 � y 2 � 1 � x 0 � y 3 \ \ \ \

� x 3 � 3 x 2 y � 3x y 2 � y 3

Expand each of the following polynomials.

1. (x � y ) 4 2. (x � 1 ) 5

3. (x � 3 ) 4 4. (x � 2 ) 3

5. (x � 1 ) 9 6. (x � 2y ) 5

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12


Ready to Go On? Chapters 1–2 InterventionA. Expressions, Equations, and

Inequalities

1. 30 2. 3 3. 2.7 4. 3 } 2 5. 81 6. 4.41

7. 18 8. 26 9. 14 10. y 1 4

} 6 11. 2q2 2 4

12. 6w 1 8 13. 12 } q ≤ 5 14. 2h 1 5 5 23

15. 12 < w 2 4 ≤ 20

Check

1. 2 2. 4 3. 7 4. 66 5. 8 6. 6 7. 4(2y) 1 8 8. 6x2 2 15 9. b 1 12 5 2b 10. 10 ≤ 3q ≤ 15 11. sometimes 12. always

B. Problem Solving

1. Yes 2. No 3. No 4. No 5. Yes 6. Yes 7. You know: The short trail is 2 km long and the long trail is 5 km long. You ride 3 laps on the short trail and x laps on the long. The total distance you ride is 21 km. You want to fi nd out: How many laps on the long trail do you need to ride a total of 21 km? 8. Number of laps on the short trail p Length of the short trail (km) 1 Number of laps on the long trail p Length of the long trail (km) 5 Total distance (km)

Check

1. No 2. Yes 3. Yes 4. Yes 5. Yes 6. No 7. You know: Jim’s grandmother walks down the hall 90 yards, across 20 yards, up the hall 90 yards, and across 20 yards. She walks the length of the hall 9 times and she walks the width of the hall x times. She wants to walk a total of 970 yards. You want to fi nd out: How many times does she need to walk the width of the hall to walk a total of 970 yards.Verbal model:Number of times she walks the length of the hall p Length of the hall (yd) 1 Number of times she walks the width of the hall p Width of the hall (yd) 5 Total distance (yd)

C. Representations of Functions

1. Domain: 0, 1, 2, 3; Range: 1, 3, 5, 7 2. Domain: 25, 210, 215, 220; Range: 1, 2, 3, 4

3. Domain: 26, 24, 21, 0; Range: 0, 100, 400,

600 4. Yes 5. Yes 6. Yes 7. y 5 x }

2

8. y 5x 2 4 9. y 5 x 1 9 10. y 5 2x

11. Input, x 3 6 9 12 15

Output, y 2 4 6 8 10

Range: 2, 4, 6, 8, 10

12. Input, x 25 24 23 22 21

Output, y 26.1 25.1 24.1 23.1 22.1

Range: 26.1, 25.1, 24.1, 23.1, 22.1

13. Input, x 1 2 4 7 9

Output, y 3 1 23 29 213

Range: 213, 29, 23, 1, 3

14. Input, x 20 30 40 50 60

Output, y 21 } 2

31 }

2

41 }

2

51 }

2

61 }

2

Range: 21

} 2 ,

31 }

2 ,

41 }

2 ,

51 }

2 ,

61 }

2

15. Input, x 2 5 6 8 9

Output, y 16 19 20 22 23

Range: 16, 19, 20, 22, 23

16. Input, x 23 21 4 8 11

Output, y 15 5 220 240 255

Range: 255, 240, 220, 5, 15

17. y

5

7.5

10

12.5

15

17.5

10 12.5 15 17.5 7.5 5 2.5

x 2.5

Answers

Copyright © by Holt McDougal. All rights reserved.190 Holt McDougal Algebra 1190 Holt McDougal Algebra 1

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LAH_A1_11_FL_RTGO_AK_190-214.indd 190LAH_A1_11_FL_RTGO_AK_190-214.indd 190 3/1/09 1:26:04 AM3/1/09 1:26:04 AM

18. y

10

15

20

25

30

35

20 25 30 3515105

x5

19. y

5

7.5

10

12.5

15

17.5

10 12.515 17.57.552.5

x2.5

20. y

5

7.5

10

12.5

15

17.5

10 12.515 17.57.552.5

x2.5

21. y

4

6

8

10

12

14

8 10 12 14642

x2

22. y

5

7.5

10

12.5

15

17.5

10 12.515 17.57.552.5

x2.5

Check

1. Domain: 7, 10, 14, 17; Range: 23, 35, 51, 63 2. Domain: 210, 28, 25, 24; Range: 90, 54, 15, 6

3. Domain: 23, 22, 21, 0; Range: 2 3 } 8 , 2

1 } 4 , 2

1 } 8 , 0

4. No 5. No 6. Yes 7. y 5 5x 8. y 5 22x

9. y 5 x 2 7 10. y 5 3 }

2 x

11. Input, x 23 21 2 5 6

Output, y 22 14 2 210 214

Range: 214, 210, 2, 14, 22

12. Input, x 10 12 14 16 18

Output, y 13 } 2 8

19 }

2 11

25 }

2

Range: 13

} 2 , 8,

19 }

2 , 11,

25 }

2

13. Input, x 1 9 13 19 23

Output, y 5 } 4

21 }

4

29 }

4

41 }

4

49 }

4

Range: 5 }

4 ,

21 }

4 ,

29 }

4 ,

41 }

4 ,

49 }

4

14. y

4

6

8

10

12

14

8 10 12 14642

x2

15. y

20

30

40

50

60

70

40 50 60 70302010

x10

Answers continued


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ey


Answers continued

16. y

4

6

8

10

12

14

8 10 12 14642

x2

D. Operations

1. 6.2 2. 2809 3. 20.25 4. 2 45

} 8 5. 6 }

11

6. 8 1 } 7 7. 0.4 8. 50 9. 1.6 10. 9 11.

23 } 5

12. 10 8 }

9 13. 25 14. 26 15. 4 16. 26

17. 22 18. 29 19. 6.1 20. 22 21. 2134

22. 7 5 }

12 23. 210

3 }

4 24. 7.4 25. 23.5

26. 219.2 27. 291 28. 7 3 } 5 29. 2

56 } 15

30. 155 31. 230 32. 2 33

} 2 33. 2 25

} 47

34. 144 35. 2182 36. 21.84 37. 2 1 } 6

38. 1 39. 4 } 3 40. 2

5 } 7 41. 2 }

19 42. 2

32 } 175

43. 23 44. 2 18 } 5 45. 81 46. 2

2 } 5 47. 4 }

15

48. 2 30

} 47 49. 220 50. 63 51. 9 52. 0

53. 2 54. 630

Check

1. 16; 16; 2 1 } 16 2. 27

3 }

10 ; 7

3 }

10 ;

10 }

73

3. 0.3; 0.3; 2 10

} 3 4. 116 5. 2 3 } 8 6. 2

59 } 15

7. 27 8. 70 9. 8 10. 2 15

} 2 11. 69

12. 27.2 13. 3 14. 23 15. 10

16. A ø B 5 {0, 1, 2, 3, 4, 5, 6}; A ù B 5 {2, 4}

E. Properties and Real Numbers

1. Integer 2. Rational 3. Rational 4. Whole 5. Rational 6. Whole

7. 1.23, 3 }

2 , 1

2 }

3 , Ï

}

3 8. 21.9, 2 Ï}

0.04 , 0.08, Ï}

2

9. Ï}

6 , 6.01, 6.1, 6 1 }

6 10. Identity

11. Commutative 12. Associative 13. Inverse 14. Associative15. Commutative 16. Commutative 17. Negative one 18. Associative

19. Identity 20. Zero 21. Negative one 22. 228 2 7v 23. 2c2 + c 24. 215m + 5m2 25. 9t 2 72 26. 6 + 8r 27. 15p + 21p2

Check

1. Irrational, rational, whole, rational; 8 }

3 , Ï

}

8 , 3, 4.1

2. Rational, rational, rational, integer; 29.1,

29.02, 2 Ï}

9 , 2 2 } 9

3. Whole, rational, rational, integer; 21.1,2 Ï}

1 ,

20.01, 0 }

1

4. Associative property of addition 5. Negative one property of multiplication 6. Additive inverse 7. Multiplicative identity 8. Distributive property 9. Commutative property of addition 10. Commutative property of multiplication 11. Associative property of multiplication 12. Distributive property 13. Zero property 14. Distributive property 15. Additive identity 16. 5h + 10 17. 24d 2 3d2 18. 232f 2 2 80f

Ready to Go On? Chapters 3–4 InterventionA. Solving Equations in One Variable

1. 21 2. 11 3. 4 4. 20 5. 8 6. 227 7. 2

8. 23 9. 224 10. 2 10

} 7 11. 9 12. 25

13. 3 14. 26 15. 2 }

13 16. All real numbers

17. 220 18. No solution

Check

1. 5.3 2. 2.2 3. 215 4. 2 6 } 5 5. 10.5 6. 7

7. 2 }

9 8. 110 9. 162 10. 6 11. 15

12. 220 13. 211 14. 2 }

3 15.

5 }

2

16. 0 17. All real numbers 18. 210

B. Proportion and Percent Problems

1. 2 }

3 2.

1 } 5 3.

1 }

2 4.

12 }

37 5.

2 }

3 6.

27 }

10 7. 14

8. 60 9. 10 10. 5 11. 6 12. 12 13. 57 14. 27 15. 180 16. 845 17. 15 18. 7 19. 20.4 20. 125 21. 80% 22. 18% 23. 3.824. 68 25. 1825 26. 75 27. 25% 28. 33.3% 29. 22.77 30. 200


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Answers continued

Check

1. 2 } 5 2.

3 }

4 3.

1 }

14 4. 15

1 }

3 5. 30 6. 25

7. 52 8. 163 9. 9 10. 14.44 11. 150 12. 85% 13. 600 14. 65% 15. 4.59 16. Increase; 67% 17. Decrease; 56%18. $20.25 19. $82.50


1. P 5 I } rt 2. r 5 Ï

}

A

} π 3. V 5 E 2 F 1 2

4. w 5 V

} lh

5. h 5 3V

} πr2 6. b1 5

2A }

h 2 b2

7. V 5 m

} d ; 64 cm3 8. d 5

40 }

Ï} s ; 4 miles

9. h 5 n2

} 64

1 2r; 46 ft 10. y 5 2x 2 10

11. y 5 2 4 } 3 x 2

8 }

3 12. y 5 2

1 } 9 x 2 3

13. y 5 8 }

3 x 1 16 14. y 5 2

25 } 2 x 1 75

15. y 5 2 1 } 2 x 2

3 } 2

Check

1. l 5 P

} 4 2. h 5

V } πr2 3. a 5 2s 2 b 2 c

4. l 5 S } πr

2 r 5. v 5 16t 1 h 2 c

} t

6. w 5 S 2 2lh

} 2l 1 2h

7. 3 ohms

8. n 5 s }

180 1 2; 15 sides 9. c 5

1 }

(1 2 e)2 ; 16

10. y 5 1 }

2 x 2 2 11. y 5 2

3 } 7 x 1 2

12. y 5 27x 1 7 }

3 13. y 5 2

4 } 5 x 2

9 } 5

14. y 5 2 8 } 3 x 2

2 }

3 15. y 5 6x 1 9

D. Graphing Linear Equations

1–6.

1

2

3

4y

1 x21222324 2 3 421

22

23

24

B

C A

D

FE

1. Quadrant I 2. x-axis 3. Quadrant II 4. y-axis 5. Quadrant III 6. Quadrant IV

7. Yes, it is a solution. 8. Yes, it is a solution. 9. No, it is not a solution. 10. No, it is not a solution. 11. Yes, it is a solution. 12. No, it is not a solution.

13.

1

2

3

4

5

6

7y

1 x21222324 2 3 421

14.

1

2

3y

1 x21222324 2 3 421

22

23

24

25

15.

1

2

3

4

5y

1 x21222324 2 3 421

22

23

16.

1

2

3y

1 x21222324 2 3 421

22

23

24

25


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Answers continued

17.

1

2

3

4

5

6y

1 x21222324 2 3 421

22

18.

1

2

3

4y

1 x21222324 2 3 421

22

23

24

19.

1

2

3

4y

1 x2122232425262721

22

23

24

20.

1

2

3

4y

1 x21222324 2 3 421

22

23

24

21.

1

2

3

4y

1 x21222324 2 3 421

22

23

24

22.

1

2

3

4y

1 x21222324 2 3 421

22

23

24

23.

1

2

3

4y

1 x2122 2 3 4 5 621

22

23

24

24.

1

2y

1 x21222324 2 3 421

22

23

24

25

26

Check

1–6.

2

4

6

8y

2 x22242628 4 6 822

24

26

28

B

C

A

F

ED

1. Quadrant IV 2. Quadrant III 3. y-axis 4. Quadrant I 5. x-axis 6. Quadrant II 7. Yes, it is a solution. 8. No, it is not a solution. 9. No, it is not a solution. 10. Yes, it is a solution. 11. No, it is not a solution. 12. Yes, it is a solution.


An

sw

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Key


Answers continued

13.

1

2

3

4

5

6y

1 x21222324 2 3 421

22

14. y

1 x21222324 2 3 421

22

23

24

25

26

27

28

15.

1

2

3

4y

1 x21222324 2 3 421

22

23

24

16. y

1 x21222324 2 3 421

22

23

24

25

26

27

28

17. 1

y

1 x21222324 2 3 421

22

23

24

25

26

27

18.

1

2

3

4y

1 x2 3 4 5 6 7 821

22

23

24

19.

1

2

3

4y

1 x21222324 2 3 421

22

23

24

20.

1

2

3

4y

1 x2122232425262721

22

23

24

21.

1

2y

1 x21222324 2 3 421

22

23

24

25

26


1. x-intercept: 26, y-intercept: 26

2. x-intercept: 2 2 } 3 , y-intercept:

8 }

3

3. x-intercept: 2, y-intercept: 18 4. x-intercept: 0,

y-intercept: 0 5. x-intercept: 2 1 } 4 , y-intercept:

3 }

2

6. x-intercept: 1 }

6 , y-intercept: 2

1 }

2 7. 21

8. Undefi ned 9. 2 1 }

8 10. 0 11. 2

4 }

3 12. 5


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Answers continued

13.

1

2

3

4

5

6

7

8y

1 x21 2 3 4 5 6 7

14.

1

2y

1 x212223242526 221

22

23

24

25

26

15.

1

2 y

1 x 21 22 23 24 2 3 4 21

22

23

24

25

26

16.

1

2y

1 x21222324 2 3 421

22

23

24

25

26

17.

1

2y

1 x21222324 2 3 421

22

23

24

25

26

18.

1

2

3

4

5

6

7

8y

1 x21222324 2 3 4

19. Yes; 2 7 }

8 20. No 21. Yes;

5 }

9 22. Yes; 2

1 }

47

23. Yes; 21 24. No 25. y 5 2 1 }

3 x; 24

26. y 5 2x; 64 27. y 5 2 1 }

2 x; 2 9

28. y 5 2 }

9 x; 6 29. y 5 2

7 } 5 x; 2140

30. y 5 1 }

2 x; 37

Check

1. x-intercept: 2 1 }

4 , y-intercept:

3 }

2 2. x-intercept:

3, y-intercept: 21 3. x-intercept: 1.25,

y-intercept: 5 4. 2 9 }

11 5. 0 6. 21

7.

1

2

3

4

5y

1 x21222324 2 3 421

22

23

when x 5 6, y 5 7

8.

1

2

3 y

1 x 21 22 23 24 2 3 4 21

22

23

24

25

when x 5 6, y 5 22


An

swer

Key


Answers continued

9.

1

2y

1 x21222324 2 3 421

22

23

24

25

26

when x 5 6, y 5 2

10. Yes; 2 4 } 5 11. No 12. Yes;

7 }

4

13. y 5 2 5 } 2 x; 250 14. y 5 3x; 129

15. y 5 2 3 } 2 x; 296


A. Writing Linear Equations

1. y 5 6x 2 4 2. y 5 2x 1 3 3. y 5 3 } 5 x 2 5

4. y 5 2 } 5 x 2 3 5. y 5 24x 1 5

6. y 5 2 1 } 3 x 2 2 7. y 5

4 }

3 x 1 6

8. y 5 2 1 } 4 x 1

11 }

4 9. y 5 26x 1 22

10. y 5 3 }

2 x 2

21 }

2 11. y 5 2

2 } 3 x 1 4

12. y 5 2x 2 2 13. y 5 2 }

3 x 2 3

14. y 5 2 1 } 4 x 1 2 15. y 5

1 }

2 x 2

5 }

2

16. y 5 3 }

4 x 1

1 }

4 17. y 5 2x 2 2

18. y 5 2 1 } 3 x 1

2 }

3 19. y 1 1 5

2 }

3 (x 2 3)

20. y 2 0 5 2 1 }

4 (x 2 4) 21. y 1 4 5

1 }

2 (x 1 3)

22. y 2 1 5 3 } 4 (x 2 1) 23. y 2 3 5 2(x 1 5)

24. y 2 2 5 2 1 }

3 (x 1 4) 25. 24x 1 y 5 3

26. 3x 1 y 5 2 27. 23x 1 2y 5 328. 4x 1 3y 5 21 29. 2x 1 y 5 2330. 3x 1 4y 5 1

Check

1. y 5 4x 1 3 2. y 5 22x 1 1 3. y 5 5 }

2 x 2 4

4. y 5 2 1 }

3 x 2 5 5. y 5

1 } 5 x 2

17 } 5

6. y 2 7 5 2 3 }

4 (x 1 2) 7. y 2 5 5 24(x 2 1)

8. y 5 2 1 } 2 x 1 2; x 1 2y 5 4 9. y 5 2

4 } 3 x 2 3;

4x 1 3y 5 29 10. It does represent an arithmetic sequence.; $96

B. Parallel and Perpendicular Lines

1. a and c are parallel. 2. d and e are perpendicular. 3. h and j are parallel. g is

perpendicular to h and j. 4. y 5 4 }

3 x 1 3

5. y 5 2 1 }

4 x 1 3 6. y 5 26x 1 15

7. y 5 3 }

2 x 2 3 8. y 5 2

2 }

3 x 9. y 5 2x 1 8

10. y 5 2 2 } 3 x 2 4 11. y 5

4 }

3 x 1 9

12. y 5 1 }

8 x 1

19 }

4 13. y 5 23x 1 12

14. y 5 3 }

2 x 15. y 5 2

1 } 2 x 2 6

Check

1. a and b are parallel. c is perpendicular to a and b. 2. d and e are perpendicular.

3. h and j are parallel. 4. y 5 3 }

8 x 2 2

5. y 5 2 2

} 3 x 1 1 6. y 5 2x 1 8

7. y 5 4 }

3 x 1 10 8. y 5 2

1 }

4 x 2 2

9. y 5 25x 1 17 10. y 5 23x 1 10

11. y 5 3 }

2 x 2 5 12. y 5 2

1 } 2 x 2 5

13. y 5 2 2 }

3 x 2 2 14. y 5

4 }

3 x 2 5 15. y 5

1 }

4 x

C. Linear models

1. No correlation 2. Negative correlation3. Positive correlation

4.

1

2

3

4

y

1 x21 2 3 4 5 6 721

22

5

6

positive correlation


An

swer K

ey


Answers continued

5.

1

2

x2122232425 1 2 321

22

23

24

25

26

y

negative correlation

6. Answers may vary. Sample answer: y 5 4 } 5 x 2 4

7. Answers may vary. Sample answer: y 5 5 }

2 x 2 4

8. y

x1 2 3 4

1

2

3

4

21

22

23

24

21222324

y 5 20.73x 1 0.5; 22.4

9. y

1 x2 3 4 5 6 7 821

22

23

24

2

1

3

4

y 5 0.63x 2 1.94; 0.58

10. y

x1 2 3 4

1

2

3

4

21

22

23

24

21222324

y 5 2 2 } 3 x; 24 2 }

3

11.

x1212223 2 3 4 521

22

23

24

25

1

2

3y

y 5 x 2 2.7; 4.3

Check

1. No correlation 2. Positive correlation3. Negative correlation

4a–c.

Days of rain2 4 6 8 10 12 14

Num

ber o

f sal

es

708090

100110120130

t

y

00

4b. Positive correlation 4c. y 5 6x 1 574d. y 5 5.85x 1 57.5 4e. 81 sales; 116 sales

D. Graphing Inequalities

1. 70 71 72 73 74 75 76 77

2. �31 �30 �29 �28 �27 �26 �25 �24

3. 6 7 85 9 10 11 12

4. 162160 164 166

5. x > 22 6. x ≤ 210 7. x ≥ 6 8. x < 259. 23 < x < 0

�1 0 1 2 3�2�3�4

10. x ≤ 2 or x ≥ 9

2 3 4 5 6 7 8 9


An

sw

er

Key


Answers continued

11. 21 ≤ x < 6

�1 0 1 2 3 4 5 6

12. x < 55 or x ≥ 60

54 55 56 57 58 59 60 61

13.

1

2

3

4

5

6 y

1 x 21 22 23 24 2 3 4 21

22

14. y

x2 3 4

1

2

3

2122232421

1

23

24

25

22

15. y

x2 3 4

1

2

3

4

2122232421

1

23

24

22

16. y

x2 3 4

1

2

3

4

2122232421

1

23

24

22

17.

1

2

x2122232425 1 2 321

22

23

24

25

26

y

18.

2

4

6

8

10

12

14y

2 x22242628 4 6 822

Check

1. 10 15 205 25 30 35 40

2. 7 8 9 10

3. x < 4 4. x ≥ 25

5. 213 ≤ x < 210;

�14 �12 �10 �8

6. x ≤ 0 or x > 6;

0 1 2 3 4 5 6 7

7.

x2122232425 1 2 321

22

23

24

25

26

y

1

2


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Answers continued

8. y

x2 3 4

1

2

3

4

2122232421

1

23

24

22

9. y

x2

1

2

3

2122232421

1

23

24

25

22

43

E. Solving Inequalities

1. p > 3.3

0 1 2 3

2. m ≤ 7

4 5 6 7 8 9 10 11

3. k > 233

�35 �34 �33 �32 �31 �30 �29�36

4. t ≥ 2 31

} 5

�6 �5 �4�7

5. r < 3.4

0 1 2 3 4 5 6 7

6. m ≤ 40

36 37 38 39 40 41 4342

7. x < 1 }

4

0 1 2 3

8. d ≥ 27

22 23 24 25 26 27 2928

9. c ≥ 26

�7 �6 �5 �4 �3 �2�9 �8

10. z < 29

�9�10�11 �8 �7 �6 �5 �4

11. w < 11.5

9 10 11 12

12. n ≤ 240

�41 �40 �39 �38 �37 �36 �35�42

13. x > 4

0 1 2 3 4 5 6 7

14. t ≥ 1 }

2

0 1 2 3

15. q > 20

14 15 16 18 20 22 24

16. v ≤ 2

0 1 2 3 4 5 6 7

17. n ≥ 22

�7 �6 �5 �4 �3 �2 �1 0

18. a < 2 1 }

2

�1 0 111 2

19. 3 < a < 5

0 1 2 3 4 5 6 7

20. 26 ≤ x ≤ 12

�8 0 8 16


An

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Key


Answers continued

21. 26 ≤ p ≤ 23

�7 �6 �5 �4 �3 �2 �1 0

22. 22 < n ≤ 21

�5 �4 �3 �2 �1 0 1 2

23. 29 < w < 23

�9 �8 �7 �6 �5 �4 �3 �2

24. 21 < j < 2

�3 �2 �1 0 1 2 3 4

25. a ≤ 26 or a > 23

�7 �6 �5 �4 �3 �2 �1 0

26. n < 23 or n ≥ 4

�3 �2 �1 0 1 2 3 4

27. y < 25 or y > 21

�7 �6 �5 �4 �3 �2 �1 0

28. k ≤ 24 or k > 3 } 2

�5 �4 �3 �2 �1 0 1 2

29. b ≤ 1 or b ≥ 2

0 1 2 3 4 5 6 7

30. All real numbers

�3 �2 �1 0 1 2 3

Check

1. y > 22.4

�3 �2 �1�4

2. b ≤ 7

0 1 2 3 4 5 6 7

3. n > 230

�32 �30 �28 �26

4. k ≥ 2 10

} 3

�3 �2�4

5. d < 23.8

�4 �3 �2�5

6. a ≤ 51

46 47 48 49 50 51 52

7. x < 2

0 1 2 3 4 5 6 7

8. x ≥ 28

24 25 26 27 28 29 30

9. c ≥ 24

�7 �6 �5 �4 �3 �2 �1 0

10. z < 224

�26 �24 �22 �20

11. y < 41.4

41.0 41.2 41.4 41.6

12. b ≤ 2480

�500 �480 �460 �440

13. x > 5

0 1 2 3 4 5 6 7

14. s ≥ 2 }

11

0 1 2 3

15. p > 1 }

4

0 0.5 1 1.5

16. r ≤ 5

0 1 2 3 4 5 6 7

17. m ≤ 1

0 1 2 3 4 5 6 7


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Answers continued

18. z > 2

0 1 2 3 4 5 6 7

19. 3 < g < 6

0 1 2 3 4 5 6 7

20. 221 ≤ x ≤ 14

�15 �10 �5 50 10 15�20

21. 27 ≤ a ≤ 21

�7 �6 �5 �4 �3 �2 �1 0

22. 21 < n ≤ 2 1 } 3

�1 0�2

23. 242 < t ≤ 214

�38 �30 �22 �14

24. 214 < j < 210

�16 �14 �12 �10

25. x ≤ 6 or x > 10

4 5 6 77 8 9 1110

26. x ≤ 28 or x > 21

�8 �7 �6 �5 �4 �3 �2 �1

27. All real numbers

�3 �2 �1 0 1 2 3

28. x < 26 or x ≥ 32

26 28 30 32

29. x < 4 or x > 16

2 4 6 8 10 12 1614

30. x ≤ 3 or x ≥ 4

0 1 2 3 4 5 6 7

F . Absolute Value Equations and Inequalilties

1. x 5 12 or x 5 212 2. y 5 2.4 or y 5 22.4

3. b 5 2 }

3 or b 5 2

2 } 3 4. m 5 44 or m 5 244

5. r 5 6 5 }

8 or r 5 26

5 }

8 6. p 5 0.9 or p 5 20.9

7. x 5 0 or x 5 26 8. c 5 0 or c 5 4

9. p 5 27

} 4 or p 5 2

13 } 4 10. t 5 4 or t 5

4 }

3

11. z 5 1 or z 5 4 12. a 5 5 }

6 or a 5

25 }

6

13. 5 < x < 9

3 4 5 6 7 8 9 10

14. a ≤ 1 }

2 or a ≥

7 }

2

0 1 2 3

15. n < 216 or n > 6

�20 �16 �12 �8 �4 0 4 8

16. t ≤ 2 1 } 3 or t ≥ 2

2 } 9

23

0� 13�

17. 2 5 } 2 < q < 2

3210�1�2�3�4

18. 0 ≤ f ≤ 4 } 5

10

Check

1. a 5 14 or a 5 214 2. c 5 9.7 or x 5 29.7

3. x 5 4 }

9 or x 5 2

4 }

9 4. n 5 38 or n 5 238

5. s 5 1 6 } 7 or s 5 21

6 } 7 6. q 5 0.3 or q 5 20.3

7. x 5 21 or x 5 27 8. u 5 2 4 } 3 or u 5

8 }

3

9. q 5 43

} 2 or q 5 2

37 }

2 10. t 5 2

1 } 2 or t 5 2

5 }

4

11. y 5 2 3 } 4 or y 5

9 }

4 12. r 5 265 or r 5 95


An

sw

er

Key


Answers continued

13. 233 < x < 103

�40 0 20�20 40 60 80 100

14. a ≤ 2 3 } 7 or a ≥ 1

�1 0 1 2

15. t < 263 or t > 57

�60 �40 �20�80 0 20 40 60

16. t ≤ 20.85 or t ≥ 20.65

�0.8 �0.6 �0.4�1

17. 2 26

} 3 < b < 22

} 3

�8 �6 �4 �2 0 2 4 6 8�10

18. 21 ≤ k ≤ 2

�3 �2 �1 0 1 2 3 4

Ready to Go On? Chapters 7–8 InterventionA. Solving Linear Systems by Graphing

1. yes 2. yes 3. no 4. no 5. no 6. yes 7. (4, 3) 8. (3, 3) 9. (22, 3) 10. (2, 21) 11. (21, 3) 12. (3, 23) 13. (1, 4) 14. (2, 2) 15. (5, 22) 16. no solution; the lines are parallel 17. infi nitely many solutions; both equations represent the same line 18. no solution; the lines are parallel 19. no solution; the lines are parallel 20. infi nitely many solutions; both equations represent the same line 21. infi nitely many solutions; both equations represent the same line

Check

1. yes 2. no 3. yes 4. (2, 21) 5. (23, 22) 6. (4, 1) 7. (2, 0) 8. no solution; the lines are parallel 9. (21, 21) 10. (5, 2) 11. (3, 25) 12. infi nitely many solutions; both equations represent the same line

B. Solving Linear Systems Using Algebra

1. (1, 7) 2. (0, 23) 3. (22, 1) 4. (21, 22) 5. (2, 1) 6. (21, 1) 7. (3, 2) 8. (21, 4)9. (2, 25) 10. (26, 27) 11. (21, 22)

12. (0, 3) 13. (24, 2) 14. (2, 2) 15. (4, 3)16. (2, 22) 17. (24, 22) 18. (3, 6)

Check

1. (4, 2) 2. (1, 6) 3. (23, 22) 4. (23, 22)

5. (21, 1) 6. (1, 2) 7. (3, 2) 8. (21, 21) 9. (22, 4) 10. subtracting equations; (1, 4) 11. substitution; (2, 4) 12. multiplying fi rst; (22, 1) 13. 5 times on the Ferris wheel and 7 times on the roller coaster

C. Solving Systems of Linear Inequalities

1.

x 2 3 4

y

1

2 3

4

2324 22 21 1 21

22

23

24

2.

x 2 3 4

y

1

2 3

4

1

22

23

24

23 24 22 21 21

3.

x

y

1

2 3

4

22

23

24

23 24 22 21 21

2 3 4 1


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ey


Answers continued

4.

x

y

1

2 3

4

22

23

24

23 24 22 21 21

2 3 4 1

5.

x

y

24 21

2 3 4 1

1

2 3

4

22

23

24

23 22 21

6.

x

y

1

2 3

4

21

22

23

24

2 3 4 1 23 22 2124

7.

x

y

24 21

1

2 3

4

22

23

24

23 22 21 2 3 4 1

8.

x

y

21

1

2 3

4

22

23

24

2 3 4 1 23 22 2124

9.

x

y

1

23

4

22

23

24

23 22 21 2 3 412421

10.

x

y

1

2 3

4

22

23

24

23 22 21 2 3 4 1 24 21

11.

x

y

2

4

6

8

22

24

26

28

4 6 8 2 26 24 2228

12.

x

y

21

1

23

4

22

23

24

23 22 21 2 3 4124

13. y ≤ 24x, y > 24x 2 3 14. y < 23x + 3, y > 3x 2 1 15. y > 22x 2 1, y < 3, x ≤ 4


An

sw

er

Key


Answers continued

Check

1.

x 21

1

2 3

4

22

23

24

26 25 24 23 22 21 1 27

y

2.

x

y

21

1

2 3

4

22

23

24

23 22 21 2 3 4 1 24

3.

x

y 1

2 3 4 1 23 22 2124

22

23

24

25

27

26

21

4.

x

y

21

1

2 3

4

22

23

24

23 22 21 2 3 4 1 24

5.

x

y

21

1

2 3

4

22

23

24

23 22 21 2 3 4 1 24

6.

x

y

21

1

2 3

4

22

23

24

23 22 21 2 3 4 1 24

7. y > 22x 2 5, y < 2x 2 1 8. y < 2, x > 22, x < 21

9. y ≤ 23x 1 6, y < 2 1 } 2 x 2 2, y < 3x 2 6

D. Properties of Exponents

1. 89 2. 513 3. (26)12 4. d17 5. r20 6. k6 7. 1018 8. (26)16 9. t15 10. b16 11. (p 1 5)14 12. (h 2 1)45 13. 56 p 46 14. 27g3h3 15. 36c2d2 16. 16p4 17. 2125t3 18. 264a2 19. 186 20. 232 21. (225)2 22. 412

23. n4 24. w2 25. 23

} 53 5

8 }

125

26. 52

} 82 5

25 }

64 27.

1 }

p8 28. r6

} s6 29. 2

u9

} v9

30. 2 32

} b5 31.

1 }

112 5 1 }

121 32. 1

33. 1 }

43 5 64 34. 1 }

(23)4 5 1 }

81

35. 62 5 36 36. 1 37. 3 38. 3 39. 1 }

6

Check

1. 710 2. (28)13 3. y29 4. 615 5. (24)21 6. s12 7. (a + 11)32 8. 813 p 213 9. 2(5)7j7k7 = 278,125j7k7 10. 313

11. v9 12. 2 33

} 43 5 2

27 } 64 13.

16 }

h2

14. 2 p15

} q15 15. 1 16. 2

1 }

25 5 2 1 } 32

17. 72 5 49 18. 1 19. 8 20. 4 21. 1 } 7


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Answers continued

E. Scientifi c Notation

1. 4.2635 3 1013 2. 5.82 3 10-11

3. 7.3009 3 104 4. 7,810,000,000,000

5. 0.00000005403 6. 603.4 7. 2.9304 3 1012 8. 1.17649 3 1047 9. 1.8662 3 1026 10. 8.0 3 1011 11. 2.0 3 10215 12. 8.416533573 3 1017

Check

1. 0.00000005026 2. 2.707 3 1025 3. 6.180903 3 1014 4. 1.001001001001 3 1023 5. 311,000,000,000,000 6. 0.00945824

7. 3.816 3 105 8. 2.56 3 1024 9. 2.5748 3 1029 10. 8.75 3 106 11. 4.0 3 10222 12. 6.81472 3 1020

F. Exponential Functions

1.

1 x2324 22 21 2 3 4

y

1

23

4

5

6

7

8 ; The domain is all

real numbers. The range is all positive real numbers.

2.

1 x2324 22 21 2 3 4

y

1

23

4

5

6

7



3.

1 x2324 22 21 2 3 4

y

1

23

4

5

6

7



4.

1 x2324 22 21 2 3 4

y

1

23

4

5

6

7



5.

1 x2324 22 21 2 3 4

y

1

23

4

5

6

7



6.

1 x2324 22 21 2 3 4

y

1

23

4

5

6

7



7a. P 5 245,000(1.018)t 7b. 292,849 people8a. h 5 3.75(1.043)t 8b. 5.48 in.


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Answers continued

9.

1 x2324 22 21 2 3 4

y

1

23

4

5

6

8

7

; The domain is all real numbers. The range is all positive real numbers.

10.

1 x2324 22 21 2 3 4

y

1

23

4

5

6

7

8 ; The domain is

all real numbers. The range is all positive real numbers.

11.

1 x2324 22 21 2 3 4

y

1

23

4

5

6

7

8 ; The domain is


12.

1 x2324 22 21 2 3 4

y

1

23

4

5

6

7

8 ; The domain is


13.

1 x2324 22 21 2 3 4

y

1

23

4

5

6

7

8 ; The domain is


14.

1 x2324 22 21 2 3 4

y

1

23

4

5

6

7

8 ; The domain is


15a. E 5 95(0.75)t 15b. 23 essays16a. T 5 475(0.92)t 16b. 136°

Check

1.

1 x2324 22 21 2 3 4

y

1

23

4

5

6

7



2.

1 x2324 22 21 2 3 4

y

1

23

4

5

6

7



3.

1 x2324 22 21 2 3 4

y

1

23

4

5

6

7




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Answers continued

4.

1 x2324 22 21 2 3 4

y

1

23

4

5

6

7



5.

1 x2324 22 21 2 3 4

y

1

23

4

5

6

7


real numbers. The range is all positive realnumbers.

6.

1 x2324 22 21 2 3 4

y

1

23

4

5

6

7



7a. V 5 5000(1.058)t 7b. $63,2168a. W 5 45(0.975)t 8b. 2 pounds

Ready to Go On? Chapters 9–10 InterventionA. Adding, Subtracting, and

Multiplying Polynomials

1. Yes; 1st degree monomial 2. No; variable exponent 3. Yes; 4th degree binomial4. Yes; 5th degree trinomial 5. No; negative exponent 6. Yes; 3rd degree polynomial7. x3 1 2x2 2 x 1 2 8. 23x2 2 5x 1 59. 3x4 1 2x3 2 4x2 1 2x 1 210. 4x3 1 3x2 2 x 1 3 11. 2x2 1 4x 2 812. 5x2 1 x 2 3 13. 8b5 1 4b4 2 12b3

14. 4x7 2 10x6 2 2x5 1 12x4

15. 2y3 2 7y2 1 13y 2 5

16. 2z3 1 5z2 214z 1 317. 2a3 2 13a2 1 11a 1 618. 6x3 2 5x2 1 6x 1 8 19. 4m2 1 20m 1 25

20. z2 2 49 21. 9x2 2 24x 1 16 22. 4x2 2 1623. 4s2 1 4st 1 t2 24. 25x2 2 4

Check

1. Not a polynomial; negative exponent 2. Yes; 1st degree binomial 3. Yes; 3rd degree polynomial 4. No; variable exponent 5. Yes; 5th degree trinomial 6. Yes; 2nd degree trinomial 7. 2x3 1 x2 1 4x 1 2 8. 2x2 1 6x 2 3 9. 4x2 1 2x 2 3 10. 3x2 2 4x 2 611. 6z5 2 12z3 1 15z2

12. 15x3 2 2x2 2 36x 2 713. 4b6 2 8b5 2 4b3 14. 9y2 2 1615. k2 2 18k 1 81 16. 2x3 1 7x2 2 19x 1 617. 4p2 2 9r2 18. 10x2 2 13x 2 319. 9y2 1 30y 1 25

B. Factoring Polynomials

1. 26, 2 2. 1, 8 3. 24, 23 4. 0, 4

5. 27, 0 6. 2 3 }

4 , 0 7. 0,

7 } 5 8. 0,

1 }

2

9. 0, 1 } 5 10. (z 1 4)(z 1 3) 11. (n 2 7)(n 2 1)

12. (m 1 4)(m 2 6) 13. (y 2 2)(y 2 3) 14. (t 2 3)(t 1 5) 15. (x 1 5)(x 1 1) 16. (2x 2 5)(2x 1 1) 17. (2y 2 9)(y 2 1) 18. 2(2z 1 3)(2z 2 5) 19. (3m 2 4)(m 2 2)20. 2(2x 2 3)(x 2 7) 21. (3t 2 4)(t 1 3)22. (3x 1 1)(3x 2 1) 23. (2s 1 1)2

24. (2m 2 3n)2 25. (4t 1 3)(4t 2 3)26. (5z 2 2)2 27. (x 1 4y)2 28. (2t 1 3)(t2 2 5) 29. (m 2 5)(2m 1 n)30. (3x2 1 1)(x 2 7) 31. (x 1 2y)(x 2 4)

Check

1. 4y(y2 2 5) 2. 5m2(2m2 1 3)3. (t 2 9)(t 1 2) 4. (x 2 3)(x 1 9)5. (s 2 7)(4s 1 1) 6. (5y 1 2)(2y 1 1)7. (4x2 1 y)(4x2 2 y) 8. (5s 1 2t)(5s 2 2t)9. (m 2 3n)2 10. (2x 1 6y)2

11. (x 2 3)(x 1 4y) 12. (t 2 2)(t 1 3s)13. 23, 0 14. 25, 4 15. 22, 8

16. 25, 1 }

3 17.

1 }

2 ,

3 }

2 18. 28, 8 19. 2

2 } 3 ,

2 }

3

20. 2 5 }

3 21.

1 }

2


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Answers continued

C. Graphing Quadratic Functions

1.

1

2

3

4

5

6

7

8y

1 x21222324 2 3 4

y 5 4x2

Same vertex, (0, 0), and axis of symmetry, x 5 0. The graph is narrower than the graph of y 5 x2

— it is a vertical stretch (by a factor of 4) of the graph of y 5 x2.2. y

1 x21222324 2 3 4

y 5 2 x213

28

27

26

25

24

23

22

21

Same vertex, (0, 0), and axis of symmetry, x 5 0. The graph is wider than the graph of y 5 x2 — it

is a vertical shrink (by a factor of 1 }

3 ) with a

refl ection in the x-axis of the graph of y 5 x2.

3.

1

2

3

4

5

6

7

8y

1 x21222324 2 3 4

y 5 x212

Same vertex, (0, 0), and axis of symmetry, x 5 0. The graph is wider than the graph of y 5 x2 — it

is a vertical shrink (by a factor of 1 }

2 ) of the graph

of y 5 x2.

4.

y 5 x2 1 11

2

3

4

5

6

7

8y

1 x21222324 2 3 4

Graph also opens up, has the same axis of symmetry, x 5 0. The vertex (0, 1), is different. The graph of y 5 x2 1 1 is a vertical translation (of 1 unit up) of the graph of y 5 x2.

5.

y 5 x2 2 2

1

2

3

4

5

6y

1 x21222324 2 3 421

22

Graph also opens up, has the same axis of symmetry, x 5 0. The vertex (0, 22), is different. The graph of y 5 x2 2 2 is a vertical translation (of 2 units down) of the graph of y 5 x2.

6.

y 5 x2 1 4

1

2

3

4

5

6

7

8y

1 x21222324 2 3 4

Graph also opens up, has the same axis of symmetry, x 5 0. The vertex (0, 4), is different. The graph of y 5 x2 1 4 is a vertical translation (of 4 units up) of the graph of y 5 x2.7. x 5 21, (21, 8) 8. x 5 4, (4, 29)9. x 5 2, (2, 4) 10. x 5 22, (22, 23)

11. x 5 3, (3, 3) 12. x 5 3 }

2 , 1 3 }

2 ,

19 }

2 2


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Answers continued

13.

y 5 2x2 2 8x 1 3

1

2

3y

1 2 x2122 3 4 5 621

22

23

24

25

14.

y 5 23x2 1 6x 2 4

y

1 x21222324 2 3 4

28

27

26

25

24

23

22

21

15.

1

2

3

4

5y

x212223242526272821

22

23

16. Maximum value, 3 17. Minimum value, 2618. Minimum value, 4 19. Maximum value, 520. Maximum value, 22 21. Minimum value, 24

Check

1.


vertex (0, 0)y

1 x21222324 2 3 4

28

27

26

25

24

23

22

21

2.

1

2

3

4

5

6y

1 x21222324 2 3 421

22


vertex (0, 22)

3.


vertex (1, 3)

1

2

3

4

5

6

7

8y

1 x21222324 2 3 4

4.

1

2

3

4y

1 x2122 3 4 5 621

22

23

24vertex (2, 24)


5.

1

2

3y

x2123242526272821

22

23

24

25

vertex (22, 3)


6. y

1 x2123242526 2

28

27

26

25

24

23

22

21

vertex (22, 23)



An

swer

Key


Answers continued

7. Minimum value, 7 8. Maximum value, 26

9. Minimum value, 2 1 } 8 10. Minimum value, 27

11. Maximum value, 5 1 }

2 12. Maximum value, 7

D. Solving Quadratic Equations

1. 22, 1 2. No solution 3. 22 4. 3 5. No solution 6. 23, 2 7. 0 8. 63 9. No solution 10. No solution 11. 0 12. 64 13. 26, 2 14. 23, 1 15. 25, 716. 28, 24 17. 23, 9 18. 213, 2319. 25.32, 1.32 20. 21, 1.67 21. 21.78, 0.2822. Factoring; 2x2 2 x 2 6 can be factoredeasily 23. Factoring; x2 2 6x can be factoredeasily 24. Quadratic formula; 4x2 1 2x 2 9 cannot be factored 25. Square roots; write in the form x2 5 d 26. Quadratic formula;x2 2 3x 2 6 cannot be factored27. Completing the square; a 5 1 and b is even28. Two solutions 29. No solution30. Two solutions 31. One solution32. No solution 33. One solution

Check

1. 4 2. No solution 3. 23, 4 4. 62 5. No solution 6. 0 7. 22, 10 8. 213, 19. 2, 12 10. 0.23, 1.43 11. 22.16, 1.1612. 21.83, 3.83 13. No solution 14. Two solutions 15. One solution


1. Linear function 2. Exponential function3. Quadratic function 4. Linear function5. Quadratic function 6. Exponential function7. Quadratic function; y 5 0.2x2 8. Linear function; y 5 3x 2 2 9. Linear function; y 5 2x 1 3 10. Quadratic function; y 5 3x2

Check

1. Quadratic function 2. Linear function 3. Exponential function 4. Linear function5. Exponential function 6. Quadratic function7. Linear function; y 5 2x 2 1 8. Quadratic function; y 5 4x2 9. Quadratic function;y 5 0.1x2 10. Linear function; y 5 4x 1 2

Ready to Go On? Chapters 11–12 InterventionA. Radical Functions

1. y

x1 2 3 4 6 7 85

1

2

3

4

5

6

7

8

y 5 5 x�

The domain is x ≥ 0. The range is y ≥ 0. The graph of y 5 5 Ï}

x is a vertical stretch (by a factor of 5) of the graph of y 5 Ï}

x .

2. y

x1 2 3 4 6 7 85

3

2

1

4

5

6

y 5 20.8 x�

21

22

The domain is x ≥ 0. The range is y ≤ 0. The graph of y 5 20.8 Ï}

x is a vertical shrink (by a factor of 0.8) with a refl ection in the x-axis of the graph of y 5 Ï}

x .

3. y

x1 2 3 4 6 7 85

1

2

3

4

5

6

7

8

y 5 x 1 1�

The domain is x ≥ 0. The range is y ≥ 1. The graph of y 5 Ï}

x 1 1 is a vertical translation (of 1 unit up) of the graph of y 5 Ï}

x .


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Answers continued

4. y

x1 2 3 4 6 7 85

1

2

3

4

5

6

7

8

y 5 x 2 3�

The domain is x ≥ 3. The range is y ≥ 0. The graph of y 5 Ï

}


x .

5. 4 Ï}

3 6. Ï

}

11 } 5 7.

Ï}

30x } 5 8.

Ï}

15 }

x2 9. 2 Ï

}

3x } z

10. Ï}

30a }

3a 11. 3 Ï

} 7 12. 4y Ï

} y 13. 2z Ï

} 7y

14. 4 Ï

}

3x }

3x 15.

2 Ï}

5 }

3x 16.

5 Ï}

b }

2b2 17. 7 Ï}

6 2 Ï}

3

18. 17 Ï}

2 19. 5 Ï}

3 1 6 20. 3 Ï}

3

21. 2 Ï}

15 2 14 22. 2 1 3 Ï}

6 23. 16 24. 57

25. 4 }

9 26. 3 27. 16 28. 6 29. 3 30. 6

31. 3, 4

Check

1. y

x1 2 3 4 6 7 85

y 5 2 x�

1

2

3

4

5

6

7

8

The domain is x ≥ 0. The range is y ≥ 0. The graph of y 5 2 Ï}

x is a vertical stretch (by a factor of 2) of the graph of y 5 Ï}

x .

2. y

x1 2 3 4 6 7 85

y 5 23 x�

21

22

23

24

25

26

27

28

The domain is x ≥ 0. The range is y ≤ 0. The graph of y 5 23 Ï}

x is a vertical stretch (by a factor of 3) with a refl ection in the x-axis of the graph of y 5 Ï}

x .

3. y

x1 2 3 4 6 7 85

y 5 x 1 4�

1

2

3

4

5

6

7

8

The domain is x ≥ 0. The range is y ≥ 4. The graph of y 5 Ï}

x 1 4 is a vertical translation (of 4 units up) of the graph of y 5 Ï}

x .

4. y

x1 2 3 4 6 7 85

1

2

3

4

5

6

7

8

y 5 x 2 2�

The domain is x ≥ 2 . The range is y ≥ 0. The

graph of y 5 Ï}


x .

5. 3 Ï}

11 6. Ï

}

13 }

8 7.

Ï}

17y }

2z2 8. 2x2 Ï}

3x 9. Ï

} 7x }

3y

10. 3xy Ï}

5x 11. 2x4 Ï}

11 12. 4 Ï

}

2y } 5z

13. 9 2 3 Ï}

3 14. 24 Ï}

2 15. 5 Ï}

7

16. 18 1 2 Ï}

21 17. 2 Ï}

3 18. 22 1 5 Ï}

30

19. 25 20. 102 21. 1 }

4 22. 5 23. 9

24. 8 25. 5 26. 1, 2 27. 4

B. Radicals in Geometry

1. 2 in., 7 in. 2. 3 ft, 7 ft 3. Ï}

5 m, 2 Ï}

5 m,

4. 2 Ï}

13 units 5. Ï}

82 units 6. 3 Ï}

5 units

7. 3 Ï}

2 units 8. 5 units 9. Ï}

13 units 10. 1, 511. 23, 7 12. 27, 3 13. (21, 5) 14. (5, 3)


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Key


Answers continued

15. 1 2 1 } 2 , 2

3 } 2 2 16. (0, 22) 17. (2, 21)

18. 1 1 1 }

2 , 3

1 }

2 2

Check

1. 5 in., 6 in. 2. 4 ft, 12 ft 3. 5 units

4. Ï}

34 units 5. 2 Ï}

5 units 6. 22, 2 7. 5, 21

8. (22, 0) 9. (1, 21) 10. 1 2 1 }

2 , 4

1 }

2 2

C. Operations on Rational Expressions

1. 2x2 2 3x 1 1 2. 3y 3 1 y 2 2 4y 3. 5y2 1 3y 2 10 4. 4z3 2 z2 2 5z

5. y 2 3 6. b 2 4 7. a 1 3 1 7 }

3a 2 1

8. 2x 1 3 2 2 }

x 2 4 9. 0 10. 23

11. 25, 5 12. 4 13. 21, 4 14. 015. No excluded values 16. 3

17. x 2 2

} 3(x 2 4)

18. 2z(z 1 5)

} z 1 2

19. x(x 2 1)

} 4(x 1 6)

20. w(w 1 3)

} 2(w 1 1)

21. 36x3

22. (b2 1 2)(b 2 2)

23. 5(y 1 3)(y 2 1)(y 1 1)

24. 8b 1 15

} 20b2 25.

20x2 2 9x 1 27 }}

4x(x 2 3)

26. 5z 2 7 }}

(z 2 5)(z 1 3)(z 2 2)

Check

1. a2 1 6a 2 3 2. 4x3 1 6x2 1 2x 3. 4m2 2 m 2 3

4. 7y3 2 3y2 1 9 5. n 1 6 6. 2x 1 1

7. 2b 2 1 2 4 }

2b 1 3 8. 4y 1 1 1

6 } y 2 5

9. 0 10. No excluded values 11. 21, 1

12. No excluded values 13. a 1 2

} a 1 4

14. (y 2 3)(y 2 1)

}} y(y 1 2)

15. 2b(b 2 5)

} b 2 2

16. x 2 2

} 2x(x 1 1)

17. 75y3

18. 2(m 2 2)(m 1 2)(m 1 5)

19. 3x(x2 1 3x 1 3) 20. 25x2 2 9

} 60x3

21. 3m3 1 7m 1 14

}} 3m2(m 1 2)

22. 3b 1 2

}} (b 2 4)(b 1 4)(b 2 2)


1. y 5 212

} x , 23 2. y 5 8 } x , 2 3. y 5

26 } x , 2

3 } 2

4. y 5 14

} x , 7 }

2 5. y 5

236 } x , 29 6. y 5

227 } x , 2

27 } 4

7. y

x1 2 3 4

1

2

3

4

2121222324

22

23

24

The graph of y 5 3 } x is a vertical stretch of

the graph of y 5 1 } x . The domain and range are all

nonzero real numbers.

8. y

x1 2 3 4

1

2

3

4

2121222324

22

23

24

The graph of y 5 21

} 2x

is a vertical shrink and a

refl ection in the x-axis of the graph of y 5 1 } x .

The domain and range are all nonzero real numbers.

9. 1y

1 x21222324 2 3 421

22

23

24

25

26

27

The graph of y 5 1 } x 2 3 is a vertical translation

(of 3 units down) of the graph of y 5 1 } x . The

domain is all real numbers except 0. The range is all real numbers except 23.


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Answers continued

10.

1

2

3

4y

1 2 x21222324252621

22

23

24


x 1 2 is a horizontal translation

(of 2 units to the left) of the graph of y 5 1 } x .

The domain is all real numbers except 22. The range is all real numbers except 0.11. 25, 1 12. 24, 3 13. 28, 22 14. 2

15. 21, 4 16. 24, 3 17. 22, 1 }

2

Check

1. y 5 8 } x , 4 2. y 5

210 } x , 25 3. y 5

212 } x , 26

4. y 5 24

} x , 12

5. y

x1 2 3 4

1

2

3

4

2121222324

22

23

24

The graph of y 5 4 } x is a vertical stretch of the

graph of y 5 1 } x . The domain and range are all

nonzero real numbers.

6. y

x1 2 3 4

1

2

3

4

2121222324

22

23

24

The graph of y 5 21

} 3x

is a vertical shrink and

a refl ection in the x-axis of the graph of y 5 1 } x .

The domain and range are all nonzero real numbers.

7.

1

2y

1 x21222324 2 3 421

22

23

24

25

26

The graph of y 5 1 } x 2 2 is a vertical translation

(of 2 units down) of the graph of y 5 1 } x . The

domain is all real numbers except 0. The range is all real numbers except 22.

8.

1

2

3

4y

1 x21222324252621

22

23

24


x 1 3 is a horizontal translation

(of 3 units to the left) of the graph of y 5 1 } x .

The domain is all real numbers except 23. The range is all real numbers except 0.

9. 26, 4 10. 21, 8 11. 2, 6 12. 2 13. 7, 814. 23 15. 24, 2


An

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Key



1. 14 2. 15 3. 12 4. 49 5. 6 6. –29

7a. 3375 cm3 7b. 631 cm3 8. B 9. B 10. A

11. A 12. No 13. Yes 14. Yes 15. No

16. You know the number of magnets you have to sell, the cost per magnet, the amount you spend on the booth and set up, and the amount of profi t you want to make; You want to fi nd the price per magnet so that if you sell half the magnets you make a profi t of $200.

17. 50p 2 [100(0.60) 1 50] 5 100; You should sell each magnet for $4.20. 18a. 7r 1 5c

18b. 7r 1 5c ≤ 35 18c. Yes; 7(2) 1 5(3) 5 29, which is ≤ 35. 19. m 5 2n 20. y 5 4x21. C

22. x 22 0 2 4

y 25 21 3 7

23. x 22 21 1 2

y 10 7 1 22

24a. p

n

5

4

3

2

1

67

8

6 7 851 2 3 4

(1, 2)

(2, 4)

(3, 6)

(4, 8)

24b. p 5 2n 24c. 140 25. B26. Domain: 1, 2, 3, 4, 5; Range: 1, 2, 3, 427. Domain: 2, 6, 8, 10, 12; Range: 2, 6, 10, 1228a. s 5 2.5w

28b. s

w

10

8

6

4

2

1214

16

6 7 851 2 3 4

(1, 2.5) (2, 5)

(3, 7.5)

(4, 10)

29. 2a 5 3; | a | 5 3 30. 2a 5 20.2; | a | 5 0.2

31. 2a 5 Ï}

5 ; | a | 5 Ï}

5 32. 2 5 }

24 33. 1.8

34. 217 35. 211.7 36. 12 37. 1 }

6 38. ab

39. 28t 40. 12 41. 2 1 }

15 42.

2 } 75 43. 227

44. 11 45. 212 46. 68 47a. 2 5 } 11

47b. The inverse property of multiplication states

a p 1 } a 5 1. Since 22 1

} 5 5 2 11

} 5 , 1 } a 5 2

5 } 11 . 48. A

49. B 50. 6 51a. 8 ft 51b. Ï}

61 is between

Ï}

49 and Ï}

64 . So, it is between 7 and 8, and iscloser to 8. 52. Yes; Any whole number can be expressed as the quotient of itself divided by 1.

53. C 54. 26.9, 2 21

} 4 , |27 |, Ï

}

50

55. 24, 2|23.2|, 2 Ï}

3 , 3 }

2 56. D

57. 1 }

3 m 1 0 5

1 }

3 m 58. C 59. Associative

property of multiplication; inverse property of multiplication 60. 8x3 2 10x2

61. 212n2 1 18n 62. 2x 1 x2 2 2xy


1. a 5 19 2. h 5 8 3. z 5 12 4. b 5 29

5. x 5 5 6. m 5 12 7. w 5 10 8. a 5 3.3

9. x 5 5 10. d 5 53

} 4 11. g 5 2

3 }

4 12. v 5

1 } 5

13. A 14. D 15. C 16a. 3 }

2 16b.

2 } 5 17. B

18. w 5 8 } 5 19. b 5 18 20. x 5 12 21. h 5 21

22. v 5 1 }

2 23. y 5 4 24a. 2

1 }

4 cups

24b. 3 3 }

8 cups 25. D 26. A 27. 49.5%

28. B 29. C 30. $30 31a. x 5 b } a 2 c

31b. x 5 5 }

8 32a. x 5

a 1 d }

c 2 b 32b. x 5 7

33a. C

} 2r

5 π 33b. 3.14 34. B

35. y 5 2 1 } 2 x 1

3 }

8 36. y 5 2x 1 1

37. y 5 5 } x 38. y 5 22x 2 3

Answers


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39–42. y

x

2(23, 1)(2, 0)

(4, 24)(0, 22)22

24

26

28

4

6

8

24 222628 2 4 6 8

43. D

44. x 0 3 6 9

y 13 9 5 1

y

x

12

10

8

6

4

2

14

16

18

2100

3 4 5 6 7 8 9

45–48. y

x

2

22

24

26

28

4

6

8

24 222628 2 4 6 8

x 5 25 x 5 2

y 5 8

y 5 24

49. x-intercept: 16; y-intercept: 26 50. x-intercept: 210; y-intercept: 6 51. x-intercept: 215; y-intercept: 6 52. C 53. D 54. A 55a. y 5 x 2 1

56a. y 5 2 1 } 3 x 1 2 57a. y 5 5x 1 15

55b–57b. y

x2

22

24

26

28

4

6

8

24 222628 2 4 6 8

y 5 5x1 15

3y 5 2 x 1 21

y 5 x2 1

58. C 59. Not a direct variation. 60. Direct

variation; 3 }

4 61. Not a direct variation.

62. Direct variation; 21 63. B

64a. y 5 5 }

3 x 64b. y 5 225

65a. c 5 0.03m 65b. $9

Ready to Go On? Chapters 5–6 Quiz1. y 5 23x 1 2 2. C 3. C 4. y 5 22x 1 2

5. A 6. y 1 9 5 2 13

} 11 (x 2 5) 7. 3x 1 y 5 13

8a. p } q 8b. 2px 1 qy 5 qp 9. C

10. y 5 3 }

4 x 2

15 }

4 11. B 12. y 5 2

5 } 2 x 1 10

13. C 14. Positive correlation 15. Negative correlation 16. No correlation 17. Negative correlation 18. B

19a.

4030Years since 1930

Aver

age

Leng

th (i

n m

inut

es o

ver 1

00)

2010 6050 70 80 90

2010

30

50

70

40

60

8090

t

m

00

19b. Answers may vary. Sample: m 5 2 }

3 t 1 31

19c. About 184 minutes; t 5 2010 2 1930 5 80,

m 5 2 } 3 (80) 1 31 < 84,

runtime 5 84 1 100 5 184

Answers continued


An

sw

er

Key


20a.

10Years (in years since 1980)

Popu

latio

n

5 15 20

800

700

900

1000

1100

t

p

00

20b. Answers may vary. Sample: p 5 212t1 1055 20c. About 911; t 5 1992 2 1980 5 12, p 5 212(12) 1 1055 5 911

21. �3�4�5 �2 �1 0 3 4 51 2

22. �3�4�5 �2 �1 0 3 4 51 2

23a. y ≥ 2 1 }

3 x 1 30;

4030Cups Sold

T-sh

irts

Sol

d

2010 6050 70 80 90

2010

30

50

70

40

60

8090

x

y

00

23b. No; the point (20, 18) is not in the feasible region. 23c. 30; the number of T-shirts the team must sell if they do not sell any cups.24. A 25. B 26. 23 < x ≤ 4

�3�4�5 �2 �1 0 3 4 51 2

27. x ≥ 1 or x < 0

�3�4�5 �2 �1 0 3 4 51 2

28. y

x

2

22

24

26

28

4

6

8

24 222628 2 4 6 8

29. y

x

2

22

24

26

28

4

6

8

24 222628 2 4 6 8

30. y

x

2

22

24

26

28

4

6

8

24 222628 2 4 6 8

31. y

x

2

22

24

26

28

4

6

8

24 222628 2 4 6 8

32. x > 27 33. a ≤ 2 13

} 2 34. x > –3

35. p ≥ 26 36. ƒ ≤ 219 37. All real numbers

38. No solution 39. q ≥ 5 }

2

40. 2 > b ≥ 25 41. 3 < x < 6

42. c < 2 11

} 2 or c ≤ 26 43. n ≤ 28 or n > 4

Answers continued


An

sw

er K

ey


44. C 45a. 5 < p ≤ 10 45b. Neither; 14 is not a solution to p ≤ 5 (position for an A) nor 5 < p ≤ 10 (position for a B). 46. x 5 213 or x 5 13 47. b 5 28 or b 5 2

48. m 5 2 7 } 2 or m 5

15 }

2 49. No solution

50. d 5 26 or d 5 0 51. No solution 52. 22 < z < 2 53. g ≥ 4 or g ≤ 22

54. �3�4�5 �2 �1 0 3 4 51 2

55. B 56. B

Ready to Go On? Chapters 7–8 Quiz1. C 2a. 9j 2 4m 5 72; j 2 12 5

1 }

2 (m 2 12)

2b. j

m

10

210

220

230

240

20

30

40

220210230240 10 20 30 40

2c. Mari will be 41 and James will be 29.3. y

x

2

22

24

26

28

4

6

8

24 222628 2 4 6 8

(23, 21)

4. A 5. The graph of each equation in the system is the same line. 6. Infi nitely many solutions

7. x 5 21, y 5 2 8. x 5 2 5 } 4 , y 5 22

9. p 5 3 } 7 , q 5

64 }

21 10. x 5 1, y 5 21

11. m 5 2 8 } 3 , n 5 2

17 } 9 12. a 5 2

13 } 6 , b 5 2

10 }

3

13. x 5 3, y 5 4 14. No solution 15. A16a. d 5 111.2h 2 25.6; d 5 112h 2 2816b. 308 miles; 5.5 hours 17. C

18. y

x

2

22

24

26

28

4

6

8

24 222628 2 4 6 8

19. B20. y

x

1

2

21

22

23

24

3

41 2 3 6 7 8

21. B 22. y ≥ 2 x 1 2; y < 1; y > x 2 4

23a. w 2 3h ≤ 0; w 1 1.75h < 9 23b. Yes; h 5 1.5 and w 5 2.75 is a solution of both inequalities.

24. (23)5 25. 1 }

a5 26. x18 27. 2 243p5

} q10

28. 4z3

} 5w 29. 2 }

a11 30. 1 31. 4

32. 1 }

8 33. C 34. A 35. B 36. A

37. D

38. B 39. 1.008 3 108 40. 3.38 3 102

41. 4.41 3 10214 42. 3 3 1028

43. y

x

1

2

3

4

5

6

7

8

222324 1 2 3 421

Domain: all x; Range: y ≥ 044. B 45a. p 5 9.8 3 106(1.015)t

45b. 1.14 3 107

Answers continued


An

sw

er

Key


46. y

x

1

2

3

4

5

6

7

8

222324 1 2 3 421

Domain: all x; Range: y ≥ 047. D 48a. y 5 16,000(0.858)x 48b. $4699

Ready to Go On? Chapters 9–10 Quiz1. C 2. D 3. Yes; 6th degree trinomial 4. 10y5 1 5y3 2 11y2 1 14 5. B 6. 2x2 2 2x 2 3 7. A8. 220a6 2 4a5 1 28a3

9. 10p2 2 18pq 2 4q2 10. d3 2 d2 2 5d 1 2 11a. Entire lawn: 18x2 1 42x 2 16; Watered lawn: 4πx2 2 4πx 1 π 11b. 5.44x2 1 54.56x 2 19.14 12. C

13. 9x2 1 6xy 1 y2 14. B 15. x 5 1 }

2 or x 5 23

16. D 17. x 5 0 or x 5 2 18. n 5 0

or n 52 1 } 2 19. D 20. (d 2 6)(d 1 4)

21. (k 2 7)(k 1 8) 22. A 23. (5y 1 2) (y 2 1)

24. 2(3z 1 4)(z 2 1) 25. D 26a. 24(4t 1 1) (t 2 5) 26b. Set each factor equal to 0 and solve for t. 27. (r 1 5)(r 2 5) 28. D 29. A 30. (2x2 1 3)(x 1 3)

31. y

x

1

21

22

23

24

2

3

4

22 212324 1 2 3 4

32. y

x

2

22

24

26

28

4

6

8

24 222628 2 4 6 8

33. B 34. A

35. y

x

2

22

24

26

28

4

6

8

24 222628 2 4 6 8

36. y 5 x2 2 5x 1 2 37. D 38a. 18 ft 38b. 100 ft

39. y

x22

26

24

28

210

212

214

216

218

2

6

4

8

10

12

14

16

18

24 222628212210214216218 2 4 6 8 10 12 14 16 18

(2, 0) (�1.5, 0)

40. p 5 6 7 }

3 41. d 5 66 42. No solution

43. h 5 10 6 Ï}

79 44. y 5 1 }

6 or y 5

8 }

3

Answers continued


An

sw

er K

ey


45. x 5 3 6 Ï}

23 46. D 47. A 48. A49. B 50. B 51. Quadratic; The second differences are equal. 52. B

53. y 5 1 }

2 x2 1 3x 1 5

Ready to Go On? Chapters 11–12 Quiz1. y

x

2

22

24

26

28

4

6

8

24 222628 2 4 6 8

2. D 3. A 4. 7 Ï}

3 5. 2 Ï}

3 } 11

6. 6x2 Ï}

2

7. Ï}

15c } 5c 8. 11 Ï}

3 9. 2112 Ï}

7 10. D 11. C

12. x 5 16 13. x 5 24 14. x 5 1 15. x 5 416. C 17. 2 in., 6 in. 18. A 19. c 5 7 or c 5 3 20. B 21. B 22. 3x2 1 2x 2 5

23. x 2 17 24. C 25. x 2 1 1 6 }

2x 2 3

26. D 27. B 28. C 29. B

30. (w 2 3)(w 1 3)

}} w 1 2

31. 3x 1 1 1 10 }

x 2 4

32. 5a 1 3 2 6 }

5a 2 3 33. B 34. 2 }

v 1 8

35. x(3x 1 26)

}} (x 2 3)(x 1 4)

36. 22x2 2 5x 1 1 }} (x 1 1)2 (x 2 3)

37. 56c2 1 3c 2 12 }} 7c(c 2 4)

38. D 39. A

40a. Yes; t 5 36

} s 40b. 4.3 ft/sec

41. y

x

2

22

24

26

28

4

6

8

24 222628 2 4 6 8

42. x 5 218 43. m 5 24 or m 5 7

44. c 5 11

} 2 or c 5 29 45. 4 6 3 Ï

}

3

Answers continued


An

sw

er

Key



Ready to Go On? Chapter 1 Cumulative Review

1. 72 2. 1.3 3. 7 4. 16 5. 32 6. 3 }

8 7. 25

8. 20 9. 23 10. 2 11. 9x 12. 15 1 y

13. z2 2 6 14. 3w 2 11 15. 7x 5 42

16. y 1 17 ≤ 36 17. 5(z 1 3) < 45

18. yes 19. no 20. no 21. yes

22. I 5 Prt; $140 23. d 5 rt; 292.5 mi

24. domain: 2, 3, 4, 5, 6; range: 3, 8, 13, 18, 23

25. domain: 1, 4, 7, 10, 13; range: 1, 2, 3, 4, 5

26. Input 11 13 15 17 20

Output 4 6 8 10 13

range: 4, 6, 8, 10, 13

27. Input 2 3 5 7

Output 2 6 14 22

range: 2, 6, 14, 22

28. 1.55 p 1 Number of boxes of screws 2 1 1.05 p 1 Number of boxes

of nails 2 5 Total cost; $9.90

29.

x

y

O 1 2 3

5

4 5 6

10

15

20

25 30.

x

y

O 2 4 6

1

8 10 12

2

3

4

5

6

7

8

9

31. y 5 2x 1 1; domain: 0, 1, 2, 3, 4; range: 1, 3, 5, 7, 9 32. y 5 9 2 2x; domain: 0, 1, 2, 3, 4; range: 9, 7, 5, 3, 1


1. 19 2. 28 3. 11 4. 11 5. 9 6. 1 }

2 7. 29

8. 3x 2 4 9. 8y 2 11 10. x 1 15 5 7

11. 7y < y2 12. no 13. yes 14. no

15. I 5 11.50s 1 13.75l; $149.25

16. y 5 x 2 4; domain: 5, 6, 7, 8, 9; range: 1, 2, 3,

4, 5 17. y 5 1 }

2 x; domain: 0, 2, 4, 6, and 8; range:

0, 1, 2, 3, and 4

18.

x

y

O 1 2 3

2

4 5

4

6

8

10

12

14

19. 20.6, 2 1 }

3 , 0,

1 }

2 , 1.5

20. 23.5, 2 Ï}

8 , 2 Ï}

4 , Ï}

9 , 3.2 21. 21, 21

22. 0.75, 0.75 23. 2 2 } 7 ,

2 } 7 24. 215 25. 21.4

26. 24 27. 211.1 28. 1.2 29. 2 1 }

16 30. 266

31. 26 32. 7 33. 6 34. 227 35. 2 3 } 7

36. Inverse property of addition

37. Associative property of addition

38. Identity property of multiplication

39. Commutative property of multiplication

40. 26x 2 54 41. 23y2 1 33y 42. 9 }

2 z 2 27

43. 23x 2 12 44. 12y 2 1 }

2 45. 23z + 1

46. 21 47. 2 48. If a number is irrational, then it is a real number.; true 49. If a number is an interger, then it is a whole number.; false; Sample Answer: 21 is an integer, but it is not a whole number. 50. 212 51. 66 52. 4 53. 219


1. 267 2. 11 3. 35 4. 41 5. 8 6. 13 7. 3

8. 12 9. 16 10. 3x 2 11 11. y2

} 6 ≥ 20

12. 17 1 2z 5 32 13. 0.75n 1 2.99m; $82.30

14. y 5 x 2 9 15. y 5 23x

16. real number, irrational number

17. real number, rational number

18. real number, rational number, integer

19. real number, rational number

20. 221 21. 22.5 22. 18.9 23. 1 }

10 24. 224

25. 9 26. 26 27. 23 28. 34 29. 28m 2 16

30. 211x 1 4 31. 23x 1 8 32. 2x 2 7

33. x 5 263 34. y 5 6 35. z 5 29

36. a 5 24 37. b 5 50 38. c 5 75

39. p 5 212 40. m 5 2 41. w 5 14

} 13

42. 57.5 mi 43. 20% 44. 36% 45. 16.8

Answers

An

sw

er K

ey



46. 105% 47. 56% 48. y 5 23x 1 12

49. y 5 5x 2 7 50. y 5 5 }

2 x 2 2

51. a. l 5 V

} wh

b. 21 in.


1. 15.625 ft3 2. 55 3. 42 4. 26 5. 35 6. 8

7. 18 8. 5x

} 7 9. y3 2 4 < 8y 10. 6(z 1 3) 5 27

11. no 12. yes 13. no 14. 211.3 15. 22 3 }

10

16. 272 17. 2156 18. 230 19. 9 }

11

20. $9.80 21. 26.2, 2 Ï}

36 , 0.5, Ï}

2 , 2

22. 23.5, 2 Ï}

6 , 2 1 } 3 , Ï

}

3 , 2.1 23. a 5 26

24. k 5 233 25. x 5 36 26. q 5 6 27. p 5 6

28. m 5 3 29. x 5 48 30. y 5 5 31. z 5 3

32. 100 ft; 60 ft 33. 4.2 34. 40% 35. 35%

36. 405 37. 44%

38. 39.

�6

x

y

2

�2�4�6�8 4 6 8

4

�2

�4

�8

�6

x

y

2

2

�2�4�6�8 4 6 8

4

8

�2

�4

�8

40. 41. 8 }

9 42. 23 43. 21

�6

x

y

2

�2�4�6�8 4 6 8

4

8

�4

�8

6

44. m 5 25; b 5 8 45. m 5 2 1 }

2 ; b 5 5

46. m 5 4 }

3 ; b 5 28 47. yes;

5 } 7 48. no

49. yes; 2 3 }

2 50. a. s 5

1 } 5 C b. 7g


1. 7 2. 18 3. 2 2 } 9 4. 224 5. 3 6. 6

7. 12 pieces 8. 7 9. 28.4 10. 1 }

2 11. 260

12. 12 13. 49 14. 23x 2 12 15. 22b 2 45

16. 26a 2 2 17. 29 18. 25 19. 3 20. 10

21. 0.5 22. 14 23. 6 books 24. 4.4 25. 56

26. 9 27. 12 28. $375 29. y 5 2ax 1 1

} b

30. x 5 2y

} 3z

31. 1 }

2 32. 3 33. 2

1 } 8

34. m 5 22, b 5 7 35. m 5 26, b 5 10

36. m 5 5 }

4 , b 5 0 37. yes,

1 } 7 38. no 39. yes,

3 }

4

40. 41.

x

y

O

1

2

3

2122

25

24

22

2324 3 4

x

y

O

1

2

2122

25

26

24

22

2324 3

42.

x

y

O

1

3

4

222

24

22

23

2324 3 4

43. 16

44. 12 45. y 5 8x 2 3 46. y 5 1 }

2 x 1 7

47. y 5 22x 1 1 48. y 5 3

49.

x

y

O

2

4

6

8

10

12

14

16

21 3 4 5

50.

x

y

O

1

2

3

4

5

21 3 4 5

y 5 1 }

2 x 1 2.5

y 5 2x 1 5.5


1. 75 2. 24 3. 24 4. yes 5. no 6. yes

7. 28 8. 238 9. 27 10. 77 11. 60

12. 212 13. 13 14. 23 15. 616 16. 60.2

17. 12 18. 9 19. 3 20. 20 21. 1 }

2 22. 24

23. 1 }

2 24. 6.5 25. no solution 26. 48 pages

27. 40% 28. 52.5 29. 925 students

Answers continued

An

swer

Key



30. a 5 2by 1 c

} x 31. y 5 abx 2 bc

32–34.

x

y

1

3

5

1�1�1

�3

�3 3

A(3, 6)

B(22, 24)

C (0, 23)

32. Quadrant I 33. Quadrant III 34. on y-axis 35. 5 36.

1 } 5 37. 0

38.

x

y

1

3

�1

�3

�3 �1 3

39.

x

y

1

3

�1

�3

�3�5 �1 1

40.

x

y

1

�1�3 �1 1

41. 4 42. 23

43. 4h 5 6w; 18 thousand gallons

44. y 5 2x 1 5 45. y 5 3 }

4 x 2 6

46. y 5 25x 1 3 47. 7x 1 y 5 14

48.

x

y

8

24

40

62 10

The graph shows a positive correlation. So, as x increases y increases.

49. x < 5; 0 1 2 3 4 5 621

50. x ≥ 2; 0 1 2 3 4 5 621

51. x < 1; 0 1 2 3 421

52. 1 < x < 6; 0 1 2 3 4 5 621

53. x ≤ 2 or x ≥ 7; 0 1 2 3 4 5 6 7 8

54. 1 }

2 < x < 5

1 }

2 ;

0 1 2 3 4 5 621

12

125

55.

x

y1

�1

�3

�3 �1 3

56. 5w 1 8s ≥ 300;

w

s

10

20

30

40

40 50 6030201000

Ho

urs

as a

sw

imm

ing

in

stu

cto

r

Hours washing cars

Summer Work


1. 9 2. 14 3. 12 4. 73 5. 8 }

27 6.

5 }

2

7. x 0 3 5 7 10

y –5 1 5 9 15

range: 25, 1, 5, 9, 15

8. x 0 2 4 6 8

y 3 4 5 6 7

range: 3, 4, 5, 6, 7

9. 2 10. 2 3 } 4 11. 4.5 12. 230 13. 27

14. 2 14

} 27 15. 7 16. 1 17. 9 18. 12 in.; 18 in.

19. 40% 20. 60.5 21. 16% 22. 260% 23. 3

24. 21

25.

x

y

1

3

12123

23

3

26.

x

y

1

3

12121

23 3

27. y 5 2 5 } 2 x 2 3 28. y 5 2x 2 3

29. y 5 2x 2 12 30. y 5 3x 31. 23x 1 4y 5 1

32. y < 20 33. x ≥ 25 34. z > 5

35. 22 ≤ x < 3 36. no solution

37. y > 1 }

2 or y < 2

7 } 2

38. 8p 1 6m ≤ 350;

8162432404854

16 24 32 40 48 54 62800

Sw

imm

ing

po

ol

Movies

Summer Activities

Answers will vary.

Answers continued

An

sw

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ey



39. (2, 23) 40. (1, 3) 41. (24, 22) 42. one solution 43. infi nitely many 44. no solution

45.

x

y

1

3

1212321

3

46.

x

y

1

3

5

21 5


1. 17 2. 8 3. 55 4. 4 } 7 5. 24 6. 48 7. 26 ft

8. 25.2 9. 2 19

} 24 10. 48 11. 2 8 } 11 12. 2

1 } 49

13. 2 2 } 3 14. 2x 2 4 15. 27x 1 63

16. 26z2 2 5z 17. 23 18. 24 19. 2 20. 18

} 13

21. 67.6 km 22. m 5 1; b 5 23

23. m 5 5 }

3 ; b 5 210 24. m 5 2

1 } 2 ; b 5 22

25. 6 26. 213 27. y 2 3 5 1 }

4 (x 2 6)

28. y 1 3 5 2 3 } 4 (x 1 3) 29. y 5 21(x 2 1)

30. a. 2x 1 y 5 4 b. 2x 1 2y 5 27

31. x < 8; 8765432 9 10

32. y ≤ 12; 111098765 12 13

33. x ≤ 6; 8765432 9 10

34. 25 ≤ x < 23;

28 27 26 25 24 23 22 21 0

35. x > 22 or x ≤ 23;

25 24 23 22 21 0 1 2 3

36. 2 2 } 3 < x < 2;

23 22 21 0 1 2 3 4 5

232

37. yes 38. no 39. yes 40. (13, 2)

41. (2, 23) 42. (23, 24)

43.

21212323 x

y

21

3

1

44.

11 772121

23

3

1

y

x

45.

22222626 6

26

22

2

y

x

46. 56 47. m14 48. 10x3

} yz

49. 6.72 3 1029

50. 3.5 3 1010

51.

2123 1 3

3

5

7

y

x

domain: all real numbers; range: all positive real numbers


1. 8x 5 32 2. 3( y 1 8) 3. 10 2 z2 < 18

4.

2226 2 6

6

2

10

14

18

22

x

y

5. 7 6. 211 7. 210 8. 21 9. 3 10. 22

11. 6 12. 4 13. 9 14. 56%

15. m 5 4 } 3 , b 5 21 16. m 5

5 } 3 , b 5 210

17. m 5 6, b 5 23 18. yes; 2 19. no

20. yes; 1 }

4 21. y 5 5x 2 2 22. y 5 x 1 5

23. y 5 1 } 2 x 1 2 24. y 5 7x 2 3

25. x < 27 26. x < 2 27. x > 7.5 or x < 26.5

28.

22222626 62

26

22

2

6

y

x

29.

21212323 31

1

3

5

y

x

Answers continued

An

sw

er

Key



30. (2, 8) 31. (21, 7) 32. (22, 7) 33. no solution 34. one solution 35. infi nitely many solutions 36. (23)11 37. 2121x6

38. 1 }

4m6n4 39. 128a3b3 40. 1.9321 3 107

41. 0.000321

42.

123 21 3

3

5

7

x

y

domain: all real numbers; range: all positive real numbers

43. 16h3 1 2h2 2 7h 44. 14k2 2 10k 1 3

45. w3 1 7w2 1 4w 2 12 46. 6y2 2 5y 2 21

47. 25x2 2 90x 1 81 48. (k 2 12)(k 2 3)

49. 3m(m 1 6)(m 2 2) 50. 22(a 1 6)(a 2 6)


1. 47 2. 12 3. 12 4. 1 }

16 5. 45 6. 16

7. 2 1 } 2 8. 4.3 9. 23.3 10.

1 }

9 11. 26

12. 1 }

121 13. 23 14. 214 15. 3 16. 4.5 pints

17. 60% 18. m 5 7; b 5 2 19. m 5 2 } 3 ; b 5 3

20. m 5 9 } 5 ; b 5 29

21.

x

y

3 51

3

5

1

22. x

y

3121

23

23. y 5 2 1 } 2 x 1 6 24. y 5 23x 1 7

25. y 5 2x 26. y 5 2 1 } 3 x 2 3

27. x < 213;

022 2242628210212

213

214216

28. x < 16;

16 1812 148 104 620

29. x ≥ 211;

022 2242628210212

211

214216

30. 1 < x ≤ 4;

021 54321

31. 21.5 < x < 4;

02122 4321

21.5

32. no solution 33. (4, 6) 34. (3, 24)

35. (2, 6)

36.

x

y

3123 2121

37.

x

y

3123 21

3

1

38. 63 39. 236 40. p24 41. 16y8

} 9x6

42. 1.521 3 109 43. 4.75 3 105

44. 1.656 3 1022 45. 3x2 1 31x 2 22

46. 49y2 2 42y 1 9 47. 5z3 1 39z2 2 2z 2 24

48. 0, 4 } 7 49. 22, 0, 2 50. 28, 7

51. x 5 12 in.; length 5 56 in.; height 5 24 in. 52. two solutions 53. one solution 54. no solution


1. 36 2. 5 }

3 3. 23 4. 65 mi/h 5. 12 6. 27

7. 2108 8. 40 9. 29 10. 4 }

15 11. 0.75

12. 42 13. 23 14. 5 h 52 min 15–18.

x

y

1

1

2123 3

3

5

23

A (3, 5)

D (4, 23)B (22, 21)

C (0, 3)

15. Quadrant I 16. Quadrant III 17. y-axis

18. Quadrant IV 19. 2 20. 0 21. 2 4 }

3

22. 24x 1 5y 5 18 23. 8x 1 y 5 219

Answers continued

An

sw

er K

ey



24. f (m) 5 55 1 0.25m 25. x < 22

26. 22 ≤ x < 4 27. not possible 28. (11.5, 6.5)

29. (22, 22) 30. (21, 23) 31. 1224

32. r14 33. 3k8

34.

x

3

5

12123

y ; domain: all real numbers; range: all positive real numbers

35.

x

3

1

121 3

y ; domain: all real numbers; range: all positive real numbers

36. 35x2 2 x 2 12 37. 16x2 2 104xy 1 169y2

38. 8x3 2 14x2 1 17x 2 21 39. (x 2 9)(x 1 9)

40. 2(3y 2 7)(2y 1 3) 41. (7z2 2 3)(2z 1 5)

42. 26, 6 43. 213.14, 1.14 44. 20.79, 1.08

45. quadratic function: y 5 x2 2 x 2 12

46.

x

3

5

12123 3

y

y 5 1 2 x

y 5 x

; domain: all real numbers ≥ 22; range: all positive real numbers

The graph of y 5 Ï}

x 1 2 is a horizontal translation (of 2 units to the left) of the graph of y 5 Ï

}

x .

47. 6x3 Ï}

2 48. Ï

}

30 }

2y 49. 24 2 6 Ï

}

6

50. x 5 4 51. x 5 31 52. x 5 70

53. d 5 4 Ï}

13 , (5, 1) 54. d 5 Ï}

37 , (25, 22.5)


1. 8(13 1 y) 2. 11 2 m2 ≤ 2m 3. 44 4. 18

5. 23 6. 23

} 18

7. 90 8. 1 9. 25 10. 21

11. 23 12. 360% 13. 360 14. 2 3 } 2 15. 2

16.

2226210 2

2

10

22x

y 17. 22 2 6 10

22

26

210

x

y

18.

21 1 3

1

21

x

y

19. y 5 22x 1 8 20. y 5 2x 2 3

21. 0.5 ≥ x; 02122 72 3 4 5 61

0.5

22. x > 26; 23242526272829 22 21 0

23. 20.8 ≤ x ≤ 2.8;

21 0 1 2 3 6 7 8

2.820.8

2223

24. 10 miles or less 25. (10, 9) 26. (5, 6)

27. (23, 4) 28. 212 29. 3t4

} s2

30.

q3

} 125s3

31. 2.68 × 1023 32. 71,900 33. 196 2 121b2

34. 4x2 2 25x 2 56 35. 7x3 1 42x2 1 63x

36. (x 1 7)(2x 2 1) 37. 3y( y 2 5)( y 1 5)

38. (5z2 2 2)(z 1 2) 39. 23, 3 40. 4, 7

41. 0.27, 3.73 42. 25.46, 1.46 43. 20.48, 3.48

44. 21.17, 0.62 45. 25 cm 46. a 5 15

47. c 5 30.5 48. b 5 56 49. y 5 40

} x ; 20

50. y 5 9 } x ; 4.5 51. 2r 1 3 1

5 }

r 1 8

52. 3t 2 2 1 3 }

2t 2 5

Answers continued

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1

Number PuzzlesGiven the following scenarios, fill in the blank with one of these numbers.

6.5 500 �4 1 �75 ��

17 �5 18 11 __ 9

5 __ 6

��

15 96 ___ 4

4.5 � ��

11 0 2 __ 3

300 �12

1. When you square my value and subtract it from 23, the result is the square root of 36. I am an irrational number. What is my number?

2. If you cube my value and divide it by the quotient of 150 and 6, the result is my value. I am not a natural number, but I am a rational number. What is my number?

3. When you take my value and multiply it by �8, the result is an integer greater than �220. If you take the result and divide it by the sum of �10 and 2, the result is my value. I am a rational number. What is my number?

4. If you take my absolute value and subtract the square of 16, the result is a double-digit number whose prime factorization is 11 � 2 � 2. I am a natural number. What is my number?

5. When you add 5 to my value and subtract 1 1 __ 2

, the result is twice

the square root of 25. I am a terminating decimal. What is my number?

6. If you take 3 to the value of my power, the result is a non-terminating, non-repeating decimal. If you take 3 and raise it to the power of the absolute value of my value, the result is 81. I am an integer. What is my number?

7. When you divide my value by the absolute value of �6, the result is a fraction that when simplified has a numerator that is the first natural number and a denominator that is a perfect square. If you take the square root of the reciprocal of this fraction, the result is 3. I am a real number. What is my number?

��

17

�5

96 ___ 4

300

6.5

�4

2 __ 3

LAH_A1_11_FL_RTGO_Ch 01_178.indd 178 2/20/09 11:35:15 AM

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2

Profit-Loss-Revenue

An important concept in business is the ability to make a profit. Profit is equal to the amount of sales minus the cost of production. If the sales are greater than the cost, the business makes a profit. If the sales are less than the cost, the business is losing money.

Use the information below to answer each question.

A manufacturer of compact-disc players sells them to a retailer for $45 each. It costs the manufacturer $200 plus $25 each to produce the compact-disc player.

1. Write a function, s, to represent the total amount of sales of


2. Write a function, c, to represent the total cost of producing the


3. Graph the functions s and c on the same coordinate grid.

x

y

2 6 10 14 18

500450400350300250200150100

50

4 86 12 16 20

Am

ou

nt,

$

Number of Compact-Disc Players

4. For what dollar amount is the sales and the cost equal?

5. For what value of n is the sales and the cost equal?

6. Write an inequality that represents the value(s) of n for which the cost is more than the sales.

7. Write an inequality that represents the value(s) of n for which the manufacturer makes a profit.

s � 45n

c � 200 � 25n

s � c � 450n � 10

0 � n � 10 or 0 � n � 9

n � 11 or n � 10


Diophantine Equations

Diophantus (about 200-284) is known to some as the ‘father of algebra’.

He studied primarily the solutions of algebraic equations and the theory of

numbers.

One type of equation he studied has the form ax � by � c where a, b, and

c are all integers and the solutions to the equation (x, y) are also integers.

These types of equations are now known as Diophantine Equations. They

can be quite difficult to solve and many times the only way to solve them

is by guessing and checking.

Solve each Diophantine Equation. Find at least one pair of positive integers for x and y that make the equation true.

1. 3x � 4y � 12 2. �2x � 3y � �9

a. Solve the equation for y. a. Solve the equation for y.

b. What number must x be divisible by? b. What number must x be divisible by?

Why? Why?

c. Find at least one solution. c. Find at least one solution.

3. x � 2y � 10 4. �4x � y � 15

5. 8x � 19y � 100 6. 3x � 7y � 35

7. �5x � 11y � 30 8. 3x � 4y � 32

9. 7x � y � 14 10. �3x � 5y � 9

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3

y � 3 � 3 __ 4 x y � �3 � 2 __

3 x

4; So the result is an integer. 3; So the result is an integer.

(4, 0)

(30, 10)

(6, 1)

(15, 75)

(60, 20) (7, 2)

(5, 5) (4, 5)

(3, 7) (2, 3)

LAH_A1_11_FL_RTGO_180.indd 180 8/22/09 10:12:09 AM

Triangle Inequalities

The Triangle Inequality Theorem states that “For any triangle, the sum of

the lengths of any two sides is greater than the length of the third side”.

This inequality defines the existence of a triangle. There is a theorem

in geometry that determines whether a given triangle is a right triangle,

obtuse triangle, or acute triangle.

A right triangle has exactly one 90 degree angle.

An obtuse triangle has exactly one angle greater than 90 degrees.

An acute triangle has no angle with a measure greater than or equal to 90 degrees.

In a triangle with sides a, b, and c with c being the longest side:

If c 2 � a 2 � b 2 , the triangle is obtuse.

If c 2 � a 2 � b 2 , the triangle is a right triangle.

If c 2 � a 2 � b 2 , the triangle is acute.

Determine whether the triangles, with these given side lengths, are acute, right, or obtuse.

1. 2, 3, 4 2. 3, 4, 5

3. 6, 6, 7 4. 7, 20, 24

5.

12

13 5

6.

7

7 7

7.

8

11 6

8.

5

6 4

9. The longest side of an acute triangle measures 12 inches. One of the shorter

sides is 7 inches. Express the length of the third side as an inequality.

10. The two shorter sides of an obtuse triangle measure 5 cm and 10 cm. Express

the length of the third side as an inequality.

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4

Obtuse Right

Acute Obtuse

Right

Obtuse

Acute

Acute

��

95 � x � 12

x � 11 or x � 15

LAH_A1_11_FL_RTGO_181.indd 181 8/22/09 10:22:01 AM

Copyright © by Holt McDougal. 227 Holt McDougal Algebra 1All rights reserved.


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5

Onto FunctionsA relation is a function if for every number in the domain, there is one and only one number in the range. Another way to describe a function is to determine if it is onto. A function is onto if and only if all the numbers of the range are paired with all of the numbers in the domain.

For example, the function in Table 1 is described as onto because each number in the range has a number from the domain assigned to it.

The function in the Table 2 is not onto because the number 7 in the range does not have a number in the domain assigned to it.

Determine if each function can be described as onto.

1. 2.

3. 4.

Domain

Range 6 7 8 9 10

1 2 3 4 5

Domain

Range 6 7 8 9 11

1 2 3 4 5

Domain

Range 0 2 4 6 8

2 4 6 8 10 Domain

Range 1 2

1 2 4 8

0 3 6 9 12

Domain

Range 6 7 8 9 10

5 10 15 20 25 Domain

Range 3 4 5 6 7

1 3 5 7 9

Table 1

Table 2

Not onto Onto

Onto Not onto

LAH_A1_11_FL_RTGO_Ch 05_182.indd 182 2/20/09 2:34:44 PM

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6

Intercept–Intercept FormLook at the graph of the equation 1__

2x �1__3y � 1 and

1___�4x �

1__5y � 1 shown on the right. The y-intercept for the

first equation is 3, so you know that the point (0, 3) lies on the line. Another point containing 0 also lies on the line; it is the point (2, 0). The x-coordinate of the point at which the line crosses the x-axis is called the x-intercept.

You can see that for the second equation, the y-intercept is 5.

What is the x-intercept of the second equation?

An equation in the form 1__ax �

1__by � 1 is in intercept-intercept form. In this

form, a is the x-intercept and b is the y-intercept.

Look back at the equation 1___�4x �

1__5y � 1. Since it is in

intercept-intercept form, you know the following relationships.

1__a �

1___�4

1__b �

1__5 How would you determine a and b?

a � �4 b � 5

You can use the intercept-intercept for to determine the slope of the line. Recall that slope equals rise over run. You can use the two intercepts to count the rise and run on the graph. Applying the definition of slope, you can calculate the slope using the intercepts a and b.

m �y2 � y1_______x2 � x1

�0�b_____

a � 0 � �b__a

What is the slope of the line described by the equation 1__2x �

1__3y � 1?

m �

For each equation, determine the x-intercept, the y-intercept and the slope of the line.

1. x �1__6y � 1 2. x � y � 1 3. x � y � 1

4. 1__2x �

1__6y � 1 5. x � y � 2 6. x � 2y � 2

3

5

1

–3

–5

x

y

1 3 5–1–3–5

x + y = 112

13x + y = 11

–415

�4

Sample answer: Use cross products.

�3___2

a � 1, b � 6, m � �6 a � 1, b � �1, m � 1 a � 1, b � 1, m � �1

a � 2, b � 6, m � �3 a � 2, b � 2, m � �1 a � 2, b � �1, m � 1__2

LAH_A1_11_FL_RTGO_183.indd 183 2/24/09 7:26:46 PM

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7

Using a Matrix to Represent a System of EquationsA matrix is a rectangular array of numbers enclosed in a single set of brackets. If each equation in a system of equations is written in standard form, you can represent the system with a matrix equation. The matrix equation is made up of three matrices; one for the coefficients on the variables x and y, one for the variables x and y, and one for the constants.

For example, the system of equations {3x � y � 6x � y � �2

is represented by the matrix

equation [3 �11 1 ][ x

y ] � [ 6�2] .

Determine which system of equations represents the correct matrix equation.

1. {x � y � 8x � y � 2

a. [3 �16 2 ][ x

y ] � [ 4�8]

2. { 3x � y � 46x � 2y � �8

b. [1 �12 3 ][ x

y ] � [29] 3. { x � y � 2

2x � 3y � 9c. [�5 8

10 3][ xy ] � [21

15] 4. { �5x � 8y � 21

10x � 3y � 15d. [1 1

1 �1][ xy ] � [82]

Create a matrix equation for each system of equations.

5. { x � 5y � 02x � 3y � 7

6. {4x � 3y � 193x � 4y � 8

7. {5x � 3y � 124x � 5y � 17

8. {x � y � 7x � y � 9

9. {12x � 9y � 11412x � 7y � 82

10. {2x � 3y � �4x � 3y � 7

11. {1__2x � y � 12

x �1__5y � 10

12. {2__3x �

1__3y � �9

1__4x �

3__4y � 16

13. {1.2x � 1.6y � 2.4�0.8x � 0.2y � �1.2

d

a

b

c

[1 �52 �3][ x

y ] � [07] [4 3

3 �4] [ xy ] � [19

8 ] [5 3

4 �5] [ xy ] � [12

17]

[1 11 �1][ x

y ] � [79] [12 �9

12 7 ] [ x

y ] � [114 82

] [2 �31 3

] [ xy ] � [�4

7 ]

� 1__2

1

1 1__5

� [ xy ] � � 12

10� � 2__3

1__3

1__4

3__4

� [ xy ] � � �9

16 � � 1.2 �1.6

�0.8 0.2� [ xy ] �

� 2.4�1.2�


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8

Describing Geometric Regions with a System of Inequalities

Describe the shaded region of the graph by writing a system of inequalities consisting of three different linear inequalities. To write this system, follow these steps:

1. Determine the slope and y-intercept of each line.

2. Write inequalities in slope-intercept form.

3. If the line is solid use either � or �. If the line is dashed, use either � or �.

4. If the shaded region is “above” the line use the symbol �, and if the shaded region is “below” the line, use the symbol �.

For example, the system of inequalities that describes the region below is {x � �2y � �4

y � �3___2

x � 1

.

x

y5

–5

–5 5

Write a system of inequalities that describes each region.

1.

x

y5

–5

–5 5

2.

x

y5

–5

–5 5

3.

x

y5

–5

–5 5

{y � 3__2

x � 3

y � �7__4x � 3

y � �4

{y � 3 y � �3y � x � 4 y � �x � 4y � �x � 4 y � x � 4

{x � �4 x � 3 y � �x � 5y � 4 y � �3 y � x � 4y � �x � 5 y � x � 6




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9

Fourth Degree TrinomialsSometimes it is possible to write a trinomial of the fourth degree, a4

� a2b2� b4,

as a difference of two squares and then factor.

Example: Factor 4 a4� 21 a2b2

� 9 b4.

Step I Find the square roots of the first and last terms.

��

4a4� 2 a2 �

� 9b4� 3 b2

Step II Find twice the product of the square roots from the terms in Step 1.

2(2 a2)(3b2) � 12 a2b2

Step III Split the middle term of the trinomial into two parts. One part is either the answer from the Step II or its opposite. The other part should be the opposite of a perfect square.

�21a2b2� �12a2b2

� 9 a2b2

Step IV Rewrite the trinomial as the difference of two squares and then factor.

4 a4� 21 a2b2

� 9 b4� (4 a4

� 12 a2b2� 9 b4) � 9 a2b2

� � 2a2� 3 b2 �2

� 9 a2b2

� [(2 a2� 3 b2) � 3ab] [(2 a2

� 3 b2) � 3ab]

� (2 a2� 3ab � 3 b2)(2a2

� 3ab � 3 b2)

Factor each trinomial.

1. 16 d4� 7 d2

� 1

2. p4� p2

� 1

3. 4 x4� 13 x2

� 1

4. 4 x4� 9 x2y2

� 16 y4

5. 9r 4� 26r 2s2

� 25 s4

6. 4 a4� 5 a2c2

� 25 c4

(4d2 � d � 1)(4 d2 � d � 1)

(p2 � p � 1)( p2 � p � 1)

(2x2� 3x � 1)(2 x2

� 3x � 1)

(2x2� 5xy � 4 y2 )(2 x2

� 5xy � 4 y2 )

(3r2� 2rs � 5 s2 )(3 r2

� 2rs � 5 s2 )

(2a2� 5ac � 5 c2 )(2 a2

� 5ac � 5 c2 )


Graphing Circles by Completing the SquareCompleting the square can be used to graph circles. The general equation for a circle with its center at the origin is x2

� y2� r 2, where r is the

radius of the circle. The general equation of a circle with its center translated from the origin is (x � h)2

� (y � k)2� r 2. An equation

representing a circle can be transformed into the sum of two squares.

Example: x2� 14x � y2

� 6y � 49 � 0(x2

� 14x � ) � ( y2� 6y � ) � �49

(x2� 14x � 49) � ( y2

� 6y � 9) � �49 � 49 � 9(x � 7 )2

� (y � 3 )2� 9

(x � 7 )2� (y � 3 )2

� 32

The center of the circle is (7, �3) and the radius is 3.

The circle is shown at the right.

Complete the square on the following equations. Identify the center and radius of the circle and then graph.

1. x2� 8x � y2

� 2y � 13 � 0 2. x2� 6x � y2

� 4y � 12 � 0

y

x

–4

–2

–6

2

2 4 6 8

–2–4

y

x2

–2

–4

4

2 4

Center: Center:

Radius: Radius:

3. x2� y2

� 10y � 75 � 0 4. x2� 8x � y2

� 84 � 0

–8–16

y

x8

–8

–16

16

8 16 –8–16

y

x8

–8

–16

16

8 16

Center: Center:

Radius: Radius:

y

x

–4

–2

–6

2

2 4 6 8 10

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10

(4, �1)

2

(�3, �2)

1

(0, �5)10

(4, 0)

10

LAH_A1_11_FL_RTGO_Ch 10_187.indd 187 2/20/09 2:34:50 PM

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11

Geometric Series

A series is the sum of a sequence of terms. In other words, a series is a list of numbers with addition operations between them. Series can be finite, meaning that there is an end, or infinite. A geometric series is a sequence of numbers such that the quotient of any two successive numbers form a common ratio.

A geometric series can be written as Sn � � k�0

n

ark where r � 0 is the common ratio

and a is a scale factor. The Greek letter sigma, �, represents the sum of each term in the sequence. For example:

S4 � � k�0

4

2(�2)k

� 2(�2)0� 2(�2)1

� 2(�2)2� 2(�2)3

� 2(�2)4

� 2 � (�4) � 8 � (�16) � 32 � 22

Find the value of each geometric series.

1. S3 � � k�0

3

3(�3)k 2. S4 � � k�0

4

2 � 1__2 �k

3. S4 � � k�0

4

4 � �1__2 �k

4. S3 � � k�0

31__2

(4)k

5. S3 � � k�0

3

�1� �1__3 �k

6. S4 � � k�0

4

9 � 2__3 �k

7. S5 � � k�0

5

4(�1)k 8. S4 � � k�0

41__3

(�3)k

�60 37__8

421__2

23__4

�20___27 234__

9

0 201__3


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12

Pascal’s TrianglePascal’s Triangle is a geometric arrangement of numbers. These numbers represent the binomial coefficients. That is, they represent the coefficients of the terms of the expansion of (x � y)n. The first seven rows of Pascal’s Triangle look like this.

Row 0 1

Row 1 1 1

Row 2 1 2 1

Row 3 1 3 3 1

Row 4 1 4 6 4 1

Row 5 1 5 10 10 5 1

Row 6 1 6 15 20 15 6 1

Notice that each number is the sum of the two numbers above it.

For example what two numbers were added to get 10 in the 5th row?

What are the numbers for the 7th row?

As an example, find (x � y)3.

Look at row 3, what are the coefficients of the expansion?

The first term of the expansion starts with the highest power of x, namelyx3, and the lowest power of y, namely y0

� 1. The power of x increases by 1 for each successive term and the power of y increases by 1 for each successive term.

(x � y)3� 1 � x3

� y0� 3 � x2

� y1� 3 � x1

� y2� 1 � x0

� y3\ \ \ \

� x3� 3 x2y � 3xy2

� y3

Expand each of the following polynomials.

1. (x � y)4 2. (x � 1 )5

3. (x � 3 )4 4. (x � 2 )3

5. (x � 1 )9 6. (x � 2y)5

4 and 6

1, 7, 21, 35, 35, 21, 7, 1

1, 3, 3, 1

x 4 � 4 x3y � 6 x2y2 � 4x y 3 � y4 x 5 � 5 x4� 10 x3

� 10 x2� 5x � 1

x 4 � 12 x3� 54 x2

� 108x � 81 x 3 � 6 x2 � 12x � 8

x 9 � 9 x8 � 36 x7 � 84 x6 � 126 x5 �

126x4 � 84 x3 � 36 x2 � 9x � 1 x 5 � 10 x4y � 40 x3y3

� 80x2y3 �

80x y 4 � 32 y5




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Holt McDougal Florida Larson Algebra 1 - rjssolutions.comrjssolutions.com/echs/Algebra1Files/resource_index/ready_to_go.pdf · Holt McDougal Florida Larson Algebra 1 ... A. Solving