CHAPTER 1 Linear Equations and Inequalities · 48 CHAPTER 1 Linear Equations and Inequalities In Words In the equation the expression ... x - 2b += 4 1 3 x - 2 = 4 1 3 x - 2 = 4

Linear Equations and Inequalities

Did you know that the amount of income tax that we pay to the federalgovernment can be found by solving a linear equation? See Problems103 and 104 in Section 1.1.

OUTLINEPart I: Linear Equations and Inequalities in One Variable1.1 Linear Equations in One Variable1.2 An Introduction to Problem Solving1.3 Using Formulas to Solve Problems1.4 Linear Inequalities in One Variable

Putting the Concepts Together (Sections 1.1–1.4)

Part II: Linear Equations and Inequalities in Two Variables1.5 Rectangular Coordinates and Graphs of

Equations1.6 Linear Equations in Two Variables1.7 Parallel and Perpendicular Lines1.8 Linear Inequalities in Two Variables

Chapter 1 Activity: Pass the PaperChapter 1 ReviewChapter 1 TestCumulative Review Chapters R–1

The Big Picture: Putting It TogetherIn Chapter R, we reviewed skills learned in earlier courses.These skills will be used throughout the text and shouldalways be kept fresh in your mind.

We now begin our discussion of algebra in earnest. Theword “algebra” is derived from the Arabic word al-jabr.The word is part of the title of a ninth-century work,“Hisâb al-jabr w’al-muqâbalah,” written during the goldenage of Islamic science and mathematics by Muhammad ibnMûqâ al-Khowârizmî.

The word al-jabr means “restoration,” a reference to thefact that, if a number is added to one side of an equation,then it must also be added to the other side in order to“restore” the equality. The title of the work “Hisâb al-jabrw’al-muqâbalah” means “the science of restoring and can-celing.”Today, algebra has come to mean a great deal more.

The material in Chapter 1 is presented in two parts.Part I reviews linear equations and inequalities in one vari-able, and Part II reviews linear equations and inequalitiesin two variables.

1

47

CH

AP

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48 CHAPTER 1 Linear Equations and Inequalities

In WordsIn the equation the expression

is the left side of the equation and 0is the right side. In this equation only the leftside contains the variable, y. In the equation

the expressionis the left side of the equation and

is the right side. In this equationboth sides have expressions that contain thevariable, x.

-2x + 104x + 54x + 5 = -2x + 10,

2y + 52y + 5 = 0,

PART I: L INEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE

1.1 Linear Equations in One VariablePreparing for Linear EquationsBefore getting started, take this readiness quiz. If you get a problem wrong, go back to the sectioncited and review the material.

P1. Determine the additive inverse of 5. [Section R.3, p. 21]P2. Determine the multiplicative inverse of [Section R.3, pp. 23–24]

P3. Use the Reduction Property to simplify [Section R.3, pp. 25–26]

P4. Find the Least Common Denominator of and [Section R.3, pp. 27–28]

P5. Use the Distributive Property to remove the parentheses: [Section R.3, pp. 29–30]

P6. What is the coefficient of [Section R.5, p. 41]P7. Simplify by combining like terms: [Section R.5, pp. 41–43]P8. Evaluate the expression when [Section R.5, pp. 40–41]

P9. Is in the domain of Is in the domain? [Section R.5, pp. 43–44]

x = -32

x + 3?x = 3

x = -2-51x + 32 - 841y - 22 - y + 5

-4x?61z - 22

512

.38

15

# 5x.-3.

OBJECTIVES1 Determine Whether a Number

Is a Solution to an Equation

2 Solve Linear Equations

3 Determine Whether an EquationIs a Conditional Equation,Identity, or Contradiction

1 Determine Whether a Number Is a Solution to an EquationAn equation in one variable is a statement made up of two expressions that are equal,and in the statement, at least one of the expressions contains the variable. The expres-sions are called the sides of the equation. Examples of equations in one variable are

In this section, we will concentrate on solving linear equations in one variable.

2y + 5 = 0 4x + 5 = -2x + 10 3z + 2

= 9

DEFINITIONA linear equation in one variable is an equation that has one unknown and theunknown is written to the first power. Linear equations in one variable can bewritten in the form

where a and b are real numbers and a Z 0.

ax + b = 0

The following are all examples of linear equations in one variable because they can bewritten in the form with a little algebraic manipulation.

Because an equation is a statement, it can be either true or false, depending uponthe value of the variable. Any value of the variable that results in a true statement iscalled a solution of the equation. When a value of the variable results in a true state-ment, we say that the value satisfies the equation. To determine whether a number sat-isfies an equation, we replace the variable with the number and determine whether theleft side of the equation equals the right side of the equation—if it does, then we havea true statement and the number substituted is a solution.

4x - 3 = 12 23y +

15

=215

-0.73p + 1.23 = 1.34p + 8.05

ax + b = 0

Preparing for...Answers P1.

P2. P3. x P4. 24 P5.

P6. P7. P8.P9. Yes; No

-133y - 3-4

6z - 12-13

-5

Section 1.1 Linear Equations in One Variable 49

EXAMPLE 1 Determining Whether a Number Is a Solution to a Linear Equation

Determine if the following numbers are solutions to the equation

(a) (b)

Solution

(a) Let in the equation and simplify.

Simplify:

Because the left side of the equation does not equal the right side of theequation, we do not have a true statement. Therefore, is not asolution.

(b) Let in the equation and simplify.

Simplify:True

Because the left side of the equation equals the right side of the equation,we have a true statement.Therefore, is a solution to the equation.x = 3

6 = 6

3122 � -6 + 12

313 - 12 � -2132 + 12

31x - 12 = -2x + 12

x = 3

x = 5

12 Z 2

3142 � -10 + 12

315 - 12 � -2152 + 12

31x - 12 = -2x + 12

x = 5

x = 3x = 5

31x - 12 = -2x + 12

Quick 1. The equation is a equation in one variable. The

expressions are called of the equation.

2. The values of the variable that result in a true statement are called .

In Problems 3–5, determine which of the given numbers are solutions to the equation.

3.

4.

5. -31z + 22 = 4z + 1; z = -3, z = -1, z = 2

3x + 2 = 2x - 5; x = 0, x = 6, x = -7

-5x + 3 = -2; x = -2, x = 1, x = 3

3x + 5 and 2x - 33x + 5 = 2x - 3

2 Solve Linear EquationsTo solve an equation means to find ALL the solutions of the equation. The set of allsolutions to the equation is called the solution set of the equation.

One method for solving equations algebraically requires that a series of equivalentequations be developed from the original equation until a solution results.

DEFINITIONTwo or more equations that have precisely the same solutions are called equivalentequations.

But how do we obtain equivalent equations? The first method we introduce for obtain-ing an equivalent equation is called the Addition Property of Equality.

In WordsThe symbol is used to indicate that we areunsure whether the left side of the equationequals the right side of the equation.

�

Work SmartThe directions solve, simplify,and evaluate are different! We solveequations. We simplify algebraicexpressions to form equivalentalgebraic expressions. We evaluatealgebraic expressions to find thevalue of the expression for a specificvalue of the variable.


For example, if then (we added 2 to both sides of the equation).Because is equivalent to the Addition Property can be used to add areal number to each side of an equation or subtract a real number from each side of anequation. You will use this handy property a great deal in algebra.

A second method that results in an equivalent equation is called the MultiplicationProperty of Equality.

a + 1-b2,a - bx + 2 = 3 + 2x = 3,

In WordsThe Addition Property says that whateveryou add to one side of an equation, you mustalso add to the other side.

In WordsThe Multiplication Property says thatwhenever you multiply one side of an equationby a nonzero expression, you must alsomultiply the other side by the same nonzeroexpression.

For example, if then Remember, the quotient is equivalent

to the product so dividing by some number b is really multiplying by the multi-

plicative inverse of b, So, the Multiplication Property can be used to multiply or

divide each side of the equation by some nonzero quantity.

1b

.

a # 1b

,

a

b

15

# 5x =15

# 30.5x = 30,

Work SmartThe number in front of the variableexpression is the coefficient.For example, the coefficient inthe expression 2x is 2.

ADDITION PROPERTY OF EQUALITY

The Addition Property of Equality states that for real numbers a, b, and c,

if a = b, then a + c = b + c

MULTIPLICATION PROPERTY OF EQUALITY

The Multiplication Property of Equality states that for real numbers a, b, and cwhere

if a = b, then ac = bc

c Z 0,

EXAMPLE 2 Using the Addition and Multiplication Properties to Solve a Linear Equation

Solve the linear equation:

Solution

The goal in solving any linear equation is to get the variable by itself with a coefficientof 1, that is, to isolate the variable.

Addition Property of Equality; add 2 to both sides:

Simplify:

Multiplication Property of Equality; multiply both sides by 3:

Simplify:

Check

Let x � 18 in the original equation:

True

Because satisfies the equation, the solution of the equation is 18, or the solutionset is 5186.x = 18

4 = 4

6 - 2 � 4

13

(18) - 2 � 4

13x - 2 = 4

x = 18

3a13xb = 3 # 6

13x = 6

a13x - 2b + 2 = 4 + 2

13x - 2 = 4

13x - 2 = 4


Quick In Problems 12–14, solve each linear equation.Be sure to verify your solution.

12.

13.

14. 2w + 8 - 7w + 1 = 3w - 1 + 2w - 5

4b + 3 - b - 8 - 5b = 2b - 1 - b - 1

2x + 3 + 5x + 1 = 4x + 10

Often, we must combine like terms or use the Distributive Property to eliminateparentheses before we can use the Addition or Multiplication Properties. Remember,when solving linear equations, our goal is to get all terms involving the variable on oneside of the equation and all constants on the other side.

EXAMPLE 3 Solving a Linear Equation by Combining Like TermsSolve the linear equation:

Solution

Combine like terms:Subtract 6y from both sides:

Add 2 to both sides:

Divide both sides by 2:

Check

Let y 5 in the original equation:

True

Because satisfies the equation, the solution of the equation is 5, or the solutionset is 556.y = 5

38 = 38

15 - 2 + 25 � 10 + 5 + 20 + 3

3152 - 2 + 5152 � 2152 + 5 + 4152 + 3= 3y - 2 + 5y = 2y + 5 + 4y + 3

y = 5

2y2

=102

2y = 10

2y - 2 + 2 = 8 + 2

2y - 2 = 8

8y - 2 - 6y = 6y + 8 - 6y

8y - 2 = 6y + 8

3y - 2 + 5y = 2y + 5 + 4y + 3

3y - 2 + 5y = 2y + 5 + 4y + 3

Quick 6. What does it mean to solve an equation?

7. State the Addition Property of Equality.

8. State the Multiplication Property of Equality.

In Problems 9–11, solve each equation and verify your solution.

9. 10. 11. 5y + 1 = 2-4a - 7 = 13x + 8 = 17

EXAMPLE 4 Solving a Linear Equation Using the Distributive PropertySolve the linear equation:

Solution

Use the Distributive Propertyto remove parentheses:

Combine like terms:Add 2x to both sides:

Subtract 12 from both sides: 6x = -6

6x + 12 - 12 = 6 - 12

6x + 12 = 6

4x + 12 + 2x = -2x + 6 + 2x

4x + 12 = -2x + 6

4x + 12 = x - 3x + 6

41x + 32 = x - 31x - 22

41x + 32 = x - 31x - 22



Check

Let in the original equation:

True

Because satisfies the equation, the solution of the equation is or thesolution set is 5-16. -1,x = -1

8 = 8

8 � -1 + 9

4122 � -1 - 31-32 41-1 + 32 � -1 - 31-1 - 22x = -1

41x + 32 = x - 31x - 22 x = -1

6x6

=-66

We now summarize the steps for solving a linear equation. Bear in mind that it ispossible that one or more of these steps may not be necessary when solving a linearequation.

Quick In Problems 15–18, solve each linear equation. Be sure to verify yoursolution.

15.

16.

17. 41x + 32 - 8x = 31x + 22 + x-21x - 42 - 6 = 31x + 62 + 4

41x - 12 = 12

18. 51x - 32 + 31x + 32 = 2x - 3

SUMMARY STEPS FOR SOLVING A LINEAR EQUATIONStep 1: Remove any parentheses using the Distributive Property.

Step 2: Combine like terms on each side of the equation.

Step 3: Use the Addition Property of Equality to get all variables on one side ofthe equation and all constants on the other side.

Step 4: Use the Multiplication Property of Equality to get the coefficient of thevariable to equal 1.

Step 5: Check your answer to be sure that it satisfies the original equation.

Linear Equations with Fractions or DecimalsA linear equation that contains fractions can be rewritten (transformed) into an equiv-alent equation without fractions by multiplying both sides of the equation by the LeastCommon Denominator (LCD) of all the fractions in the equation.

EXAMPLE 5 How to Solve a Linear Equation That Contains Fractions

Solve the linear equation:

Step-by-Step Solution

Before we follow the summary steps, we rewrite the equation without fractions by multiplying both sides of the equationby the Least Common Denominator (LCD). The LCD is 20, so we multiply both sides of the equation by 20 and obtain

20 # ay + 14

+y - 2

10 b = 20 # ay + 7

20 b

y + 14

+y - 2

10 =y + 7

20


Step 1: Remove all parentheses usingthe Distributive Property.

Use the Distributive Property:

Divide out common factors:Use the Distributive Property: 5y + 5 + 2y - 4 = y + 7

51y + 12 + 21y - 22 = y + 7

20 # y + 14

+ 20 # y - 210

= 20 # y + 720

20 # ay + 14

+y - 2

10 b = 20 # ay + 7

20 b

Step 2: Combine like terms on eachside of the equation.

7y + 1 = y + 7

Step 3: Use the Addition Property ofEquality to get all variables on oneside of the equation and allconstants on the other side.

Subtract y from both sides:

Subtract 1 from both sides: 6y = 6

6y + 1 - 1 = 7 - 1

6y + 1 = 7

7y + 1 - y = y + 7 - y

Step 4: Use the Multiplication Propertyof Equality to get the coefficient onthe variable to equal 1.


y = 1

6y6

=66

Step 5: Check: Verify the solution.


Rewrite each rational number with

True 8

20 =

820

1020

+-220

� 820

24# 55

+-110

# 22

� 820

LCD = 20:

24

+-110

� 820

1 + 1

4 +

1 - 210

� 1 + 720

y = 1

y + 1

4 +y - 2

10=y + 7

20

Now we can follow Steps 1–5 for solving a linear equation.

Because satisfies the equation, the solution of the equation is 1, or the solutionset is 516.y = 1

Quick In Problems 19–22,solve each linear equation.Be sure to verify your solution.

19. 20.

21. 22.4x + 3

9 -

2x + 12

=16

x + 2

6 + 2 =

53

3x4

-512

=5x6

3y2

+y

6 =

103

When decimals occur in a linear equation, we can rewrite (transform) the equationinto an equivalent equation that does not have a decimal. We use the same techniquethat we used in equations with fractions. The idea behind the procedure is to multiplyboth sides of the equation by a power of 10 so that the decimals are removed. For


example, because multiplying 0.7 by 10 “eliminates” the decimal since

Because multiplying 0.03 by 100 “eliminates” the

decimal.

0.03 =3

100,1010.72 = 10 # 7

10 = 7.

0.7 =710

,

EXAMPLE 6 Solving a Linear Equation That Contains DecimalsSolve the linear equation:

Solution

We want to rewrite the equation so that the equivalent equation does not contain adecimal. This is done by multiplying both sides of the equation by 10. Do you seewhy? Each of the decimals is written to the tenths position, so multiplying by 10 will“eliminate” the decimal.

Use the Distributive Property:

Subtract 3x from both sides:



Check


True

Because satisfies the equation, the solution of the equation is 3, or the solutionset is 536.x = 3

1.1 = 1.1

1.5 - 0.4 � 0.9 + 0.2

0.5132 - 0.4 � 0.3132 + 0.2x = 3 0.5x - 0.4 = 0.3x + 0.2

x = 3

2x2

=62

2x = 6

2x - 4 + 4 = 2 + 4

2x - 4 = 2

5x - 4 - 3x = 3x + 2 - 3x

5x - 4 = 3x + 2

1010.5x2 - 1010.42 = 1010.3x2 + 1010.22 1010.5x - 0.42 = 1010.3x + 0.22

0.5x - 0.4 = 0.3x + 0.2

3 Determine Whether an Equation Is a Conditional Equation,Identity, or Contradiction

All of the linear equations that we have studied thus far have had one solution. Whileit is tempting to say that all linear equations must have one solution, this statement isnot true in general. In fact, linear equations may have either one solution, no solution,or infinitely many solutions. We give names to the type of equation depending uponthe number of solutions that the linear equation has.

The equations that we have solved thus far are called conditional equations.

Quick In Problems 23–25, solve each linear equation. Be sure to verify yoursolution.

23. 24.

25. 0.41y + 32 = 0.51y - 420.07x - 1.3 = 0.05x - 1.10.2t + 1.4 = 0.8

DEFINITIONA conditional equation is an equation that is true for some values of the variableand false for other values of the variable.


For example, the equation

is a conditional equation because it is true when and false for every other realnumber x.

x = 3

x + 7 = 10

DEFINITIONAn equation that is false for every value of the variable is called a contradiction.

For example, the equation

is a contradiction because it is false for any value of x. Contradictions are identifiedthrough the process of creating equivalent equations. For example, if we subtract 3xfrom both sides of we obtain which is clearly false. Contra-dictions have no solution and therefore the solution set is empty. We express thesolution set of contradictions as either � or 5 6.

8 = 6,3x + 8 = 3x + 6,

3x + 8 = 3x + 6

DEFINITIONAn equation that is satisfied for every choice of the variable for which both sidesof the equation are defined is called an identity.

In WordsConditional equations are true for somevalues of the variable and false for others.Contradictions are false for all values of thevariable. Identities are true for all allowedvalues of the variable.

For example,

is an identity because any real number x satisfies the equation. Just as with contradic-tions, identities are recognized through the process of creating equivalent equations.For example, if we combine like terms in the equationwe obtain which is true no matter what value of x we choose.Therefore, the solution set of linear identities is the set of all real numbers. We expressthe solution of linear identities as either 5x | x is any real number6 or �.

3x + 11 = 3x + 11,2x + 3 + x + 8 = 3x + 11

2x + 3 + x + 8 = 3x + 11

EXAMPLE 7 Classifying a Linear EquationSolve the linear equation State whether the equa-tion is an identity, contradiction, or conditional equation.

Solution

As with any linear equation, our goal is to get the variable by itself with a coefficient of 1.

Use the Distributive Property:Combine like terms:

Add 3x to both sides:

The last statement states that This is a false statement, so the equation is a con-tradiction. The original equation is a contradiction and has no solution. The solutionset is or 5 6.�

9 = 5.

9 = 5

-3x + 9 + 3x = -3x + 5 + 3x

-3x + 9 = -3x + 5

3x + 9 - 6x = 5x + 5 - 8x

31x + 32 - 6x = 51x + 12 - 8x

31x + 32 - 6x = 51x + 12 - 8x.

Work SmartIn the solution to Example 7 weobtained the equation

You may recognize at this point thatthe equation is a contradiction andstate the solution set as or { }.�

-3x + 9 = -3x + 5


Solution

Use the Distributive Property:Combine like terms:

At this point it should be clear that the equation is true for all realnumbers x. So, the original equation is an identity and its solution set is all real numbersor

Had we continued to solve the equation in Example 8 by subtracting 8x from bothsides of the equation, we would obtain

The statement is true for all real numbers x, so the solution set of the originalequation is all real numbers.

14 = 14

14 = 14 8x + 14 - 8x = 8x + 14 - 8x

5x | x is any real number} or �.

8x + 14 = 8x + 14

8x + 14 = 8x + 14

-4x + 8 + 12x + 6 = 8x + 14

-41x - 22 + 314x + 22 = 214x + 72

Quick26. Identify the three classifications of equations. Explain what each classification

means.

In Problems 27–30, solve the equation and state whether it is an identity,contradiction, or conditional equation.

27.

28.

29.

30. -31z + 12 + 21z - 32 = z + 6 - 2z - 15

-4x + 2 + x + 1 = -41x + 22 + 11

31x - 22 = 2x - 6 + x41x + 22 = 4x + 2

1.1 EXERCISES1–30. are the Quick s that follow each EXAMPLE

Building SkillsIn Problems 31–36, determine which of the numbers are solutionsto the given equation. See Objective 1.

31.

32.

33.

34.

35.

36.

In Problems 37–58, solve each linear equation. Be sure to verifyyour solution. See Objective 2.

37. 38. 8x - 6 = 183x + 1 = 7

31t + 12 - t = 4t + 9; t = -3, t = -1, t = 2

41x - 12 = 3x + 1; x = -1, x = 2, x = 5

6x + 1 = -2x + 9; x = -2, x = 1, x = 4

5m - 3 = -3m + 5; m = -2, m = 1, m = 3

-4x - 3 = -15; x = -2, x = 1, x = 3

8x - 10 = 6; x = -2, x = 1, x = 2

43. 44.

45. 46.

47.

48.

49. 50.

51. 52.

53.

54.2x + 1

3 -

6x - 14

= -5

12

4x + 39

-2x + 1

2 =

16

3x2

+x

6 = -

53

4y5

-1415

=y

3

41z - 22 = 1231x + 22 = -6

-6x + 2 + 2x + 9 + x = 5x + 10 - 6x + 11

5x + 2 - 2x + 3 = 7x + 2 - x + 5

-5z + 3 = -3z + 13m + 4 = 2m - 5

-7t - 3 + 5t = 11-3w + 2w + 5 = -4

39. 40. -6x - 5 = 135x + 4 = 14

41. 42. 8y + 3 = 54z + 3 = 2

EXAMPLE 8 Classify a Linear EquationSolve the linear equation State whether theequation is an identity, contradiction, or conditional equation.

-4(x - 2) + 3(4x + 2) = 2(4x + 7).

Work Smart: Study SkillsSelected problems in the exercisesets are identified by a symbol.For extra help, view the workedsolutions to these problems on thebook’s CD Lecture Series.


79.

80.

81.

82.

83.

84.

85.

86.

87.

88.

89.

90.

Applying the Concepts91. Find a such that the solution set of is

92. Find a such that the solution set of is

93. Find a such that the solution set ofis the set of all real numbers.

94. Find a such that the solution set of is the set of all real numbers.

In Section R.5 we introduced the domain of a variable.The domainof a variable is the set of all values that a variable may assume. Re-call that division by zero is not defined, so any value of the variablethat results in division by zero must be excluded from the domain.In Problems 95–100, determine which values of the variable mustbe excluded from the domain.

95. 96.

97. 98.

99. 100.

101. Interest Suppose you have a credit card debt of $2000.Last month, the bank charged you $25 interest

on the debt.The solution to the equation

represents the annual interest rate on the creditcard, r. Find the annual interest rate on the creditcard.

25 =200012

# r

-2x + 741x - 32 + 2

6x - 2

31x + 12 - 6

2x3x + 1

3x + 74x - 3

-35x + 8

5

2x + 1

2x + 31x + 12 ax + 3 =

31x - 12 a1x - 12 =

576. ax + 6 = 20

5-36. ax + 3 = 15

15

12a - 52 - 4 =12

1a + 42 -710

14

1x - 42 + 3 =13

12x + 62 -56

0.51x + 32 = 0.21x - 62-0.81x + 12 = 0.21x + 42-0.8y + 0.3 = 0.2y - 3.7

0.3x - 1.3 = 0.5x - 0.7

z - 46

-2z + 1

9 =

13

m + 14

+56

=2m - 1

12

13z - 81z + 12 = 21z - 32 + 3z

4b - 31b + 12 - b = 51b - 12 - 5b

71x + 22 = 51x - 22 + 2(x + 12)

41p + 32 = 31p - 22 + p + 1855.

56.

57.

58.

In Problems 59–74, solve the equation. Identify each equation as anidentity, contradiction, or conditional equation. Be sure to verifyyour solution. See Objective 3.

59.

60.

61.

62.

63.

64.

65.

66.

67.

68.

69.

70.

71.

72.

73.

74.

Mixed PracticeIn Problems 75–90, solve each linear equation and verify thesolution. State whether the equation is an identity, contradiction,or a conditional equation.

75. 76.

77.

78. -5x + 5 + 3x + 7 = 5x - 6 + x + 12

4a + 3 - 2a + 4 = 5a - 7 + a

4y + 5 = 77y - 8 = -7

451y - 42 + 3 =

231y + 12 +

415

1312x - 32 + 2 =

561x + 32 -

1112

0.91z - 32 - 0.21z - 52 = 0.41z + 12 + 0.3z - 2.1

0.41z + 12 - 0.7z = -0.1z + 0.7 - 0.2z - 0.3

3x + 14

-7x - 4

2 =

263

2x + 12

-x + 1

5 =

2310

r

2 + 21r - 12 =

5r2

+ 4

3p -p

4 =

11p4

+ 1

z - 24

+2z - 3

6 = 7

x

4 +

3x10

= -3320

81w + 22 - 3w = 71w + 22 + 2(1 - w)

21y + 12 - 31y - 22 = 5y + 8 - 6y

101x - 12 - 4x = 2x - 1 + 41x + 124m + 1 - 6m = 21m + 32 - 4m

51s + 32 = 3s + 2s

41x + 12 = 4x

0.12y - 5.26 = 0.05y + 1.25

0.14x + 2.23 = 0.09x + 1.98

0.3z + 0.8 = -0.1

0.5x - 3.2 = -1.7


102. How Much Do I Make? Last week, before taxes,you earned $539 after working 26 hours at your reg-ular hourly rate and 6 hours at time-and-a-half. Thesolution to the equation representsyour regular hourly rate, x. Determine your regularhourly rate.

103. Paying Your Taxes You are single and just deter-mined that you paid $3727.50 in federal incometaxes in 2008. The solution to the equation

represents theamount x that you earned in 2008. Determine howmuch you earned in 2008. (SOURCE: Internal Revenue

Service)

104. Paying Your Taxes You are married and just deter-mined that you paid $10,187.50 in federal incometaxes in 2008. The solution to the equation

represents0.251x - 65,1002 + 8962.5010,187.50 =

0.151x - 80252 + 802.503727.50 =

26x + 9x = 539

the amount x that you and your spouse earned in2008. Determine how much you and your spouseearned in 2008. (SOURCE: Internal Revenue Service)

Explaining the Concepts105. Explain the difference between and

. In general, what is the difference be-tween an algebraic expression and an equation?

106. Explain the difference between the directions“solve” and “simplify.”

107. Make up a linear equation that has one solution.Make up a linear equation that has no solution. Makeup a linear equation that is an identity. Comment onthe differences and similarities in making up eachequation.

4(x + 1) = 24(x + 1) - 2

1.2 An Introduction to Problem Solving

Work SmartWe learned in the last section thatan equation is a statement in whichtwo algebraic expressions,separated by an equal sign, are equal.

Table 1 Words That Translate into an Equal Signis yields are equals is equivalent to

was gives results in is equal to

Preparing for Problem SolvingBefore getting started, take the following readiness quiz. If you get a problem wrong, go back tothe section cited and review the material.

In Problems P1–P6, express each English phrase as an algebraic expression. [Section R.5, pp. 39–40]

P1. The sum of 12 and a number z.P2. The product of 4 and x.P3. A number y decreased by 87.P4. The quotient of z and 12.P5. Four times the sum of x and 7.P6. The sum of four times a number x and 7.

OBJECTIVES1 Translate English Sentences into

Mathematical Statements

2 Model and Solve DirectTranslation Problems

3 Model and Solve MixtureProblems

4 Model and Solve UniformMotion Problems

1 Translate English Sentences into Mathematical StatementsIn English, a complete sentence must contain a subject and a verb, so expressions or“phrases” are not complete sentences. For example,“Beats me!” is an expression, not acomplete sentence, because it does not contain a subject. The expression “5 more thana number x” does not contain a verb and therefore is not a complete sentence either.The statement “5 more than a number x is 18” is a complete sentence because it con-tains a subject and a verb, so we can translate it into a mathematical statement. In math-ematics, statements can be represented symbolically through equations. In English,statements can be true or false. For example, “The moon is made of green cheese” is afalse statement, while “Humans are mammals” is a true statement. Mathematical state-ments can be true or false as well—we call them conditional equations.

You may want to look back at page 39 in Section R.5 for a review of key words thattranslate into mathematical symbols. Table 1 provides a summary of words that typi-cally translate into an equal sign.

Let’s look at some examples where we translate English sentences into mathemati-cal statements.


P2. 4x P3. P4.

P5. P6. 4x + 741x + 72z

12 y - 87

12 + z

Section 1.2 An Introduction to Problem Solving 59

EXAMPLE 1 Translating English Sentences into Mathematical StatementsTranslate each of the following sentences into a mathematical statement. Do not solvethe equation.

(a) Five more than a number x is 20.

(b) Four times the sum of a number z and 3 is 15.

(c) The difference of x and 5 equals the quotient of x and 2.

Solution

(a)

(b) The expression that reads “Four times the sum of” tells us we need tofind the sum first and then multiply this result by 4.

(c)

x

2 =x - 5

The difference of x and 5 ('''')''''* equals (')'* the quotient of x and 2 (''''')'''''*

15=41z + 32Four times the sum of a number z and 3 (''''''')'''''''*

is " 15 "

20=x + 5

5 more than a number x (''''')''''* is " 20 "

QuickIn Problems 1–5, translate each English statement into a mathematical statement.Do not solve the equation.

1. The sum of x and 7 results in 12.

2. The product of 3 and y is equal to 21.

3. Two times the sum of 3 and x is equivalent to the product of 5 and x.

An Introduction to Problem Solving and Mathematical ModelsEvery day we encounter various types of problems that must be solved. Problemsolving is the ability to use information, tools, and our own skills to achieve a goal. Forexample, suppose Kevin wants a glass of water, but he is too short to reach the sink.Kevin has a problem.To solve the problem, he finds a step stool and pulls it over to thesink. He uses the step stool to climb on the counter, opens the kitchen cabinet and pullsout a cup. He then crawls along the counter top, turns on the faucet, and proceeds to fillhis cup with water. Problem solved!

Of course, this is not the only way that Kevin could solve the problem. Can youthink of any other solutions? Just as there are various approaches to solving life’severyday problems, there are many ways to solve problems using mathematics. How-ever, regardless of the approach, there are always some common aspects in solvingany problem. For example, regardless of how Kevin ultimately ends up with his cupof water, someone must get a cup from the cabinet and someone must turn on thefaucet.

One of the purposes of learning algebra is to be able to solve certain types of prob-lems. To solve these problems, we will need techniques that can help us translate theverbal description of the problem into an equation that can be solved. The process oftaking a verbal description of a problem and developing a mathematical equation thatcan be used to solve the problem is mathematical modeling.

4. The difference of x and 10 equals the quotient of x and 2.

5. Three less than a number y is five times y.

Work SmartThe English statement “The sum offour times a number z and 3 is 15”would be expressed mathematicallyas Do you see thedifference between this statementand the one in Example 1(b)?

4z + 3 = 15.


Mathematical modeling begins with a problem. The problem is then summarized asa verbal description. The verbal description is then translated into the languageof mathematics. This translation results in an equation that can be solved (the mathe-matical problem).The solution must be checked against the mathematical problem (theequation) and the verbal description.This entire process is called the modeling process.We call the equation that is developed the mathematical model. See Figure 1.

Verbaldescription

Language ofmathematics

Mathematicalproblem

Solution

Check

Check

Check

Realproblem

Figure 1

Not all models are mathematical. In general, a model is a way of representing re-ality through graphs, pictures, small-scale reproductions, equations, or even verbaldescriptions. Because the world is an extremely complex place, we often need tosimplify information when we develop a model. For example, a map is a model ofour road system. Maps don’t show all the details of the system such as trees, build-ings, or potholes, but they do a good job of describing how to get from point A topoint B. Mathematical models are similar in that we often make assumptions regard-ing our world in order to make the mathematics more manageable. The fewerassumptions that are made, the more complicated the mathematics becomes in themodel.

It is difficult to give a step-by-step approach for solving problems because eachproblem is unique in some way. However, because there are common links to manytypes of problems, we can categorize problems. In this text, we will present five cate-gories of problems.

Five Categories of Problems

1. Direct Translation—problems where we must translate from English intoMathematics by using key words in the verbal description

2. Mixture—problems where two or more quantities are combined in some fashion

3. Geometry—problems where the unknown quantities are related through geomet-ric formulas

4. Uniform Motion—problems where an object travels at a constant speed

5. Work Problems—problems where two or more entities join forces to complete a job

We will present strategies for solving these categories of problems throughout the text.In this section, we will concentrate on direct translation problems, mixture problems,and uniform motion problems.

Regardless of the type of problem we must solve, there are certain steps that can beused to assist in solving the problem. On the next page we provide you with a series ofsteps that should be followed when developing any mathematical model. As weproceed through this course, and in future courses, you will use the techniques thatyou have studied in this course to solve more complicated problems, but the approachremains the same.


STEPS FOR SOLVING PROBLEMS WITH MATHEMATICAL MODELS

Step 1: Identify What You Are Looking ForRead the problem very carefully, perhaps two or three times. Identifythe type of problem and the information that we wish to learn from theproblem. It is fairly typical that the last sentence in the problem indicateswhat it is we wish to solve for.

Step 2: Give Names to the UnknownsAssign variables to the unknown quantities in the problem. Remember,you can use any letter to represent the unknown(s) when you make yourmodel. Choose a variable that is representative of the unknown quantity.For example, use t for time.

Step 3: Translate the Problem into the Language of MathematicsRead the problem again. This time, after each sentence is read, determineif the sentence can be translated into a mathematical statement or expres-sion in terms of the variables identified in Step 2. It is often helpful to createa table, chart, or figure. When you have finished reading the problem, ifnecessary, combine the mathematical statements or expressions into anequation that can be solved.

Step 4: Solve the Equation(s) Found in Step 3Solve the equation for the variable.

Step 5: Check the Reasonableness of Your AnswerCheck your answer to be sure that it makes sense. If it does not, go backand try again.

Step 6: Answer the QuestionWrite your answer in a complete sentence.

Work SmartWhen you solve a problem bymaking a mathematical model,check your work to make sure youhave the right answer. Typically,errors can happen in two ways.� One type of error occurs if you

correctly translate the problem intoa model but then make an errorsolving the equation. This type oferror is usually easy to find.

� However, if you misinterpret theproblem and develop an incorrectmodel, then the solution youobtain may still satisfy yourmodel, but it probably will not bethe correct solution to the originalproblem. We can check for thistype of error by determiningwhether the solution isreasonable. Does your answermake sense? Always be sure thatyou are answering the questionthat is being asked.

2 Model and Solve Direct Translation ProblemsLet’s look at solving direct translation problems, which are problems that can be set upby reading the problem and translating the verbal description into a mathematicalequation.

EXAMPLE 2 Consecutive IntegersThe sum of three consecutive odd integers results in 45. Find the integers.

Solution

Step 1: Identify This is a direct translation problem. We are looking for three oddintegers. The odd integers are 1, 3, 5, and so on.

Step 2: Name If we let x represent the first odd integer, then is the next oddinteger, and is the third odd integer.

Step 3: Translate Since we know that their sum is 45, we have

The Model= 45x + 4 2

+x + 2 2

+x #

x + 4x + 2

First Integer Second Integer Third Integer

Step 4: Solve We now proceed to solve the equation.

x = 13

3x = 39

3x + 6 = 45

x + x + 2 + x + 4 = 45

Combine like terms:Subtract 6 from both sides:



Step 5: Check Since x represents the first odd integer, the remaining two oddintegers are and It is always a good idea to make sure youranswer is reasonable. Since we know we have the right answer!

Step 6: Answer the Question The three consecutive odd integers whose sum is45 are 13, 15, and 17.

13 + 15 + 17 = 45,13 + 4 = 17.13 + 2 = 15

Quick In Problems 6 and 7, translate each English statement into an equationand solve the equation.

6. The sum of three consecutive even integers is 60. Find the integers.

7. The sum of three consecutive integers is 78. Find the integers.

EXAMPLE 3 How Much Do I Make per Hour?Before taxes, Marissa earned $725 one week after working 52 hours. Her employerpays time-and-a-half for all hours worked in excess of 40 hours. What is Marissa’shourly wage?

Solution

Step 1: Identify This is a direct translation problem. We are looking for Marissa’shourly wage.

Step 2: Name Let represent Marissa’s hourly wage.

Step 3: Translate We know that Marissa earned $725 by working 40 hours at herregular wage and 12 hours earning 1.5 times her regular wage. For each hour thatMarissa works for the first 40 hours, she earns dollars and for each hour sheworks for the next 12 hours, she earns 1.5 dollars. Therefore, her total salary is

725 2

= 1211.5w2 4 +40w

2

ww

w

Regular Earnings Overtime Earnings Total Earnings

The Model

Step 4: Solve

w = 12.50

58w 58

=725 58

58w = 725

40w + 18w = 725

40w + 1211.5w2 = 725

Simplify:Combine like terms:


Step 5: Check We believe that Marissa’s hourly wage is $12.50. So, for the first40 hours she earned and for the next 12 hours she earned

Her total salary was This checkswith the information presented in the problem.

Step 6: Answer the Question Marissa’s hourly wage is $12.50.

$500 + $225 = $725.1.51$12.502 1122 = $225.$12.501402 = $500

Quick 8. Before taxes, Melody earned $735 one week after working 46 hours. Her

employer pays time-and-a-half for all hours worked in excess of 40 hours.What is Melody’s hourly wage?

9. Before taxes, Jim earned $564 one week after working 30 hours at his regularwage, 6 hours at time-and-a-half on Saturday, and 4 hours at double-time on Sunday. What is Jim’s hourly wage?


Solution

Step 1: Identify This is a direct translation problem. We are looking for the numberof minutes for which the two plans cost the same.

Step 2: Name Let m represent the number of long-distance minutes used in themonth.

Step 3: Translate The monthly fee for MCI is $6.99 plus $0.04 for each minute used.So, if one minute is used, the fee is dollars. If two minutes areused, the fee is dollars. In general, if m minutes are used, themonthly fee is dollars. Similar logic results in the monthly fee forSprint being dollars. We want to know the number of minutes forwhich the cost for the two plans will be the same, which means we need to solve

The Model

Step 4: Solve

104 = m 1.04 = 0.01m

1.04 + 0.04m = 0.05m

6.99 + 0.04m = 5.95 + 0.05m

6.99 + 0.04m = 5.95 + 0.05m

Cost for MCI = Cost for Sprint

5.95 + 0.05m6.99 + 0.04m6.99 + 0.04122 = 7.07

6.99 + 0.04112 = 7.03

Subtract 5.95 from both sides:Subtract 0.04m from both sides:

Multiply both sides by 100:

Step 5: Check We believe the cost of the two plans will be the same if 104 minutesare used. The cost of MCI’s plan will be The cost ofSprint’s plan will be They are the same!

Step 6: Answer the Question The cost of the two plans will be the same if 104 minutesare used.

5.95 + 0.0511042 = $11.15.6.99 + 0.0411042 = $11.15.

EXAMPLE 4 Choosing a Long-Distance CarrierMCI has a long-distance phone plan that charges $6.99 a month plus $0.04 per minuteof usage. Sprint has a long-distance phone plan that charges $5.95 a month plus $0.05per minute of usage. For how many minutes of long-distance calls will the costs for thetwo plans be the same? (SOURCE: MCI and Sprint)

Quick 10. You need to rent a moving truck. You have identified two companies that rent

trucks. EZ-Rental charges $35 per day plus $0.15 per mile. Do It YourselfRental charges $20 per day plus $0.25 per mile. For how many miles will thecost of renting be the same?

11. You need a new cell phone for emergencies only. Company A charges $12 permonth plus $0.10 per minute, while Company B charges $0.15 per minute withno monthly service charge. For how many minutes will the monthly cost bethe same?

There are many types of direct translation problems. One specific type of directtranslation problem is a “percent problem.”Typically, these problems involve discountsor markups that businesses use in determining their prices.

Percent means divided by 100 or per hundred. We use the symbol % to denote

percent, so 45% means 45 out of 100 or or 0.45. In applications involving

percents, we often encounter the word “of,” as in 20% of 100. Remember that theword “of” translates into multiplication in mathematics so 20% of 100 means

So, 20% of 100 is 20.

20% # 100 or 0.20 # 100

45100


When dealing with percents and the price of goods, it is helpful to remember thefollowing:

Net Price + Markup = Gross Price Original Price - Discount = Sale Price

EXAMPLE 5 Discounted PriceSuppose that you have just entered your favorite clothing store and find thateverything in the store is marked at a discount of 40% off. If the sale price of a suitis $144, what was the original price?

Solution

Step 1: Identify This is a direct translation problem involving percents. We wish tofind the original price of the suit.

Step 2: Name Let p represent the original price.

Step 3: Translate We know that the original price minus the discount will give us thesale price. We also know that the sale price is $144, so

The discount was 40% off of the original price so that the discount is 0.40p.Substituting into the equation we obtain

The Model

Step 4: Solve We now solve for p, the original price.

p = 240

0.60p0.60

=1440.60

0.60p = 144

p - 0.40p = 144

p - 0.40p = 144

p - discount = 144,

p - discount = 144

Combine like terms:

Divide both sides by 0.60:

Simplify:

Step 5: Check If the original price of the suit was $240, then the discount would beSubtracting $96 from the original price results in a sale price of

$144. This agrees with the information in the problem.

Step 6: Answer the Question The original price of the suit was $240.

0.41$2402 = $96.

Work SmartYou could eliminate the decimals bymultiplying both sides of theequation by 10.

Quick12. What is 40% of 100?

13. 8 is 5% of what?

14. 15 is what percent of 20?

15. Suppose that you have just entered your favorite clothing store and find thateverything in the store is marked at a discount of 30% off. If the sale price of ashirt is $21, what was the original price?

16. A Milex Tune-Up automotive facility marks up its parts 35%. Suppose thatMilex charges its customers $1.62 for each spark plug it installs. What is Milex’s cost for each spark plug?

Interest is money paid for the use of money. The total amount borrowed is called theprincipal. The principal can be in the form of a loan (an individual borrows from thebank) or a deposit (the bank borrows from the individual). The rate of interest,expressed as a percent, is the amount charged for the use of the principal for a givenperiod of time, usually on a yearly (that is, per annum) basis.


SIMPLE INTEREST FORMULA

If a principal of P dollars is borrowed for a period of t years at an annual interestrate r, expressed as a decimal, the interest I charged is

Interest charged according to this formula is called simple interest.

I = Prt

EXAMPLE 6 Computing Credit Card InterestSuppose that Yolanda has a credit card balance of $2800. Each month, the credit cardcharges 14% annual simple interest on any outstanding balances. What is the interestthat Yolanda will be charged on this loan after one month? What is Yolanda’s creditcard balance after one month?

Solution

We wish to know the interest I charged on the loan. Because the interest rate is givenas an annual rate, the length of time that the money is borrowed must be expressed

in years. Since we are talking about 1 month, the length of time is of a year so that

The outstanding balance, or principal, is The annual interest

rate is Substituting into we obtain

Yolanda will owe the amount borrowed, $2800, plus accrued interest, $32.67, for atotal of $2800 + $32.67 = $2832.67.

I = 1$28002 10.142a 112b = $32.67

I = Prt,r = 14% = 0.14.

p = $2800.t =1

12.

112

Quick17. is money paid for the use of money. The total amount borrowed is

called the .

18. Suppose that Dave has a car loan of $6500.The bank charges 6% annual simpleinterest.What is the interest charge on Dave’s car loan after 1 month?

19. Suppose that you have $1400 in a savings account. The bank pays 1.5% annualsimple interest. What would be the interest paid after 6 months? What is the balance in the account?

3 Model and Solve Mixture ProblemsMixture problems are problems in which two or more items are combined to form athird item. There are a number of different types of mixture problems, but they all fol-low a basic approach to solving the problem. In solving mixture problems rememberthe following idea:

One type of mixture problem is the so-called interest problem.

Portion from Item A + Portion from Item B = Whole or Total

EXAMPLE 7 Financial PlanningKevin has $15,000 to invest. His goal is to obtain an overall annual rate of return of9%, or $1350 annually. His financial advisor recommends that he invest some of themoney in corporate bonds that pay 12% and the rest in government-backed Treasurybonds paying 4%. How much should be placed in each investment in order for Kevinto achieve his goal?


Solution

Step 1: Identify This is a mixture problem involving simple interest. Kevin needs toknow how much to place into corporate bonds and how much to place into Treasurybonds in order to earn $1350 in interest.

Step 2: Name Let b represent the amount invested in corporate bonds, so thatrepresents the amount invested in Treasury bonds.

Step 3: Translate We organize the information given in Table 2.

$15,000 - b

Use the Distributive Propertyto remove parentheses:


Divide both sides by 0.08:

Step 5: Check It appears that Kevin should invest $9375 in corporate bonds andin Treasury bonds. The simple interest earned each year on

the corporate bonds is ($9375)(0.12)(1) The simple interest earned eachyear on the Treasury bonds is The total interest earned is

Kevin wanted to earn 9% annually and ($15,000)(0.09)(1)This agrees with the information presented in the problem.

Step 6: Answer the Question Kevin will invest $9375 in corporate bonds and in Treasury bonds.

$5625

$1350.=$1125 + $225 = $1350.

1$5625210.042112 = $225.$1125.=

$15,000 - $9375 = $5625

Table 2Principal

$Rate%

TimeYr

Interest$

Kevin wants to earn 9% each year on his principal of $15,000. Using the simpleinterest formula, Kevin wants to earn each year in interest.The total interest will be the sum of the interest from the corporate bonds and theTreasury bonds.

The Model

Step 4: Solve

b = 9375

0.08b = 750

0.08b + 600 = 1350

0.12b + 600 - 0.04b = 1350

0.12b + 0.04115,000 - b2 = 1350

0.12b + 0.04115,000 - b2 = 1350

= $1350Interest from corporate bonds + Interest from Treasury bonds

$15,00010.092112 = $1350

Quick20. Sophia has recently retired and requires an extra $5400 per year in income.

She has $90,000 to invest and can invest in either an Aaa-rated bond that pays5% per annum or a B-rated bond paying 9% per annum. How much should beplaced in each investment in order for Sophia to achieve her goal?

Often, new blends are created by mixing two quantities. For example, a chef mightmix buckwheat flour with wheat flour to make buckwheat pancakes. Or a coffee shopmight mix two different types of coffee to create a new coffee blend.

Work SmartRemember that mixtures can includeinterest (money), solids (nuts), liquids(chocolate milk), and even gases(Earth’s atmosphere). “Mixture” is asimplifying assumption.

Corporate Bond b 0.12 1 0.12b

Treasury Bond 0.04 1

Total 15,000 0.09 1 0.09115,0002 = $1350

0.04115,000 - b215,000 - b

21. Steve has $25,000 to invest and wishes to earn an overall annual rate of returnof 8%. His financial advisor recommends that he invest some of the money in a5-year CD paying 4% per annum and the rest in a corporate bond paying 9%per annum. How much should be placed in each investment in order for Steve to achieve his goal?


Table 3Price

$/Pound

� �

Brazilian$8 per pound

Blend$9 per pound

Sumatra$12 per pound

Figure 2

EXAMPLE 8 Blending CoffeesThe manager of a coffee shop wishes to form a new blend of coffee. She wants to mixSumatra beans, known for their strong, distinctive taste, that sell for $12 per poundwith milder Brazilian beans that sell for $8 per pound to get 50 pounds of the newblend. The new blend will sell for $9 per pound and there will be no difference inrevenue from selling the new blend versus selling the beans separately. See Figure 2.How many pounds of the Sumatra and Brazilian beans are required?

Solution

Step 1: Identify This is a mixture problem. We want to know the number of poundsof Sumatra beans and the number of pounds of Brazilian beans that are required inthe new blend.

Step 2: Name Let s represent the number of pounds of Sumatra beans that arerequired. So, will be the number of pounds of Brazilian coffee required.

Step 3: Translate We are told that there is to be no difference in revenue betweenselling the Sumatra and Brazilian separately versus the blend. This means that if theblend contains one pound of Sumatra and one pound of Brazilian, we should collect

because that is how much we would collect if we sold thebeans separately.

We set up Table 3.

$12112 + $8112 = $20

50 - s

Number ofPounds Revenue

Sumatra 12 s 12s

Brazilian 8

Blend 9 50 91502 = 450

8150 - s250 - s

In general, if the mixture contains s pounds of Sumatra beans, we should collect$12s. If the mixture contains pounds of Brazilian beans, we should collect$8( ). If the mixture sells for $9 per pound and we make 50 pounds of the blend,then we should collect for the blend.$91502 = $450

50 - s50 - s

$9 # 50 5 = 8 # 150 - s2 + 12 # s

= aPrice per pound of blend

b aPounds of blend

b aPrice per pound of Sumatra

b aPounds of Sumatra

b + aPrice per pound of Brazilian

b aPounds of Brazilian

bh h

Revenue from Sumatra Revenue from Brazilian Revenue from Mixture

We have the equation

The Model12s + 8150 - s2 = 450


Step 4: Solve Solve the equation for s.

s = 12.5

4s = 50

4s + 400 = 450

12s + 400 - 8s = 450

12s + 8150 - s2 = 450Use the Distributive Property toremove the parentheses:



Step 5: Check It appears that we should mix 12.5 pounds of Sumatra withpounds of Brazilian. The 12.5 pounds of Sumatra beans would

sell for and the 37.5 pounds of Brazilian beans would sell forthe total revenue would be which equals the

revenue obtained from selling the blend. This checks with the information presentedin the problem.

Step 6: Answer the Question The manager should mix 12.5 pounds of Sumatra beanswith 37.5 pounds of Brazilian beans to make the blend.

$150 + $300 = $450,$8137.52 = $300;$12112.52 = $150

50 - 12.5 = 37.5

Quick 22. Suppose that you want to blend two teas in order to obtain 10 pounds of the

new blend that sells for $3.50 per pound.Tea A sells for $4.00 per pound and Tea Bsells for $2.75 per pound.Assuming that there will be no difference in revenuefrom selling the new blend versus selling the tea separately,determine the numberof pounds of each tea that will be required in the blend.

23. “We’re Nuts!” sells cashews for $6.00 per pound and peanuts for $1.50 perpound. The manager has decided to make a “trail mix” that combines thecashews and peanuts. She wants the trail mix to sell for $3.00 per pound andthere should be no loss in revenue from selling the trail mix versus selling thenuts alone. How many pounds of cashews and peanuts are required to create 30 pounds of trail mix?

4 Model and Solve Uniform Motion ProblemsObjects that move at a constant velocity are said to be in uniform motion. When theaverage velocity of an object is known, it can be interpreted as its constant velocity.For example, a car traveling at an average velocity of 40 miles per hour is in uniformmotion.

UNIFORM MOTION FORMULA

If an object moves at an average speed r, the distance d covered in time t is givenby the formula

d = rt

In WordsThe uniform motion formula states thatdistance equals rate times time.

EXAMPLE 9 Uniform MotionRoger and Bill decide to have a 10-mile race. Roger can run at an average speed of12 miles per hour while Bill can run at an average speed of 10 miles per hour.To “even things up,” Roger agrees to give Bill a head start of 0.15 hour. When willRoger catch up to Bill?

Solution

Step 1: Identify This is a uniform motion problem.We wish to know the numberof hours it will take for Roger to catch up to Bill if each man runs at his average speed.


Step 2: Name Let t represent the number of hours it takes for Roger to catch up toBill. Since Bill is given a head start of 0.15 hour, Bill will have been running for

hours.

Step 3: Translate Figure 3 illustrates the situation. We will set up Table 4.

t + 0.15

START

t t � 0.15

RogerBill

FINISH

Figure 3

Table 4Rate, mph Time, hours Distance, miles

Roger 12 t 12t

Bill 10 101t + 0.152t + 0.15

The distance that Roger will travel is 12t, since Roger’s average velocity is 12 milesper hour and he runs for t hours. The distance that Bill will travel is sinceBill’s average velocity is 10 miles per hour and he runs for hours. We wantthe distance that each travels to be equal, so we set up the following model:

The Model

Step 4: Solve We wish to solve for t:

t =1.52

= 0.75

2t = 1.5

12t = 10t + 1.5

12t = 101t + 0.152

12t = 101t + 0.152 Distance Roger runs = Distance Bill runs

t + 0.15101t + 0.152

Use the Distributive Property:Subtract 10t from both sides:


Step 5: Check It appears that it will take Roger 0.75 hour to catch up to Bill. After0.75 hour, Roger will have traveled miles and Bill will have traveled

miles. It checks!

Step 6: Answer the Question It will take Roger 0.75 hour or 45 minutes to catch upto Bill.

1010.75 + 0.152 = 91210.752 = 9

Quick24. Objects that move at a constant velocity are said to be in .

25. A Chevrolet Cobalt left Omaha, Nebraska, traveling at an average velocity of40 miles per hour.Two hours later, a Dodge Charger left Omaha, Nebraska,traveling on the same road at an average velocity of 60 miles per hour.Whenwill the Charger catch up to the Cobalt? How far will each car have traveled?

26. A train leaves Union Station traveling at an average velocity of 50 miles perhour. Four hours later, a helicopter follows along the train tracks traveling at anaverage velocity of 90 miles per hour. When will the helicopter catch up to thetrain? How far will each have traveled?



Building Skills

27. What is 25% of 40?

28. What is 150% of 70?

29. 12 is 30% of what?

30. 50 is 90% of what?



In Problems 33–42, translate each of the following English statements into a mathematical statement.Then solve the equation.See Objective 1.

33. The sum of a number x and 12 is 20.

34. The difference between 10 and a number zis 6.

35. Twice the sum of y and 3 is 16.

36. The sum of two times y and 3 is 16.

37. The difference between and 22 equals three times .

38. The sum of x and 4 results in twice x.

39. Four times a number x is equivalent to the sum oftwo times x and 14.

40. Five times a number x is equivalent to the differenceof three times x and 10.

41. 80% of a number is equivalent to the sum of thenumber and 5.

42. 40% of a number equals the difference between thenumber and 10.

Applying the Concepts43. Number Sense Grant is thinking of two numbers.

He says that one of the numbers is twice the othernumber and the sum of the numbers is 39. What arethe numbers?

44. Number Sense Pattie is thinking of two numbers.She says that one of the numbers is 8 more than theother number and the sum of the numbers is 56.What are the numbers?

ww

47. Computing Grades Going into the final exam, whichcounts as two grades, Kendra has test scores of 84, 78,64, and 88. What score does Kendra need on the finalexam in order to have an average of 80?

48. Computing Grades Going into the final exam, whichcounts as three grades, Mark has test scores of 65, 79,83, and 68. What score does Mark need on the finalexam in order to have an average of 70?

49. Comparing Printers Jacob is trying to decide betweentwo laser printers, one manufactured by Hewlett-Packard, the other by Brother. Both have similarfeatures and warranties, so price is the determiningfactor. The Hewlett-Packard costs $180 and printingcosts are approximately $0.03 per page. The Brothercosts $230 and printing costs are approximately $0.01 per page. How many pages need to be printedfor the cost of the two printers to be the same?

50. Comparing Job Offers Maria has just been offeredtwo sales jobs. The first job offer is a base monthlysalary of $2500 plus a commission of 3% of total sales. The second job offer is a base monthly salary of $1500 plus a commission of 3.5% of total sales.For what level of monthly sales is the salary of thetwo jobs equivalent?

51. Finance An inheritance of $800,000 is to be dividedamong Avery, Connor, and Olivia in the followingmanner: Olivia is to receive of what Connor gets,while Avery gets of what Connor gets. How muchdoes each receive?

52. Sharing the Cost of a Pizza Judy and Linda agree toshare the cost of a $21 pizza based on how much eachate. If Judy ate of the amount that Linda ate, howmuch should each pay?

53. Sales Tax In the state of Colorado there is a sales tax of 2.9% on all goods purchased. If Jan buys a television for $599, what will be the final bill,including sales tax?

54. Sales Tax In the state of Texas there is a sales taxof 6.25% on all goods purchased. If Megan buys 4 compact disks for $39.83, what will be the final bill,including sales tax?

55. Markups A new 2008 Honda Accord has a list priceof $25,800. Suppose that the dealer markup on thiscar is 15%. What is the dealer’s cost?

56. Markups Suppose that the price of a new Intermedi-ate Algebra text is $95. The bookstore has a policy ofmarking texts up 30%. What is the cost of the text tothe bookstore?

57. Discount Pricing Suppose that you just received an email alert from buy.com indicating that

3�4

1�4

3�4

45. Consecutive Integers The sum of three consecutiveintegers is 75. Find the integers.

46. Consecutive Integers The sum of four consecutiveodd integers is 104. Find the integers.


66. Simple Interest Faye has a home equity loan of$70,000. The bank charges Faye 6% per annum simpleinterest. What is the interest charge on Faye’s loanafter 1 month?

67. Banking A bank has loaned out $500,000, part of it at6% per annum and the rest of it at 11% per annum. Ifthe bank receives $43,750 in interest each year, howmuch was loaned at 6%?

68. Banking Patrick is a loan officer at a bank. He has$2,000,000 to lend out and has two loan programs.His home equity loan is currently priced at 6% perannum, while his unsecured personal loan is priced at14%. The bank president wants Patrick to earn a rateof return of 12% on the $2,000,000 available. Howmuch should Patrick lend out at 6%?

69. Investments Pedro wants to invest his $25,000 bonuscheck. His investment advisor has recommended thathe put some of the money in a bond fund that yields5% per annum and the rest in a stock fund that yields9% per annum. If Pedro wants to earn $1875 eachyear from his investments, how much should he placein each investment?

70. Investments Johnny is a shrewd 8-year-old.ForChristmas,his grandparents gave him $10,000.Johnnydecides to invest some of the money in a savings accountthat pays 2% per annum and the rest in a stock fundpaying 10% per annum.Johnny wants his investmentsto yield 7% per annum.How much should Johnny putinto each account?

71. Making Coffee Suppose that you want to blend twocoffees in order to obtain 50 pounds of a new blend.If x represents the number of pounds of coffee A,write an algebraic expression that represents thenumber of pounds of coffee B.

72. Candy “Sweet Tooth!” candy store sells chocolate-covered almonds for $6.50 per pound and chocolate-covered peanuts for $4.00 per pound. The managerdecides to make a bridge mix that combines the al-monds with the peanuts. She wants the bridge mix tosell for $5.00 per pound, and there should be no lossin revenue from selling the bridge mix versus the almonds and peanuts alone. How many pounds ofchocolate-covered almonds and chocolate-coveredpeanuts are required to create 50 pounds of bridge mix?

73. Coins Bobby has been saving quarters and dimes. He opened up his piggy bank and determinedthat it contained 47 coins worth $9.50. Determine how many dimes and quarters were in the piggy bank.

74. More Coins Diana has been saving nickels anddimes. She opened up her piggy bank and determinedthat it contained 48 coins worth $4.50. Determine how many nickels and dimes were in the piggy bank.

4-gigabyte flash drives have just been discounted by40%. If the sale price of the flash drive is now $24.90,what was the original price?

58. Discount Pricing Suppose that you just received anemail alert from Kohls indicating that the fall line ofclothing has just been discounted by 30% and knitpolo shirts are now $28. What was the original priceof a polo shirt?

59. Cars A Mazda 6s weighs 20 pounds more than a Nissan Altima. A Honda Accord EX weighs70 pounds more than a Nissan Altima. The totalweight of all three cars is 10,050 pounds. How much does each car weigh? (SOURCE: Road and Track

magazine)

60. Cars A Mazda 6s requires 18 more feet to stop from80 miles per hour than a Nissan Altima. A Honda Accord EX requires 26 more feet to stop from80 miles per hour than a Nissan Altima. The combined distance that each car requires to stop from 80 miles per hour is 734 feet. What is the stopping distance required for each car? (SOURCE: Road

and Track magazine)

61. Finance A total of $20,000 is going to be split betweenAdam and Krissy with Adam receiving $3000 less thanKrissy. How much will each get?

62. Finance A total of $40,000 is going to be invested in stocks and bonds. A financial advisor recommendsthat $6000 more should be invested in stocks thanbonds. How much is invested in stocks? How much is invested in bonds?

63. Investments Suppose that your long-lost Aunt Sarahas left you an unexpected inheritance of $24,000.You have decided to invest the money rather thanspend it on frivolous purchases. Your financial advisor has recommended that you diversify by placing some of the money in stocks and some inbonds. Based upon current market conditions, she has recommended that the amount in bonds shouldequal three-fifths of the amount invested in stocks.How much should be invested in stocks? How muchshould be invested in bonds?

64. Investments Jack and Diane have $60,000 to invest.Their financial advisor has recommended that theydiversify by placing some of the money in stocksand some in bonds. Based upon current market conditions, he has recommended that the amount in bonds should equal two-thirds of the amount invested in stocks. How much should be invested in stocks? How much should be invested in bonds?

65. Simple Interest Elena has a credit card balance of$2500. The credit card company charges 14% perannum simple interest. What is the interest charge on Elena’s credit card after 1 month?


75. Gold The purity of gold is measured in karats, withpure gold being 24 karats. Other purities of goldare expressed as proportional parts of pure gold.

For example, 18-karat gold is or 75%, pure gold;

12-karat gold is or 50%, pure gold; and so on.

How much pure gold should be mixed with 12-karatgold to obtain 72 grams of 18-karat gold?

76. Antifreeze The cooling system of a car has a capacityof 15 liters. If the system is currently filled with a mix-ture that is 40% antifreeze, how much of this mixtureshould be drained and replaced with pure antifreezeso that the system is filled with a solution that is50% antifreeze?

77. A Biathlon Suppose that you have entered a 62-milebiathlon that consists of a run and a bicycle race. Dur-ing your run, your average velocity is 8 miles per hourand during your bicycle race, your average velocity is20 miles per hour.You finish the race in 4 hours.Whatis the distance of the run? What is the distance of thebicycle race?

78. A Biathlon Suppose that you have entered a 15-milebiathlon that consists of a run and swim. During yourrun, your average velocity is 7 miles per hour andduring your swim, your average velocity is 2 milesper hour. You finish the race in 2.5 hours. What is thelength of the swim? What is the distance of the run?

79. Collision Course Two cars that are traveling towardeach other are 455 miles apart. One car is traveling 10 miles per hour faster than the other car. The carspass after 3.5 hours. How fast is each car traveling?

80. Collision Course Two planes that are travelingtoward each other are 720 miles apart. One plane istraveling 40 miles per hour faster than the other. Theplanes pass after 0.75 hour. How fast is each planetraveling?

81. Boats Two boats leave a port at the same time, onetraveling north and the other traveling south. Thenorthbound boat travels at 12 miles per hour (mph)faster than the southbound boat. If the southboundboat is traveling at 25 mph, how long before they are155 miles apart?

82. Cyclists Two cyclists leave a city at the same time, onegoing east and the other going west. The westboundcyclist bikes at 3 mph faster than the eastbound cyclist.If after 6 hours they are 162 miles apart, how fast iseach cyclist riding?

83. Walking At 10:00 A.M. two people leave their homesthat are 15 miles apart and begin walking toward eachother. If one person walks at a rate that is 2 mph faster

1224

,

1824

,

than the other and they meet after 1.5 hours, how fastwas each person walking?

84. Trains At 9:00 A.M., two trains are 715 miles aparttraveling toward each other on parallel tracks. If onetrain is traveling at a rate that is 10 miles per hour(mph) faster than the other train and they meet after 5.5 hours, how fast is each train traveling?

Extending the Concepts85. Computing Average Speed On a recent trip to

Florida, we averaged 50 miles per hour. On the returntrip we averaged 60 miles per hour. What do youthink the average speed of the trip to Florida andback was? Defend your position. Algebraically,determine the average speed of the trip to Floridaand back.

86. Discount Pricing Suppose that you are the managerof a clothing store and have just purchased 100 shirtsfor $15 each.After 1 month of selling the shirts at theregular price, you plan to have a sale giving 30% offthe original selling price. However, you still want tomake a profit of $6 on each shirt at the sale price.Whatshould you price the shirts at initially to ensure this?

87. Critical Thinking Make up an applied problem that would result in the equation

88. Critical Thinking Make up an applied problem thatwould result in the equation

89. Uniform Motion Suppose that you are walkingalong the side of train tracks at 2 miles per hour.A train that is traveling 20 miles per hour in the samedirection requires 1 minute to pass you. How long isthe train?

Explaining the Concepts90. Consider the phrase “. . . there will be no difference

in revenue from selling the new blend versus sellingthe beans separately” presented in Example 8.Explain what this means.

91. How is mathematical modeling related to problemsolving?

92. Why do we make assumptions when creating mathe-matical models?

93. Name the different categories of problems presentedin this section. Name two different kinds of mixtureproblems.

94. Think of two models that you use in your everydaylife. Are any of them mathematical models? Describeyour models to members of your class.

10 + 0.14x = 50.

x - 0.05x = 60.

Section 1.3 Using Formulas to Solve Problems 73

A known relation that exists between two or more variables can be used to solvecertain types of problems. A formula is an equation that describes how two or morevariables are related.

Table 5Verbal Description Formula

Geometry: What is the area of a rectangle? The area A of a rectangle is the product of its length l and width w.

Physics: How do we measure the energy of an object in motion? Kinetic energy K is one-half the product of the mass m and the square of the velocity v.

Economics: What proportion of the population is in the labor force? The participation rate R is the sum of the number of employed E and the number of unemployed U divided by the adult population P.

Finance: How much money will I have next year? The future value Ais the product of the present value P and 1 plus the annual interest rate r.

A = P11 + r2

R =E + U

P

K =12

mv2

A = lw

Quick 1. A is an equation that describes how two or more variables are related.

In Problems 2–5, translate the verbal description into a mathematical formula.

2. The area A of a circle is the product of the number and the square of its radius r.

3. The volume V of a right circular cylinder is the product of the number thesquare of its radius r, and its height h.

4. The daily cost C of manufacturing computers is $175 times the number of com-puters manufactured x plus $7000.

5. The distance s that an object free-falls is one-half the product of acceleration dueto gravity g and the square of time t.

p,

p

Work SmartThink of a formula as another kindof mathematical model.

1.3 Using Formulas to Solve ProblemsPreparing for FormulasBefore getting started, take this readiness quiz. If you get a problem wrong, go back to the sectioncited and review the material.

In Problems P1 and P2, (a) round, and (b) truncate each decimal to the indicated number ofplaces. [Section R.2, pp. 14–15]

P1. 3.00343; 3 decimal placesP2. 14.957; 2 decimal places

OBJECTIVES1 Solve for a Variable in a Formula

2 Use Formulas to Solve Problems

Preparing for...Answers P1. (a) 3.003(b) 3.003 P2. (a) 14.96 (b) 14.95

EXAMPLE 1 Answering Questions with FormulasThe information in Table 5 gives descriptions, in words, of known relations and thecorresponding formula.

1 Solve for a Variable in a FormulaThe expression “solve for the variable” means to get the variable by itself on one sideof the equation with all other variables and constants, if any, on the other side by form-ing equivalent equations. For example, in the formula for the area of a rectangle,


the formula is solved for A because A is by itself on one side of the equationwhile all other variables are on the other side.

When solving certain problems, it becomes important for us to be able to solve for-mulas for certain variables. The steps that we follow when solving a formula for a cer-tain variable are identical to those that we followed when solving an equation.

A = lw,

Work SmartWhen working with formulas, keeptrack of the units. Verify that the unitsin your answer are reasonable.

EXAMPLE 2 Solving for a Variable in a Formula

The volume V of a cone is given by the formula where r is the radius andh is the height of the cone. See Figure 4.

(a) Solve the formula for h.

(b) Use the result from part (a) to find the height of a cone if its volume iscubic feet and its radius is 5 feet.

Solution

(a) Because we want to solve the formula for h, we want to get h by itselfand all other variables and constants on the other side of the equation.

Multiply both sides by 3:

Divide out common factors:

Divide both sides by

Divide out common factors:

If a � b, then b � a:

(b) Substituting and ft into we obtain

Example 2 presents a formula from geometry. Formulas from geometry are useful insolving many types of problems. We list some of these formulas in Table 6.

h = 6 feet

h =150p ft3

25p ft2

h =3150p ft32p15 ft22

h =3V

pr2,r = 5V = 50p ft3

h =3V

pr2

3V

pr2= h

3V

pr2=pr2h

pr2 pr 2:

3V = pr2h

3 # V = a13

pr2hb # 3 V =

13

pr2h

50p

V =13

pr2h,

Work SmartSolving for a variable is just likesolving an equation with oneunknown. When solving for avariable, treat all the other variablesas constants.

Table 6Figure Formulas Figure Formulas

Square Area:Perimeter:

s

s

P = 4sA = s2

h

r

Figure 4

Rectangle Area:Perimeter:

l

w

P = 2l + 2wA = lw


Figure Formulas Figure Formulas

Rectangular Solid Volume:Surface Area:

h

w

l

S = 2lw + 2lh + 2whV = lwh

Quick 6. The formula for the area of a circle is .

7. What is the formula for the perimeter of a rectangle?

8. The area A of a triangle is given by the formula where b is the base of the triangle and h is the height.


(b) Find the height of the triangle whose area is 10 square inches and whose baseis 4 inches.

9. The perimeter P of a parallelogram is given by the formula where a isthe length of one side of the parallelogram and b is the length of the adjacent side.

(a) Solve the formula for b.

(b) Find the length of one side of a parallelogram whose perimeter is 60 cm andwhose adjacent length is 20 cm.

P = 2a + 2b,

A =12

bh,

Cube Volume:Surface Area:

ss

s

S = 6s2V = s3

Triangle Area:

Perimeter:

ha c

b

P = a + b + c

A =12

bh

Trapezoid Area:

Perimeter:

ha c

B

b

P = a + b + c + B

A =12

h1B + b2

Parallelogram Area:Perimeter:

a

a

bb h

P = 2a + 2bA = ah

Circle Area:Circumference:

rd

C = 2pr = pdA = pr2

Sphere Volume:

Surface Area:

r

S = 4pr2

V =43pr3

Right Circular Cylinder Volume:Surface Area:

h

r

S = 2pr2 + 2prhV = pr2h

Cone Volume:

h

r

V =13

pr2h


Quick In Problems 10–13, solve for the indicated variable.

10. for P 11. for y

12. for h 13. for nS = na + 1n - 12d2xh - 4x = 3h - 3

Ax + By = CI = Prt

2 Use Formulas to Solve ProblemsFormulas are often needed in order to solve certain types of word problems. We followthe same steps to solving these problems as were presented in Section 1.2 on page 61.

EXAMPLE 4 The Perimeter of a WindowThe perimeter of a rectangular picture window is 466 inches. The length of the windowis 55 inches more than the width. See Figure 5. Find the dimensions of the window.

Solution

Step 1: Identify This is a geometry problem that requires the formula for the perime-ter of a rectangle. We want to determine the dimensions of the window, which is in theshape of a rectangle.That is, we want to determine the length and width of the window.

Step 2: Name Let represent the width. Since the length is 55 inches more than thewidth, we know that l 55.

Step 3: Translate The perimeter of a rectangle is We substitute theknown values into the formula for the perimeter of a rectangle.

The Model

Step 4: Solve

Distribute the 2:

Combine like terms:

Subtract 110 from both sides:

Divide both sides by 4: 89 = w

356 = 4w

466 = 4w + 110

466 = 2w + 110 + 2w

466 = 21w + 552 + 2w

466 = 21w + 552 + 2wP = 466; l = w + 55: P = 2l + 2w

w.P = 2l + 2

= w +w

w

w � 55

Figure 5

EXAMPLE 3 Solving for a Variable in a FormulaThe formula is a model used in economics to describe thetotal income of an economy. In the model, Y is income, C is consumption, I is invest-ment in capital, G is government spending, N is net exports, and b is a constant. Solvethe formula for Y.

Solution

We want to get all terms with Y on the same side of the equal sign.

Subtract bY from both sides:Combine like terms:

Use the Distributive Property “in reverse” to isolate Y:

Divide both sides by 1 � b:

Simplify: Y =C + I + G + N

1 - b

Y11 - b2

1 - b =C + I + G + N

1 - b

Y11 - b2 = C + I + G + N

Y - bY = C + I + G + N Y - bY = C + bY + I + G + N - bY

Y = C + bY + I + G + N

Y = C + bY + I + G + N


Step 5: Check It appears that the width of the window is 89 inches, so the length isinches.The perimeter of the window is inches. It

checks!

Step 6: Answer the Question The width of the window is 89 inches and the length is 144 inches.

There is an interesting side note to the result of Example 4. If we compute the ratio

of the length of the window to the width, we obtain Rectangles whose

dimensions form this ratio are called golden rectangles. Golden rectangles are said tohave dimensions that are “pleasing to the eye.” The golden rectangle was first con-structed by the Greek philosopher Pythagoras in the sixth century B.C. These rectan-gles are used in architecture (The Parthenon) and in art (the Mona Lisa). See Figure 6.

14489

L 1.618.

211442 + 21892 = 46689 + 55 = 144

Figure 6

Quick14. The perimeter of a rectangular pool is 180 feet. If the length of the pool is to be

10 feet more than the width, find the dimensions of the pool.

15. The opening of a rectangular bookcase has a perimeter of 224 inches. If theheight of the bookcase is 32 inches more than the width, determine the dimen-sions of the opening of the bookcase.

Solution

Step 1: Identify This is a geometry problem that requires the formula for the surface area of a cylinder. We wish to find the height of the can of soup.

Step 2: Name Let h represent the height of the can of soup.

Step 3: Translate We know that the surface area S of the can of soup is 46.5 squareinches. The radius of the can of soup is 1.375 inches. We substitute these values intothe formula for the surface area of the can.

The Model 46.5 = 2p11.37522 + 2p11.3752hS = 46.5; r = 1 .375: S = 2pr2 + 2prh

Figure 7

r � 1.375�

EXAMPLE 5 Constructing a Soup CanA can of Campbell’s soup has a surface area of 46.5 square inches. See Figure 7. Thesurface area S of a right circular cylinder is where r is the radius ofthe can and h is the height of the can. Find the height of a can of Campbell’s soup if itsradius is 1.375 inches. Round your answer to two decimal places.

S = 2pr2 + 2prh,

Dagli Orti (A)/Picture Desk, Inc./Kobal Collection


Figure 8

Work SmartRound-off error occurs whendecimals are continually roundedduring the course of solving aproblem. The more times we round,the more inaccurate the results maybe. So, do not do any arithmeticuntil the last step.

Quick 16. A can of peaches has a surface area of 51.8 square inches. The surface area S of

a right circular cylinder is where r is the radius of the canand h is the height of the can. Find the height of a can of peaches if its radius is1.5 inches. Round your answer to two decimal places.

S = 2pr2 + 2prh,

Step 4: Solve Solve for h. We will not compute any of the values until the last calculation. This is done to avoid round-off error.

Subtract fromboth sides:

Divide both sides by

We will use a calculator to evaluate this expression. Figure 8 shows the outputfrom a TI-84 Plus graphing calculator.

Rounded to two decimal places, we obtain inches.

Step 5: Check The surface area S of the can with height 4.01 inches is

Step 6: Answer the Question The height of the can is 4.01 inches rounded to twodecimal places.

= 46.5 square inches.

= 2p11.37522 + 2p11.375214.012 S = 2pr2 + 2prh

h = 4.01

46.5 - 2p11.37522

2p11.3752 = h2p11.3752:

46.5 - 2p11.37522 = 2p11.3752h 46.5 = 2p11.37522 + 2p11.3752h2p11.37522


Building SkillsIn Problems 17–20, translate the verbal description into a mathematical formula.

17. Force F equals the product of mass m and accelera-tion a.

18. The area A of a triangle is one-half the product of its

base b and its height h.

19. The volume V of a sphere is four-thirds the product ofthe number and the cube of its radius r.

20. The revenue R of selling computers is $800 times thenumber of computers sold x.

In Problems 21–32, solve the formula for the indicated variable.See Objective 1.

21. Uniform Motion Solve for r.

22. Direct Variation Solve for k.y = kx

d = rt

p

23. Algebra Solve for m.

24. Algebra Solve for m.

25. Statistics Solve for x.

26. Statistics Solve for

27. Newton’s Law of Gravitation Solve

for

28. Sequences Solve for S.

29. Finance Solve for P.

30. Bernoulli’s Equation Solve

for r.

p +12

rv2 + rgy = a

A = P + Prt

S - rS = a - ar5

m1 .

F = G

m1m2

r2

1n .E =Z # s 1n

Z =x - m

s

y = mx + b

y - y1 = m1x - x1 2


45. Finance The formula can be used to relate the future value A of a deposit of P dollarsin an account that earns an annual interest rate r(expressed as a decimal) after t years.

(a) Solve the formula for P.

(b) How much would you have to deposit today inorder to have $5000 in 5 years in a bank accountthat pays 4% annual interest?

46. Federal Taxes According to the tax code in 2007, amarried couple filing a joint income tax return thatearns over $234,600 per year is subject to having theiritemized deductions reduced. The formula

can be used to deter-mine the permitted deductions P where D representsthe amount of deductions from Schedule A and Irepresent’s the couple’s adjusted gross income.

(a) Solve the formula for I.

(b) Determine the adjusted gross income of amarried couple filing a joint return whoseallowed deductions were $24,192 and Schedule Adeductions were $25,000.

47. Supplementary Angles Two angles are supplementaryif the sum of the measures of the angles is 180°. If oneangle is 30° more than its supplement, find the mea-sures of the two angles.

48. Supplementary Angles See Problem 47. If one angleis twice the measure of its supplement, find the mea-sures of the two angles.

49. Complementary Angles Two angles are complementaryif the sum of the measures of the angles is 90°. If oneangle is more than 3 times its complement, find themeasures of the two angles.

50. Complementary Angles See Problem 49. If one angleis 30° less than twice its complement, find the mea-sures of the two angles.

51. Dimensions of a Window The perimeter of a rectan-gular window is 26 feet.The width of the window is3 feet more than the length.What are the dimensionsof the window?

10°

P = D - 0.021I - 234,6002

A = P11 + r2t 31. Temperature Conversion Solve

for F.

32. Trapezoid Solve for b.

In Problems 33–40, solve for y.

A =12h1B + b2

C =591F - 322

x�y�

x� � y� � 180�

33. 34. -4x + y = 122x + y = 13

35. 36. 4x + 2y = 209x - 3y = 15

37. 38. 5x - 6y = 184x + 3y = 13

39. 40.23x -

52y = 5

12x +

16y = 2

Applying the Concepts41. Cylinders The volume V of a right circular cylinder is

given by the formula where r is the radiusand h is the height.


(b) Find the height of a right circular cylinder whosevolume is cubic inches and whose radius is 2 inches.

42. Cylinders The surface area S of a right circular cylin-der is given by the formula wherer is the radius and h is the height.


(b) Determine the height of a right circular cylinderwhose surface area is square centimeters andwhose radius is 4 centimeters.

43. Maximum Heart Rate The modelwas developed by Londeree and Moeschberger

to determine the maximum heart rate M of an indi-vidual who is age A. (SOURCE: Londeree and Moeschberger,

“Effect of Age and Other Factors on HR max,” Research Quarterly

for Exercise and Sport, 53(4), 297–304)

(a) Solve the model for A.

(b) According to this model, what is the age ofan individual whose maximum heart rate is 160?

44. Maximum Heart Rate The modelwas developed by Miller to determine the maximumheart rate M of an individual who is age A. (SOURCE:

Miller et al., “Predicting Max HR,” Medicine and Science in Sports

and Exercise, 25(9), 1077–1081)

(a) Solve the model for A.

(b) According to this model, what is the age ofan individual whose maximum heart rate is160?

M = -0.85A + 217

206.3M = -0.711A +

72p

S = 2prh + 2pr2 ,

32p

V = pr2h,

x� y�

x� � y� � 90�


52. Dimensions of a Window The perimeter of a rectan-gular window is 120 inches. The length of the windowis twice its width. What are the dimensions of thewindow?

53. Art An artist wants to place a piece of roundstained glass into a square that is made of copperwire. See the figure. If the perimeter of the square is40 inches, what is the area of the largest circular pieceof stained glass that can fit into the copper square?

54. Art An artist wants to place two circular pieces ofstained glass into a rectangle that is made of copperwire. See the figure. The perimeter of the rectangle is36 centimeters and the length is twice the width. Findthe area of each circle assuming they are to be aslarge as possible and still fit inside the rectangle.

55. Angles in a Triangle The sum of themeasure of the interior angles in atriangle is 180°. The measure of thesecond angle is 15° more than themeasure of the first angle. The mea-sure of the third angle is 45° morethan the measure of the first angle.Find the measures of the interior angles in thetriangle.

56. Angles in a Triangle See Problem 55. The measure ofthe second angle is 3 times the measure of the firstangle. The measure of the third angle is 20° more thanthe measure of the first angle. Find the measures ofthe interior angles in the triangle.

57. Designing a Patio Suppose that you wish to build arectangular cement patio that is to have a perimeter of80 feet. The length is to be 5 feet more than the width.

(a) Find the dimensions of the patio.(b) If the building code requires that the cement be

4 inches deep, how much cement do you have topurchase?

25 ft

Water

Deck

58. Designing a Foundation You have just purchased acircular gazebo whose diameter is 12 feet. Before youhave the gazebo delivered, you must lay a cementfoundation to place the gazebo on.

(a) Find the area of the base of the gazebo.(b) If the building code requires that the cement be

4 inches deep, how much cement do you have topurchase?

Extending the Concepts59. Critical Thinking Suppose that you have just purchased

a swimming pool whose diameter is 25 feet.You decideto build a deck around the pool that is 3 feet wide.

(a) What is the area of the deck?

8 ft

3 ft

(b) The building code requires that the pool must beenclosed in a fence. How much fence is requiredto encircle the pool?

(c) If the fence costs $25 per linear foot to install,how much will the fence cost?

60. Critical Thinking Suppose that you wish to install awindow whose dimensions are given in the figure.

z�x�

y�

x� � y� � z� � 180�

(a) What is the area of the opening of the window?

(b) What is the perimeter of the window?(c) If glass costs $8.25 per square foot, what is the

cost of the glass for the window?

61. If the radius of a circle is doubled, does the areadouble? Explain. If the length of the side of a cube isdoubled, what happens to the volume?

Section 1.4 Linear Inequalities in One Variable 81

1.4 Linear Inequalities in One VariablePreparing for Linear Inequalities in One VariableBefore getting started, take the following readiness quiz. If you get a problem wrong, go back to thesection cited and review the material.

In Problems P1–P4, replace the question mark by or to make the statement true.[Section R.2, pp. 16–17]

P1. 3 ? 6 P2. P3. P4.

P5. True or False: The inequality is called a strict inequality. [Section R.2, p. 17]P6. Use set-builder notation to represent the set

of all digits that are divisible by 3. [Section R.2, pp. 8–9]

Ú

23

? 35

12

? 0.5-3 ? -6

=6 , 7 ,

OBJECTIVES1 Represent Inequalities Using the

Real Number Line and IntervalNotation

2 Understand the Properties ofInequalities

3 Solve Linear Inequalities

4 Solve Problems Involving LinearInequalities

An inequality in one variable is a statement involving two expressions, at least one con-taining the variable, separated by one of the inequality symbols or Tosolve an inequality means to find all values of the variable for which the statement istrue. These values are called solutions of the inequality. The set of all solutions is calledthe solution set.

Ú .6 , … , 7 ,

Preparing for...Answers P1. P2.P3. P4. P5. False P6. isa digit that is divisible by 36

5x | x7=76

DEFINITIONA linear inequality in one variable is an inequality that can be written in the form

where a, b, and c are real numbers and a Z 0.

ax + b 6 c or ax + b … c or ax + b 7 c or ax + b Ú c

For example, the following are all linear inequalities involving one variable:

Before we discuss methods for solving linear inequalities, we will present three ways ofrepresenting the solution set. One of the methods for representing the solution set isthrough set-builder notation—something we are already familiar with. However, set-builder notation can be somewhat cumbersome, so we introduce a more streamlinedway to represent a solution set to an inequality using interval notation. Finally, becausewe often like to visualize solution sets, we present a method for graphing the solutionset on a real number line.

x - 4 7 9 5x - 1 … 14 8z 6 0 5x - 1 Ú 3x + 8

1 Represent Inequalities Using the Real Number Line and Interval Notation

Suppose that a and b are two real numbers and We shall use the notation

to mean that x is a number between a and b. So, the expression is equivalentto the two inequalities and Similarly, the expression is equiv-alent to the two inequalities and We define and similarly. Expressions such as or are said to be in inequality notation.

While the expression is technically correct, it is not the preferred way towrite an inequality. For ease of reading, we prefer that the numbers in the inequality gofrom smaller values to larger values. So, we would write as

A statement such as is false because there is no number x for whichand We also never mix inequalities as in

In addition to representing inequalities using inequality notation, we can useinterval notation.

2 … x Ú 3.x … 1.3 … x3 … x … 1

2 … x … 3.3 Ú x Ú 2

3 Ú x Ú 2

x Ú 5-2 6 x 6 5a 6 x … ba … x 6 bx … b.a … x

a … x … bx 6 b.a 6 xa 6 x 6 b

a 6 x 6 b

a 6 b.

Work SmartRemember that the inequalitiesand are called strict inequalities,while and are called nonstrictinequalities.

Ú…7

6


DEFINITION INTERVAL NOTATIONLet a and b represent two real numbers with

A closed interval, denoted by [a, b], consists of all real numbers x for which

An open interval, denoted by (a, b), consists of all real numbers x for which

The half-open, or half-closed, intervals are (a, b], consisting of all real numbers x forwhich and [a, b), consisting of all real numbers x for which

In each of these definitions, a is called the left endpoint and b is called the rightendpoint of the interval.

a … x 6 b.a 6 x … b,

a 6 x 6 b.

a … x … b.

a 6 b.

The symbol (read as “infinity”) is not a real number, but a notational device usedto indicate unboundedness in the positive direction. In other words, the symbol means that there is no right endpoint on the inequality. The symbol (read as“minus infinity” or “negative infinity”) also is not a real number, but a notational de-vice used to indicate unboundedness in the negative direction. The symbol meansthat there is no left endpoint on the inequality. Using the symbols and we candefine five other kinds of intervals.

-q ,q-q

-qq

q

In addition to representing inequalities using interval notation, we can represent in-equalities using a graph on the real number line. The inequality or the interval(3, ) consists of all numbers x that lie to the right of 3 on the real number line.We canrepresent these values by shading the real number line to the right of 3.To indicate that 3is not included in the set, we will agree to use a parenthesis on the endpoint. See Figure 9.

To represent the inequality or the interval graphically, we also shadeto the right of 3, but this time we use a bracket on the endpoint to indicate that 3 isincluded in the set. See Figure 10.

Table 7 summarizes interval notation, inequality notation, and their graphs.

[3, q2x Ú 3

qx 7 3

Figure 9x � 3

1 4 5320�1

1 4 5320�1

Figure 10x 3Ú

INTERVALS INCLUDING

consists of all real numbers x for which consists of all real numbers x for which consists of all real numbers x for which consists of all real numbers x for which consists of all real numbers x (or )-q 6 x 6 q1-q , q2

x 6 a1-q , a2x … a1-q , a]x 7 a1a, q2x Ú a[a, q2

ˆWork SmartThe symbols and are neverincluded as endpoints because theyare not real numbers. So, we useparentheses when or areendpoints.

q-q

-qq

Table 7

Interval Notation Inequality Notation Graph

The open interval (a, b)

The closed interval [a, b]

The half-open interval [a, b)

The half-open interval (a, b]

The interval

The interval

The interval

The interval

The interval 5x | x is a real number6(-q , q )

5x | x 6 a6(-q , a)

5x | x … a6(-q , a]

5x | x 7 a6(a, q )

5x | x Ú a6[a, q )

5x | a 6 x … b65x | a … x 6 b65x | a … x … b65x | a 6 x 6 b6

a

a

a

a

a b

a b

ba

ba


EXAMPLE 1 Using Interval Notation and Graphing InequalitiesWrite each inequality using interval notation. Graph the inequality.

(a) (b)

Solution

(a) describes all numbers x between and 4, inclusive. In in-terval notation, we write To graph we place brack-ets at and 4 and shade in between. See Figure 11.-2

-2 … x … 4,[-2, 4].-2-2 … x … 4

1 6 x … 5-2 … x … 4

Figure 11

1 4 5320�3 �2 �1

1 4 5 6320�1

Figure 12

EXAMPLE 2 Using Interval Notation and Graphing InequalitiesWrite each inequality using interval notation. Graph the inequality.

(a) (b) x Ú -3x 6 2

Solution

(a) describes all numbers x less than 2. In interval notation, we writeTo graph we place a parenthesis at 2 and then shade to

the left. See Figure 13.x 6 2,1-q , 22.x 6 2

1 4320�1�2Figure 13

Figure 14

(b) describes all numbers x greater than or equal to In intervalnotation, we write To graph we place a bracket at and then shade to the right. See Figure 14.

-3x Ú -3,[-3, q2. -3.x Ú -3

Quick 1. A(n) , denoted [a, b], consists of all real numbers x for which

2. In the interval (a, b), a is called the and b is called the of the interval.

In Problems 3–6, write each inequality in interval notation. Graph the inequality.

3. 4.

5. 6.12

6 x 672

x … 3

3 … x 6 6-3 … x … 2

a … x … b.

1 320�1�2�3�4

(b) describes all numbers x greater than 1 and less than orequal to 5. In interval notation, we write (1, 5]. To graph weplace a parenthesis at 1 and a bracket at 5 and shade in between. See Figure 12.

1 6 x … 5,1 6 x … 5


Figure 15

1 3 4 520�1�2�3

Figure 16

EXAMPLE 4 Using Inequality Notation and Graphing InequalitiesWrite each interval in inequality notation involving x. Graph the inequality.

(a) (b)

Solution

(a) The interval consists of all numbers x for which

See Figure 17 for the graph.

(b) The interval consists of all numbers x for which SeeFigure 18 for the graph.

x 6 1.1-q , 12

x Ú32

.c32

, q b

1-q , 12c32

, q b

Figure 17 Figure 18

1 4320�1�2�3�41 4 5 6 7320�1 3__2

Quick In Problems 7–10, write each interval as an inequality. Graph the inequality.

7. (0, 5] 8.

9. 10. a -q , 83

d15, q21-6, 02

2 Understand the Properties of InequalitiesConsider the inequality If we add 3 to both sides of the inequality, the expressionon the left becomes 5 and the expression on the right becomes 8. Since we can seethat adding the same quantity to both sides of an inequality does not change the sense,or direction, of the inequality.This result is called the Addition Property of Inequalities.

5 6 8,2 6 5.

In WordsThe Addition Property states that thedirection of the inequality does not changewhen the same quantity is added to each sideof the inequality.

ADDITION PROPERTY OF INEQUALITIES

For real numbers a, b, and c,

If a 6 b, then a + c 6 b + c If a 7 b, then a + c 7 b + c

For example, since we have that or In addition, sincewe have that or

Because is equivalent to , the Addition Property of Inequalities canalso be used to subtract a real number from each side of an inequality without chang-ing the direction of the inequality.

a + (-b)a - b1 7 -3.3 + 1-22 7 -1 + 1-223 7 -1,

6 6 9.2 + 4 6 5 + 42 6 5,

1 3 4 5 6 720�1

EXAMPLE 3 Using Inequality Notation and Graphing InequalitiesWrite each interval in inequality notation involving x. Graph the inequality.

(a) (b) (1, 5)

Solution

(a) The interval consists of all numbers x for which See Figure 15 for the graph.

(b) The interval (1, 5) consists of all numbers x for which SeeFigure 16 for the graph.

1 6 x 6 5.

-2 … x 6 4.[-2, 42

[-2, 42


We’ve seen what happens when we add a real number to both sides of an inequality.What happens when we multiply both sides by a nonzero constant? Let’s see.

Consider the inequality Multiply both sides of the inequality by 2. The ex-pression on the left side of the inequality becomes and the expression on theright becomes Certainly so the direction of the inequality did notchange.

Again consider the inequality Now multiply both sides of the inequality byThe expression on the left side of the inequality becomes and the

expression on the right becomes Because we see that thedirection of the inequality is reversed.

These results are true in general and lead us to the Multiplication Properties ofInequalities.

-6 7 -10-2152 = -10.-2132 = -6-2.

3 6 5.

6 6 10,2152 = 10.2132 = 6

3 6 5.

In WordsThe Multiplication Property states that if bothsides of an inequality are multiplied by apositive real number, the direction of theinequality is unchanged. If both sides ofan inequality are multiplied by a negativereal number, the direction of the inequality isreversed.

MULTIPLICATION PROPERTIES OF INEQUALITIES

Let a, b, and c be real numbers.

If a 7 b and if c 6 0, then ac 6 bc If a 6 b and if c 6 0, then ac 7 bc

If a 7 b and if c 7 0, then ac 7 bc If a 6 b and if c 7 0, then ac 6 bc

Quick 11. Write the inequality that results by adding 5 to each side of the inequality .

What property of inequalities does this illustrate?

12. Write the inequality that results by subtracting 3 from each side of the inequal-ity . What property of inequalities does this illustrate?

13. Write the inequality that results by multiplying both sides of the inequality

by . What property of inequalities does this illustrate?

14. Write the inequality that results by dividing both sides of the inequality by . What property of inequalities does this illustrate?

15. Write the inequality that results by dividing both sides of the inequality by 5. What property of inequalities does this illustrate?5x 6 30

-3-6 6 9

12

2 6 8

x + 3 7 -6

4 6 7

Because the quotient is equivalent to , the Multiplication Properties of Inequalities

can also be used to divide both sides of an inequality by a nonzero real number.

a # 1b

a

b

3 Solve Linear InequalitiesTwo inequalities that have exactly the same solution set are called equivalentinequalities. As with equations, one method for solving a linear inequality is to replaceit by a series of equivalent inequalities until an inequality with an obvious solution,such as is obtained. We obtain equivalent inequalities by applying some of thesame operations as those used to find equivalent equations.The Addition Property andMultiplication Properties form the basis for the procedures.

Although not essential, it is easier to read an inequality if the variable is placed onthe left side and the constant on the right. If the variable does end up on the right sideof the inequality, we can rewrite it with the variable on the left side using the fact that

and a 7 x is equivalent to x 6 a

a 6 x is equivalent to x 7 a

x 7 2,

In WordsIf the sides of an inequality are interchanged,the direction of the inequality reverses.


The solution using set-builder notation is The solution using interval no-tation is Figure 19 shows the graph of the solution set.15, q2. 5x | x 7 56.

EXAMPLE 5 How to Solve a Linear InequalitySolve the inequality Graph the solution set.


The goal in solving any linear inequality is to get the variable by itself with a coefficient of 1.

3x - 2 7 13.

Step 1: Isolate the term containing thevariable.

Add 2 to both sides (Addition Property):

3x 7 15

3x - 2 + 2 7 13 + 2

3x - 2 7 13

Step 2: Get a coefficient of 1 on thevariable.

Divide both sides by 3 (Multiplication Property):

x 7 5

3x3

7153

Figure 19

1 4 5 6 7320�1

Quick In Problems 16–19, solve each linear inequality. Express your solutionusing set-builder notation and interval notation. Graph the solution set.

EXAMPLE 6 Solving Linear InequalitiesSolve the inequality:

Solution

x … -4

-4x-4

…16-4

-4x Ú 16

x - 5x Ú 5x + 16 - 5x

x Ú 5x + 16

x - 4 + 4 Ú 5x + 12 + 4

x - 4 Ú 5x + 12

x - 4 Ú 5x + 12



Divide both sides by �4. Don’t forget tochange the direction of the inequality:

The solution using set-builder notation is The solution using intervalnotation is See Figure 20 for the graph of the solution set.1-q , -4].

5x | x … -46.Figure 20

10�1�2�3�4�5�6�7


20. 21.

22. -5x + 12 6 x - 3

-2x + 1 … 3x + 113x + 1 7 x - 5

16. 17.

18. 19. -2x + 1 … 134x - 3 6 13

13x … 2x + 3 7 5


EXAMPLE 7 Solving Linear InequalitiesSolve the inequality:

Solution

Distribute the 3:Combine like terms:Add 3 to both sides:


Multiply both sides of the inequality by

The solution using set-builder notation is The solution using intervalnotation is See Figure 21 for the graph of the solution set.1-6, q2. 5x | x 7 -66.

x 7 -6

1-121-x2 7 1-126-1: -x 6 6

5x - 6x 6 6x + 6 - 6x

5x 6 6x + 6

5x - 3 + 3 6 6x + 3 + 3

5x - 3 6 6x + 3

3x - 3 + 2x 6 6x + 3

31x - 12 + 2x 6 6x + 3

31x - 12 + 2x 6 6x + 3

Figure 2110�1�2�3�4�5�6�7


EXAMPLE 8 Solving Linear Inequalities Involving Fractions

Solve the inequality:

Solution

We begin by rewriting the linear inequality without fractions by multiplying bothsides of the inequality by 6, the Least Common Denominator.

Distribute:Subtract 2 from both sides:


The solution using set-builder notation is The solution using intervalnotation is See Figure 22 for the graph of the solution set.1-8, q2. 5x | x 7 -86.

x 7 -8

4x - 3x 7 3x - 8 - 3x

4x 7 3x - 8

4x + 2 - 2 7 3x - 6 - 2

4x + 2 7 3x - 6

212x + 12 7 31x - 22 6 # a2x + 1

3 b 7 6 # ax - 2

2 b

2x + 1 3

7x - 2

2

Figure 2280 2 4 6�2�4�6�8�10

23.

24.

25. 7 - 21x + 12 … 31x - 52-21x + 12 Ú 41x + 3241x - 22 6 3x - 4


When solving applications involving linear inequalities, we use the same steps forsetting up applied problems that we introduced in Section 1.2 on page 61.

4 Solve Problems Involving Linear InequalitiesWhen you are confronted with a word problem, one of the first things that you need todo is look for key words that tip you off as to the type of word problem that it is. Thereare certain phrases that frequently occur in problems that lead to linear inequalities.We list some of these phrases for you in Table 8.

Table 8Phrase Inequality Phrase Inequality

At least No more than

No less than At most

More than Fewer than

Greater than Less than 67

67

…Ú

…Ú

EXAMPLE 9 Comparing Credit CardsChase Bank has offered you two different credit card options. The Southwest rewardscard charges an annual fee of $39 plus 12.90% simple interest on all outstanding bal-ances. The Marriott rewards card charges an annual fee of $30 plus 14.15% simple in-terest on all outstanding balances. What annual balance results in the Southwest cardcosting less than the Marriott card? (SOURCE: chase.com)

Solution

Step 1: Identify We want to know the credit card balance for which the Southwestcredit card costs less than the Marriott rewards card. The phrase “costs less” impliesthat this is an inequality problem.

Step 2: Name Let b represent the credit card balance on each card.

Step 3: Translate Each card charges an annual fee plus simple interest. So, for eachcard the cost will be “annual fee � interest.”

Recall from Section 1.2 that simple interest is found using the formula

where I is the interest charged, P is the balance on the credit card, r is the annual interest rate, and t is time.

In this problem, we let b represent the credit card balance. The annual interestrate r will either be 0.129 (for Southwest) or 0.1415 (for Marriott). Because we arediscussing annual cost, we have that t � 1.

I = Prt


26. 27.

28.12

1x + 32 713

1x - 42

25

x +3

106

12

3x + 1

5 Ú 2


Since we want to know what balance results in Southwest costing less than Marriott,we have the following inequality:

Southwest Credit Card Credit Card CostCost 6 MarriottAnnual Fee

for SouthwestInterest Charged

for SouthwestAnnual Feefor Marriott

Interest Chargedfor Marriott

� � The Model 0.1415b2

30#

0.129b4

39#

Step 4: Solve Solve the inequality for b.

b 7 720

-0.0125b 6 -9

0.129b 6 -9 + 0.1415b

39 + 0.129b 6 30 + 0.1415b

Subtract 39 from both sides:Subtract 0.1415b from both sides:

Divide both sides by �0.0125.Don’t forget to reverse the

inequality symbol:

Step 5: Check If the balance is $720, then the annual cost for Southwest is 39 �0.129(720) � $131.88. The annual cost for Marriott is 30 � 0.1415(720) � $131.88.For a balance greater than $720, say $750, the annual cost for Southwest is 39 �0.129(750) � $135.75. The annual cost for Marriott is 30 � 0.1415(750) � $136.13.

Step 6: Answer the Question If the annual balance is greater than $720, then theSouthwest card offers a better deal than the Marriott card.

Quick 29. You have just received two credit card applications in the mail. The card from

Bank A has an annual fee of $25 and charges 9.9% simple interest. The cardfrom Bank B has no annual fee, but charges 14.9% simple interest. For whatannual balance will the card from Bank A cost less than the card from Bank B?

30. Suppose the daily revenue from selling x boxes of candy is given by the equa-tion The daily cost of operating the store and making the candy isgiven by the equation For how many boxes of candy will revenueexceed costs? That is, solve R 7 C.

C = 8x + 96.R = 12x.

1–30. are the Quick s that follow each EXAMPLE

Building SkillsIn Problems 31–38,write each inequality using interval notation.Graph the inequality.See Objective 1.

31. 32.

33. 34.

35. 36. x 6 0x Ú 6

-8 6 x … 1-4 … x 6 0

1 6 x 6 72 … x … 10

1.4 EXERCISES

37. 38.

In Problems 39–46, write each interval as an inequality involving x.Graph each inequality. See Objective 1.

39. (1, 8) 40.

41. 42.

43. 44.

45. 46. (-q , 8][3, q)

(2, q)(-q , 5)

[1, 4)(-5, 1]

[-2, 3]

x Ú -52

x 632


In Problems 47–54, fill in the blank with the correct inequalitysymbol. State which property of inequalities is being utilized. SeeObjective 2.

47. If , then 10.

48. If , then 11.

49. If , then 15.

50. If , then 9.

51. If , then 4.

52. If , then 2.

53. If , then .

54. If , then .

In Problems 55–92, solve each linear inequality. Express your solu-tion using set-builder notation and interval notation. Graph thesolution set.See Objective 3.

55. 56.

57. 58.

59. 60.

61. 62.

63. 64.

65.

66.

67.

68.

69.

70.

71.

72.

73.

74.

75.

76.

77.3x + 1

4 6

12

2.3x - 1.2 7 1.8x + 0.4

0.5x + 4 … 0.2x - 5

-31x + 42 + 5x 6 41x + 32 - 14

41x + 12 - 2x Ú 51x - 22 + 2

31x - 22 + 5 7 41x + 12 + x

31x - 32 6 21x + 423x + 4 Ú 5x - 8

-3x + 1 6 2x + 11

8x + 3 Ú 5x - 9

6x + 5 … 3x + 2

-6x - 5 6 13

-3x + 1 7 13

5x - 4 … 163x + 2 7 11

38x 6

916

4

15 x 7

85

-8x 7 32-7x 6 21

4x Ú 206x 6 24

x + 6 6 9x - 4 … 2

-5x-6x 6 30

-5x-2x Ú 10

25x

25x + 6 … 8

2x2x + 5 … 9

x4x 7 36

x13x 7 5

2x2x - 5 7 6

xx - 3 6 7

78.

79.

80.

81.

82.

83.

84.

85.

86.

87.

88.

89.

90.

91.

92.

Mixed PracticeIn Problems 93–106, solve each linear inequality. Express your solution using set-builder notation and interval notation. Graph thesolution set.

93. 94.

95. 96.

97.

98.

99.

100. 51y + 72 6 61y + 4241x + 22 … 31x - 224x + 3 Ú -6x - 2

13x - 5 7 10x - 6

-5x 6 30-3a … -21

y - 5 Ú 7y + 8 7 -7

x

12 Úx

2 -

2x + 14

23x 6

1412x + 32

25x +

310

612

4x - 33

6 3

5613x - 22 -

23

14x - 12 6 -29

12x + 52

23

14x - 12 -49

1x - 42 75

12 12x + 32

71x + 22 - 412x + 32 6 -2[5x - 21x + 32] + 7x

413x - 12 - 51x + 42 Ú 3[2 - 1x + 32] - 6x

-312x + 12 … 2[3x - 21x - 52]-51x - 32 Ú 3[4 - 1x + 42]

23

-56

x 7 2

35

- x 753

1313x + 52 6

161x + 42

121x - 42 7

3412x + 12

2x - 3 3

743


115. Payload Restrictions An Airbus A320 has a maximum payload of 45,686 pounds. Suppose that on a flight from Chicago to New Orleans,United Airlines sold out the flight at 179 seats.Assuming that the average passenger weighs150 pounds, determine the weight of the luggageand other cargo that the plane can carry. (SOURCE:

Airbus)

116. Moving Trucks A 15-foot moving truck from Bud-get costs $39.95 per day plus $0.65 per mile. If yourbudget only allows for you to spend at most $125.75,what is the number of miles you can drive? (SOURCE:

Budget)

117. Health Benefits The average monthly benefit B, indollars, for individuals on disability is given by theequation where t is the num-ber of years since 1990. In what year will the monthlybenefit B exceed $1000? That is, solve

118. Health Expenditures Total private health expendi-tures H, in billions of dollars, are given by the equa-tion where t is the number of yearssince 1990. In what year will total private health ex-penditures exceed $1 trillion ($1000 billion)? That is,solve

119. Commissions Susan sells computer systems. Her annual base salary is $34,000. She also earns a com-mission of 1.2% on the sale price of all computer systems that she sells. For what value of the com-puter systems sold will Susan’s annual salary be atleast $100,000?

120. Commissions Al sells used cars for a Chevy dealer.His annual base salary is $24,300. He also earns acommission of 3% on the sale price of the cars thathe sells. For what value of the cars sold will Al’s an-nual salary be more than $60,000? If used cars at thisparticular dealership sell for an average of $15,000,how many cars does Al have to sell to meet hissalary goal?

121. Supply and Demand The quantity demanded ofcustom monogrammed shirts is given by the equa-tion The quantity supplied of cus-tom monogrammed shirts is given by the equation

where p is the price of a shirt. Forwhat prices will quantity supplied exceed quantitydemanded, thereby resulting in a surplus of shirts?That is, solve

122. Supply and Demand The quantity supplied of digital cameras is given by the equation

. The quantity demanded of digitalcameras is given by the equation where p is the price of a camera. For what prices willquantity demanded exceed quantity supplied,

D = 1800 - 12p,S = -2800 + 13p

S 7 D.

S = -200 + 10p,

1000 - 20p.D =

26t + 411 7 1000.

H = 26t + 411,

B 7 1000.

B = 19.25t + 585.72,

101.

102.

103.

104.

105.

106.

Applying the Concepts107. Find the set of all x such that the sum of twice x and

5 is at least 13.

108. Find the set of all x such that the difference between3 times x and 2 is less than 7.

109. Find the set of all z such that the product of 4 and zminus 3 is no more than 9.

110. Find the set of all y such that the sum of twice y and3 is greater than 13.

111. Computing Grades In order to earn an A in Mr.Ruffatto’s Intermediate Algebra course, Jackiemust earn at least 540 points.Thus far, Jackie hasearned 90, 83, 95, and 90 points on her four exams.The final exam, which counts as 200 points, is rapidlyapproaching. How many points does Jackie needto earn on the final to earn an A in Mr. Ruffatto’sclass?

112. Computing Grades In order to earn anA in Mrs.Padilla’s IntermediateAlgebra course,Mark mustobtain an average score of at least 90.On his first fourexams Mark scored 94,83,88,and 92.The final examcounts as two test scores.What score does Mark needon the final to earn anA in Mrs.Padilla’s class?

113. McDonald’s Suppose that you have ordered onemedium order of French Fries and one 16-ouncetriple-thick chocolate shake from McDonald’s.Thefries have 22 grams of fat and the shake has 17 gramsof fat. Each McDonald’s hamburger has 10 grams offat. How many hamburgers can you order and stillkeep the total fat content of the meal to no more than 69grams? (SOURCE: McDonald’s Corporation)

114. Burger King Suppose that you have ordered onemedium onion ring and one 16-ounce chocolateshake from Burger King. The onion rings have 16 grams of fat and the shake has 8 grams of fat.Each cheeseburger has 21 grams of fat. How manycheeseburgers can you order and still keep the totalfat content of the meal to no more than 87 grams?(SOURCE: Burger King)

b

3 +

56

61112

x

2 +

34

Ú38

3[1 + 21x - 42] Ú 3x + 3

2[4 - 31x + 12] … -4x + 8

215 - x2 - 3 … 4 - 5x

314 - 3x2 7 6 - 5x


thereby resulting in a shortage of cameras? That is,solve

Extending the Concepts123. Solve the linear inequality

124. Solve the linear inequality-31x - 22 + 7x 7 212x + 52.

31x + 22 + 2x 7 51x + 12.

D 7 S.Explaining the Concepts

travels at 45 miles per hour. How long will it takethe two cars to be 255 miles apart?

9. Solve for y.

10. Solve the formula for r.

11. The volume of a right circular cylinder is given bythe formula where r is the radius of thecylinder and h is the height of the cylinder.


(b) Use the result from part (a) to find the heightof a right circular cylinder with volume

and radius

12. Write the following inequalities in interval nota-tion. Graph the inequality.

r = 7 inches.V = 294p inches3

V = pr2h,

A = P + Prt

3x - 2y = 4

These problems cover important concepts from Sections 1.1through 1.4. We designed these problems so that you can re-view the chapter so far and show your mastery of the con-cepts. Take time to work these problems before proceedingwith the next section. The answers to these problems are lo-cated at the back of the text on page AN-4.

1. Determine which, if any, of the following aresolutions to

(a) (b)

In Problems 2 and 3, solve the equation.

2.

3.

Determine if the equation is an identity, a contradiction, or aconditional equation.

4.

In Problems 5 and 6, translate the English statement into amathematical statement. Do not solve the equation.

5. The difference of a number and 3 is two more than half the number.

6. The quotient of a number and 2 is less than the number increased by 5.

7. Mixture Two acid solutions are available to achemist. One is a 20% nitric acid solution and theother is a 40% nitric acid solution. How much ofeach type of solution should be mixed together toform 16 liters of a 35% nitric acid solution?

8. Travel Two cars leave from the same location andtravel in opposite directions along a straight road.One car travels 30 miles per hour while the other

5 - 21x + 12 + 4x = 61x + 12 - 13 + 4x2

73

x +45

=5x + 12

15

312x - 12 + 6 = 5x - 2

x = 1x = -3

512x - 32 + 1 = 2x - 6

PUTTING THE CONCEPTS TOGETHER (SECTIONS 1.1–1.4)

125. Write a brief paragraph that explains the circum-stances under which the direction, or sense, of aninequality changes.

126. Explain why the inequality is false.

127. Explain why we never mix inequalities as in4 6 x 7 7.

5 6 x 6 1

(a)

(b)

13. Write the interval in inequality notation involvingx. Graph the inequality.

(a)

(b)

In Problems 14–16, solve the inequality and graph thesolution set on a real number line.

14.

15.

16.

17. Birthday Party A recreational center offers a chil-dren’s birthday party for $75 plus $5 for each child.How many children can Logan invite to his birth-day party if the budget for the party is no morethan $125?

x - 9 … x + 312 - x2-3 7 3x - 1x + 522x + 3 … 4x - 9

1-3, 1]

1-q , -1.5]

2 6 x … 5

x 7 -3

Section 1.5 Rectangular Coordinates and Graphs of Equations 93

PART I I : L INEAR EQUATIONS AND INEQUALITIES IN TWO VARIABLES

Preparing for Rectangular Coordinates and Graphs of EquationsBefore getting started, take this readiness quiz. If you get a problem wrong, go back to the sectioncited and review the material.

P1. Plot the following points on the real

number line: [Section R.2, pp. 15–16]

P2. Determine which of the following are solutionsto the equation (a) (b) (c) [Section 1.1, pp. 48–49]

P3. Evaluate the expression for thegiven values of the variable.(a) (b) (c) [Section R.5, pp. 40–41]

P4. Solve the equation for y. [Section 1.3, pp. 73–76]P5. Evaluate [Section R.3, pp. 19–20]ƒ -4 ƒ .

3x + 2y = 8x = -3x = 2x = 0

2x2 - 3x + 1x = -7x = -3x = 0

3x - 51x + 22 = 4.

-2, 4, 0, 12

.

OBJECTIVES1 Plot Points in the Rectangular

Coordinate System

2 Determine Whether an OrderedPair Is a Point on the Graph ofan Equation

3 Graph an Equation Using thePoint-Plotting Method

4 Identify the Intercepts from theGraph of an Equation

5 Interpret Graphs

1 Plot Points in the Rectangular Coordinate SystemRecall from Section R.2 that we locate a point on the real number line by assigning it asingle real number, called the coordinate of the point. See Preparing for Problem P1.When we graph a point on the real number line, we are working in one dimension.When we wish to work in two dimensions, we locate a point using two real numbers.

We begin by drawing two real number lines that intersect at right (90°) angles. One ofthe real number lines is drawn horizontal, while the other is drawn vertical. We call thehorizontal real number line the x-axis, and the vertical real number line is the y-axis.The point where the x-axis and y-axis intersect is called the origin, O. See Figure 23.

�1�2�3�4 1 2 3 4 x

4

3

2

1

�1

�2

�3

�4

O

yFigure 23

The origin O has a value of 0 on the x-axis and the y-axis. Points on the x-axis to theright of O are positive real numbers; points on the x-axis to the left of O are negativereal numbers. Points on the y-axis that are above O are positive real numbers; points onthe y-axis that are below O are negative real numbers. In Figure 23 we label the x-axis“x” and the y-axis “y.” Notice that an arrow is used at the end of each axis to denote thepositive direction. We do not use an arrow to denote the negative direction.

The coordinate system presented in Figure 23 is called a rectangular or Cartesian coordinate system, named after René Descartes (1596–1650), a French mathematician,philosopher, and theologian. The plane formed by the x-axis and y-axis is often re-ferred to as the xy-plane, and the x-axis and y-axis are called the coordinate axes.

We can represent any point P in the rectangular coordinate system by using anordered pair (x, y) of real numbers. If we travel x units to the right of the y-axis;x 7 0,

Preparing for...Answers

P1.

�1�2�3 0 1 2 3 4 5 61__2

P2. (a) No (b) No (c) Yes P3. (a) 1

(b) 3 (c) 28 P4. P5. 4y = -32x + 4

1.5 Rectangular Coordinates and Graphs of Equations


if we travel units to the left of the y-axis. If we travel y units above thex-axis; if we travel units below the x-axis.The ordered pair (x, y) is also calledthe coordinates of P.

ƒy ƒy 6 0,y 7 0,ƒx ƒx 6 0,

The origin O has coordinates (0, 0). Any point on the x-axis has coordinates of theform (x, 0), and any point on the y-axis has coordinates of the form (0, y).

If (x, y) are the coordinates of a point P, then x is called the x-coordinate, or abscissa,of P and y is called the y-coordinate or ordinate, of P.

If you look back at Figure 23, you should notice that the x- and y-axes divide theplane into four separate regions or quadrants. In quadrant I,both the x- and y-coordinateare positive; in quadrant II, x is negative and y is positive; in quadrant III, both x and yare negative; and in quadrant IV, x is positive and y is negative. Points on the coordi-nate axes do not belong to a quadrant. See Figure 24.

Figure 24

�1�2�3�4 1 2 3 4 x

4

3

2

1

�1

�2

�3

�4

O

y

Quadrant IIx 0, y � 0

Quadrant Ix � 0, y � 0

Quadrant IIIx 0, y 0

Quadrant IVx � 0, y 0

EXAMPLE 1 Plotting Points in the Rectangular Coordinate System andDetermining the Quadrant in which the Point Lies

Plot the points in the xy-plane. Tell which quadrant each point is in.

(a) (b) (c)

(d) (e)

Solution

Before we plot the points, we draw a rectangular or Cartesian coordinate system. SeeFigure 25(a). We now plot the points.

(a) To plot point from the origin O, we travel 3 units to the right andthen 2 units up. Label the point A. See Figure 25(b). Point A is in quadrant Ibecause both x and y are positive.

A13, 22,

E1-2, 02D13, -42C1-1, -32B1-2, 42A13, 22

B (�2, 4)

C (�1, �3)

E (�2, 0)

D (3, �4)

A (3, 2)

(b)

�1 1 2 3 4�2�3�4�1

2

1

O

3

4

y

x

�2

�3

�4

(a)

�1 1 2 3 4�2�3�4�1

2

1

O

3

4

y

x

�2

�3

�4

Figure 25


(b) To plot point from the origin O, we travel 2 units to the left andthen 4 units up. Label the point B. See Figure 25(b). Point B is in quadrant II.

(c) See Figure 25(b). Point C is in quadrant III.

(d) See Figure 25(b). Point D is in quadrant IV.

(e) See Figure 25(b). Point E is not in a quadrant because it lies on the x-axis.

B1-2, 42,

2 Determine Whether an Ordered Pair Is a Point on the Graph of an Equation

In Section 1.1, we solved linear equations in one variable. The solution was either asingle value of the variable, the empty set, or all real numbers. We will now look atequations in two variables. Our goal is to learn a method for representing the solutionto an equation in two variables.

Quick 1. The point where the x-axis and y-axis intersect in the Cartesian coordinate

system is called the .

2. True or False: If a point lies in quadrant III of the Cartesian coordinate system,then both x and y are negative.

In Problems 3 and 4, plot each point in the xy-plane. Tell in which quadrant or onwhich coordinate axis each point lies.

3. (a)

(b)

(c)

(d) D1-4, -32C10, -32B14, -22A15, 22

DEFINITIONAn equation in two variables, say x and y, is a statement in which the algebraic expres-sions involving x and y are equal.The expressions are called sides of the equation.

DEFINITIONThe graph of an equation in two variables x and y is the set of all points whosecoordinates, (x, y), in the xy-plane satisfy the equation.

Since an equation is a statement, it may be true or false, depending upon the values ofthe variables. Any values of the variable that make the equation a true statement aresaid to satisfy the equation.

For example, the following are all equations in two variables.

The first equation is satisfied when and since It isalso satisfied when and In fact, there are infinitely many choices of xand y that satisfy the equation However, there are some choices of x and ythat do not satisfy the equation For example, and does notsatisfy the equation because (that is, ).

When we find a value of x and y that satisfies an equation, it means that the orderedpair (x, y) represents a point on the graph of the equation.

9 Z 632 Z 4 + 2y = 4x = 3x2 = y + 2.

x2 = y + 2.y = 2.x = -2

32 = 7 + 2.y = 7x = 3x2 = y + 2

x2 = y + 2 3x + 2y = 6 y = -4x + 5

In WordsThe graph of an equation is a geometric wayof representing the set of all points that makethe equation a true statement.

4. (a)

(b)

(c)

(d) D16, 12C13, -22B1-4, 02A1-3, 22


Quick 5. True or False: The graph of an equation in two variables x and y is the set of all

points whose coordinates, (x, y), in the Cartesian plane satisfy the equation.

6. Determine if the following coordinates represent points that are on the graph of

(a) (b) (c)

7. Determine if the following coordinates represent points that are on the graph of

(a) (1, 4) (b) (c) 1-3, 1221-2, -12y = x2 + 3.

a32

, -94b12, -2212, -32

2x - 4y = 12.

EXAMPLE 2 Determining Whether a Point Is on the Graph of an Equation

Determine if the followingcoordinates representpoints thatareonthegraphof

(a) (2, 0) (b) (c)

Solution

(a) For (2, 0), we check to see if satisfies the equation

LetTrue

The statement is true when and so the point whose coordinatesare (2, 0) is on the graph.

(b) For we have

Let

False

The statement is false, so the point whose coordinates are is noton the graph.

(c) For we have

Let

True

The statement is true when and so the point whose coordinates

are is on the graph.a12

, -92b

y = -92

,x =12

6 = 6

122

� 6

32

+92

� 6

3a12b - a -

92b � 6x =

12

, y = -92

:

3x - y = 6

a12

, - 92b ,

11, -225 = 6

5 = 6

3 + 2 � 6

3112 - 1-22 � 6x = 1, y = -2:

3x - y = 6

11, -22,y = 0,x = 2

6 = 6

3122 - 0 � 6x = 2, y = 0:

3x - y = 6

3x - y = 6.x = 2, y = 0

a12

, -92b11, -22

3x - y = 6.

For the remainder of the course we will say “the point (x, y)” rather than “the pointwhose coordinates are (x, y)” for the sake of brevity.


3 Graph an Equation Using the Point-Plotting MethodOne of the most elementary methods for graphing an equation is the point-plottingmethod. With this method, we choose values for one of the variables and use the equa-tion to determine the corresponding values of the remaining variable. If x and y are thevariables in the equation, it does not matter whether we choose values of x and use theequation to find the corresponding y or choose y and find x. Convenience will deter-mine which way we go.

EXAMPLE 3 How to Graph an Equation by Plotting PointsGraph the equation by plotting points.


y = -2x + 4

Step 1: We want to find all points (x, y) that satisfy theequation. To determine these points we choose values of x(do you see why?) and use the equation to determine thecorresponding values of y. See Table 9.

Table 9x (x, y)

0 (0, 4)

1 (1, 2)

2 (2, 0)

3 (3, -2) -2(3) + 4 = -2

-2(2) + 4 = 0

-2(1) + 4 = 2

-2(0) + 4 = 4

(-1, 6) -2(-1) + 4 = 6-1

(-2, 8) -2(-2) + 4 = 8-2

(-3, 10) -2(-3) + 4 = 10-3

y = -2x + 4

Figure 26

The graph of the equation shown in Figure 26(b) does not show all the points thatsatisfy . For example, in Figure 26(b) the point is part of thegraph of but it is not shown. Since the graph of can beextended as far as we please, we use arrows on the ends of the graph to indicate thatthe pattern shown continues. It is important to show enough of the graph so that any-one who is looking at it will “see” the rest of it as an obvious continuation of what isthere. This is called a complete graph.

y = -2x + 4y = -2x + 4,18, -122y = -2x + 4

2 4�4 �2 x

8

6

10

4

2

�2

y

(b)

(2, 0)

(1, 2)

(0, 4)(�1, 6)

(3, �2)

(�2, 8)

(�3, 10)

2 4�4 �2 x

8

6

10

4

2

�2

y

(a)

Step 2: We plot the ordered pairs listed in the thirdcolumn of Table 9 as shown in Figure 26(a). Nowconnect the points to obtain the graph of theequation (a line) as shown in Figure 26(b).


EXAMPLE 4 Graphing an Equation by Plotting PointsGraph the equation by plotting points.

Solution

Table 10 shows several points on the graph.

y = x2

Table 10x (x, y)

0 (0, 0)

1 (1, 1)

2 (2, 4)

3 (3, 9)

4 (4, 16)y = (4)2 = 16

y = (3)2 = 9

y = (2)2 = 4

y = (1)2 = 1

y = (0)2 = 0

(-1, 1)y = (-1)2 = 1-1

(-2, 4)y = (-2)2 = 4-2

(-3, 9)y = (-3)2 = 9-3

(-4, 16)y = (-4)2 = 16-4

y = x2

(4, 16)(�4, 16)

(3, 9)(�3, 9)

(2, 4)(�2, 4)

(1, 1)(�1, 1)

(0, 0) 64�2 x

4

�4

8

12

16

20

�4�6

y

(a)

(4, 16)(�4, 16)

(3, 9)(�3, 9)

(2, 4)(�2, 4)

(1, 1)(�1, 1)

(0, 0) 64�2 x

4

�4

8

12

16

20

�4�6

y

(b)

Quick In Problems 8–10, graph each equation using the point-plotting method.

8. 9. 10. y = x2 + 32x + 3y = 8y = 3x + 1

Work SmartNotice we use a different scale onthe x- and y-axis in Figure 27.

Work SmartExperience will play a huge role indetermining which x-values to choosein creating a table of values. For thetime being, start by choosing valuesof x around as in Table 10.x = 0

Figure 27

In Figure 27(a), we plot the ordered pairs listed in Table 10. In Figure 27(b), we connect the points in a smooth curve.

Two questions that you might be asking yourself right now are “How do I know howmany points are sufficient?” and “How do I know which x-values (or y-values) I shouldchoose in order to obtain points on the graph?” Often, the type of equation we wish tograph indicates the number of points that are necessary. For example, we will learn inthe next section that if the equation is of the form then its graph is a lineand only two points are required to obtain the graph (as in Example 3). Other times,more points are required. At this stage in your math career, you will need to plot quitea few points to obtain a complete graph. However, as your experience and knowledgegrow, you will learn to be more efficient in obtaining complete graphs.

y = mx + b,


EXAMPLE 5 Graphing the Equation Graph the equation by plotting points.

Solution

Because the equation is solved for x, we will choose values of y and use the equationto find the corresponding values of x. See Table 11. We plot the ordered pairs listed inTable 11 and connect the points in a smooth curve. See Figure 28.

x = y2

x � y2

Table 11y x � y2 (x, y)

0 (0, 0)

1 (1, 1)

2 (4, 2)

3 (9, 3)32 = 9

22 = 4

12 = 1

02 = 0

(1, -1)(-1)2 = 1-1

(4, -2)(-2)2 = 4-2

(9, -3)(-3)2 = 9-3

Figure 28

x

y

62�2 4

�2

�4

2

4

8 10

(9, 3) (4, 2)

(0, 0)

(1, �1) (4, �2)

(9, �3)

(1, 1)

Quick In Problems 11 and 12, graph each equation using the point-plottingmethod.

11. 12. x = 1y - 122x = y2 + 2

4 Identify the Intercepts from the Graph of an EquationOne of the key components that should be displayed in a complete graph is theintercepts of the graph.

DEFINITIONThe intercepts are the coordinates of the points, if any, where a graph crosses ortouches the coordinate axes. The x-coordinate of a point at which the graph crossesor touches the x-axis is an x-intercept, and the y-coordinate of a point at which thegraph crosses or touches the y-axis is a y-intercept.

Figure 29

x-intercepts

y-intercept

y

x

touches

crosses

Work SmartIn order for a graph to be complete,all of its intercepts must bedisplayed.

EXAMPLE 6 Finding Intercepts from a GraphFind the intercepts of the graph shown in Figure 30. What are the x-intercepts? Whatare the y-intercepts?

Figure 30

(1, 0)

(3.8, 0)

(0, 2)

(�3, 0)

y

x5

5

�5

�5

See Figure 29 for an illustration. Notice that an x-intercept exists when and a y-intercept exists when x = 0.

y = 0


Solution

The intercepts of the graph are the points

The x-intercepts are and 3.8. The y-intercept is 2.

In Example 6, you should notice the following: If we do not specify the type of in-tercept (x- versus y-), then we report the intercept as an ordered pair. However, if wespecify the type of intercept, then we only need to report the coordinate of the inter-cept. For x-intercepts, we report the x-coordinate of the intercept (since it is under-stood that the y-coordinate is 0); for y-intercepts, we report the y-coordinate of theintercept (since the x-coordinate is understood to be 0).

-3, 1,

1-3, 02, 10, 22, 11, 02, and 13.8, 02

Quick 13. The points, if any, at which a graph crosses or touches a coordinate axis are

called _____________.

(1, 0) (6.7, 0)

(0, –0.9)

(�5, 0)

y

x63�3�6

�3

5 Interpret GraphsGraphs play an important role in helping us to visualize relationships that existbetween two variables or quantities. We have all heard the expression “A picture isworth a thousand words.” A graph is a “picture” that illustrates the relationshipbetween two variables. By visualizing this relationship, we are able to see important in-formation and draw conclusions regarding the relationship between the two variables.

EXAMPLE 7 Interpret a GraphThe graph in Figure 31 shows the profit P for selling x gallons of gasoline in an hourat a gas station. The vertical axis represents the profit and the horizontal axis repre-sents the number of gallons of gasoline sold.

Figure 31

100 200 300 400 500 600 700 8000

100

�100

200

�200

300

400

500

600

Number of Gallons

Pro

fit (

$)

(375, 565)

P

x

14. True or False:The graph of an equation must have at least one x-intercept.

15. Find the intercepts of the graph shown in the figure. What are the x-intercepts?What are the y-intercepts?


(a) What is the profit if 150 gallons of gasoline are sold?

(b) How many gallons of gasoline are sold when profit is highest? What isthe highest profit?

(c) Identify and interpret the intercepts.

Solution

(a) Draw a vertical line up from 150 on the horizontal axis until we reach thepoint on the graph. Then draw a horizontal line from this point to thevertical axis. The point where the horizontal line intersects the verticalaxis is the profit when 150 gallons of gasoline are sold. The profit fromselling 150 gallons of gasoline is $200.

(b) The profit is highest when 375 gallons of gasoline are sold. The highestprofit is $565.

(c) The intercepts are , (100, 0), and (750, 0). For : If theprice of gasoline is too high, demand for gasoline (in theory) will be 0gallons. This is the explanation for selling 0 gallons of gasoline. Thenegative profit is due to the fact that the company has $0 in revenue(since it didn’t sell any gas), but had hourly expenses of $200. For(100, 0): The company sells just enough gas to pay its bills. This can bethought of as the break-even point. For (750, 0): The 750 gallons soldrepresents the maximum number of gallons that the station can pumpand still break even.

(0, -200)(0, -200)

Quick 16. The graph shown represents the cost C (in thousands of dollars) of refining

x gallons of gasoline per hour (in thousands). The vertical axis represents thecost and the horizontal axis represents the number of gallons of gasolinerefined.

(a) What is the cost of refining 250 thousand gallons of gasoline per hour?

(b) What is the cost of refining 400 thousand gallons of gasoline per hour?

(c) In the context of the problem, explain the meaning of the graph ending at700 thousand gallons of gasoline.

(d) Identify and interpret the intercept.

100 200 300 400 500 600 700 800Number of Gallons (thousands)

x0

200

400

600

800

1000

Cos

t (th

ousa

nds

of $

)

C



Building Skills17. Determine the coordinates of each of the points

plotted. Tell in which quadrant or on what coordinateaxis each point lies. See Objective 1.

FE

A

C

D

B

642�2 x

�2

�4

�6

2

4

6

�4�6

y

18. Determine the coordinates of each of the pointsplotted. Tell in which quadrant or on what coordinateaxis each point lies. See Objective 1.

In Problems 19 and 20, plot each point in the xy-plane.Tell inwhich quadrant or on what coordinate axis each point lies. See Objective 1.

19.

F1-4, 12E10, 32D11, -62C15, 02B1-2, -62A13, 52

21.

(a) (1, 2)

(b)

(c)

(d) a- 32

, 3b1-4, 421-2, 32

2x + 5y = 12

20.

F10, -52E11, 22D1-6, -22C12, -52B1-6, 02A1-3, 12

In Problems 21–26, determine whether the given points are on thegraph of the equation. See Objective 2.

22.

(a) (1, 7)

(b) (0, 6)

(c)

(d) a32

, 4b1-3, 102

-4x + 3y = 18

23.

(a)

(b) (3, 10)

(c) (0, 1)

(d) 12, -32

1-2, -152y = -2x2 + 3x - 1 24.

(a) (2, 2)

(b) (3, 8)

(c)

(d) (0, 0)

1-3, -182

y = x3 - 3x

25.

(a) (1, 4)

(b) (4, 1)

(c)

(d) (0, 3)

1-6, 92

y = ƒx - 3 ƒ 26.

(a) (0, 1)

(b) (1, 1)

(c)

(d) a232

, 12b

a12

, 12b

x2 + y2 = 1

In Problems 27–54, graph each equation by plotting points. See Objective 3.

27. 28.

29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

39. 40.

41. 42.

43. 44.

45. 46.

47. 48.

49. 50.

51. 52.

53. 54. x = y2 + 2x = y2 - 1

x2 + y = 5x2 - y = 4

y = x3 - 2y = x3 + 1

y = -x3y = x3

y = - ƒx ƒy = ƒx - 1 ƒ

y = ƒx ƒ - 2y = ƒx ƒ

y = -2x2 + 8y = 2x2 - 8

y = x2 - 2y = -x2

3x + y = 92x + y = 7

y = -12

x + 2y =12

x - 4

y = -4x + 2y = -3x + 1

y = x - 2y = x + 3

y = -13

xy = -12

x

y = 2xy = 4x

F

E

A

CD

B

642�2 x

�2

�4

�6

2

4

6

�4�6

y


Applying the Concepts59. If (a, 4) is a point on the graph of what is

a?

60. If is a point on the graph of

what is a?

61. If (3, b) is a point on the graph of what is b?

62. If is a point on the graph of

what is b?

63. Area of a Window Bob Villa wishes to put a new win-dow in his home. He wants the perimeter of the win-dow to be 100 feet. The graph above and to the rightshows the relation between the width, x, of the open-ing and the area of the opening.

y = -2x2 + 3x + 1,

1-2, b2y = x2 - 2x + 1,

y = -3x + 5,1a, -22

y = 4x - 3,

56.

x

y

3

3

�3

�3 x

y

3�3

3

�3

57.

55.

x

y

3�3

�4

3

(a) What is the area of the opening if the width is10 feet?

(b) What is the width of the opening in order for areato be a maximum? What is the maximum area ofthe opening?


64. Projectile Motion The graph below shows the height,in feet, of a ball thrown straight up with an initialspeed of 80 feet per second from an initial height of96 feet after t seconds.

1 2 3 4 5 60

50

100

150

200

Time (seconds)

Hei

ght

(2.5, 196)

(0, 96)

H

t

(a) What is the height of the object after 1.5 seconds?

(b) At what time is the height a maximum? What isthe maximum height?


65. Cell Phones We all struggle with selecting a cellular phone provider. The graph below shows the relation between themonthly cost of a cellular phone and the number of minutes used, m, when using the Sprint PCS 2000-minute plan.(SOURCE: SprintPCS.com)

10 20 30 40 500

200

400

600

800

A

Width (feet)

Are

a (s

quar

e fe

et) (25, 625)

x

58.

x

y

3�3

�9

12

Cos

t per

Mon

th (

$)

Anytime Minutes per Month

4000

3000

2000

1000

02000 4000 6000 8000 10,000 12,000 14,000 16,000

(2000, 100)(0, 100)

(15120, 3380)

(8000, 1600)

In Problems 55–58, the graph of an equation is given. List the inter-cepts of the graph. See Objective 4.


72. Make up an equation that contains the points (0, 3),(1, 3), and Compare your equation with thoseof your classmates. How many are the same?

Explaining the Concepts73. Explain what is meant by a complete graph.

74. Explain what the graph of an equation represents.

75. What is the point-plotting method for graphing anequation?

76. What is the y-coordinate of a point that is an x-intercept? What is the x-coordinate of a point that isa y-intercept?

The Graphing CalculatorJust as we have graphed equations using point plotting,the graphing calculator also graphs equations by plottingpoints. Figure 32 shows the graph of and Table 12shows points on the graph of using a TI-84 Plusgraphing calculator.

y = x2y = x2

1-4, 32.

Figure 32

�4 40

16

Table 12

(a) What is the cost of talking for 200 minutes in amonth? 500 minutes?

(b) What is the cost of talking 8000 minutes in amonth?

(c) Identify and interpret the intercept.

66. Wind Chill It is 10° Celsius outside. The wind is calmbut then gusts up to 20 meters per second. You feelthe chill go right through your bones. The followinggraph shows the relation between the wind chilltemperature (in degrees Celsius) and wind speed (inmeters per second).

10 200

4

8

�4

12

Win

d C

hill

(20, �3.7)

(10, 0)

(0, 10)

�C

v(m/s)

(a) What is the wind chill if the wind is blowing 4 meters per second?

(b) What is the wind chill if the wind is blowing 20 meters per second?


67. Plot the points (4, 0), (4, 2), and Describe the set of all points of the form (4, y) wherey is a real number.

68. Plot the points (4, 2), (1, 2), (0, 2), and Describe the set of all points of the form (x, 2) wherex is a real number.

Extending the Concepts69. Draw a graph of an equation that contains two

x-intercepts, and 3. At the x-intercept thegraph crosses the x-axis; at the x-intercept 3, thegraph touches the x-axis. Compare your graph withthose of your classmates. How are they similar? Howare they different?

70. Draw a graph that contains the points(0, 3), and (1, 5). Compare your graph with

those of your classmates. How many of the graphsare straight lines? How many are “curved”?

71. Make up an equation that is satisfied by the points (2, 0), (4, 0), and (1, 0). Compare your equation withthose of your classmates. How are they similar? Howare they different?

1-1, 12, 1-3, -12,

-2,-2

1-3, 22.

14, -62.14, -32,

In Problems 77–84, use a graphing calculator to draw a completegraph of each equation. Use the TABLE feature to assist in select-ing an appropriate viewing window.

77. 78.

79. 80.

81. 82.

83. 84. y = -x3 + 3xy = x3 - 6x + 1

y - x2 = -15y + 2x2 = 13

y = 2x2 - 4y = -x2 + 8

y = -5x + 8y = 3x - 9

Section 1.6 Linear Equations in Two Variables 105

DEFINITIONA linear equation in two variables is an equation of the form

where A, B, and C are real numbers. A and B cannot both be 0.

Ax + By = C

1.6 Linear Equations in Two VariablesPreparing for Linear Equations in Two VariablesBefore getting started, take this readiness quiz. If you get a problem wrong, go back to the sectioncited and review the material.

P1. Solve: [Section 1.1, pp. 49–50]P2. Solve: [Section 1.1, pp. 49–50]P3. Solve for y: [Section 1.3, pp. 73–76]

P4. Evaluate: [Section R.4, pp. 34–35]

P5. Evaluate: [Section R.4, pp. 34–35]

P6. Distribute: [Section R.3, pp. 29–30]-2(x + 3)

-7 - 2-5 - (-2)

5 - 2-2 - 4

3x - 2y = 102x + 5 = 133x + 12 = 0

OBJECTIVES1 Graph Linear Equations Using

Point Plotting

2 Graph Linear Equations UsingIntercepts

3 Graph Vertical and HorizontalLines

4 Find the Slope of a Line GivenTwo Points

5 Interpret Slope as an AverageRate of Change

6 Graph a Line Given a Point andIts Slope

7 Use the Point-Slope Form of aLine

8 Identify the Slope and y-Intercept of a Line fromIts Equation

9 Find the Equation of a LineGiven Two Points

1 Graph Linear Equations Using Point PlottingIn Section 1.5 we discussed how to graph any equation using the point-plottingmethod. Remember, the graph of an equation is the set of all ordered pairs (x, y) suchthat the equation is a true statement.

We are now going to learn methods for graphing a specific type of equation, called alinear equation in two variables.

When a linear equation is written in the form we say that the linearequation is in standard form.

Some examples of linear equations in standard form are

The graph of a linear equation is a line. Let’s graph a linear equation using the point-plotting method.

3x - 4y = 9 12

x +23

y = 4 3x = 9 -2y = 5

Ax + By = C,


P2. P3. P4.

P5. 3 P6. -2x - 6

-12

y =32x - 5{4}

{-4}

EXAMPLE 1 Graphing a Linear Equation Using the Point-Plotting Method

Graph the linear equation:

Solution

To graph the linear equation, we choose various values for x and then use the equa-tion to find the corresponding values of y. For this equation, we will let and 1.

y = 5 y = 7

2y = 10 2y = 14

-4 + 2y = 6 -8 + 2y = 6

41-12 + 2y = 6x = -1:41-22 + 2y = 6x = -2:

x = -2, -1, 0,

4x + 2y = 6

Divide both sides by 2: Divide both sides by 2:

Add 4 to both sides:Add 8 to both sides:

is on the graph.(-2, 7) is on the graph.(-1, 5)


Table 13 shows the points that are on the graph of We plot the orderedpairs and (1, 1) in the Cartesian plane and connect the points ina straight line. See Figure 33.

1-2, 72, 1-1, 52, 10, 32, 4x + 2y = 6.

Subtract 4 from both sides: y = 1 y = 3

2y = 2 2y = 6

4 + 2y = 6 0 + 2y = 6

4112 + 2y = 6x = 1:4102 + 2y = 6x = 0:

Divide both sides by 2: Divide both sides by 2:

(0, 3) is on the graph. (1, 1) is on the graph.

Table 13

7

5

0 3

1 1 11, 1210, 321-1, 52-1

1-2, 72-2

1x, y2yx

4 x

6

�4

y

(1, 1)

(0, 3)

(�1, 5)

(�2, 7)

�3

3

Figure 33

One of the problems with using the point-plotting method to graph equations isdetermining how many points need to be plotted before we obtain a complete graph.Based on the results of Example 1, we can see that only two points are required toobtain a complete graph of a linear equation.To guard against making an error, however,you should plot at least three points.

Work SmartIn Example 1, we chose to pick x-values and use the equation to find the corresponding y-values;however, we could also have chosena y-value and used the equation tofind the corresponding x-value.

x-intercepts

y-intercept

y

x

touches

crosses

Figure 34

Quick

2 Graph Linear Equations Using InterceptsIn Section 1.5, we said any complete graph should display the intercepts, if any. Recall,the intercepts of the graph of an equation are the points, if any, where the graph crossesor touches the coordinate axes. See Figure 34.

Now we will explain how to find the intercepts algebraically. From Figure 34 it isapparent that an x-intercept exists when the value of y is 0 and that a y-intercept existswhen the value of x is 0. This leads to the following procedure for finding intercepts.

Work SmartWe recommend that you find at leastthree points to be sure your graph iscorrect. Also, remember that acomplete graph is a graph thatshows all the interesting features ofthe graph, such as its intercepts.

PROCEDURE FOR FINDING INTERCEPTS

� To find the x-intercept(s), if any, of the graph of an equation, let in theequation and solve for x.

� To find the y-intercept(s), if any, of the graph of an equation, let in theequation and solve for y.

x = 0

y = 0

In Problems 3–5, graph each linear equation using the point-plotting method.

1. A(n) is an equation of the form , where , , andare real numbers. and cannot both be 0.

2. The graph of a linear equation is called a .

BACBAAx + By = C

3. 4. 5. -6x + 3y = 1212

x + y = 2y = 2x - 3


The procedure given can be used to find the intercepts of any type of equation. Let’suse this procedure to find the intercepts of a linear equation.

EXAMPLE 2 Graphing a Linear Equation by Finding Its InterceptsGraph the linear equation by finding its intercepts.

Solution

To find the y-intercept, we let and solve the equation for y.

y = 6

2y = 12

0 + 2y = 12

3102 + 2y = 12

3x + 2y = 12x = 0

3x + 2y = 12

Let :


x = 0

The y-intercept is 6, so the point (0, 6) is on the graph of the equation.To find the x-intercept, we let and solve the equation for x.

x = 4 3x = 12

3x + 0 = 12

3x + 2102 = 12

3x + 2y = 12y = 0

Let


y = 0:

The x-intercept is 4, so the point (4, 0) is on the graph of the equation.We obtain one additional point on the graph by letting (or any other value of

x besides 0 or 4), and find y to be 3. Table 14 shows the points that are on the graph ofWe plot the points (0, 6), (4, 0), and (2, 3). Connect the points in a

straight line and obtain the graph in Figure 35.3x + 2y = 12.

x = 2

Table 14x y (x, y)

0 6 (0, 6)

4 0 (4, 0)

2 3 (2, 3)3 x�3 6

y

6

(2, 3)

(4, 0)

(0, 6)

�3

3

Figure 35

Work SmartLinear equations in one variablehave no solution, one solution, orinfinitely many solutions. Becausethe procedure for finding interceptsof linear equations in two variablesresults in a linear equation in onevariable, linear equations can haveno x-intercepts, one x-intercept, orinfinitely many x-intercepts. Thesame applies to y-intercepts.

Quick 6. True or False: To find the x-intercept(s), if any, of the graph of an equation,

let in the equation and solve for .

In Problems 7 and 8, graph each linear equation by finding its intercepts.

7. 8. 4x - 5y = 20x + y = 4

xy = 0

EXAMPLE 3 Graphing a Linear Equation by Finding Its InterceptsGraph the linear equation by finding its intercepts.

Solution

x + 3y = 0

To find the y-intercept, we let andsolve the equation for y.

To find the x-intercept, we let andsolve the equation for x.x + 3y = 0

y = 0x + 3y = 0

x = 0

y = 0

3y = 0

0 + 3y = 0Let


x = 0:

x = 0

x + 0 = 0

x + 3102 = 0Let y = 0:

The y-intercept is 0, so the point (0, 0) ison the graph of the equation.

The x-intercept is 0, so the point (0, 0)is on the graph of the equation.


Because both the x- and y-intercepts are 0, we will find two additional points onthe graph of the equation. By letting we find that By letting we find that We plot the points (0, 0), and connect the pointsin a straight line and obtain the graph in Figure 36.

13, -12,1-3, 12,y = 1.x = -3,y = -1.x = 3,

Work SmartWe chose x � �3 and x � 3, toavoid fractions. This makes plottingthe points easier.

5 x�5

y

5

�5

(�3, 1)

(3, �1)

(0, 0)

Figure 36

x�2 42

y

3

�3

(3, 2)(3, 1)

(3, �1)(3, �2)

Figure 37x = 3

From Example 3, we learn that any equation of the form whereand has only one intercept at (0, 0).Therefore, to graph equations of this

form, we find two additional points on the graph.B Z 0,A Z 0

Ax + By = 0,

Quick In Problem 9, graph the equation by finding its intercepts.

9. 3x - 2y = 0

3 Graph Vertical and Horizontal LinesIn the equation of a line, we said that A and B cannot both be zero.But what if or B = 0?A = 0

Ax + By = C,

EXAMPLE 4 Graphing a Vertical LineGraph the equation using the point-plotting method.

Solution

Because the equation can be written as we know that the graphis a line. When you look at the equation it should be clear to you that nomatter what value of y we choose, the corresponding value of x is going to be 3.Therefore, the points (3, 0), (3, 1), and (3, 2) are all points on theline. See Figure 37.

13, -22, 13, -12,x = 3,

1x + 0y = 3,x = 3

x = 3

Based on the results of Example 4, we can write the equation of a vertical line:

EQUATION OF A VERTICAL LINE

A vertical line is given by an equation of the form

where a is the x-intercept.x = a

Now let’s look at equations that lead to graphs that are horizontal lines.


EXAMPLE 5 Graphing a Horizontal LineGraph the equation using the point-plotting method.

Solution

Because the equation can be written as we know that thegraph is a line. In looking at the equation it should be clear that no matterwhat value of x we choose, the corresponding value of y is going to be Therefore,the points and are all points on the line.See Figure 38.

Based on the results of Example 5, we can generalize:

12, -221-2, -22, 1-1, -22, 10, -22, 11, -22, -2.y = -2,

0x + 1y = -2,y = -2

y = -2

EQUATION OF A HORIZONTAL LINE

A horizontal line is given by an equation of the form

where b is the y-intercept.

y = bx�4 4

y

3

�3(�2, �2)

(�1, �2) (1, �2)

(2, �2)

Figure 38y = -2

4 Find the Slope of a Line Given Two PointsUp to this point, we have given you a linear equation and asked you to graph the line.Now we want to discuss an interesting property of linear equations. Look back atTable 13 and Figure 33 from Example 1. Notice that for each 1-unit increase in x, thevalue of y decreases by 2. For example, as x increases from to 0, y decreases from5 to 3; as x increases from 0 to 1, y decreases from 3 to 1. This property of linear equa-tions is referred to as slope.

-1

Consider the staircase drawn in Figure 39(a). If we draw a line at the top of eachriser on the staircase (in blue), we can see that each step contains exactly the samehorizontal run and the same vertical rise. We call the ratio of the rise to the run theslope of the line. It is a numerical measure of the steepness of the line. For example,suppose that the staircase in Figure 39(a) has a run of 7 inches and a rise of 6 inches.

Then the slope of the line is If the run of the stair is increased to

10 inches, while the rise remains the same, then the slope of the line is

See Figure 39(b). If the run is decreased to 4 inches and the rise

remains the same, then the slope of the line is See Figure 39(c).riserun

=6 inches4 inches

.

riserun

=6 inches10 inches

.

riserun

=6 inches7 inches

.

Run � 7 inches

Rise � 6 inches

Run � 10 inches Run � 4 inches

(a) Slope � 6__7 (c) Slope � 6__

4(b) Slope � 6___10

Rise � 6 inches Rise � 6 inches

Figure 39

Quick In Problems 10–12, graph each equation.

10. 11. 12. -3x + 4 = 1y = -4x = 5


In WordsSlope is the change in y divided by the changein x.

We can define the slope of a line using rectangular coordinates.

DEFINITIONLet and be two distinct points. If the slope mof the nonvertical line L containing P and Q is defined by the formula

If then L is a vertical line and the slope m of L is undefined (since this results in division by 0).x1 = x2 ,

m =y2 - y1

x2 - x1, x1 Z x2

x1 Z x2 ,Q = 1x2 , y22P = 1x1 , y12

The accepted symbol for the slope of a line is m. It comes from the French wordmonter, which means to ascend or climb. Figure 40(a) provides an illustration of theslope of a nonvertical line; Figure 40(b) illustrates a vertical line.

x1

y1

y2

x2

Rise � y2 � y1

Q � (x2, y2) L

P � (x1, y1)

Run � x2 � x1

y2 � y1__x2

�

x1

(a) Slope of L is m �

x1

y1

y2 Q � (x1, y2)

P � (x1, y1)

L

(b) Slope of L is undefined;L is vertical

Figure 40

From Figure 40(a) we can see that the slope m of a nonvertical line may beviewed as

We can also write the slope m of a nonvertical line as

The symbol is the Greek letter delta, which comes from the Greek word diaphora,which means “difference.” In mathematics, we read the symbol as “change in.” So the

notation is read “change in y divided by change in x.”

The slope m of a nonvertical line measures the amount that y changes (the verticalchange) as x changes from to (the horizontal change). The slope m of a verticalline is undefined since it results in division by zero.

x2x1

¢y¢x

¢¢

m =y2 - y1

x2 - x1=

Change in yChange in x

=¢y¢x

m =y2 - y1

x2 - x1=

RiseRun

Comments Regarding the Slope of a Nonvertical Line1. Any two different points on the line can be used to compute the slope of the line

shown in Figure 41. The slope m of the line L is given by

This result is due to the fact that the two triangles formed in Figure 41 are similar(the measure of the angles is the same in both triangles). Therefore, the ratios ofthe sides are proportional.

2. The slope of a line may be computed from to or fromQ to P because

m =y2 - y1

x2 - x1=

-1y1 - y22-1x1 - x22 =

y1 - y2

x1 - x2

Q = 1x2 , y22P = 1x1 , y12

m =y2 - y1

x2 - x1 or m =

y4 - y3

x4 - x3

y

x

P � (x1, y1)

Q � (x2, y2)

S � (x4, y4)

R � (x3, y3)

x4 � x3

y4 � y3

x2 � x1

y2 � y1

L

Figure 41

Work SmartIt doesn’t matter whether wecompute the slope of the line frompoint P to Q or from point Q to P.


EXAMPLE 6 Finding and Interpreting the Slope of a LineFind and interpret the slope of the line containing the points (3, 6) and

Solution

We plot the points and and draw a linethrough the points as shown in Figure 42. The slope of the line drawn in Figure 42 is

We could also compute the slope as

m =y1 - y2

x1 - x2=

6 - 23 - 1-22 =

45

m =y2 - y1

x2 - x1=

2 - 6-2 - 3

=-4-5

=45

Q = 1x2 , y22 = 1-2, 22P = 1x1 , y12 = 13, 62

1-2, 22.

2 4 x�4 �2

y

6

2

4

Q � (�2, 2)

P � (3, 6)

Run � x � 5

Rise � y � 4

Figure 42

A slope of can be interpreted as: For every 5-unit increase in x, y will increase by

4 units. Or for every 5-unit decrease in x, y will decrease by 4 units.

45

Quick

13. If a line is vertical, then its slope is .

14. On a line, for every 10-foot run there is a 4-foot rise.The slope of the line is .

15. True or False: If and are two distinct points with, the slope m of the nonvertical line L containing P and Q is defined by

the formula

16. True or False: If the slope of a line is then if x increases by 2, y will increase by 1.

In Problems 17–20, find and interpret the slope of the line containing the points.

17. (0, 3); (3, 12) 18.

19. (3, 2); 20. 1-2, 42; 1-2, -121-3, 221-1, 32; 13, -42

12

,

m =x2 - x1

y2 - y1, y1 Z y2.

x1 Z x2

Q = (x2, y2)P = (x1, y1)

EXAMPLE 7 Finding Slopes of Different Lines Each of Which Contains (3, 5)

Find the slope of the lines and containing the following pairs of points.Graph the lines in the Cartesian plane.

L4 : P13, 52 Q413, 02 L3 : P13, 52 Q315, -22 L2 : P13, 52 Q216, 52 L1 : P13, 52 Q115, 82

L4L1 , L2 , L3 ,


2 4 6 x�2

y

6

8

2

�2

4

(5, 8)

(3, 5)

L1

(a)

Figure 43

2 4 6 x�2

y

6

8

2

�2

4 (3, 5) (6, 5)

L2

(b)

2 4 6 x�2

y

6

8

2

�2

4(3, 5)

(5, �2)

L3

(c)

2 4 6 x�2

y

6

8

2

�2

4(3, 5)

(3, 0)

L4

(d)

Relying on Figure 43, we have the following properties of slope:

PROPERTIES OF SLOPE

� When the slope of a line is positive, the line slants upward from left to right,as shown by in Figure 43(a).

� When the slope of a line is zero, the line is horizontal, as shown by inFigure 43(b).

� When the slope of a line is negative, the line slants downward from left to right,as shown by in Figure 43(c).

� When the slope of a line is undefined, the line is vertical, as shown by inFigure 43(d).

L4

L3

L2

L1

Quick21. True or False: If the slope of a line is negative, then the line slants downward

from left to right.

22. Find the slope of the lines and containing the following pairs ofpoints. Graph all four lines on the same Cartesian plane.

L3 : P11, 32 Q31-3, 72 L4 : P11, 32 Q41-4, 32 L1 : P11, 32 Q116, 42 L2 : P11, 32 Q211, 82

L4L1 , L2 , L3 ,

Solution

Let and denote the slopes of the lines and respectively.Then

undefined

The graphs of the four lines are shown in Figure 43.

m3 =-2 - 55 - 3

=-72

= -72 m4 =

0 - 53 - 3

=-50

m1 =8 - 55 - 3

=32 m2 =

5 - 56 - 3

=03

= 0

L4 ,L1 , L2 , L3 ,m4m1 , m2 , m3 ,

5 Interpret Slope as an Average Rate of ChangeThe slope m of a nonvertical line measures the amount that y changes as x changesfrom to The slope of a line is also called the average rate of change of y withrespect to x.

In applications, we are often interested in knowing how the change in one variablemight impact some other variable. For example, if your income increases by $1000, howmuch will your spending (on average) change? Or, if the speed of your car increases by10 miles per hour, how much (on average) will your car’s gas mileage change?

x2 .x1


EXAMPLE 8 Slope as an Average Rate of ChangeA strain of E. coli Beu 397-recA441 is placed into a Petri dish at 30° Celsius andallowed to grow. The data shown in Table 15 are collected. The population ismeasured in grams and the time in hours. The population growth is shown inFigure 44.

Table 15Time (hours), x Population (grams), y

0 0.09

1 0.12

2 0.16

3 0.22

4 0.29

5 0.392 4 60

0.050.100.150.200.250.300.350.400.45

Time (hours)

Growth of E-coli

Pop

ulat

ion

(gra

ms)

Figure 44

SOURCE: Dr. Polly Lavery, Joliet Junior College

(a) Compute and interpret the average rate of change in the population between 0 and 1 hour.

(b) Compute and interpret the average rate of change in the population between 3 and 4 hours.

(c) Based upon your results to parts (a) and (b), do you think that the population grows linearly? Why?

Solution(a) To find the average rate of change, we compute the slope of the line

between the points (0, 0.09) and (1, 0.12).

The population of E. coli was growing at the rate of 0.03 gram per hourbetween 0 and 1 hour.

(b) We want to compute the slope of the line between the points (3, 0.22) and (4, 0.29).

The population of E. coli was growing at the rate of 0.07 gram per hourbetween 3 and 4 hours.

(c) The population is not growing linearly because the average rate of change(slope) is not constant. In fact, because the average rate of change is increasing as time passes, the population is growing more rapidly over time.

m = average rate of change =0.29 - 0.22

4 - 3= 0.07 gram per hour

m = average rate of change =0.12 - 0.09

1 - 0= 0.03 gram per hour

Quick

23. The data to the left represent the total revenue that would be received from selling x bicycles at Gibson’s Bicycle Shop.

(a) Plot the ordered pairs (x, y) on a graph and connect the points with straightlines.

(b) Compute and interpret the average rate of change in the revenue between 0 and 25 bicycles sold.

(c) Compute and interpret the average rate of change in the revenue between 102 and 150 bicycles sold.

(d) Based upon your results to parts (a), (b), and (c), do you think that the revenuegrows linearly? Why?

Number ofBicycles, x

TotalRevenue, y

0 0

25 28,000

60 45,000

102 53,400

150 59,160


6 Graph a Line Given a Point and Its SlopeOne reason we care so much about slope is that it can be used to help graph lines.

EXAMPLE 9 Graphing a Line Given a Point and Its SlopeDraw a graph of the line that contains the point (1, 2) and has a slope of

(a) 3 (b)

Solution

(a) Because the we have that This means

that if x increases by 1 unit, then y will increase by 3 units. So, if we start at(1, 2) and move 1 unit to the right and then 3 units up, we end up at thepoint (2, 5). We then draw a line through the points (1, 2) and (2, 5) toobtain the graph of the line. See Figure 45.

(b) Because the we have that This means

that if x increases by 2 units, then y will decrease by 3 units. So, if we start at(1, 2) and move 2 units to the right and then 3 units down, we end up at thepoint We then draw a line through the points (1, 2) and toobtain the graph of the line. See Figure 46.

It is perfectly acceptable to set so that we move left

2 units from (1, 2) and then up 3 units. We would then end up at which is also on the graph of the line as indicated in Figure 46.

1-1, 52,¢y¢x

= -32

=3-2

13, -1213, -12.

-32

=-32

=¢y¢x

.slope =RiseRun

=¢y¢x

,

3 =31

=¢y¢x

.slope =RiseRun

=¢y¢x

,

-32

2 4 x�2 6

y

6

2

�2

4(2, 5)

(1, 2)

Run � 1

Rise � 3

Figure 45

2 4 x�2 6

y

6

�2

(�1, 5)

(3, �1)

(1, 2)

Run � 2Run � �2 Rise � �3

Rise � 3

Figure 46 Quick24. Draw a graph of the line that contains the point and has a slope of

(a) (b) (c) 0-413

1-1, 32

7 Use the Point-Slope Form of a LineA second reason we care so much about slope is that it can be used to help us find theequation of a line. Suppose that L is a nonvertical line with slope m containing thepoint See Figure 47.1x1 , y12.

For any other point (x, y) on L, we know from the formula for the slope of a line that

Multiplying both sides by we can rewrite this expression as

y - y1 = m1x - x12x - x1 ,

m =y - y1

x - x1

Some otherpointGiven

pointy � y1

(x, y) L

(x1, y1)

x � x1

x

yFigure 47


EXAMPLE 10 Using the Point-Slope Form of an Equation of a LineFind the equation of a line whose slope is 3 and that contains the point Graph the line.

Solution

Because we are given the slope and a point on the line, we use the point-slope form ofa line with and

y - 5 = 31x + 22 y - 5 = 31x - 1-222

y - y1 = m1x - x121x1 , y12 = 1-2, 52.m = 3

1-2, 52.

y1 = 5: x1 = -2,m = 3,

See Figure 48 for a graph of the line.

POINT-SLOPE FORM OF AN EQUATION OF A LINE

An equation of a nonvertical line with slope m that contains the point is

y - y1 = m1x - x12

1x1 , y12Slope

T

Given Pointcc

Quick In Problems 25–28, find an equation of the line with the given properties.Graph the line.

25.

26.

27.

28. m = 0, 1x1 , y12 = 14, -22m =

13

, 1x1 , y12 = 13, -42m = -4, 1x1 , y12 = 1-2, 32m = 2, 1x1 , y12 = 13, 52

2 x�4 �2

y

6

4

8

2

�2

(�1, 8)

(�2, 5)

y � 3

x � 1

Figure 48

8 Identify the Slope and y-Intercept of a Line from Its EquationIf we solve the equation in Example 10 for y, we obtain the following:

y = 3x + 11

y - 5 = 3x + 6

y - 5 = 31x + 22Distribute the 3 to remove parentheses:

Add 5 to both sides of the equation:

The coefficient of x, 3, is the slope of the line and the y-intercept of the line isWhen an equation is written in the form we say the

equation is in slope-intercept form.y = mx + b,y = 3(0) + 11 = 11.

SLOPE-INTERCEPT FORM OF AN EQUATION OF A LINE

An equation of a line L with slope m and y-intercept b is

y = mx + b

EXAMPLE 11 Finding the Slope and y-Intercept of a Line from Its Equation

Write the equation in slope-intercept form. Find the slope m and y-intercept b of the line. Graph the line.

x - 3y = 9


Solution

To put the equation in slope-intercept form, we solve the equation for y.

y =13

x - 3

y =-x + 9

-3

-3y = -x + 9 x - 3y = 9

y = mx + b,

Subtract x from both sides of the equation:

Divide both sides of the equation by :

Divide into both terms in the numerator:-3

-3

Comparing to we see that the coefficient of is the

slope, and the y-intercept is We graph the line by plotting a point at We then use the slope to find an

additional point on the graph by moving right 3 units and up 1 unit from the pointto the point Draw a line through the two points and obtain the graph

shown in Figure 49.13, -22.10, -32

10, -32.-3.

x, 13

,y = mx + b,y =13

x - 3

Quick In Problems 29–32, find the slope and y-intercept of each line. Graph the line.

2 x�4 �2 4 6 8 10

y

2

�2

�4

4

(0, �3)

(3, �2)Rise � 1

Run � 3

Figure 49

29. 30.

31. 32. 7x + 3y = 03x - 2y = 7

6x + 2y = 83x - y = 2

9 Find the Equation of a Line Given Two PointsWe know that two points are all that is needed to graph a line. If we are given twopoints, we can find an equation of the line through the points by first finding the slopeof the line and then using the point-slope form of a line.

EXAMPLE 12 How to Find an Equation of a Line from Two PointsFind the equation of a line through the points and If possible, write the equation in slope-intercept form.Graph the line.


12, -52.1-1, 42

Step 1: Find the slope of the line containing thepoints.

Let and Substitute these values into the formula for the slope of a line.

m =y2 - y1

x2 - x1=

-5 - 42 - 1-12 =

-93

= -3

1x2 , y22 = 12, -52.1x1 , y12 = 1-1, 42

Step 2: Use the point-slope form of a line to findthe equation.

With and we have

y - 4 = -3(x + 1)

y - 4 = -31x - 1-122 y - y1 = m1x - x12

y1 = 4,m = -3, x1 = -1,

Step 3: Solve the equation for y. y = -3x + 1

y - 4 = -3x - 3Distribute the �3:Add 4 to both sides:

Work SmartIn Step 2 of Example 12, we choseto use and but wecould also have used and

Choose the values of xand y that make the algebra easiest.y1 = -5.

x1 = 2y1 = 4,x1 = -1


x�2�4 2

y

3

�3

(�3, 2)

(�3, �2)

Figure 51

The slope of the line is and the y-intercept is 1. See Figure 50 for the graph.-3

x�2 42

y

4

�4(2, �5)

(1, �2)

(�1, 4)

(0, 1) Run = 1Rise = �3

Figure 50

Quick In Problems 33–35, find the equation of the line containing the givenpoints. If possible, write the equation in slope-intercept form. Graph the line.

33. (1, 3); (4, 9) 34. (2, 2)

35. (-4, 6); (3, 6)

1-2, 42;

EXAMPLE 13 Finding an Equation of a Line from Two PointsFind the equation of a line through the points and If possible, writethe equation in slope-intercept form. Graph the line.

Solution

Let and Substitute these values into the for-mula for the slope of a line.

undefined

The slope is undefined, so the line is vertical. The equation of the line is See Figure 51 for the graph.

x = -3.

m =y2 - y1

x2 - x1 =

-2 - 2-3 - 1-32 =

-40

1x2 , y22 = 1-3, -22.1x1 , y12 = 1-3, 22

1-3, -22.1-3, 22

Work SmartIf you plot the points first, it will beclear that the line through the pointsis vertical.

The presentation of lines dealt with two types of problems.

1. Given an equation, classify and graph it. (See Examples 1–5.)

2. Given a graph, or information about a graph, find its equation. (See Examples 10,12, and 13.)

This text deals with both types of problems.

Work Smart: Study SkillsTo determine which equation of aline to use, ask yourself “Whatinformation do I know?”

If you know (1) the slope and apoint that isn’t the y-intercept, thenuse the point-slope form (2) the slopeand the y-intercept, then use theslope-intercept form (3) two points,then use the slope formula with thepoint-slope form (4) and if the slopeis undefined, use the vertical line.

Quick36. Find an equation of the line containing the points (3, 2) and If possible,

write the answer in slope-intercept form. Graph the line.13, -42.

SUMMARY EQUATIONS OF LINESForm of Line Formula Comments

Horizontal Line Graph is a horizontal line (slope is 0) with y-intercept b.

Vertical Line Graph is a vertical line (undefined slope) with x-intercept a.

Point-slope Useful for finding the equation of a line given a point and a slope or two points.

Slope-intercept Useful for quickly determining the slope andy-intercept of the line.

Standard Straightforward to find the x- and y-intercepts.Ax + By = C

y = mx + b

y - y1 = m1x - x12

x = a

y = b



Building SkillsIn Problems 37–44, graph each linear equation by plotting points.See Objective 1.

37. 38.

39. 40.

41. 42.

43. 44.

In Problems 45–54, graph each linear equation by finding its intercepts. See Objective 2.

45. 46.

47. 48.

49. 50.

51. 52.

53. 54.

In Problems 55–60, graph each linear equation. See Objective 3.

55. 56.

57. 58.

59. 60.

In Problems 61–64, (a) find the slope of the line and (b) interpretthe slope. See Objective 4.

61. 62.

3y + 20 = -102y + 8 = -6

y = 6y = 1

x = 5x = -5

-32

x +34

y = 023

x -12

y = 0

4x + 3y = 02x + y = 0

14

x +15

y = 213

x -12

y = 1

-4x + 3y = 245x - 3y = 15

-2x + y = 43x + y = 6

-7x + 3y = 95x - 3y = 6

2x -32

y = 1023

x + y = 6

-5x + y = 103x + 2y = 12

2x - y = -8x - 2y = 6

In Problems 65–76, plot each pair of points and graph the linecontaining them. Determine the slope of the line. See Objective 4.

65. (0, 0); (1, 5)

66. (0, 0);

67.

68.

69.

70.

71.

72.

73.

74.

75.

76.

In Problems 77–86, graph the line containing the given point and having slope m. Do not find the equation of the line.See Objective 6.

77. 78.

79. 80.

81. 82.

83. 84.

85. 86. m is undefined;

In Problems 87 and 88, the slope and a point on a line are given.Use the information to find three additional points on the line.Answers may vary.

87. 88.

In Problems 89–92, find an equation of the line. Express your answer in slope-intercept form. See Objective 7.

89. 90.

m = -23

; 11, -32m =52

; 1-2, 32

1-5, 22m = 0; 11, 22m = -

12

; 13, 32m = -32

; 12, 52m =

43

; 1-2, -52m =13

; 1-3, 42m = -4; 1-1, 52m = -2; 1-3, 12m = 2; 1-1, 42m = 3; 11, 22

a73

, 52b ; a13

9,

134b

a12

, 53b ; a9

4,

116b

14, 12; 14, -32110, 22; 110, -321-3, 12; 12, 121-3, 22; 14, 2211, -42; 1-1, 321-2, 32; 13, 7213, -12; 1-2, 1121-2, 32; 11, -62

1-2, 52

(0, 0)(3, 4)

x

y

�3 3

�4

4

(0, 0) x

y

(3, �2)

�4 4

�4

4

x

y

(2, �4)

(�1, 4)

�4 4

�5

5

x

y

(4, 2)

(�2, �3)�4 4

�4

4

63. 64.

x

y

(�4, 1)

(5, �3)

x

y

(�2, 0)(4, 4)


In Problems 115–126, find the slope and y-intercept of each line.Graph the line.See Objective 8.

115.

116.

117.

118.

119.

120.

121.

122.

123.

124.

125.

126.

Applying the Concepts

127. Find an equation for the x-axis.

128. Find an equation for the y-axis.

129. Maximum Heart Rate The data below representthe maximum number of heartbeats that a healthy individual should have during a 15-second interval of time while exercising for different ages.

y = -4

x = 3

2x - 5y - 10 = 0

x - 4y - 2 = 0

3x + 6y = 12

4x + 2y = 8

-3x + y = 1

2x + y = 3

y = -7x

y = -4x

y = 3x + 2

y = 2x - 1

91. 92.

Age, x

Maximum Number of

Heartbeats, y

20 50

30 47.5

40 45

50 42.5

60 40

70 37.5

x

y

(�4, 3)

(1, 3)

x

y

(2, 4)

(2, �2)

In Problems 93–102, find an equation of the line with the givenslope and containing the given point. Express your answer in slope-intercept form, if possible. See Objective 7.

93.

94.

95.

96.

97.

98.

99.

100.

101. m undefined; (6, 1)

102.

In Problems 103–114, find an equation of the line containing thegiven points. Express your answer in slope-intercept form, if possible. See Objective 9.

103. (0, 0); (5, 7)

104. (0, 0);

105. (3, 2); (4, 7)

106. (1, 3); (3, 7)

107.

108. (1, 6)

109. (�1, 5)

110.

111. (1, 3);

112.

113. (2, 4);

114. (3, 1); 13, -421-4, 42

1-5, 12; 11, -121-3, -72

1-3, -42; 11,-421-1, -32;1-3, 12;1-2, 12; 15, -22

14, -32

m = 0; (3, -2)

m = -43

; 11, -32m = -

54

; 1-2, 42m =

12

; 12, 12m =

43

; 13, 22m = 4; 12, -12m = -3; 1-1, 12m = -1; 10, 02m = 2; 10, 02

(a) Plot the ordered pairs (x, y) on a graph andconnect the points with straight lines.

(b) Compute and interpret the average rate of changein the maximum number of heartbeats between 20 and 30 years of age.

(c) Compute and interpret the average rate of changein the maximum number of heartbeats between 50 and 60 years of age.

(d) Based upon your results to parts (a), (b), and (c), do you think that the maximum number ofheartbeats is linearly related to age? Why?

SOURCE: American Heart Association


130. Raisins The following data represent the weight (ingrams) of a box of raisins and the number of raisinsin the box.


132. U.S. Population The following data represent thepopulation of the United States between 1930 and2000.


Age, x Average Income, y

20 $10,469

30 $31,161

40 $40,964

50 $43,627

60 $40,654

70 $21,784

Year, x Population, y

1930 123,202,624

1940 132,164,569

1950 151,325,798

1960 179,323,175

1970 203,302,031

1980 226,542,203

1990 248,709,873

2000 281,421,906

(b) Compute and interpret the average rate ofchange in the number of raisins between42.3 and 42.5 grams.

(c) Compute and interpret the average rate ofchange in the number of raisins between42.7 and 42.8 grams.

(d) Based upon your results to parts (a), (b), and (c),do you think that the number of raisins is linearlyrelated to weight? Why?

Weight(in grams), x

42.3 82

42.5 86

42.6 89

42.7 91

42.8 93

SOURCE: Jennifer Maxwell, student atJoliet Junior College

Number of Raisins, y

131. Average Income An individual’s income varies withage. The following data show the average income ofindividuals of different ages in the United States for2005.


(b) Compute and interpret the average rate ofchange in average income between 20 and 30 years of age.

(c) Compute and interpret the average rate ofchange in average income between 50 and 60 years of age.

(d) Based upon your results to parts (a), (b), and (c), do you think that average income is linearlyrelated to age? Why?

133. Measuring Temperature The relationship betweenCelsius (°C) and Fahrenheit (°F) degrees for meas-uring temperature is linear. Find an equation relat-ing °C and °F if 0°C corresponds to 32°F and 100°Ccorresponds to 212°F. Use the equation to find theCelsius measure of 60°F.

134. Building Codes As a result of the Americans withDisabilities Act (ADA, 1990), the building codestates that access ramps must have a slope not

steeper than Interpret what this result means.112

.

x

y

x

y

(b) Compute and interpret the average rate ofchange in population between 1930 and 1940.

(c) Compute and interpret the average rate ofchange in population between 1990 and 2000.

SOURCE: Statistical Abstract, 2008

(d) Based upon your results to parts (a), (b), and (c),do you think that population is linearly relatedto the year? Why?

SOURCE: U.S. Census Bureau

135. Which of the following equations might have thegraph shown? (More than one answer is possible.)

(a)(b)(c)(d)(e)

136. Which of the following equations might have thegraph shown? (More than one answer is possible.)

(a)(b)

(c)

(d)(e) -2x + y = -4

4x + 3y = -5

y = -23x - 3

y = -x + 2y = 2x - 5

-3x + 2y = -43x - 2y = 4y = 2x + 3y = -2x + 3y = 3x - 1

Section 1.7 Parallel and Perpendicular Lines 121

Explaining the Concepts137. Name the five forms of equations of lines given in

this section.

138. What type of line has one x-intercept, but no y-intercept?

139. What type of line has one y-intercept, but nox-intercept?

140. What type of line has one x-intercept and one y-intercept?

141. Are there any lines that have no intercepts? Explainyour answer.

142. Exploration Graph and on the same Cartesian

plane. What pattern do you observe? In general,describe the graph of

143. Exploration Graph and onthe

same Cartesian plane. What pattern do you observe?In general, describe the graph of with

144. Exploration Graph and

on the same Cartesian plane. What pattern do you

y = -2xy = -12

x, y = -x,

a 7 0.y = ax

y = 2xy =12

x, y = x,

y = 2x + b.

y = 2x - 4y = 2x + 7,y = 2x, y = 2x + 3,

observe? In general, describe the graph of with

The Graphing Calculator145. To see the role that the slope m plays in the graph of

a linear equation graph the followinglines on the same screen.

State some general conclusions about the graph offor Now graph

State some general conclusions about the graph offor

146. To see the role that the y-intercept b plays in thegraph of a linear equation graph thefollowing lines on the same screen.

State some general conclusions about the graph ofy = 2x + b.

Y3 = 2x + 5 Y4 = 2x - 4

Y1 = 2x Y2 = 2x + 2

y = mx + b,

m 6 0.y = mx + b

Y1 = -12x + 2 Y2 = -2x + 2 Y3 = -6x + 2

m Ú 0.y = mx + b

Y3 = 2x + 2 Y4 = 6x + 2

Y1 = 0x + 2 Y2 =12

x + 2

y = mx + b,

a 6 0.y = ax

Work SmartThe words “if and only if” given inthe definition mean that there aretwo statements being made.

If two nonvertical lines areparallel, then their slopes are equaland they have different y-intercepts.

If two nonvertical lines haveequal slopes and different y-intercepts, then they are parallel.

1.7 Parallel and Perpendicular LinesPreparing for Parallel and Perpendicular LinesBefore getting started, take this readiness quiz. If you get a problem wrong, go back to the sectioncited and review the material.

P1. Determine the reciprocal of 3. [Section R.3, pp. 23–24]

P2. Determine the reciprocal of [Section R.3, pp. 23–24]-35

.

OBJECTIVES1 Define Parallel Lines

2 Find Equations of Parallel Lines

3 Define Perpendicular Lines

4 Find Equations of PerpendicularLines

1 Define Parallel LinesWhen two lines (in the Cartesian plane) do not intersect (that is, they have no points incommon), they are said to be parallel.

PARALLEL LINES

Two nonvertical lines are parallel if and only if their slopes are equal and they havedifferent y-intercepts. Vertical lines are parallel if they have different x-intercepts.

Figure 52(a) on the next page shows nonvertical parallel lines. Figure 52(b) on thenext page shows vertical parallel lines.


P2. -53

13


Because the lines have the same slope, but different y-intercepts, the lines are parallel.

32

,

(b) Solve for y:

Add 3x to both sides:


Divide each term in the numerator by 2:

The slope of is and the

y-intercept is 3.

32

L1

y =32

x + 3

y =3x + 6

2

2y = 3x + 6

-3x + 2y = 6L1 Solve for y:Subtract 6x from

both sides:

Divide bothsides by �4:

Divide each term inthe numerator by �4:

The slope of is and the

y-intercept is -2.

32

L2

y =32

x - 2

y =-6x + 8

-4

-4y = -6x + 8

6x - 4y = 8L2

Quick1. Two lines are parallel if and only if they have the same and

different .

In Problems 2–4, determine whether the two lines are parallel.

2. L2 : y = -3x - 3 L1 : y = 3x + 1

4. L2 : 6x + 10y = 10 L1 : -3x + 5y = 10

3. L2 : -8x - 4y = 12L1 : 6x + 3y = 3

Figure 52Parallel lines

x

y

(a)

x

y

(b)

EXAMPLE 1 Determining Whether Two Lines Are ParallelDetermine whether the given lines are parallel.

(a)

L2 : 6x + 2y = 12

L1 : 4x + y = 8 (b)

L2 : 6x - 4y = 8

L1 : -3x + 2y = 6

Solution

To determine whether two lines are parallel, we determine the slope and y-interceptof each line by putting the equation of the line in slope-intercept form. If the slopesare the same, but the y-intercepts are different, then the lines are parallel.

(a) ] Solve L1 for y:Subtract 4x

from both sides:

The slope of is and they-intercept is 8.

-4L1

y = -4x + 8

4x + y = 8

Because the lines have different slopes, they are not parallel.

Solve for y:Subtract 6x from

both sides: Divide both sides

by 2: Divide each term inthe numerator by 2:

The slope of is and they-intercept is 6.

-3L2

y = -3x + 6

y =-6x + 12

2

2y = -6x + 12

6x + 2y = 12L2


2 Find Equations of Parallel LinesNow that we know how to identify parallel lines, let’s discuss how to find the equationof a line that is parallel to a given line.

Step 1: Find the slope of the given line byputting the equation in slope-intercept form. Subtract 4x from both sides:

Divide both sides by 2: y = -2x + 1

2y = -4x + 2

4x + 2y = 2

The slope of the line is �2.

Step 2: Use the point-slope form of a line withthe given point and the slope found in Step 1 tofind the equation of the parallel line.

m = -2, x1 = -2, y1 = 3: y - 3 = -21x - 1-222 y - y1 = m1x - x12

Step 3: Put the equation in slope-interceptform by solving for y. Distribute the �2:

Add 3 to both sides: y = -2x - 1

y - 3 = -2x - 4

y - 3 = -21x + 22

EXAMPLE 2 How to Find the Equation of a Line That Is Parallel to a Given LineFind an equation for the line that is parallel to and contains the point Graph the lines in theCartesian plane.


1-2, 32.4x + 2y = 2

The line parallel to containing is Notice that theslopes of the two lines are the same, but the y-intercepts are different. Figure 53 showsthe graph of the parallel lines.

y = -2x - 1.(-2, 3)4x + 2y = 2

3 x�3

y

�4

4

4x � 2y � 2

y � �2x � 1

(�2, 3)

Figure 53

Quick In Problems 5 and 6, find the equation of the line that contains the givenpoint and is parallel to the given line. Write the line in slope-intercept form. Graphthe lines.

5. (5, 8); 6. 3x + 2y = 101-2, 42;y = 3x + 1

3 Define Perpendicular LinesWhen two lines intersect at a right angle (90°), they are said to be perpendicular. See Figure 54.

We use the slopes of the lines to determine whether two lines are perpendicular.

x

90�

y

Figure 54Perpendicular lines

PERPENDICULAR LINES

Two nonvertical lines are perpendicular if and only if the product of their slopes isAlternatively, two nonvertical lines are perpendicular if their slopes are negative

reciprocals of each other.Any vertical line is perpendicular to any horizontal line.-1.

Work SmartIf and are negativereciprocals of each other, then

m1 =-1m2

.

m2m1


Solution

(a) The slope of is The slope of is Because the product of the slopes, the lines are notperpendicular. Notice that the slopes are not negative reciprocals of eachother.

(b) The slope of is The slope of is Because the

product of the slopes is the lines are perpendicular.

Notice that the slopes are negative reciprocals of each other.

m1#m2 =

23# a - 3

2b = -1,

m2 = -32

.L2m1 =23

.L1

m1m2 = 4 # 1-42 = -16 Z -1,m2 = -4.L2m1 = 4.L1

EXAMPLE 3 Finding the Slope of a Line Perpendicular to a Given Line

Find the slope of any line perpendicular to a line whose slope is

Solution

The negative reciprocal of is Any line whose slope is will be

perpendicular to the line whose slope is because -45# a5

4b = -1.

54

-45

-154

= -1 # 45

= -45

.54

54

.

Quick7. Two lines are perpendicular if and only if the product of their slopes is .

8. Find the slope of any line perpendicular to a line whose slope is -3.

EXAMPLE 4 Determining Whether Two Lines Are PerpendicularDetermine whether the given lines are perpendicular.

(a)

L2 : y = -4x - 3

L1 : y = 4x + 1 (b)

L2 : y = -32

x + 2

L1 : y =23

x - 5

Quick In Problems 9–11, determine whether the given lines are perpendicular.

9.

L2 : y = -15

x - 4

L1 : y = 5x - 3

11.L2: 3x - 6 = 0L1: 2y + 4 = 0

10.

L2 : x - 4y = 2

L1 : 4x - y = 3

4 Find Equations of Perpendicular LinesNow that we know how to find the slope of a line perpendicular to a second line, wecan find the equation of a line perpendicular to a second line.

EXAMPLE 5 How to Find the Equation of a Line Perpendicular to a Given LineFind an equation of the line that is perpendicular to the line and contains the point Write theequation in slope-intercept form. Graph the two lines.

14, -12.2x + 5y = 10


The equation of the line perpendicular to through is

Figure 55 shows the graphs of the two lines.y =52x - 11.

(4, -1)2x + 5y = 10

Step 1: Find the slope of the given line byputting the equation in slope-intercept form. Subtract 2x from both sides:

Divide both sides by 5: y = -25

x + 2

5y = -2x + 10

2x + 5y = 10

Step 2: Find the slope of the perpendicularline.

The slope of the perpendicular line is the negative reciprocal of

which is 52

.-25

,

Step 3: Use the point-slope form of a line withthe given point and the slope found in Step 2 tofind the equation of the perpendicular line. m =

52

, x1 = 4, y1 = -1: y - 1-12 =52

1x - 42 y - y1 = m1x - x12

The slope of the line is -25

.

Step 4: Put the equation in slope-intercept formby solving for y.

Distribute the

Subtract 1 from both sides:

52

:

y =52

x - 11

y + 1 =52

x - 10

y + 1 =52

1x - 42

x�4

y

4

�4

�8

�12

4

2x � 5y � 10

x � 11y � 5__2

(5, 0)

(0, 2)

Figure 55

Quick In Problems 12–14, find the equation of the line that contains the givenpoint and is perpendicular to the given line. Write the line in slope-intercept form.Graph the lines.

12. 13.

14. (3, -2); x = 3

1-3, -42; 3x - 4y = 81-4, 22; y = 2x + 1




Building SkillsIn Problems 15–18, a slope of a line is given. Determine (a) theslope of any line parallel to the line whose slope is given and (b) theslope of any line perpendicular to the line whose slope is given. SeeObjectives 1 and 3.

15. 16.

17. 18. m � 0

In Problems 19–26, determine whether the given linear equationsare parallel, perpendicular, or neither. See Objectives 1 and 3.

m = -56

m = -85

m = 5

29.

L is perpendicular to

30.


31.


32.


In Problems 33–46, find an equation of the line with the given prop-erties. Express your answer in slope-intercept form. Graph the lines.See Objectives 2 and 4.

33. Parallel to through the point (3, 1)


35. Perpendicular to through the point (2, 3)

36. Perpendicular to through the point (4, 1)

37. Parallel to through the point


39. Perpendicular to through the point (1, 3)x = 1

x = -2

1-1, -32y = 1

y = 4x + 3

y = -2x + 1

y = -3x + 1

y = 2x + 3

y = 3

x

y

3

L

�3

2

�2

y � 3(0, 3)

x = -2

x

y

3

L

�3

3

�3

x � �2

(0, 2)(�2, 0)

y =23

x + 1

x

y

3

L

�3

3

�3

x � 1y � 2__3

(0, 1)

(�4, 4)

y = -2x - 1

x

y

3

L

�3

3

�3

y � �2x � 1

(0, 2)

19. 20.

y = -13

x - 5

y = 3x - 1y = 5x - 7y = 5x + 4

21. 22.6x + 2y = 9

-3x - y = 32x - 8y = 38x + y = 12

23. 24.5x + 6y = 3

10x - 3y = 5x + 2y = 6

-4x + 2y = 12

25. 26.

2x +23

y = 1

12

x -32

y = 3

x -13

y =53

-x +13

y =13

In Problems 27–32, find an equation of the line L. Express youranswer in slope-intercept form. See Objectives 2 and 4.

27.

L is parallel to

28.

L is parallel to y = -12

x + 1

x

y

3

L

�3

3

�3

x � 1y � � 1__2

(2, 0)

(0, �1)

y =32

x

x

y

3

L

�3

3

�3 (0, �3)

3__2y � x


(a) Plot the points in a Cartesian plane. Connect thepoints to form a quadrilateral.

(b) Verify that the quadrilateral is a parallelogram byshowing that the opposite sides are parallel.

56. Geometry In geometry, we learn that a parallelogramis a quadrilateral in which both pairs of opposite sidesare parallel. Given the points

and

(a) Plot the points in a Cartesian plane. Connect thepoints to form a quadrilateral.

(b) Verify that the quadrilateral is a parallelogram byshowing that the opposite sides are parallel.

Extending the Concepts57. Find A so that is perpendicular to

58. Find B so that is perpendicular to

59. The figure shows the graph of two parallel lines.Which of the following pairs of equations might havesuch a graph?

(a)

(b)

(c)

(d)

(e)

60. The figure shows the graph of two perpendicularlines. Which of the following pairs of equations mighthave such a graph?

(a)

(b)

(c)

(d)

(e)

Explaining the Concepts61. If two nonvertical lines have the same x-intercept, but

different y-intercepts, can they be parallel? Explainyour answer.

62. Why don’t we say that a horizontal line is perpendicu-lar to a vertical line if they have slopes that are nega-tive reciprocals of each other?

2y + x = 2 x - 2y = 6 -3x + 4y = -2

3x + 4y = 5 3x - 2y = 5 2x + 3y = -2

x + 2y = 1 -2x + y = 3

y = -13

x - 2

y = 3x + 4

3x - 3y = 9 x - y = 3

-2x + y = 2 -2x + y = 5 x - 2y = -3 x - 2y = 4 y = 2x + 1 y = 2x + 3 y = -x - 1

y = x + 3

2x - 3y = 8.-6x + By = 3

4x + y = 3.Ax + 4y = 12

D = 1-1, 32,B = 14, 12, C = 15, 52, A = 1-2, -12,

40. Perpendicular to through the point




44. Perpendicular to through thepoint



Mixed PracticeIn Problems 47–52, two points on and two points on aregiven. Plot the points in the Cartesian plane and draw a linethrough the points. Compute the slope of the line containing thesepoints and determine whether the lines are parallel, perpendicular,or neither.

L2L1

13, -123x + y = 1

1-2, -325x + 2y = 1

12, -32 -2x + 5y - 3 = 0

1-4, 12 4x + 3y - 1 = 0

1-4, 322x + y = 5

3x - y = 2

12, -42y = 8

47. (1, 2); (6, 5) 48. (1, 1); (4, 3)

L2 : 1-1, 32; 13, -32L1 :

L2 : 1-2, 32; 11, -22L1 :

49. 50.

(4, 8); (2, 5)L2 :

L1 : 1-3, 02; 10, 22L2 : 1-1, -62; 1-4, 12L1 : 1-2, 42; 11, -32

51. (0, 5); (1, 3) 52.(0, 4); (8, 2)L2 :

L1 : 11, -32; 15, -42L2 : 1-4, 62; 10, 142L1 :

Applying the Concepts53. Geometry Given the points

and

(a) Plot the points in a Cartesian plane. Connect thepoints to form a triangle.

(b) Verify that the triangle is a right triangle byshowing that the line segment isperpendicular to the line segment andtherefore forms a right angle.

54. Geometry Given the points and

(a) Plot the points in a Cartesian plane. Connect thepoints to form a triangle.

(b) Verify that the triangle is a right triangle byshowing that the line segment isperpendicular to the line segment andtherefore forms a right angle.

55. Geometry In geometry, we learn that a parallelogramis a quadrilateral in which both pairs of opposite sidesare parallel. Given the points

and D = 13, 52,C = 18, 62, A = 12, 22, B = 17, 32,

ACAB

C = 1-5, 32, A = 1-2, -22, B = 13, 12,BC

AB

C = 12, 62, A = 11, 12, B = 14, 32,

x

y

x

y


1.8 Linear Inequalities in Two VariablesPreparing for Linear Inequalities in Two VariablesBefore getting started, take this readiness quiz. If you get a problem wrong, go back to the sectioncited and review the material.

P1. Determine whether satisfies the inequality [Section 1.4, pp. 85–88]

P2. Solve the inequality: [Section 1.4, pp. 85–88]-4x - 3 7 93x + 1 Ú 7.

x = 4

OBJECTIVES1 Determine Whether an Ordered

Pair Is a Solution to a LinearInequality

2 Graph Linear Inequalities

3 Solve Problems Involving LinearInequalities

In Section 1.4, we solved inequalities in one variable. In this section, we discuss linearinequalities in two variables.

1 Determine Whether an Ordered Pair Is a Solution to a Linear Inequality

Linear inequalities in two variables are inequalities in one of the forms

Preparing for...Answers P1. Satisfies

P2. or (- q, -3){x |x 6 -3}

where A, B, and C are real numbers and A and B are not both zero.If we replace the inequality symbol with an equal sign, we obtain the equation of a

line, The line separates the xy-plane into two regions, called half-planes. See Figure 56.

Ax + By = C.

Ax + By 6 C Ax + By 7 C Ax + By … C Ax + By Ú C

x

y

Ax � By � C

Figure 56

A linear inequality in two variables x and y is satisfied by an ordered pair (a, b) if,when x is replaced by a and y is replaced by b, a true statement results.

EXAMPLE 1 Determining Whether an Ordered Pair Is a Solution to a Linear Inequality In Two Variables

Determine which of the following ordered pairs are solutions to the linear inequality

(a) (2, 4) (b) (c) (1, 3)

Solution

(a) Let and in the inequality. If a true statement results, then (2, 4)is a solution to the inequality.

False

Because 10 is not less than 7, the statement is false, so (2, 4) is not a solution tothe inequality.

10 6? 7

6 + 4 6? 7

3122 + 4 6? 7x = 2, y = 4:

3x + y 6 7

y = 4x = 2

1-3, 123x + y 6 7.

Section 1.8 Linear Inequalities in Two Variables 129

(b) Let and in the inequality. If a true statement results, thenis a solution to the inequality.

True

Because is less than 7, the statement is true, so is a solution tothe inequality.

(c) Let and in the inequality.

True

Because 6 is less than 7, the statement is true, so (1, 3) is a solution to the inequality.

6 6? 7

3 + 3 6? 7

3112 + 3 6? 7x = 1, y = 3:

3x + y 6 7

y = 3x = 1

1-3, 12-8

-8 6? 7

-9 + 1 6? 7

31-32 + 1 6? 7x = -3, y = 1:

3x + y 6 7

1-3, 12 y = 1x = -3

Work SmartTo review the distinction betweenstrict and nonstrict inequalities, turnback to page 17.

Step 1: We replace the inequality symbol with an equal signand graph the corresponding line. If the inequality is strict ( or ), graph the line as a dashed line. If the inequality isnonstrict ( or ), graph the line as a solid line.

We replace with to obtain We graphthe line using a dashed linebecause the inequality is strict. See Figure 57(a) on thenext page.

3x + y = 7 1y = -3x + 723x + y = 7.=6

Ú…76

Quick1. If we replace the inequality symbol in with an equal sign, we

obtain the equation of a line, The line separates the xy-planeinto two regions, called .

2. Determine which of the following ordered pairs are solutions to the linearinequality

(a) (4, 1) (b) (c) (2, 3) (d) (0, 1)1-1, 22-2x + 3y Ú 3.

Ax + By = C.Ax + By 7 C

2 Graph Linear InequalitiesNow that we know how to determine whether an ordered pair is a solution to a linearinequality in two variables, we are prepared to graph linear inequalities in two variables.A graph of a linear inequality in two variables x and y consists of all points (x, y) whosecoordinates satisfy the inequality.

The graph of any linear inequality in two variables may be obtained by graphing theequation corresponding to the inequality, using dashes if the inequality is strict ( or�) and a solid line if the inequality is nonstrict ( or ). This graph will separate thexy-plane into two half-planes. In each half-plane either all points satisfy the inequalityor no points satisfy the inequality. So the use of a single test point is all that is requiredto obtain the graph of a linear inequality in two variables.

Ú…

EXAMPLE 2 How to Graph a Linear Inequality in Two VariablesGraph the linear inequality:


3x + y 6 7


Step 2: We select any test point that is not on the lineand determine whether the test point satisfies theinequality. When the line does not contain the origin, itis usually easiest to choose the origin, (0, 0), as thetest point.

Test Point: (0, 0):

TrueBecause 0 is less than 7, the point (0, 0) satisfies the

0 6? 7

3102 + 0 6? 7

3x + y 6 7

x�3

y

3

�3

3

6

9

3x � y � 7

(0, 7)

Figure 57

x�3

y

3

�3

3

6

9

3x � y � 7

(0, 0)

(0, 7)

Only one test point is needed to obtain the graph of the inequality. Why? Considerthe inequality presented in Examples 1 and 2. We notice that doesnot satisfy the inequality, while and do satisfy the inequality. Noticethat point A is not in the shaded region of Figure 57(b), while points B and C are in theshaded region. So, if a point does not satisfy the inequality, then none of the points in thehalf-plane containing that point satisfy the inequality. If a point does satisfy the inequal-ity, then all the points in the half-plane containing the point satisfy the inequality.

Below we summarize the steps for graphing a linear inequality in two variables.

C11, 32B1-3, 12 A12, 423x + y 6 7

inequality. Therefore, we shade the half-plane containingthe point (0, 0). See Figure 57(b). The shaded regionrepresents the solution to the linear inequality.

STEPS FOR GRAPHING A LINEAR INEQUALITY IN TWO VARIABLES

Step 1: Replace the inequality symbol with an equal sign and graph the resultingequation. If the inequality is strict ( or �), use dashes to graph the line; ifthe inequality is nonstrict ( or ) use a solid line. The graph separates the xy-plane into two half-planes.

Step 2: Select a test point P that is not on the line (that is, select a test point in oneof the half-planes).

(a) If the coordinates of P satisfy the inequality, then shade the half-planecontaining P.

(b) If the coordinates of P do not satisfy the inequality, then shade the half-plane that does not contain P.

Ú…

Work SmartAn alternative to using test points isto solve the inequality for y. If theinequality is of the form y or shade above the line. If theinequality is of the form y or ,shade below the line.

y …6

y Ú ,7

Quick3. True or False: The graph of a linear inequality is a line.

4. True or False: In a graph of a linear inequality in two variables with a strictinequality, the line separating the two half-planes should be dashed.

In Problems 5 and 6, graph each linear inequality.

5. 6. 6x - 3y … 15y 6 -2x + 3

(a) (b)


EXAMPLE 3 Graphing a Linear Inequality in Two Variables

Graph the linear inequality:

Solution

We replace the inequality symbol with an equal sign to obtain We graph the

line using a solid line because the inequality is nonstrict. See Figure 58(a).

Now, we select any test point that is not on the line and determine whether thetest point satisfies the inequality. Because the line contains the origin, we decide touse (2, 0) as the test point.

Test Point: (2, 0):

False

Because 0 is not greater than 1, the point (2, 0) does not satisfy the inequality.Therefore,we shade the half-plane that does not contain (2, 0). See Figure 58(b).The shaded regionrepresents the solution to the linear inequality.

Notice the inequality is of the form so we shade above the line.y Ú ,

0 Ú? 1

0 Ú?12# 2

y Ú12

x

y =12

x

y =12

x.

y Ú12

x

3 x�3

y

�3

3

(a)

Figure 58

3 x�3

y

�3

3

(b)

(2, 0)

Work SmartDo not use (0, 0) as a test point forequations of the form because the graph of this equationcontains the origin.

Ax + By = 0

Quick7. Graph the linear inequality: 2x + y 6 0

3 Solve Problems Involving Linear InequalitiesLinear inequalities involving two variables can be used to solve problems in areas suchas nutrition, manufacturing, or sales. Let’s look at an application of linear inequalitiesfrom nutrition.

EXAMPLE 4 Saturated Fat IntakeRandy really enjoys Wendy’s Junior Cheeseburgers and Biggie French Fries.However, he knows that his intake of saturated fat during lunch should not exceed16 grams. Each Junior Cheeseburger contains 6 grams of saturated fat and eachBiggie Fries contains 3 grams of saturated fat. (SOURCE: wendys.com)

(a) Write a linear inequality that describes Randy’s options for eating atWendy’s. That is, write an inequality that represents all the combinationsof Junior Cheeseburgers and Biggie Fries that Randy can order.

(b) Can Randy eat 2 Junior Cheeseburgers and 1 Biggie Fry during lunchand stay within his allotment of saturated fat?

(c) Can Randy eat 3 Junior Cheeseburgers and 1 Biggie Fry during lunchand stay within his allotment of saturated fat?

Solution

(a) We are going to use the first three steps in the problem-solving strategygiven in Section 1.2 on page 61 to help us develop the linear inequality.


Step 1: Identify We want to determine the number of Junior Cheeseburgersand Biggie Fries Randy can eat while not exceeding 16 grams of saturated fat.

Step 2: Name the Unknowns Let x represent the number of Junior Cheeseburg-ers that Randy eats and let y represent the number of Biggie Fries Randy eats.

Step 3: Translate If Randy eats one Junior Cheeseburger, then he will con-sume 6 grams of saturated fat. If he eats two, then he will consume 12 gramsof saturated fat. In general, if he eats x Junior Cheeseburgers, he will con-sume 6x grams of saturated fat. Similar logic for the Biggie Fries tells us thatif Randy eats y Biggie Fries, he will consume 3y grams of saturated fat. Thewords “cannot exceed” imply a inequality. Therefore, a linear inequalitythat describes Randy’s options for eating at Wendy’s is

The Model

(b) Letting and we obtain

True

Because the inequality is true, Randy can eat 2 Junior Cheeseburgers and1 Biggie Fry and remain within the allotment of 16 grams of saturated fat.

(c) Letting and we obtain

False

Because the inequality is false, Randy cannot eat 3 Junior Cheeseburgers and1 Biggie Fry and remain within the allotment of 16 grams of saturated fat.

21 …? 16

6132 + 3112 …? 16

y = 1,x = 3

15 …? 16

6122 + 3112 …? 16

y = 1,x = 2

6x + 3y … 16

…

Quick8. Avery is on a diet that requires that he consume no more than 800 calories for

lunch. He really enjoys Wendy’s Chicken Breast filet and Frosties. Each ChickenBreast filet contains 430 calories and each Frosty contains 330 calories.

(a) Write a linear inequality that describes Avery’s options for eating at Wendy’s.(b) Can Avery eat 1 Chicken Breast filet and 1 Frosty and stay within his

allotment of calories?(c) Can Avery eat 2 Chicken Breast filets and 1 Frosty and stay within his

allotment of calories?

1.8 EXERCISES

Building SkillsIn Problems 9–12, determine whether the given points are solutionsto the linear inequality. See Objective 1.

9.

(a) (0, 1)(b)(c) 18, -121-2, 42x + 3y 6 6 10.

(a)(b)(c) 1-5, 4211, -3212, -122x + y 7 -3

11.

(a)(b) (0, 2)(c) (0, 3)

1-4, 22-3x + 4y Ú 12 12.

(a)(b) (3, 0)(c) 11, 2211, 02

2x - 5y … 2

In Problems 13–32, graph each inequality. See Objective 2.

13. y 7 3 14. y 6 -2

15. x Ú -2 16. x 6 7

17. y 6 5x 18. y Ú23

x

1–8. are the Quick s that follow each EXAMPLE

19. y 7 2x + 3 20. y 6 -3x + 1

21. y …12

x - 5 22. y Ú -43

x + 5

23. 3x + y … 4 24. -4x + y Ú -5

25. 2x + 5y … -10 26. 3x + 4y Ú 12

28. -5x + 3y 6 3027. -4x + 6y 7 24


Extending the ConceptsIn Problems 37–40, determine the linear inequality whose graph isgiven.

Applying the Concepts33. Nutrition Sammy goes to McDonald’s for lunch. He

is on a diet that requires that he consume no morethan 150 grams of carbohydrates for lunch. Sammyenjoys McDonald’s Filet-o-Fish sandwiches andFrench fries. Each Filet-o-Fish has 45 grams of car-bohydrates and each small order of French fries has40 grams of carbohydrates. (SOURCE: McDonald’s Corp.)

(a) Write a linear inequality that describes Sammy’soptions for eating at McDonald’s.

(b) Can Sammy eat two Filet-o-Fish and one order offries and meet his carbohydrate requirement?

(c) Can Sammy eat three Filet-o-Fish and one order of fries and meet his carbohydraterequirement?

34. Salesperson Juanita sells two different computermodels. For each Model A computer sold shemakes $45 and for each Model B computer soldshe makes $65. Juanita set a monthly goal of earningat least $4000.

(a) Write a linear inequality that describes Juanita’soptions for making her sales goal.

(b) Will Juanita make her sales goal if she sells 50Model A and 28 Model B computers?

(c) Will Juanita make her sales goal if she sells 41Model A and 33 Model B computers?

35. Production Planning Acme Switch Company is asmall manufacturing firm that makes two differentstyles of microwave switches. Switch A requires 2 hours to assemble, while Switch B requires 1.5 hoursto assemble. Suppose that there are at most 80 hoursof assembly time available each week.

(a) Write a linear inequality that describes Acme’soptions for making the microwave switches.

(b) Can Acme manufacture 24 of Switch A and 41 ofSwitch B in a week?

(c) Can Acme manufacture 16 of Switch A and 45 ofSwitch B in a week?

36. Budget Constraints Johnny can spend no more than$3.00 that he got from his grandparents. He goes tothe candy store and wants to buy gummy bears thatcost $0.10 each and suckers that cost $0.25 each.

(a) Write a linear inequality that describes Johnny’soptions for buying candy.

(b) Can Johnny buy 18 gummy bears and 5 suckers? (c) Can Johnny buy 19 gummy bears and

4 suckers?

37. 38.

3 x�3

y

�3

3

(0, 1)

(2, �3)

3 x�3

y

�3

3(2, 2)

(0, �2)

39. 40.

3 x�3

y

�2

3

(2, 1)

(�1, �3)

3 x�3

y

�3

3(�2, 3)

(2, �1)

Explaining the Concepts41. If an ordered pair does not satisfy a linear inequality,

then we shade the side opposite the point. Explainwhy this works.

42. Explain why we cannot use a test point that lies onthe line separating the two half-planes when deter-mining the solution to a linear inequality involvingtwo variables.

The Graphing CalculatorGraphing calculators can be used to graph linear inequali-ties in two variables. Figure 59 shows the result of Exam-ple 2 using a TI-84 Plus graphing calculator. Consult yourowner’s manual for specific keystrokes.

In Problems 43–54, graph the inequalities using a graphing calculator.

43. 44.

45. 46.

47. 48.

49. 50.

51. 52.

53. 54. 3x + 4y Ú 122x + 5y … -10

-4x + y Ú -53x + y … 4

y Ú -43

x + 5y …12

x - 5

y 6 -3x + 1y 7 2x + 3

y Ú23

xy 6 5x

y 6 -2y 7 3

30.x

3-y

4… 129.

x

2+y

36 1

31.23

x -32

y Ú 2 32.54

x -35

y … 2

�2 6

10

�10

Figure 59


YOU SHOULD BE ABLE TO... EXAMPLE REVIEW EXERCISES

CHAPTER 1 ReviewSection 1.1 Linear Equations

KEY CONCEPTS

� Linear Equation in One VariableAn equation equivalent to one of the form where a and b are real numbers with

� Addition Property of Equality

For real numbers a, b, and c, if then

� Multiplication Property of Equality

For real numbers a, b, and c where if then ac = bc.a = b,c Z 0,

a + c = b + c.a = b,

a Z 0.ax + b = 0,

KEY TERMS

Equation in one variableSides of the equationSolutionSatisfiesSolve an equationSolution setEquivalent equationsConditional equationContradictionIdentity

Focus: Solving equations and inequalitiesTime: 15 minutesGroup size: 4Materials needed: One blank piece of notebook paper per groupmember

To the right are two equations and two inequalities. In this activityyou will work together to solve these problems by following theprocedure below. Be sure to read through the entire proceduretogether before beginning the activity so all group members willunderstand the procedure.

Procedure

1. Write one of the problems shown at the top of yourpaper. Be sure that each group member chooses adifferent problem.

2. Two lines below the original problem, write out thefirst step for solving the problem.

3. Fold the top of the paper down to cover the originalproblem, leaving your first step visible.

4. Pass your paper to a different group member. Youmight want to arrange your seats so that the paperscan be passed around in a circle.

5. Continue solving the problems one step at a time,covering the step above yours, and passing the paperto the next group member, until all problems aresolved.

6. As a group, discuss the solutions and decide whetheror not they are correct. If any of the solutions areincorrect, solve them correctly together.

Problems

(a)

(b)

(c)

(d) 4(x + 1) - 2x Ú 5x - 11

31x - 42 - 5x 7 2x + 12

31x - 22 + 8 = 5x - 21x - 124 - 1x - 32 = -10 + 51x + 12

CHAPTER 1 Activity: Pass the Paper

5–18

Examples 7 and 8 5–14

1–4Example 1

2 Solve linear equations (p. 49)

1 Determine whether a number is a solution to anequation (p. 48)

Examples 2 through 6

3 Determine whether an equation is a conditionalequation, identity, or contradiction (p. 54)

Section 1.2 An Introduction to Problem SolvingKEY CONCEPTS

� Simple Interest Formulawhere I is interest, P is principal, r is the per annum

interest rate expressed as a decimal, t is time in years

� Uniform Motion Formula

where d is distance, r is average speed, t is timed = rt,

I = Prt,

Chapter 1 Review 135

In Problems 1–4, determine which of the numbers, if any,are solutions to the given equation.

1.

2.

3.

4.

In Problems 5–14, solve the linear equation. State whetherthe equation is an identity, contradiction, or conditionalequation.

5.

6.

7.

8.

9.

10.

11.

12.x

3+

2x5

=x - 20

15

2y + 34

-y

2= 5

312r + 12 - 5 = 91r - 12 - 3r

-21x - 42 = 8 - 2x

7x + 5 - 8x = 13

2x + 5x - 1 = 20

-4 = 8 - 3y

2w + 9 = 15

w - 73

-w

4 = -

76

; w = -14, w = 7

y = -2, y = 0

4y - 11 - y2 + 5 = -6 - 213y - 52 - 2y;

-1 - 4x = 213 - 2x2 - 7; x = -2, x = -1

3x - 4 = 6 + x; x = 5, x = 6

13.

14.

In Problems 15 and 16, determine which values of the vari-able must be excluded from the domain.

15.

16.

17. State Income Tax A resident of Missouri completesher state tax return and determines that she paid$2370 in state income tax in 2007. The solution to theequation representsher Missouri taxable income x in 2007. Solve theequation to determine her Missouri taxable income.(SOURCE: Missouri Department of Revenue)

18. Movie Club The DVD club to which you belongoffers unlimited DVDs at $10 off the regular price ifyou buy 1 at the regular price. You purchase 5 DVDsthrough this offer and spend $69.75 (not includingtax and shipping). The solution to the equation

represents the regular clubprice x for a DVD. Solve the equation to determinethe regular club price for a DVD.

x + 41x - 102 = 69.75

2370 = 0.061x - 90002 + 315

6x - 5 61x - 12 + 3

8 2x + 3

2.1w - 312.4 - 0.2w2 = 0.913w - 52 - 2.7

0.21x - 62 + 1.75 = 4.25 + 0.113x + 102

KEY TERMS

Problem solvingMathematical modelingModeling processMathematical modelDirect translationInterest

PrincipalRate of interestSimple interestMixture problemsUniform motion


1 Translate English sentences into mathematical statements (p. 58) Example 1 19–22

2 Model and solve direct translation problems (p. 61) Examples 2 through 6 23–28

3 Model and solve mixture problems (p. 65) Examples 7 and 8 29–32

4 Model and solve uniform motion problems (p. 68) Example 9 33–34

In Problems 19–22, translate each of the following Englishstatements into a mathematical statement. Do not solve theequation.

19. The sum of three times a number and 7 is 22.

20. The difference of a number and 3 is equivalent to the

quotient of the number and 2.

21. 20% of a number equals the difference of the numberand 12.

22. The product of six and a number is the same as 4 lessthan twice the number.

For Problems 23 and 24, translate each English statementinto a mathematical statement. Then solve the equation.

23. Shawn is 8 years older than Payton and the sum of theirages is 18. What are their ages?

24. The sum of five consecutive odd integers is 125. Findthe integers.


Section 1.3 Using Formulas to Solve ProblemsKEY CONCEPTS

� Geometry Formulas (see pages 74–75)

In Problems 35–40, solve for the indicated variable.

35. Solve for x.

36. Solve for C.

37. Solve for W.

38. Solve for

39. Solve for T.

40. Solve for W.S = 2LW + 2LH + 2WH

PV = nRT

m2 .r = m1v1 + m2v2

P = 2L + 2W

F =95

C + 32

y =kx

In Problems 41–44, solve for y.

41.

42.

43.

44.

45. Temperature Conversions To convert temperaturesfrom Fahrenheit to Celsius, we can use the formula.

25x +

13y = 8

48x - 12y = 60

-5x + 4y = 10

3x + 4y = 2


1 Solve for a variable in a formula (p. 73) Examples 2 and 3 35–44

2 Use formulas to solve problems (p. 76) Examples 4 and 5 45–52

25. Computing Grades Logan is in an elementary statis-tics course and has test scores of 85, 81, 84, and 77. Ifthe final exam counts the same as two tests, whatscore does Logan need on the final to have an aver-age of 80?

26. Home Equity Loans On January 15, 2008, Bank ofAmerica offered a home equity line of credit at a rateof 6.24% annual simple interest. If Cherie has such acredit line with a balance of $3200, how much interestwill she accrue at the end of 1 month?

27. Discounted Price Suppose that REI sells a 0° sleep-ing bag at the discounted price of $94.50. If this pricerepresents a discount of 30% off the original sellingprice, find the original price.

28. Minimum Wage On July 24, 2008, the federal mini-mum wage was increased 12% to $6.55. Determine thefederal minimum wage prior to July 24, 2008.

29. Making a Mixture CoffeeAM sells chocolate-coveredblueberries for $10.95 per pound and chocolate-covered strawberries for $13.95 per pound. The com-pany wants to sell a mix of the two that would sellfor $12.95 per pound with no loss in revenue. Howmany pounds of each treat should be used to make12 pounds of the mix?

30. A Sports Mix The Candy Depot sells baseball gum-balls for $3.50 per pound and soccer gumballs for$4.50 per pound. The company wants to sell a “sports

mix” that sells for $3.75 per pound with no loss in rev-enue. How many pounds of each gumball type shouldbe included to make 10 pounds of the mix?

31. Investments Angie received an $8000 bonus andwants to invest the money. She can invest part of themoney at 8% simple interest with a moderate riskand the rest at 18% simple interest with a high risk.She wants an overall annual return of 12% but doesnot want to risk losing any more than necessary. Howmuch should she invest at 18% to reach her goal?

32. Antifreeze A 2008 Chevrolet Malibu has an enginecoolant system capacity of 7.5 quarts. If the system iscurrently filled with a mixture that is 30% antifreeze,how much of this mixture should be drained and re-placed with pure antifreeze so that the system is filledwith a mixture that is 50% antifreeze?

33. Road Trip On a 300-mile trip to Chicago, Josh drovepart of the time at 60 miles per hour and the remain-der of the trip at 70 miles per hour. If the total triptook 4.5 hours, for how many miles did Josh drive ata rate of 60 miles per hour?

34. Uniform Motion An F15 Strike Eagle near New YorkCity and an F14 Tomcat near San Diego are about2200 miles apart and traveling towards each other.The F15 is traveling 200 miles per hour faster than theF14 and the planes pass each other after 50 minutes.How fast is each plane traveling?

KEY TERMS

FormulaGolden rectangleSupplementary anglesComplementary angles


If the melting point for platinum is

3221.6°F, convert this temperature to degrees Celsius.

46. Angles in a Triangle The measure of each congruentangle in an isosceles triangle is 30 degrees larger thanthe measure of the remaining angle. Determine themeasures of all three angles.

47. Window Dimensions The perimeter of a rectangularwindow is 76 feet. The window is 8 feet longer than itis wide. Find the dimensions of the window.

48. Long-Distance Phone Calls A long-distance tele-phone company charges a monthly fee of $2.95 and aper-minute charge of $0.04. The monthly cost for longdistance is given by where x is thenumber of minutes used.

(a) Solve the equation for x.

(b) How many full minutes can Debbie use in onemonth on this plan if she does not want tospend more than $20 in long distance in onemonth?

49. Concrete Rick has 80 cubic feet of concrete to pourfor his new patio. If the patio is rectangular and Rickwants it to be 12 feet by 18 feet, how thick will thepatio be if he uses all the concrete?

C = 2.95 + 0.04x

C =59

1F - 322. 50. Right Circular Cones The lateralsurface area for a frustum of aright circular cone is given by

where s is theslant height of the frustum, R isthe radius of the base, and r is theradius of the top.(a) Solve the equation for r.

(b) If the frustum of a right circular cone has a lateralsurface area of square feet, a slant height of2 feet, and a base whose radius is 3 feet, what isthe radius of the top of the frustum?

51. Heating Bills On January 15, 2008, the winter energycharge C for Illinois Power residential service wascomputed using the formula forx kilowatt hours (kwh).(a) Solve the equation for x.

(b) How many kwh were used if the winter energycharge was $206.03? Round to the nearest wholenumber.

52. Supplementary and Complementary Angles Thesupplement of an angle and the complement ofthe same angle sum to 150°. What is the measure ofthe angle?

C = 7.48 + 0.08674x

10p

(R + r)A = pss

R

r

Section 1.4 Linear InequalitiesKEY CONCEPTS

� Linear Inequality in One VariableAn inequality of the form or where a, b, and c are real numbers with

� Interval Notation versus Inequality Notation

a Z 0.ax + b Ú c,ax + b 6 c, ax + b … c, ax + b 7 c,

INTERVAL NOTATION INEQUALITY NOTATION GRAPH

The open interval (a, b)

The closed interval [a, b]

The half-open interval [a, b)

The half-open interval (a, b]

The interval

The interval

The interval

The interval

The interval 5x | x is a real number61-q , q)

a5x | x 6 a61-q , a)

a5x | x … a61-q , a]

a5x | x 7 a61a, q2

a5x |x Ú a6[a, q2

ba5x | a 6 x … b6

ba5x | a … x 6 b6

a b5x | a … x … b6

a b5x | a 6 x 6 b6

KEY TERMS

Solve an inequalitySolutionsSolution setInterval notationClosed intervalOpen intervalHalf-open or half-closed

intervalLeft endpointRight endpointEquivalent inequalities



� Addition Property of InequalitiesFor real numbers a, b, and c

If then

If then a + b 7 b + c.a 7 b,

a + c 6 b + c.a 6 b,

� Multiplication Properties of Inequalities

For real numbers a, b, and c

If and if then

If and if then

If and if then

If and if then ac 6 bc.c 6 0,a 7 b

ac 7 bc.c 6 0,a 6 b

ac 7 bc.c 7 0,a 7 b

ac 6 bc.c 7 0,a 6 b

1 Represent inequalities using the real number lineand interval notation (p. 81)

53–56Examples 1 through 4

2 Understand the properties of inequalities (p. 84) 57–58

3 Solve linear inequalities (p. 85) 59–68Examples 5 through 8

4 Solve problems involving linear inequalities (p. 88) 69–72Example 9

In Problems 53 and 54, write each inequality using intervalnotation and graph the inequality.

53.

54.

In Problems 55 and 56, write each interval as an inequalityinvolving x and graph the inequality.

55.

56.

In Problems 57 and 58, use the Addition Property and/orMultiplication Property to find a and b.

57. If then

58. If then

In Problems 59–68, solve each linear inequality. Express yoursolution using set-builder notation and interval notation.Graph the solution set.

59.

60.

61.

62.

63.

64. 21x + 12 + 1 7 21x - 2231p - 22 + 15 - p2 7 2 - 1p - 32-7x - 8 6 -22

-7 … 31h + 12 - 8

2 6 1 - 3x

3x + 12 … 0

a 6 3x + 5 6 b.-2 6 x 6 0,

a … 2x - 3 … b.5 … x … 9,

[-1, 321-q , 4]

x 7 -2

2 6 x … 7

65.

66.

67.

68.

69. Octoberfest The German Club plans to rent a hallfor their annual Octoberfest banquet. The hall costs$150 to rent plus $7.50 for each person who attends.If the club does not want to spend more than $600for the event, how many people can attend thebanquet?

70. Car Rentals A Ford Taurus at Enterprise Rent-a-Carrents for $43.46 per day. You receive 150 free milesper day but are charged $0.25 per mile for any addi-tional miles. How many miles can you drive per day,on average, and not exceed your daily budget of$60.00?

71. Fund Raising A middle school band sells $1 candybars at a carnival to raise money for new instru-ments. The band pays $50.00 to rent a booth andmust pay the candy company $0.60 for each barsold. How many bars must the band sell to bemaking a profit?

72. Movie Club A DVD club offers unlimited DVDsfor $9.95 if you purchase one for $24.95. How manyDVDs can you purchase without spending more than$72.00?

25y - 20 7

23y + 12

-49

w +7

126

536

0.03x + 0.10 7 0.52 - 0.07x

51x - 12 - 7x 7 212 - x2


Section 1.5 Rectangular Coordinates and Graphs of EquationsKEY CONCEPTS

� Graph of an Equation in Two VariablesThe set of all ordered pairs (x, y) in the xy-plane that satisfy the equation

� Intercepts

The points, if any, where a graph crosses or touches thecoordinate axes

KEY TERMS

x-axisy-axisOriginRectangular or Cartesian

coordinate systemxy-planeCoordinate axesOrdered pairCoordinatesx-coordinatey-coordinateAbscissaOrdinate

QuadrantsEquation in two

variablesSides SatisfyGraph of an equation

in two variablesPoint-plotting methodComplete graphInterceptx-intercepty-intercept


1 Plot points in the rectangular coordinate system (p. 93) Example 1 73, 74

2 Determine whether an ordered pair is a point on the graphof an equation (p. 95)

75, 76Example 2

3 Graph an equation using the point-plotting method (p. 97) 77–82Examples 3 through 5

4 Identify the intercepts from the graph of an equation (p. 99) 83Example 6

5 Interpret graphs (p. 100) 84Example 7

In Problems 73 and 74, plot each point in the same xy-plane.Tell in which quadrant or on what coordinate axis eachpoint lies.

73.

C(0, 4)

E(1, 0)

F(4, 3)

D1-5, 12

B1-1, -32A12, -42

In Problems 77–82, graph each equation by plotting points.

77. 78.

79. 80.

81. 82.

In Problem 83, the graph of an equation is given. List theintercepts of the graph.

83.

x

y

�4

�4

4

4

x = y2 + 1y = x3 + 2

y = ƒx + 2 ƒ - 1y = -x2 + 4

2x + y = 3y = x + 2

74. A(3, 0)

B(1, 5)

F10, -52E15, -22D1-1, 42C1-3, -52

In Problems 75 and 76, determine whether the given pointsare on the graph of the equation.

75.(a) (3, 1)

(b)

(c) (4, 0)

(d) a13

, -3b

12, -123x - 2y = 7 76.

(a)

(b) (1, 1)

(c)

(d) a12

, 32b

1-2, 1621-1, 32

y = 2x2 - 3x + 2


84. Cell Phones A cellular phone company offers a planfor $40 per month for 3000 minutes with additional min-utes costing $0.05 per minute.The graph below showsthe monthly cost, in dollars, when x minutes are used.

(a) If you talk for 2250 minutes in a month, howmuch is your monthly bill?

(b) Use the graph to estimate your monthly bill ifyou talk for 12 thousand minutes.

Minutes Used (in thousands)

2000

1500

1000

500

090 18 27 36 45

(44.64, 2122)

(3, 40)(0, 40)

Cos

t per

Mon

th (

$)

Section 1.6 Linear Equations in Two VariablesKEY CONCEPTS

� Standard Form of a Linewhere A, B, and C are real numbers. A and B are not both 0.

� Finding Intercepts

x-intercept(s): Let in the equation and solve for x

y-intercept(s): Let in the equation and solve for y

� Equation of a Vertical Line

where a is the x-intercept

� Equation of a Horizontal Line

where b is the y-intercept

� Slope of a Line

Let and be two distinct points. If the slope mof the nonvertical line L containing P and Q is defined by the formula

If then L is a vertical line and the slope m of L is undefined(since this results in division by 0).

� When the slope of a line is positive, the line slants upward from left to right.

� When the slope of a line is negative, the line slants downward from left to right.

� When the slope of a line is zero, the line is horizontal.

� When the slope of a line is undefined, the line is vertical.

� Point-slope Form of a Line

An equation of a nonvertical line of slope m that contains the point is

� Slope-intercept Form of a Line

An equation of a nonvertical line with slope m and y-intercept b is y = mx + b.

y - y1 = m1x - x12.1x1 , y12

x1 = x2 ,

m =y2 - y1

x2 - x1, x1 Z x2

x1 Z x2 ,Q = 1x2 , y22P = 1x1 , y12

y = b,

x = a,

x = 0

y = 0

Ax + By = C,

KEY TERMS

Linear equationStandard formLineVertical lineHorizontal lineRunRiseSlopeUndefined slopeAverage rate of change


2 Graph linear equations using intercepts (p. 106) Examples 2 and 3 89–92

3 Graph vertical and horizontal lines (p. 108) Examples 4 and 5 93–95

4 Find the slope of a line given two points (p. 109) Examples 6 and 7 96–99

5 Interpret slope as an average rate of change (p. 112) Example 8 100

6 Graph a line given a point and its slope (p. 114) Example 9 101–104

7 Use the point-slope form of a line (p. 114) Example 10 105–108

8 Identify the slope and y-intercept of a line from its equation (p. 115) Example 11 113–114

9 Find the equation of a line given two points (p. 116) Examples 12 and 13 109–112


1 Graph linear equations using point plotting (p. 105) Example 1 85–88

In Problems 85–88, graph each linear equation by plottingpoints.

85. 86.

87. 88.

In Problems 89–92, graph each linear equation by findingits intercepts.

89. 90.

91. 92.

In Problems 93–95, graph each linear equation.

93.94.

95.

In Problems 96 and 97, (a) find the slope of the line and(b) interpret the slope.96.

97.

(�3, 1)

(5, �1)x

y

�5

5

�5

(�2, �1)

(2, 4)

x

y

5

5

�5

3x + 5 = -1

y = -8x = 4

4x + y = 834

x -12

y = 1

4x + 3y = 05x + 3y = 30

-3x + 2y = 85x - 2y = 6

x - y = -4x + y = 7

In Problems 98 and 99, plot each pair of points and deter-mine the slope of the line containing them. Graph the line.

98.

99.

100. Illinois’s Population The following data representthe population of Illinois between 1940 and 2000.

14, 52; 10, -121-1, 52; 12, -12

Year, x Population, y

1940 7,897,241

1950 8,712,176

1960 10,081,158

1970 11,110,285

1980 11,427,409

1990 11,430,602

2000 12,419,293


(b) Compute and interpret the average rateof change in population between 1940and 1950.

(c) Compute and interpret the average rate of change in population between 1980 and 1990.

(d) Compute and interpret the average rate of change in population between 1990 and 2000.

(e) Based upon the results to parts (a), (b), (c),and (d), do you think that population is linearlyrelated to the year? Why?

SOURCE: U.S. Census Bureau


In Problems 101–104, graph the line containing the point Pand having slope m. Do not find the equation of the line.

101.

102. P (3, 2)

103. m is undefined;

104.

In Problems 105 and 106, find an equation of the line.Express your answer in either slope-intercept or standardform, whichever you prefer.

105.

106.

(�1, �3)

(2, 6)

x

y

5�5

5

(�2, 3)

(6, �1)x

y

�4 4

5

�5

m = 0; P1-3, 12P12, -42

m = -23

;

m = 4; P1-1, -52

In Problems 107 and 108, find an equation of the line withthe given slope and containing the given point. Expressyour answer in either slope-intercept or standard form,whichever you prefer.

107.

108.

In Problems 109–112, find an equation of the line contain-ing the given points. Express your answer in either slope-intercept or standard form, whichever you prefer.

109. (6, 2);

110.

111.

112.

In Problems 113–114, find the slope and y-intercept of eachline. Graph the line.

113.

114. 2x + 3y = 12

y = 4x - 6

1-1, 22; 18, -1214, -12; 11, -721-2, 32; 14, 32

1-3, 52

m =35

; 1-10, -42m = -1; 13, 22

Section 1.7 Parallel and Perpendicular LinesKEY CONCEPTS

� Parallel LinesTwo lines are parallel if they have the same slope, but different y-intercepts.

� Slopes of Perpendicular Lines

Two lines are perpendicular if the product of their slopes is �1. Alternatively,two lines are perpendicular if their slopes are negative reciprocals of each other.

KEY TERMS

Parallel linesPerpendicular lines


1 Define parallel lines (p. 121) Example 1 115, 117–120

2 Find equations of parallel lines (p. 123) Example 2 121–123

3 Define perpendicular lines (p. 123) Examples 3 and 4 116, 117–120

4 Find equations of perpendicular lines (p. 124) Example 5 124–126

In Problems 115–116, the slope of a line L is

115. Determine the slope of a line that is parallel to L.

116. Determine the slope of a line that is perpendicularto L.

m = -38

.In Problems 117–120, determine whether the given pairs oflinear equations are parallel, perpendicular, or neither.

117.

118. 3x + 4y = 28 6x - 8y = 16

9x + 3y = -3 x - 3y = 9


119.

120.

In Problems 121–126, find an equation of the line with thegiven properties. Express your answer in slope-interceptform, if possible.Graph the lines.

121. Parallel to through (1, 2)y = -2x - 5

y = 2 x = 2

-6x + 3y = 0 2x - y = 3 122. Parallel to through (4, 3)

123. Parallel to through

124. Perpendicular to through (6, 2)

125. Perpendicular to through

126. Perpendicular to through 15, -42x = 2

1-3, -223x + 4y = 6

y = 3x + 7

11, -42x = -3

5x - 2y = 8

Section 1.8 Linear Inequalities in Two VariablesKEY CONCEPT

� Linear inequalities in two variables are inequalities in one of the forms

where A, B, and C are real numbers and A and B are not both zero.

Ax + By 6 C Ax + By 7 C Ax + By … C Ax + By Ú C

KEY TERMS

Half-planesSatisfiedGraph of a linear inequality

in two variables


1 Determine whether an ordered pair is a solution to a linear inequality (p. 128) Example 1 127–128

2 Graph linear inequalities (p. 129) Examples 2 and 3 129–134

3 Solve problems involving linear inequalities (p. 131) Example 4 135–136

In Problems 127 and 128, determine whether the givenpoints are solutions to the linear inequality.

127.(a)

(b)

(c) 15, -121-6, 15214, -22

5x + 3y … 15

(b) Can Ethan buy 5 movie tickets and 2 musicCDs?

(c) Can Ethan buy 2 movie tickets and 2 musicCDs?

136. Fund Raising For a fund raiser, the Math Clubagrees to sell candy bars and candles. The club’sprofit will be 50¢ for each candy bar and $2.00for each candle it sells. The club needs to earnat least $1000 in order to pay for an upcomingfield trip.(a) Write a linear inequality that describes the com-

bination of candy bars and candles that must besold.

(b) Will selling 500 candy bars and 350 candles earnenough for the trip?

(c) Will selling 600 candy bars and 400 candles earnenough for the trip?

128.(a) (2, 3)

(b)

(c) 1-1, 3215, -22

x - 2y 7 -4

In Problems 129–134, graph each inequality.

129. 130.

131. 132.

133. 134.

135. Entertainment Budget Ethan’s entertainmentbudget permits him to spend a maximum of $60 permonth on movie tickets and music CDs. Movie tick-ets cost on average $7.50 each. Music CDs average$15.00 each.(a) Write a linear inequality that describes

Ethan’s options for spending the $60maximum budget.

x 7 -82x + 3y 6 0

y Ú 53x + 4y 7 20

2x - 4y … 12y 6 3x - 2


Remember to use your Chapter Test Prep Video CD to see fully worked-outsolutions to any problems you would like to review.CHAPTER 1 TEST

1. Determine which, if any, of the following are solutionsto

(a)

(b)

2. Write the following inequalities in interval notationand graph on a real number line.

(a)

(b)

In Problems 3 and 4, translate the English statement into amathematical statement. Do not attempt to solve.

3. Three times a number, decreased by 8, is 4 more thanthe number.

4. Two-thirds of a number, increased by twice the differ-ence of the number and 5, is more than 7.

In Problems 5 and 6, solve the equation. Determine if theequation is an identity, a contradiction, or a conditionalequation.

5.

6.

In Problems 7–9, solve the inequality and graph the solutionset on a real number line.

7.

8.

9.

10. Solve

11. Computer Sales Glen works as a computer salesmanand earns $400 weekly plus 8% commission on hisweekly sales. If he wants to make at least $750 in aweek, how much must his sales be?

12. Party Costs A recreational center offers a children’sbirthday party for $75 plus $5 for each child. Howmany children were at Payton’s birthday party if thetotal cost for the party was $145?

13. Sandbox Rick is building a rectangular sandbox forhis daughter. He wants the length of the sandbox tobe 2 feet more than the width and he has 20 feet oflumber to build the frame. Find the dimensions of thesandbox.

14. Mixture Two acid solutions are available to achemist. One is a 10% nitric acid solution and theother is a 40% nitric acid solution. How much of eachtype of solution should be mixed together to form 12 liters of a 20% nitric acid solution?

7x + 4y = 3 for y.

-x + 4 … x + 3

4x + 7 7 2x - 31x - 22x + 2 … 3x - 4

7 + x - 3 = 31x + 12 - 2x

5x - 1x - 22 = 6 + 2x

3 6 x … 7

x 7 -4

x = -2

x = 6

31x - 72 + 5 = x - 4.15. Ironman Race The last leg of the Ironman competi-

tion is a 26.2-mile run. Contestant A runs at a constantrate of 8 miles per hour. If Contestant B starts the run30 minutes after Contestant A and runs at a constantrate of 10 miles per hour, how long will it take Contes-tant B to catch up to Contestant A?

16. Plot the following ordered pairs in the same xy-plane.Tell in which quadrant or on what coordinate axiseach point lies.

17. Determine whether the ordered pair is a point on thegraph of the equation

(a)

(b)

(c) (2, 9)

In Problems 18 and 19, graph the equations by plottingpoints.

18.

19.

20. Identify the intercepts from the graph below.

y = 4x2

y = 4x - 1

1-1, -321-2, 42

y = 3x2 + x - 5.

A13, -42, B10, 22, C13, 02, D12, 12, E1-1, -42, F1-3, 52

x

y

2�2�4

�2

�4

2

4

4

(0, 3)

(�3, 0) (0, 1)

21. The following graph represents the speed of a carover time.

Time (sec)

40

20

0

y

t0 10 20 30

Spee

d (m

ph) (6, 30) (24, 30)

(a) What is the speed of the car at 6 seconds?

(b) Identify and interpret the intercepts.

Cumulative Review Chapters R–1 145

In Problems 22–26, graph each linear equation, using anyappropriate method.

22. 23.

24. 25.

26.

27. Find and interpret the slope of the line containing thepoints and

28. Draw a graph of the line that contains the point

and has a slope of Do not find the equation of the line.

29. Determine whether the graphs of the following pairof linear equations are parallel, perpendicular, orneither.

In Problems 30–33, find the equation of the line with thegiven properties. Express your answer in either slope-intercept or standard form, whichever you prefer.

30. Through the point and having a slope of 4

31. Through the points (6, 1) and

32. Parallel to and through the point

33. Perpendicular to and through the point (6, 2)

34. Determine whether the given points are solutions tothe linear inequality

(a) (b) (4, 5) (c) (5, 3)13, -123x - y 7 10.

3x - y = 4

110, -12x - 5y = 15

1-3, 721-3, 12

x + 4y = -2

8x - 2y = 1

-35

.

12, -421-1, 62.15, -22

x = -7

32

x -14

y = 13x + 2y = 12

3x + 5y = 0x - y = 8

In Problems 35 and 36, graph each linear inequality.

35. 36.

37. Area of a Circle The following data show the rela-tionship between the diameter of a circle and the areaof that circle.

5x - 2y 6 0y … -2x + 1

Diameter (feet), x Area (square feet), y1 0.79

3 7.07

6 28.27

7 38.48

8 50.27

10 78.54

13 132.73


(b) Compute and interpret the average rate ofchange in area between diameter lengths of 1 and3 feet.

(c) Compute and interpret the average rate of changein area between diameter lengths of 10 and 13 feet.

(d) Based upon the results to parts (a), (b), and (c),do you think that the area of the circle is linearlyrelated to the diameter? Why?

CUMULATIVE REVIEW CHAPTERS R–1

1. Approximate each number by (i) truncating and (ii)rounding to the indicated number of decimal places.

(a) 27.2357; 3 decimal places.

(b) 1.0729; 1 decimal place.

2. Plot the points 0, and on a real number line.

In Problems 3–8, evaluate the expressions.

3. 4.

5. 6. 1-324 -31122 -6

-3 + 4 - 7- ƒ -14 ƒ

72

-4, -52

,

7. 8.

9. Evaluate when

10. Simplify:

11. Determine if the given values are in the domain of x

for the expression

(a) (b)

12. Use the Distributive Property to remove parenthesesand then simplify:

13. Determine whether is a solution to theequation x - 12x + 32 = 5x - 1.

x = 3

31x + 22 - 412x - 12 + 8

x = 0x = -2

x + 3

x2 + x - 2 .

4a2 - 6a + a2 - 12 + 2a - 1

x = 2.3x2 + 2x - 7

23

+12

-14

5 - 211 - 423 + 5 # 3


In Problems 14 and 15, solve the equation.

14.

15.

16. Solve for y.

In Problems 17 and 18, solve the inequality and graph the solutionset on a real number line.

17.

18.

19. Plot the following ordered pairs in the same Cartesianplane.

,

In Problems 20 and 21, graph the linear equation using the methodyou prefer.

20. 21.

In Problems 22 and 23, find the equation of the line with the givenproperties. Express your answer in either slope-intercept or stan-dard form, whichever you prefer.

22. Through the points and 1-6, 10213, -22

4x - 5y = 15y = -12

x + 4

D(0, 3), E(-4, -5), F(-5, 2)

A(-3, 0), B(4, -2), C(1, 5)

51x - 32 Ú 71x - 42 + 3

x + 3 2

…3x - 1

4

2x - 5y = 6

x + 1 3

= x - 4

4x - 3 = 213x - 22 - 723. Parallel to and through the point

24. Graph

25. Computing Grades Shawn really wants an A in hisgeometry class. His four exam scores are 94, 95, 90,and 97. The final exam is worth two exam scores. Tohave an A, his average must be at least 93. For whatrange of scores on the final exam will Shawn be ableto earn an A in the course?

26. Body Mass Index The body mass index (BMI) of aperson 62 inches tall and weighing x pounds is givenby A BMI of 30 or more is considered to beobese. For what weights would a person 62 inches tallbe considered obese?

27. Supplementary Angles Two angles are supplemen-tary. The measure of the larger angle is 15 degreesmore than twice the measure of the smaller angle.Find the angle measures.

28. Cylinders Max has 100 square inches of aluminumwith which to make a closed cylinder. If the radiusof the cylinder must be 2 inches, how tall will thecylinder be? (Round to the nearest hundredth ofan inch.)

29. Consecutive Integers Find three consecutive evenintegers such that the sum of the first two is 22 morethan the third.

0.2x - 2.

x - 3y 7 12.

1-5, 72 y = -3x + 10

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CHAPTER 1 Linear Equations and Inequalities · 48 CHAPTER 1 Linear Equations and Inequalities In Words In the equation the expression ... x - 2b += 4 1 3 x - 2 = 4 1 3 x - 2 = 4