blitzer algtrig4 co08 k12rpadptn.qxd 3/8/10 6:01 pm page 1

747

Television, movies, andmagazines place great

emphasis on physical beauty.Our culture emphasizes physical

appearance to such an extent thatit is a central factor in the per ception

and judgment of others. The moder nemphasis on thinness as the ideal body

shape has been suggested as a major causeof eating disorders among adolescent women.

Cultural values of physical attractivenesschange over time. During the 1950s, actr essJayne Mansfield embodied the postwar ideal:curvy, buxom, and big-hipped. Men, too,

have been caught up in changes of how they“ought” to look. The 1960s ideal was the soft

and scrawny hippie. Today’s ideal man is toughand muscular.

Given the importance of culture in settingstandards of attractiveness, how can you establisha healthy weight range for your age and height? Inthis chapter, we will use systems of inequalities toexplore these skin-deep issues.

You’ll find a weight that fits you using the models(mathematical, not fashion) in Example 1 of Section5.5 and Exercises 77–80 in Exercise Set 5.5.

Exercises 85–86 use graphs and a formula forbody-mass index to indicate whether you are

obese, overweight, borderline overweight,normal weight, or underweight.

Blitzer_AlgTrig4_CO08_K12RPadptn.qxd 3/8/10 6:01 PM

8.1

748 Chapter 8 Systems of Equations and Inequalities

Systems of Linear Equations in Two VariablesObjectives

� Decide whether an orderedpair is a solution of a linearsystem.

� Solve linear systems bysubstitution.

� Solve linear systems byaddition.

� Identify systems that do nothave exactly one ordered-pairsolution.

� Solve problems using systemsof linear equations.

Sec t i on

� Decide whether an ordered pairis a solution of a linear system.

Researchers identified college students who generallywere procrastinators or nonprocrastinators. The

students were asked to report throughout the semes-ter how many symptoms of physical

illness they had experi-enced. Figure 8.1 shows

that by late in thesemester, all studentsexperienced increasesin symptoms. Earlyin the semester, pro-crastinators reportedfewer symptoms, but

late in the semester, as workcame due, they reported moresymptoms than their nonprocrasti-nating peers.

The data in Figure 8.1 can beanalyzed using a pair of linearmodels in two variables. The figureshows that by week 6, both groupsreported the same number ofsymptoms of illness, an average ofapproximately 3.5 symptoms pergroup. In this section, you willlearn two algebraic methods, calledsubstitution and addition, that will

reinforce this graphic observation, verifying (6, 3.5) as the point of intersection.

Systems of Linear Equations and Their SolutionsAll equations in the form are straight lines when graphed. Two suchequations are called a system of linear equations or a linear system. A solution to asystem of linear equations in two variables is an ordered pair that satisfies bothequations in the system. For example, (3, 4) satisfies the system

( is, indeed, 7.)( is, indeed, )

Thus, (3, 4) satisfies both equations and is a solution of the system.The solution can bedescribed by saying that and The solution can also be described using setnotation. The solution set to the system is —that is, the set consisting of theordered pair (3, 4).

A system of linear equations can have exactly one solution, no solution, orinfinitely many solutions. We begin with systems that have exactly one solution.

Determining Whether Ordered Pairs Are Solutionsof a System

Consider the system:

Determine if each ordered pair is a solution of the system:

a. b. 1-4, 32.14, -12

bx + 2y = 2x - 2y = 6.

EXAMPLE 1

513, 426y = 4.x = 3

- 1.3 - 43 + 4bx + y = 7

x - y = -1.

Ax + By = C

Symptoms of Physical Illnessamong College Students

9

8

7

6

5

4

3

2

Ave

rage

Num

ber

of S

ympt

oms

Week in a 16-Week Semester2 4 6 8 10 12 1614

1

Procrastinators

y

x

Nonprocrastinators

Figure 8.1Source: Gerrig and Zimbardo, Psychology andLife, 18th Edition, Allyn and Bacon, 2008

A-BLTZMC08_747-820-hr 25-09-2008 10:01 Page 748

The solution of a system of linear equa-tions can sometimes be found by graphingboth of the equations in the same rectangularcoordinate system. For a system with one solu-tion, the coordinates of the point of intersectiongive the system’s solution. For example, thesystem in Example 1,

is graphed in Figure 8.2. The solution of thesystem, corresponds to the point ofintersection of the lines.

Check Point 1 Consider the system:

Determine if each ordered pair is a solution of the system:

a. (1, 2) b. (7, 6).

Eliminating a Variable Using the Substitution MethodFinding the solution to a linear system by graphing equations may not be easy to do.For example, a solution of would be difficult to “see” as an intersectionpoint on a graph.

Let’s consider a method that does not depend on finding a system’s solutionvisually: the substitution method.This method involves converting the system to oneequation in one variable by an appropriate substitution.

A - 23 , 157

29 B

b2x - 3y = -42x + y = 4.

14, -12,

bx + 2y = 2x - 2y = 6,

Section 8.1 Systems of Linear Equations in Two Variables 749

Solution

a. We begin by determining whether is a solution. Because 4 is theand is the of we replace with 4 and

with -1.yx14, -12,y-coordinate-1x-coordinate

14, -12

true 2 = 2,

4 + 1-22 � 2

4 + 21-12 � 2

x + 2y = 2

true 6 = 6,

4 + 2 � 6

4 - 1-22 � 6

4 - 21-12 � 6

x - 2y = 6

The pair satisfies both equations: It makes each equation true.Thus, theordered pair is a solution of the system.

b. To determine whether is a solution, we replace with and with 3.y-4x1-4, 32

14, -12

true 2 = 2,

-4 + 6 � 2

-4 + 2 # 3 � 2

x + 2y = 2

false -10 = 6,

-4 - 6 � 6

-4 - 2 # 3 � 6

x - 2y = 6

The pair fails to satisfy both equations: It does not make both equationstrue.Thus, the ordered pair is not a solution of the system.

1-4, 32

Study TipWhen solving linear systems bygraphing, neatly drawn graphs areessential for determining points ofintersection.

• Use rectangular coordinategraph paper.

• Use a ruler or straightedge.

• Use a pencil with a sharppoint.

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

Point ofintersection(4, −1) givesthe commonsolution.

All points aresolutions ofx + 2y = 2.

All points aresolutions ofx − 2y = 6.

Figure 8.2 Visualizing a system’s solution

� Solve linear systems bysubstitution.

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Study TipThe equation from step 1, in whichone variable is expressed in terms ofthe other, is equivalent to one of theoriginal equations. It is often easiestto back-substitute an obtained valueinto this equation to find the value ofthe other variable. After obtainingboth values, get into the habit ofchecking the ordered-pair solution inboth equations of the system.

Study TipIn step 1, you can choose which vari-able to isolate in which equation. Ifpossible, solve for a variable whosecoefficient is 1 or to avoid workingwith fractions.

-1


Solving a System by Substitution

Solve by the substitution method:

SolutionStep 1 Solve either of the equations for one variable in terms of the other. Webegin by isolating one of the variables in either of the equations. By solving for inthe second equation, which has a coefficient of 1, we can avoid fractions.

This is the second equation in the given system.Solve for by adding to both sides.

Step 2 Substitute the expression from step 1 into the other equation. We substitutefor in the first equation.

This gives us an equation in one variable, namely

The variable has been eliminated.

Step 3 Solve the resulting equation containing one variable.

This is the equation containing one variable.

Apply the distributive property.Combine like terms.Add 15 to both sides.Divide both sides by 6.

Step 4 Back-substitute the obtained value into one of the original equations. Nowthat we have the of the solution, we back-substitute 4 for into one ofthe original equations to find Let’s use both equations to show that we obtain thesame value for in either case.x

x.yy-coordinate

y = 4 6y = 24

6y - 15 = 9 10y - 15 - 4y = 9

512y - 32 - 4y = 9

x

512y - 32 - 4y = 9.

x= 2y-3 5 x-4y=9

x2y - 3

2yx x = 2y - 3 x - 2y = -3

x

b5x - 4y = 9x - 2y = -3.

EXAMPLE 2

Solving Linear Systems by Substitution

1. Solve either of the equations for one variable in terms of the other. (If one ofthe equations is already in this form, you can skip this step.)

2. Substitute the expression found in step 1 into the other equation. This willresult in an equation in one variable.

3. Solve the equation containing one variable.4. Back-substitute the value found in step 3 into one of the original equations.

Simplify and find the value of the remaining variable.5. Check the proposed solution in both of the system’s given equations.

Using the first equation: Using the second equation:

x = 5

5x = 25

5x - 16 = 9

5x - 4142 = 9

5x - 4y = 9

x = 5

x - 8 = -3

x - 2142 = -3

x - 2y = -3

With and the proposed solution is (5, 4).

Step 5 Check. Take a moment to show that (5, 4) satisfies both given equations.The solution set is 515, 426.

y = 4,x = 5

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� Solve linear systems by addition.

TechnologyGraphic Connections

A graphing utility can be used to solve the system inExample 2. Solve each equation for graph theequations, and use the intersection feature. The utilitydisplays the solution (5, 4) as x = 5, y = 4.

y,

[–10, 10, 1] by [–10, 10, 1]

x − 2y = −3orx + 3

2y =

5x − 4y = 9or5x − 9

4y =

Check Point 2 Solve by the substitution method:

Eliminating a Variable Using the Addition MethodThe substitution method is most useful if one of the given equations has anisolated variable. A second, and frequently the easiest, method for solving alinear system is the addition method. Like the substitution method, the additionmethod involves eliminating a variable and ultimately solving an equationcontaining only one variable. However, this time we eliminate a variable byadding the equations.

For example, consider the following system of linear equations:

When we add these two equations, the are eliminated. This occurs becausethe coefficients of the 3 and are opposites (additive inverses) of eachother:

y = -2.

–2y=4The sum is an equation

in one variable.

b 3x - 4y = 11-3x + 2y = -7

-3,x-terms,x-terms

b 3x - 4y = 11-3x + 2y = -7.

b3x + 2y = 42x + y = 1.

Divide both sides by and solve for y.

-2

Now we can back-substitute for into one of the original equations to find It does not matter which equation you use; you will obtain the same value for ineither case. If we use either equation, we can show that and the solution

satisfies both equations in the system.When we use the addition method, we want to obtain two equations whose sum

is an equation containing only one variable. The key step is to obtain, for one of thevariables, coefficients that differ only in sign. To do this, we may need to multiply oneor both equations by some nonzero number so that the coefficients of one of thevariables, or become opposites. Then when the two equations are added, thisvariable is eliminated.

y,x

11, -22x = 1

xx.y-2

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Solving a System by the Addition Method

Solve by the addition method:

SolutionStep 1 Rewrite both equations in the form Both equations arealready in this form. Variable terms appear on the left and constants appear onthe right.

Step 2 If necessary, multiply either equation or both equations by appropriatenumbers so that the sum of the or the sum of the is 0. We can eliminate or Let’s eliminate Consider the terms in in eachequation, that is, and To eliminate we can multiply each term of the firstequation by and then add the equations.

b3x + 2y = 489x - 8y = -24

-3x,9x.3x

xx.y.xy-coefficientsx-coefficients

Ax � By � C.

b3x + 2y = 489x - 8y = -24.

EXAMPLE 3

Solving Linear Systems by Addition

1. If necessary, rewrite both equations in the form

2. If necessary, multiply either equation or both equations by appropriatenonzero numbers so that the sum of the or the sum of the

is 0.

3. Add the equations in step 2. The sum is an equation in one variable.

4. Solve the equation in one variable.

5. Back-substitute the value obtained in step 4 into either of the givenequations and solve for the other variable.

6. Check the solution in both of the original equations.

y-coefficientsx-coefficients

Ax + By = C.

No change "

Multiply by -3. " b -9x - 6y = -144

9x - 8y = -24

Step 3 Add the equations. Add:

Step 4 Solve the equation in one variable. We solve by dividingboth sides by

Divide both sides by

Simplify.

Step 5 Back-substitute and find the value for the other variable. We can back-substitute 12 for into either one of the given equations. We’ll use the first one.

This is the first equation in the given system.

Substitute 12 for

Multiply.

Subtract 24 from both sides.

Divide both sides by 3.

We found that and The proposed solution is (8, 12).

Step 6 Check. Take a few minutes to show that (8, 12) satisfies both of the originalequations in the system. The solution set is 518, 1226.

x = 8.y = 12

x = 8

3x = 24

3x + 24 = 48

y. 3x + 21122 = 48

3x + 2y = 48

y

y = 12

- 14. -14y

-14=

-168-14

-14.-14y = -168

-14y = -168

Study TipAlthough the addition method is alsoknown as the elimination method,variables are eliminated when usingboth the substitution and additionmethods. The name addition methodspecifically tells us that the elimina-tion of a variable is accomplished byadding two equations.

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Check Point 3 Solve by the addition method:

Some linear systems have solutions that are not integers. If the value ofone variable turns out to be a “messy” fraction, back-substitution might lead tocumbersome arithmetic. If this happens, you can return to the original system anduse the addition method to find the value of the other variable.

Solving a System by the Addition Method

Solve by the addition method:

SolutionStep 1 Rewrite both equations in the form We first arrange thesystem so that variable terms appear on the left and constants appear on the right.We obtain

b2x - 7y = -173x + 5y = 17.

Ax � By � C.

b2x = 7y - 175y = 17 - 3x.

EXAMPLE 4

b4x + 5y = 32x - 3y = 7.

Subtract from both sides of the first equation.Add to both sides of the second equation.3x

7y

Step 2 If necessary, multiply either equation or both equations by appropriatenumbers so that the sum of the or the sum of the is 0. Wecan eliminate or Let’s eliminate by multiplying the first equation by 3 and thesecond equation by

b2x - 7y = -173x + 5y = 17

-2.xy.x

y-coefficientsx-coefficients

Multiply by �2. "

Multiply by 3. " b 6x - 21y = -51

-6x - 10y = -34

Step 3 Add the equations. Add:

Step 4 Solve the equation in one variable. We solve by dividing bothsides by

Divide both sides by

Simplify.

Step 5 Back-substitute and find the value for the other variable. Back-substitutionof for into either of the given equations results in cumbersome arithmetic. Instead,let’s use the addition method on the given system in the form to findthe value for Thus, we eliminate by multiplying the first equation by 5 and thesecond equation by 7.

b2x - 7y = -173x + 5y = 17

yx.Ax + By = C

y8531

y =

8531

-31. -31y

-31=

-85-31

-31.-31y = -85

-31y = -85

Multiply by 7. "

Multiply by 5. " b10x - 35y = -85

21x + 35y = 119

Add:

Divide both sides by 31. x =

3431

31x = 34

We found that and The proposed solution is

Step 6 Check. For this system, a calculator is helpful in showing that

satisfies both of the original equations in the system. The solution set is E A3431 , 85

31 B F .

A3431 , 85

31 B

a3431

, 8531b .x =

3431

.y =

8531

A-BLTZMC08_747-820-hr 25-09-2008 10:01 53


Check Point 4 Solve by the addition method:

Linear Systems Having No Solution or Infinitely ManySolutionsWe have seen that a system of linear equations in two variables represents a pairof lines. The lines either intersect at one point, are parallel, or are identical. Thus,there are three possibilities for the number of solutions to a system of two linearequations in two variables.

b2x = 9 + 3y

4y = 8 - 3x.

� Identify systems that do not haveexactly one ordered-pairsolution.

Number of Solutions What This Means Graphically

Exactly one ordered-pair solution The two lines intersect at one point.

No solution The two lines are parallel.

Infinitely many solutions The two lines are identical.

The Number of Solutions to a System of Two Linear EquationsThe number of solutions to a system of two linear equations in two variables isgiven by one of the following. (See Figure 8.3.)

y

x

y

x

y

x

Solution

Exactly one solution No solution (parallel lines) Infinitely many solutions(lines coincide)

Figure 8.3 Possible graphs for a system of two linear equations in two variables

A linear system with no solution is called an inconsistent system. If you attemptto solve such a system by substitution or addition, you will eliminate both variables.A false statement, such as will be the result.

A System with No Solution

Solve the system:

b4x + 6y = 126x + 9y = 12.

EXAMPLE 5

0 = 12,

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b4x + 6y = 126x + 9y = 12

Add:

Multiply by �2. "

Multiply by 3. "

0= 12

–12x-18y=–24

12x+18y= 36 There are no values of xand y for which 0 = 12.No values of x and y

satisfy 0x + 0y = 12.

e

The false statement indicates that the system is inconsistent and has nosolution. The solution set is the empty set,

The lines corresponding to the two equations in Example 5 are shown inFigure 8.4. The lines are parallel and have no point of intersection.

�.0 = 12

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

6x + 9y = 12 4x + 6y = 12

Figure 8.4 The graph of an inconsistent system

Solution Because no variable is isolated, we will use the addition method. Toobtain coefficients of that differ only in sign, we multiply the first equation by 3and multiply the second equation by -2.

x

Check Point 5 Solve the system:

A linear system that has at least one solution is called a consistent system.Lines that intersect and lines that coincide both represent consistent systems. If thelines coincide, then the consistent system has infinitely many solutions, representedby every point on either line.

The equations in a linear system with infinitely many solutions are calleddependent. If you attempt to solve such a system by substitution or addition, you willeliminate both variables. However, a true statement, such as will be the result.

A System with Infinitely Many Solutions

Solve the system:

Solution Because the variable is isolated in the first equation, wecan use the substitution method. We substitute the expression for into the secondequation.

yy = 3x - 2,y

b y = 3x - 215x - 5y = 10.

EXAMPLE 6

10 = 10,

b 5x - 2y = 4-10x + 4y = 7.

DiscoveryShow that the graphs of and must be parallel lines by solvingeach equation for What is the slope and for each line? What does this mean? Ifa linear system is inconsistent, what must be true about the slopes and for thesystem’s graphs?

y-interceptsy-intercepty.

6x + 9y = 124x + 6y = 12

This statement is truefor all values of x and y.

y = 3x-2 15x-5 y =10

15x-15x+10=10

10=10

15x-5(3x-2)=10

Substitute for

The substitution results in an equation in one variable.Apply the distributive property.

Simplify.

y.3x - 2

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In our final step, both variables have been eliminated and the resulting statement,is true. This true statement indicates that the system has infinitely many

solutions.The solution set consists of all points lying on either of the coincidinglines, or as shown in Figure 8.5.

We express the solution set for the system in one of two equivalent ways:

The set of all ordered pairs(x, y) such that y = 3x − 2

The set of all ordered pairs(x, y) such that 15x − 5y = 10

{(x, y) | y=3x-2} or {(x, y) | 15x-5y=10}.

15x - 5y = 10,y = 3x - 21x, y2

10 = 10,

Study TipAlthough the system in Example 6 has infinitely many solutions, this does not mean that anyordered pair of numbers you can form will be a solution. The ordered pair must satisfyone of the system’s equations, or and there are infinitely manysuch ordered pairs. Because the graphs are coinciding lines, the ordered pairs that are solutionsof one of the equations are also solutions of the other equation.

15x - 5y = 10,y = 3x - 21x, y2

Check Point 6 Solve the system:

Functions of Business: Break-Even AnalysisSuppose that a company produces and sells units of a product. Its revenue is themoney generated by selling units of the product. Its cost is the cost of producing

units of the product.xx

x

b x = 4y - 85x - 20y = -40.

� Solve problems using systems oflinear equations.

Revenue and Cost FunctionsA company produces and sells units of a product.

Revenue Function

Cost Function

C1x2 = fixed cost + 1cost per unit produced2x

R1x2 = 1price per unit sold2x

x

The point of intersection of the graphs of the revenue and cost functions iscalled the break-even point. The of the point reveals the number ofunits that a company must produce and sell so that money coming in, the revenue, isequal to money going out, the cost. The of the break-even point givesthe amount of money coming in and going out. Example 7 illustrates the use of thesubstitution method in determining a company’s break-even point.

Finding a Break-Even Point

Technology is now promising to bring light, fast, and beautiful wheelchairs tomillions of disabled people. A company is planning to manufacture these radicallydifferent wheelchairs. Fixed cost will be $500,000 and it will cost $400 to produceeach wheelchair. Each wheelchair will be sold for $600.

a. Write the cost function, of producing wheelchairs.

b. Write the revenue function, from the sale of wheelchairs.

c. Determine the break-even point. Describe what this means.

xR,

xC,

EXAMPLE 7

y-coordinate

x-coordinate

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

y = 3x − 2

15x − 5y = 10

Figure 8.5 The graph of a systemwith infinitely many solutions

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Solution

a. The cost function is the sum of the fixed cost and variable cost.

b. The revenue function is the money generated from the sale of wheelchairs.

c. The break-even point occurs where the graphs of and intersect. Thus, wefind this point by solving the system

Using substitution, we can substitute for in the first equation:

Substitute for in

Subtract from both sides.

Divide both sides by 200.

Back-substituting 2500 for in either of the system’s equations (or functions),we obtain

The break-even point is (2500, 1,500,000). This means that the company willbreak even if it produces and sells 2500 wheelchairs. At this level, the moneycoming in is equal to the money going out: $1,500,000.

Figure 8.6 shows the graphs of the revenue and costfunctions for the wheelchair business. Similar graphs andmodels apply no matter how small or large a businessventure may be.

The intersection point confirms that the companybreaks even by producing and selling 2500 wheelchairs.Can you see what happens for The red costgraph lies above the blue revenue graph. The cost isgreater than the revenue and the business is losing money.Thus, if they sell fewer than 2500 wheelchairs, the result isa loss. By contrast, look at what happens for The blue revenue graph lies above the red cost graph. Therevenue is greater than the cost and the business is makingmoney. Thus, if they sell more than 2500 wheelchairs, theresult is a gain.

x 7 2500.

x 6 2500?

R(2500)=600(2500)=1,500,000.

We used R(x) = 600x.

x

x = 2500.

400x 200x = 500,000

y = 500,000 + 400x.y600x 600x = 500,000 + 400x

y600x

bC1x2 = 500,000 + 400x

R1x2 = 600x or by = 500,000 + 400x

y = 600x.

RC

R(x)=600x

Revenue per chair, $600, times the number of chairs sold

x

C(x)=500,000+400x

Fixed cost of$500,000

Variable cost: $400 foreach chair producedplus

5000400030002000Wheelchairs Produced and Sold

1000

Gain

Loss

$3,000,000

$2,000,000

$1,000,000

C(x) = 500,000 + 400x

R(x) = 600x

Break-even point:(2500, 1,500,000)

y

x

Figure 8.6

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Check Point 7 A company that manufactures running shoes has a fixed cost of$300,000. Additionally, it costs $30 to produce each pair of shoes. They are sold at$80 per pair.

a. Write the cost function, of producing pairs of running shoes.

b. Write the revenue function, from the sale of pairs of running shoes.


What does every entrepreneur, from a kid selling lemonade to Donald Trump,want to do? Generate profit, of course.The profit made is the money taken in, or therevenue, minus the money spent, or the cost. This relationship between revenue andcost allows us to define the profit function, P1x2.

xR,

xC,

200,000

400,000

600,000

−200,000

−400,000

−600,000

10000

3000 5000

y

x

Wheelchairs Producedand Sold

Business isin the black.

Business is in the red.

Profit functionP(x) = 200x − 500,000

Figure 8.7

The profit function for the wheelchair business in Example 7 is

The graph of this profit function is shown in Figure 8.7. The red portion lies belowthe and shows a loss when fewer than 2500 wheelchairs are sold.The businessis “in the red.” The black portion lies above the and shows a gain when morethan 2500 wheelchairs are sold. The wheelchair business is “in the black.”

x-axisx-axis

= 200x - 500,000.

= 600x - 1500,000 + 400x2

P1x2 = R1x2 - C1x2

The Profit FunctionThe profit, generated after producing and selling units of a product isgiven by the profit function

where and are the revenue and cost functions, respectively.CR

P1x2 = R1x2 - C1x2,

xP1x2,

Exercise Set 8.1

Practice Exercises

In Exercises 1–4, determine whether the given ordered pair is asolution of the system.

1. (2, 3)

bx + 3y = 11x - 5y = -13

2.

b9x + 7y = 88x - 9y = -69

1-3, 52

3. (2, 5)

b2x + 3y = 17x + 4y = 16

4. (8, 5)

b5x - 4y = 203y = 2x + 1

In Exercises 5–18, solve each system by the substitution method.

5. 6.

7. 8.

9. 10.

11. 12. b4x + 3y = 02x - y = 0

b5x + 2y = 0x - 3y = 0

bx = 3y + 7x = 2y - 1

bx = 4y - 2x = 6y + 8

b2x - 3y = -13y = 2x + 7

bx + 3y = 8y = 2x - 9

bx + y = 6y = 2x

bx + y = 4y = 3x

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In Exercises 43–46, let represent one number and let representthe other number. Use the given conditions to write a system ofequations. Solve the system and find the numbers.

43. The sum of two numbers is 7. If one number is subtracted fromthe other, their difference is Find the numbers.

44. The sum of two numbers is 2. If one number is subtractedfrom the other, their difference is 8. Find the numbers.

45. Three times a first number decreased by a second number is 1.The first number increased by twice the second number is 12.Find the numbers.

46. The sum of three times a first number and twice a secondnumber is 8. If the second number is subtracted from twicethe first number, the result is 3. Find the numbers.

Practice PlusIn Exercises 47–48, solve each system by the method of your choice.

47. 48.

In Exercises 49–50, solve each system for and expressing eithervalue in terms of or if necessary.Assume that and

49. 50.

51. For the linear function andFind and

52. For the linear function andFind and

Use the graphs of the linear functions to solve Exercises 53–54.

53. Write the linear system whose solution set is Expresseach equation in the system in slope-intercept form.

54. Write the linear system whose solution set is Express eachequation in the system in slope-intercept form.

�.

516, 226.

6 7x

y

1 2 3 4 5−1

12345

−2−3−4−5

−1−2−3

x − 3y = −6x + 3y = 12

x − 3y = 6x − y = 4

b.mf122 = -7.f1x2 = mx + b, f1-32 = 23

b.mf132 = -9.f1x2 = mx + b, f1-22 = 11

b4ax + by = 36ax + 5by = 8

b5ax + 4y = 17ax + 7y = 22

b Z 0.a Z 0b,ay,x

dx - y

3=

x + y

2-

12

x + 22

- 4 =

y + 4

3

dx + 2

2-

y + 4

3= 3

x + y

5=

x - y

2-

52

-1.

yx13. 14.

15. 16.

17. 18.

In Exercises 19–30, solve each system by the addition method.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29. 30.

In Exercises 31–42, solve by the method of your choice. Identifysystems with no solution and systems with infinitely manysolutions, using set notation to express their solution sets.

31. 32.

33. 34.

35. 36.

37. 38.

39. 40.

41. 42. b4x = 3y + 82x = -14 + 5y

b2x = 3y + 44x = 3 - 5y

cx

6-

y

2=

13

x + 2y = -3c

x

4-

y

4= -1

x + 4y = -9

b4x - 2y = 22x - y = 1

b x + 3y = 23x + 9y = 6

b2x + 5y = -43x - y = 11

b3x - 2y = -54x + y = 8

b9x - 3y = 12y = 3x - 4

by = 3x - 521x - 35 = 7y

b6x + 2y = 7y = 2 - 3x

bx = 9 - 2y

x + 2y = 13

b5x = 6y + 402y = 8 - 3x

b3x = 4y + 13y = 1 - 4x

b2x + 3y = -165x - 10y = 30

b3x - 4y = 112x + 3y = -4

b3x - 7y = 136x + 5y = 7

b4x + 3y = 152x - 5y = 1

b2x - 7y = 23x + y = -20

b x + 2y = 2-4x + 3y = 25

b3x + 2y = 143x - 2y = 10

b2x + 3y = 62x - 3y = 6

bx + y = 6x - y = -2

bx + y = 1x - y = 3

dy = -

12

x + 2

y =

34

x + 7d

y =

13

x +

23

y =

57

x - 2

b3x - 4y = x - y + 42x + 6y = 5y - 4

b2x - 3y = 8 - 2x

3x + 4y = x + 3y + 14

b 2x + 5y = 1-x + 6y = 8

b2x + 5y = -43x - y = 11

A-BLTZMC08_747-820-hr 25-09-2008 10:01 Page 759

Source: O’Sullivan and Sheffrin, Economics, Prentice Hall, 2007

64. You invested $30,000 and started a business writing greetingcards. Supplies cost 2¢ per card and you are selling each cardfor 50¢. (In solving this exercise, let represent the numberof cards produced and sold.)

An important application of systems of equations arises in connectionwith supply and demand. As the price of a product increases, thedemand for that product decreases. However, at higher prices,suppliers are willing to produce greater quantities of the product.The price at which supply and demand are equal is called theequilibrium price.The quantity supplied and demanded at thatprice is called the equilibrium quantity. Exercises 65–66 involvesupply and demand.

65. The following models describe wages for low-skilled labor.

x

Application ExercisesThe figure shows the graphs of the cost and revenue functions fora company that manufactures and sells small radios. Use theinformation in the figure to solve Exercises 55–60.

55. How many radios must be produced and sold for the companyto break even?

56. More than how many radios must be produced and sold forthe company to have a profit?

57. Use the formulas shown in the voice balloons to findDescribe what this means for the company.

58. Use the formulas shown in the voice balloons to findDescribe what this means for the company.

59. a. Use the formulas shown in the voice balloons to write thecompany’s profit function, from producing and selling

radios.

b. Find the company’s profit if 10,000 radios are producedand sold.

60. a. Use the formulas shown in the voice balloons to write thecompany’s profit function, from producing and selling

radios.

b. Find the company’s profit if 20,000 radios are producedand sold.

Exercises 61–64 describe a number of business ventures. Foreach exercise,

a. Write the cost function,

b. Write the revenue function,


61. A company that manufactures small canoes has a fixed costof $18,000. It costs $20 to produce each canoe. The sellingprice is $80 per canoe. (In solving this exercise, let repre-sent the number of canoes produced and sold.)

62. A company that manufactures bicycles has a fixed cost of$100,000. It costs $100 to produce each bicycle. The sellingprice is $300 per bike. (In solving this exercise, let representthe number of bicycles produced and sold.)

63. You invest in a new play. The cost includes an overhead of$30,000,plus production costs of $2500 per performance.A sold-out performance brings in $3125. (In solving this exercise, let represent the number of sold-out performances.)

x

x

x

R.

C.

xP,

xP,

R13002 - C13002.

R12002 - C12002.

y

x

35,000

30,000

25,000

20,000

15,000

10,000

5000

Radios Produced and Sold7006005004003002001000

C(x) = 10,000 + 30x

R(x) = 50x


p=–0.325x+5.8

Price of labor(per hour)

Millions of workersemployers will hire

p=0.375x+3

Price of labor(per hour)

Millions of available workers

Demand Model Supply Model

a. Solve the system and find the equilibrium number ofworkers, in millions, and the equilibrium hourly wage.

b. Use your answer from part (a) to complete this statement:

If workers are paid _____ per hour, there will be___ million available workers and ___ millionworkers will be hired.

c. In 2007, the federal minimum wage was set at $5.15 perhour. Substitute 5.15 for in the demand model,

and determine the millions ofworkers employers will hire at this price.

d. At a minimum wage of $5.15 per hour, use the supplymodel, to determine the millions ofavailable workers. Round to one decimal place.

e. At a minimum wage of $5.15 per hour, use your answersfrom parts (c) and (d) to determine how many morepeople are looking for work than employers are willingto hire.

66. The following models describe demand and supply for three-bedroom rental apartments.

a. Solve the system and find the equilibrium quantity andthe equilibrium price.

b. Use your answer from part (a) to complete this statement:

When rents are _____ per month, consumers willdemand _____ apartments and suppliers will offer_____ apartments for rent.

Demand Model

p=50x

Monthly rental price

Number of apartmentssupplied, in thousands

Supply Model

p=–50x+2000

Monthly rental price

Number of apartmentsdemanded, in thousands

p = 0.375x + 3,

p = -0.325x + 5.8,p

A-BLTZMC08_747-820-hr 25-09-2008 10:01 Page 760


67. In each of the 1996, 2000, and 2004 presidential elections,37% of voters believed abortion should be “always legal” or“always illegal.”The bar graph shows the breakdown of theseabortion “polars” as a percentage of voters in each election.

Abortion Polars as a Percentage of Voters

Presidential Election

10%

35%

5%

Pro-choice polar Pro-life polar

15%

20%

25%

30%

Per

cent

age

of V

oter

s

1996

12

25

2000

14

23

2004

16

21

Source: Newsweek

The data can be modeled by the following system ofequations:

In which year will the percentage of voters who are pro-choice polars be the same as the percentage of voters whoare pro-life polars? For that year, what percent will be pro-choice and what percent will be pro-life?

68. The bar graph shows the percentage of Americans for andagainst the death penalty for a person convicted of murder.

The percentage of pro-choicepolars, y, x years after 1996x+2y=50

–x+2y=24.The percentage of pro-life

polars, y, x years after 1996

e

16

79

1988

18

76

1991

16

80

1994

13

77

1995

22

71

1999

28

66

2000

100%

80%

60%

40%

20%

Are You in Favor of the Death Penaltyfor a Person Convicted of Murder?

Per

cent

age

of A

mer

ican

s

For the death penaltyAgainst the death penalty

Year

Source: Newsweek poll

The data can be modeled by the following system ofequations:

The percent, y, in favor of thedeath penalty x years after 1988

The percent, y, against thedeath penalty x years after 1988

–x+ y= 16.

13x+12y=992e

In which year will the percentage of Americans in favor ofthe death penalty be the same as the percentage of Ameri-cans who oppose it? For that year, what percent will be forthe death penalty and what percent will be against it?

69. We opened this section with a study showing that late in thesemester, procrastinating students reported more symptomsof physical illness than their nonprocrastinating peers.

a. At the beginning of the semester, procrastinators reportedan average of 0.8 symptoms, increasing at a rate of0.45 symptoms per week. Write a function that modelsthe average number of symptoms after weeks.

b. At the beginning of the semester, nonprocrastinatorsreported an average of 2.6 symptoms, increasing at a rateof 0.15 symptoms per week.Write a function that modelsthe average number of symptoms after weeks.

c. By which week in the semester did both groups reportthe same number of symptoms of physical illness? Forthat week, how many symptoms were reported by eachgroup? How is this shown in Figure 8.1 on page 748?

70. Although Social Security is a problem, some projectionsindicate that there’s a much bigger time bomb ticking inthe federal budget, and that’s Medicare. In 2000, the cost ofSocial Security was 5.48% of the gross domestic product,increasing by 0.04% of the GDP per year. In 2000, the cost ofMedicare was 1.84% of the gross domestic product, increasingby 0.17% of the GDP per year.

(Source: Congressional Budget Office)

a. Write a function that models the cost of Social Securityas a percentage of the GDP years after 2000.

b. Write a function that models the cost of Medicare as apercentage of the GDP years after 2000.

c. In which year will the cost of Medicare and Social Securitybe the same? For that year, what will be the cost of eachprogram as a percentage of the GDP? Which program willhave the greater cost after that year?

The bar graph shows the percentage of Americans who usedcigarettes, by ethnicity, in 1985 and 2005. For each of the groupsshown, cigarette use has been linearly decreasing. Use thisinformation to solve Exercises 71–72.

x

x

x

x

41%

38%

35%

32%

29%

26%

23%

24.2%

27.3%

2005

27.3%

40.0%38.0%

Per

cent

age

of A

mer

ican

s13

and

Old

er U

sing

Cig

aret

tes

Cigarette Use in the United States

Year1985

38.9%

20%

White

African American

Hispanic

Source: Department of Health and Human Services

A-BLTZMC08_747-820-hr 25-09-2008 10:01 Page 761


(Exercises 71–72 are based on the bar graph at the bottom of theprevious page.)

71. In this exercise, let represent the number of years after1985 and let represent the percentage of Americans in oneof the groups shown who used cigarettes.a. Use the data points (0, 38) and (20, 27.3) to find the

slope-intercept equation of the line that models thepercentage of African Americans who used cigarettes,

years after 1985. Round the value of the slope totwo decimal places.

b. Use the data points (0, 40) and (20, 24.2) to find theslope-intercept equation of the line that models thepercentage of Hispanics who used cigarettes, yearsafter 1985.

c. Use the models from parts (a) and (b) to find the yearduring which cigarette use was the same for AfricanAmericans and Hispanics. What percentage of eachgroup used cigarettes during that year?

72. In this exercise, let represent the number of years after1985 and let represent the percentage of Americans in oneof the groups shown who used cigarettes.a. Use the data points (0, 38.9) and (20, 27.3) to find the

slope-intercept equation of the line that models thepercentage of whites who used cigarettes, yearsafter 1985.

b. Use the data points (0, 40) and (20, 24.2) to find theslope-intercept equation of the line that models thepercentage of Hispanics who used cigarettes, yearsafter 1985.

c. Use the models from parts (a) and (b) to find the year, tothe nearest whole year, during which cigarette use wasthe same for whites and Hispanics. What percentage ofeach group, to the nearest percent, used cigarettes duringthat year?

Use a system of linear equations to solve Exercises 73–84.

Looking for Mr. Goodbar? It’s probably not a good idea if youwant to look like Mr. Universe or Julia Roberts. The graph showsthe four candy bars with the highest fat content, representinggrams of fat and calories in each bar. Exercises 73–76 are basedon the graph.

y, x

y, x

yx

y, x

my, x

yx

Candy Bars with the Highest Fat Content

14

17

12

Fat (grams) Calories

16

13

11

15

Fat (

gram

s)

Mr.Goodbar

Mounds Snickers Reese’sPeanut

Butter Cup

14.014.1

16.3

13.7 250

280

230

270

240

220

260

Cal

orie

s

Source: Krantz and Sveum, The World’s Worsts, HarperCollins, 2005

74. One Snickers bar and two Reese’s Peanut Butter Cups contain737 calories.Two Snickers bars and one Reese’s Peanut ButterCup contain 778 calories. Find the caloric content of eachcandy bar.

75. A collection of Halloween candy contains a total of five Mr.Goodbars and Mounds bars. Chew on this: The grams of fatin these candy bars exceed the daily maximum desirable fatintake of 70 grams by 7.1 grams. How many bars of each kindof candy are contained in the Halloween collection?

76. A collection of Halloween candy contains a total of12 Snickers bars and Reese’s Peanut Butter Cups. Chew onthis: The grams of fat in these candy bars exceed twice thedaily maximum desirable fat intake of 70 grams by26.5 grams. How many bars of each kind of candy arecontained in the Halloween collection?

77. A hotel has 200 rooms. Those with kitchen facilities rent for$100 per night and those without kitchen facilities rent for$80 per night. On a night when the hotel was completelyoccupied, revenues were $17,000. How many of each type ofroom does the hotel have?

78. A new restaurant is to contain two-seat tables and four-seattables. Fire codes limit the restaurant’s maximum occupancyto 56 customers. If the owners have hired enough servers tohandle 17 tables of customers, how many of each kind oftable should they purchase?

79. A rectangular lot whose perimeter is 360 feet is fenced alongthree sides. An expensive fencing along the lot’s length costs$20 per foot and an inexpensive fencing along the two sidewidths costs only $8 per foot. The total cost of the fencingalong the three sides comes to $3280. What are the lot’sdimensions?

80. A rectangular lot whose perimeter is 320 feet is fencedalong three sides. An expensive fencing along the lot’slength costs $16 per foot and an inexpensive fencing alongthe two side widths costs only $5 per foot. The total cost ofthe fencing along the three sides comes to $2140. What arethe lot’s dimensions?

81. When a crew rows with the current, it travels 16 miles in2 hours.Against the current, the crew rows 8 miles in 2 hours.Let crew’s rowing rate in still water and let rate of the current. The following chart summarizes thisinformation:

y = thex = the

73. One Mr. Goodbar and two Mounds bars contain 780 calories.Two Mr. Goodbars and one Mounds bar contain 786 calories.Find the caloric content of each candy bar.

Rate : Time � Distance

Rowing with current

x + y 2 16

Rowing against current

x - y 2 8

Find the rate of rowing in still water and the rate of thecurrent.

82. When an airplane flies with the wind, it travels 800 miles in4 hours. Against the wind, it takes 5 hours to cover thesame distance. Find the plane’s rate in still air and the rateof the wind.

A-BLTZMC08_747-820-hr 25-09-2008 10:01 Page 762


In Exercises 83–84, an isosceles triangle containing two angles withequal measure is shown. The degree measure of each triangle’sthree interior angles and an exterior angle is represented withvariables. Find the measure of the three interior angles.

83. 84.

Writing in Mathematics85. What is a system of linear equations? Provide an example

with your description.

86. What is the solution of a system of linear equations?

87. Explain how to solve a system of equations using thesubstitution method. Use and toillustrate your explanation.

88. Explain how to solve a system of equations using theaddition method. Use and toillustrate your explanation.

89. When is it easier to use the addition method rather than thesubstitution method to solve a system of equations?

90. When using the addition or substitution method, how canyou tell if a system of linear equations has infinitely manysolutions? What is the relationship between the graphs of thetwo equations?

91. When using the addition or substitution method, how canyou tell if a system of linear equations has no solution?What is the relationship between the graphs of the twoequations?

92. Describe the break-even point for a business.

Technology Exercise93. Verify your solutions to any five exercises in Exercises 5–42

by using a graphing utility to graph the two equations in thesystem in the same viewing rectangle. Then use the intersec-tion feature to display the solution.

Critical Thinking ExercisesMake Sense? In Exercises 94–97, determine whether eachstatement makes sense or does not make sense, and explainyour reasoning.

94. Even if a linear system has a solution set involving fractions,such as I can use graphs to determine if thesolution set is reasonable.

95. Each equation in a system of linear equations has infinitelymany ordered-pair solutions.

96. Every linear system has infinitely many ordered-pair solutions.

E A 811 , 43

11 B F ,

2x + 3y = 03x + 5y = -2

3x + 4y = 6y = 3 - 3x

y y

x

3x + 15

y y

x

2x − 30

97. If I know the perimeter of this rectangle and triangle, eachin the same unit of measure, I can use a system of linearequations to determine values for and

98. Write a system of equations having as a solutionset. (More than one system is possible.)

99. Solve the system for and in terms of and

100. Two identical twins can only be distinguished by thecharacteristic that one always tells the truth and the otheralways lies. One twin tells you of a lucky number pair:“When I multiply my first lucky number by 3 and mysecond lucky number by 6, the addition of the resultingnumbers produces a sum of 12. When I add my first luckynumber and twice my second lucky number, the sum is 5.”Which twin is talking?

101. A marching band has 52 members, and there are 24 in thepom-pom squad. They wish to form several hexagons andsquares like those diagrammed below. Can it be done withno people left over?

Group Exercise102. The group should write four different word problems that

can be solved using a system of linear equations in two vari-ables. All of the problems should be on different topics. Thegroup should turn in the four problems and their algebraicsolutions.

P

B

B

B

BBB

BB

P

P

PP

P

= Band Member

= Pom-pom Person

a2x + b2y = c2 .

a1x + b1y = c1

c2 :a1 , b1 , c1 , a2 , b2 ,yx

51-2, 726

y

x y + 1

y

x

y.x

Preview ExercisesExercises 103–105 will help you prepare for the material coveredin the next section.

103. If and does the ordered triplesatisfy the equation

104. Consider the following equations:

b5x - 2y - 4z = 33x + 3y + 2z = -3.

2x - y + 4z = -8?1x, y, z2z = -3,x = 3, y = 2,

Equation 1Equation 2

Use these equations to eliminate Copy Equation 1 andmultiply Equation 2 by 2. Then add the equations.

105. Write an equation involving and based on thefollowing description:

When the value of in is 4, the value ofis 1682.y

y = ax2+ bx + cx

ca, b,

z.

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