2.3 Problem Solving and Applications of Linear Equations€¦ · Problem Solving and Applications...

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Problem Solving and Applicationsof Linear Equations

2.3

2.3 OBJECTIVES

1. Apply the five-step strategy for solving anapplication

2. Solve motion problems3. Find a break-even point4. Solve a literal equation

“I have a problem!”

How often have you, or a friend, started a conversation with this statement? And, more im-portant, what do you do to solve the problem? George Polya, in his book How to Solve It,contends that there are four parts to solving any problem.

1. Understand the problem

2. Devise a plan

3. Carry out the plan

4. Look back

This approach is useful for solving any problem. One of the reasons that mathematics is arequired course in most programs is that, in a math class, you are constantly practicing theart of problem solving. To help you remember to use Polya’s approach on problems that youencounter outside of the classroom, we will consistently use a series of steps similar tothose prescribed above.

Recall that an algorithm is a series of steps that, when followed, solves a problem. Thefollowing five-step algorithm is essentially the same as Polya’s approach, but for one thing.We have divided the “devise a plan” step into two steps that are more directly relevant tosolving mathematical problems. Here is our five-step approach:

Step 1 Read the problem carefully to determine the unknown quantities.Step 2 Choose a variable to represent the unknown. Express all other

unknowns in terms of this variable.Step 3 Translate the problem to the language of algebra to form an equation. Step 4 Solve the equation, and answer the question of the original problem.Step 5 Verify your solution by returning to the original problem.

Step by Step: Solving Applications

Our first two applications fall into the category of uniform-motion problems. Uniformmotion means that the speed of an object does not change over a certain distance or time.To solve these problems, we will need a relationship between the distance traveled, repre-sented by d, the rate (or speed) of travel, r, and the time of that travel, t. In general, therelationship for the distance traveled d, rate r, and time t, is expressed as

d � r � t

Distance Rate Time

70 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

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This is the key relationship, and it will be used in all motion problems. Let’s see how itis applied in Example 1.

Solving a Motion Problem

On Friday morning Ricardo drove from his house to the beach in 4 h. In coming back onSunday afternoon, heavy traffic slowed his speed by 10 mi/h, and the trip took 5 h. Whatwas his average speed (rate) in each direction?

Step 1 We want the speed or rate in each direction.

Step 2 Let x be Ricardo’s speed to the beach. Then x � 10 is his return speed.

It is always a good idea to sketch the given information in a motion problem. Here wewould have

Step 3 Because we know that the distance is the same each way, we can write an equa-tion, using the fact that the product of the rate and the time each way must be the same.

So

Distance (going) � distance (returning)

Time � rate (going) � time � rate (returning)

4x � 5(x � 10)

Time � rate Time � rate(going) (returning)

A chart can help summarize the given information. We begin by filling in the informationgiven in the problem.

Now we fill in the missing information. Here we use the fact that d � rt to complete thechart.

Rate Time Distance

Going x 4 4x

Returning x � 10 5 5(x � 10)

Rate Time Distance

Going x 4

Returning x � 10 5

Returningx � 10 for 5 hmi

h

Goingx for 4 hmi

h

Example 1

� �

PROBLEM SOLVING AND APPLICATIONS OF LINEAR EQUATIONS SECTION 2.3 71©

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From here we set the two distances equal to each other and solve as before.

Step 4 Solve.

4x � 5(x � 10)

4x � 5x � 50

�x � �50

x � 50 mi/h

So Ricardo’s rate going to the beach was 50 mi/h, and his rate returning was 40 mi/h.

Step 5 To check, you should verify that the product of the time and the rate is the samein each direction.

Solving a Motion Problem

Katy leaves Las Vegas for Los Angeles at 10 A.M., driving at 50 mi/h. At 11 A.M. Jensenleaves Los Angeles for Las Vegas, driving at 55 mi/h along the same route. If the cities are260 mi apart, at what time will they meet?

Step 1 Let’s find the time that Katy travels until they meet.

Step 2 Let x be Katy’s time.

Then x � 1 is Jensen’s time. Jensen left 1 h later!Again, you should draw a sketch of the given information.

Step 3 To write an equation, we will again need the relationship d � rt. From this equa-tion, we can write

Katy’s distance � 50x

Jensen’s distance � 55(x � 1)

Meetingpoint

Los Angeles

(Jensen)

55 for (x � 1) h

(Katy)

50 for x h

Las Vegas

mih

mih

Example 2

NOTE x was his rate going;x � 10, his rate returning.

C H E C K Y O U R S E L F 1

A plane made a flight (with the wind) between two towns in 2 h. Returning againstthe wind, the plane’s speed was 60 mi/h slower, and the flight took 3 h. What wasthe plane’s speed in each direction?

Example 2 illustrates another way of using the distance relationship.


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As before, we can use a table to solve.

From the original problem, the sum of those distances is 260 mi, so

50x � 55(x � 1) � 260

Step 4

50x � 55(x � 1) � 260

50x � 55x � 55 � 260

105x � 55 � 260

105x � 315

x � 3 h

Finally, because Katy left at 10 A.M., the two will meet at 1 P.M. We leave the check of thisresult to you.

Rate Time Distance

Katy 50 x 50x

Jensen 55 x � 1 55(x � 1)

NOTE Be sure to answer thequestion asked in the problem.

Next, we consider an application from business. But we will need some new terminology.The total cost of manufacturing an item consists of two types of costs. The fixed cost,sometimes called the overhead, includes costs such as product design, rent, and utilities. Ingeneral, this cost is constant and does not change with the number of items produced. Thevariable cost, which is a cost per item, includes costs such as material, labor, and shipping.The variable cost depends on the number of items being produced.

A typical cost equation might be

C � 3.30x � 5000

Variable cost Fixed cost

in which total cost, C, equals variable cost times the number of items produced, x, plus thefixed cost.

The total revenue is the income the company makes. It is calculated as the product of theselling price of the item and the number of items sold. A typical revenue equation might be

R � 7.50x

Selling price Number ofper item items sold

and total revenue equals an item’s selling price times the number sold.


At noon a jogger leaves one point, running at 8 mi/h. One hour later a bicyclistleaves the same point, traveling at 20 mi/h in the opposite direction. At what timewill they be 36 mi apart?

PROBLEM SOLVING AND APPLICATIONS OF LINEAR EQUATIONS SECTION 2.3 73©

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Example 3

Finding the Break-Even Point

A firm producing DVDs finds that its fixed cost is $5000 per month and that its variable costis $3.50 per DVD. The cost of producing x DVDs is then given by

C � 3.50x � 5000

The firm can sell the units at $7.50 each, so the revenue from selling x units is

R � 7.50x

Find the break-even point.Because the break-even point is that point at which the revenue equals the cost, or

R � C, from our given equations we have

7.50x � 3.50x � 5000

Revenue Cost

Solving as before gives

4x � 5000

x � 1250

The firm will break even (no profit or loss) by producing and selling exactly 1250 DVDseach month.

�


A firm producing lawn chairs has fixed costs of $525 per week. The variable costis $8.50 per chair, and the revenue per chair is $15.50. This means that the costequation is

C � 8.50x � 525

and the revenue equation is

R � 15.50x

Find the break-even point.

The break-even point is that point at which the revenue equals the cost (the companywould exactly break even without a profit or a loss).

Let’s apply these concepts in Example 3.


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Note, in the previous example and exercise, that once we solved for r in the originalequation, we were able to easily use the result with different sets of data. You can comparethis to first substituting the known values and then having to solve for r in each caseseparately.

Solving for a specified variable also has significance in computer programming. Inusing a formula for the computation of the value of a variable, you must solve the formulafor that variable to use the formula in a program.

C H E C K Y O U R S E L F A N S W E R S

1. with the wind and 120 against 2. At 2 P.M. 3. 75 chairs

4. 7%

mi

h180

mi

h

Example 4

Solving an Interest Rate Application

Suppose a principal of $1000 is invested in a mutual fund account for 5 years. For theamount in the account to be $1400 at the end of that period, what must the interest rate be?

In Example 3 of Section 2.2, we solved the formula

A � P(1 � rt)

for r with the result

In our problem, A � $1400, P � $1000, and t � 5 years. Substituting, we have

The necessary interest rate is 8%.

r �1400 � 1000

(1000)(5)�

400

5000� 0.08

r �A � P

Pt

Let’s turn now to an application of our work in solving literal equations for a specifiedvariable.

NOTE From the given problem,we know values for A, P, and t.Two strategies are possible.

1. We can substitute the knownvalues in A � P(1 � rt) andthen solve for r.

2. Or we can solve for r andthen substitute. We haveillustrated this approach.


Suppose that a principal of $5000 is invested in a time deposit fund for 3 years. If theamount in the account at the end of that period is $6050, what was the annualinterest rate?

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Exercises

Solve the following applications.

1. Speed. On her way to a business meeting, Kim took the freeway, and the trip took3 h. Returning, she decided to take a side road, and her speed along that route

averaged 9 mi/h slower than on the freeway. If Kim’s return trip took h and the

distance driven was the same each way, find her average speed in each direction.

2. Speed. Beth was required to make a cross-country flight in training for her pilot’slicense. When she flew from her home airport, a steady 30-mi/h wind was behind her,and the first leg of the trip took 5 h. When she returned against the same wind, theflight took 7 h. Find the plane’s speed in still air and the distance traveled on each legof the flight.

3. Speed. Craig was driving on a 220-mi trip. For the first 3 h he traveled at a steadyspeed. At that point, realizing that he would be late to his destination, he increased hisspeed by 10 mi/h for the remaining 2 h of the trip. What was his driving speed foreach portion of the trip?

4. Distance. Robert can drive to work in 45 min, whereas if he decides to take the bus,the same trip takes 1 h 15 min. If the average rate of the bus is 16 mi/h slower thanhis driving rate, how far does he travel to work?

5. Time. At 9 A.M., Tom left Boston for Baltimore, traveling at 45 mi/h. One hour later,Andrea left Baltimore for Boston, traveling at 50 mi/h along the same route. If thecities are 425 mi apart, at what time did Tom and Andrea meet?

6. Time. A passenger bus left a station at 1 P.M., traveling north at an average rate of50 mi/h. One hour later, a second bus left the same station, traveling south at a rate of55 mi/h. At what time will the two buses be 260 mi apart?

7. Distance. On Tuesday, Malia drove to a conference and averaged 54 mi/h for thetrip. When she returned on Thursday, road construction slowed her average speed by9 mi/h. If her total driving time was 11 h, what was her driving time each way, andhow far away from her home was the conference?

8. Time. At 8:00 A.M., Robert left on a trip, traveling at 45 mi/h. One-half hour later,Laura discovered that Robert forgot his luggage and left along the same route,traveling at 54 mi/h, to catch up with him. When did Laura catch up with Robert?

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2.3

Name

Section Date

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

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9. Business. A firm producing gloves finds that its fixed cost is $4000 per week and itsvariable cost is $8.50 per pair. The revenue is $13.50 per pair of gloves, so that costand revenue equations are, respectively,

C � 8.50x � 4000 and R � 13.50x

Find the break-even point for the firm.

10. Business. A company that produces calculators determines that its fixed cost is$8820 per month. The variable cost is $70 per calculator: the revenue is $105 percalculator. The cost and revenue equations, respectively, are given by

C � 70x � 8820 and R � 105x

Find the number of calculators the company must produce and sell to break even.

11. Business. A firm that produces scientific calculators has a fixed cost of $1260 perweek and variable cost of $6.50 per calculator. If the company can sell the calculatorsfor $13.50, find the break-even point.

12. Business. A publisher finds that the fixed cost associated with a new paperback is$18,000. Each book costs $2 to produce and will sell for $6.50. Find the publisher’sbreak-even point.

13. Business. A firm producing flashlights finds that its fixed cost is $2400 per week,and its variable cost is $4.50 per flashlight. The revenue is $7.50 per flashlight, so thecost and revenue equations are, respectively,

C � 4.50x � 2400 and R � 7.50x

Find the break-even point for the firm (the point at which the revenue equals the cost).

ANSWERS

9.

10.

11.

12.

13.

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14. Business. A company that produces portable television sets determines that its fixedcost is $8750 per month. The variable cost is $70 per set, and the revenue is $105 perset. The cost and revenue equations, respectively, are given by

C � 70x � 8750 and R � 105x

Find the number of sets the company must produce and sell to break even.

15. Business. An important economic application involves supply and demand. Thenumber of units of a commodity that manufacturers are willing to supply, S, isrelated to the market price, p. A typical supply equation is

S � 40p � 285 (1)

(Generally the supply increases as the price increases.)The number of units that consumers are willing to buy, D, is called the demand,

and it is also related to the market price. A typical demand equation is

D � �45p � 1500 (2)

(Generally the demand decreases as the price increases.)The price at which the supply and demand are equal (or S � D) is called the

equilibrium price for the commodity. The supply and demand equations for a certainmodel portable radio are given in equations (1) and (2). Find the equilibrium price forthe radio.

16. Business. The supply and demand equations for a certain type of computer modemare

S � 25p � 2500 and D � �40p � 5300

Find the equilibrium price for the modem.

17. Business. You find a new bicycle that you like, and you plan to ride it for exerciseseveral days a week. You are also happy to find that this very model is on sale for22% off. You speak to the salesclerk, who begins writing up the sale by first addingon your state’s 7.8% sales tax. He then takes off the 22%. “No!”, you object, “Youshould take the 22% off first and then add the sales tax.” The salesclerk says he issorry, but he has been instructed to first calculate the amount of tax. Who is correct?Defend your position using algebra.

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15.

16.

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18. Density and Body Fat. You have probably heard the question, “Which weighs more,a ton of feathers or a ton of bricks?” The usual response of “bricks” is, of course,incorrect. A ton of anything weighs 2000 pounds. Why do people miss this questionso often? We know from experience that bricks are heavier than feathers. However,what we mean is that bricks have a much higher density than feathers (they weighmore per unit volume). A cubic meter of bricks weighs far more than a cubic meter offeathers.

We often compare the density of a substance to the density of water. Water has adensity of 1 g/cm3. Things denser than water (such as body muscle and bone) sink,whereas things less dense (such as body fat) float.

The concept of density is important in analyzing the significance of the percentageof body fat you have. There are many ways to measure, or estimate, your percentageof body fat. Most health clubs and many college health classes offer the opportunityto find this percentage. By multiplying percentage of body fat times your weight, youcan calculate the actual pounds of fat in your body.

However, we are looking for a more useful measure. Body density, D, can becalculated using the following formula.

D �

in which A � proportion of body fat

B � proportion of lean body tissue (1 � A)

a � density of fat body tissue in grams/cubic cm (approximately 0.9 g/cm3)

b � density of lean body tissue in grams/cubic cm (approximately 1.1 g/cm3)

Use this formula to compute the following.1. Substituting 1 � A for B, 0.9 for a, and 1.1 for b, solve the formula so that D is

expressed in terms of A.2. Find the values associated with body fat proportions of 0.1, 0.15, 0.2, 0.25, and

0.3.3. Use other resources (health professionals, health clubs, the Internet, and the

library) to find another measure of body density. How do the two methodscompare?

Answers1. 63 mi/h going; 54 mi/h returning 3. 40 mi/h for 3 h; 50 mi/h for 2 h5. 2:00 P.M. 7. 5 h going; 6 h returning; 270 mi 9. 800 pairs of gloves11. 180 calculators 13. 800 flashlights 15. $21 17.

1

A

a�

B

b

ANSWERS

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Linear Inequalities2.4

2.4 OBJECTIVES

1. Solve and graph the solution set for a linearinequality

2. Solve and graph the solution set for a compoundinequality

In Section 2.1 we defined a linear equation in one variable as an equation that could bewritten in the form

ax � b � 0

in which a and b are real numbers and a � 0.A linear inequality in one variable is defined in a similar fashion.

NOTE We can also say that asolution satisfies the inequality.

NOTE Adding the samequantity to both sides of aninequality gives an equivalentinequality.

The inequality symbol � can be replaced with any of the other inequality symbols , ,or �, so that

ax � b 0 ax � b 0 and ax � b � 0

are also linear inequalities.Fortunately your experience with linear equations in Section 2.1 provides the ground-

work for solving linear inequalities. You will see many similarities.A solution for a linear inequality in one variable is any real number that will make the

inequality a true statement when the variable is replaced by that number. The solution setfor a linear inequality is the set of all solutions.

Our strategy for solving linear inequalities is, for the most part, identical to that used forsolving linear equations. We write a sequence of equivalent inequalities to isolate the vari-able on one side of the inequality symbol.

Writing equivalent inequalities for this purpose requires two properties. First:

This addition property is similar to that seen earlier in solving equations. As before,because subtraction is defined in terms of addition, the property also allows us to subtractthe same quantity from both sides of an inequality without changing the solutions.

Our second property, dealing with multiplication, has an important difference. We beginby writing the true inequality

2 � 3

A linear inequality can be written in the form

ax � b � 0

in which a and b are real numbers and a � 0.

Definitions: Linear Inequality

If a � b, then a � c � b � c.

Rules and Properties: Addition Property of Inequalities


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Solving a Linear Inequality

Solve and graph the inequality

4x � 3 � 5 (1)

First, we add 3 to both sides.

4x � 3 � 3 � 5 � 3

4x � 8 (2)

We now divide both sides of the inequality by 4, to isolate the variable on the left.

�

x � 2 (3)

8

4

4x

4

Multiplying both sides of that inequality by the same positive number, say 3, gives

3(2) � 3(3)

6 � 9 Another true statement!

Note that the new inequality has the same sense (points in the same direction) as the origi-nal inequality.

However, if we now multiply both sides of the inequality by a negative number, say �3,we have

�3(2) � �3(3)

�6 � �9 A false statement!

To make this a true statement, we must reverse the sense of the inequality to write

�6 �9

This suggests that if we multiply both sides of an inequality by a negative number, we mustreverse the sense of the inequality to form an equivalent inequality. From this discussion wecan now state our second property.

NOTE Multiplying both sidesof an inequality by a positivenumber gives an equivalentinequality.

NOTE Multiplying both sidesof an inequality by a negativenumber and reversing the sensegive an equivalent inequality.

NOTE As in solving equations,we apply the addition propertyand then the multiplicationproperty, to isolate the variable.

Again, because division is defined in terms of multiplication, this property also allowsus to divide both sides of an inequality by the same nonzero number, reversing the sense ofthe inequality if that number is negative.

We will use these properties in solving inequalities in much the same way as we did insolving equations, with the one significant difference pointed out above.

The following examples illustrate the solution process for linear inequalities.

If a � b

then ac � bc when c is a positive number (c 0)

and

ac bc when c is a negative number (c � 0)

Rules and Properties: Multiplication Property of Inequalities

Example 1

LINEAR INEQUALITIES SECTION 2.4 81©

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NOTE Recall that theparenthesis means 2 is notincluded in the solution set.


Solve and graph the inequality.

5x � 7 22



3x � 5 � 5x � 3

Add 5 to both sides.

3x � 5 � 5 � 5x � 3 � 5

3x � 5x � 8

Subtract 5x from both sides.

3x � 5x � 5x � 5x � 8

�2x � 8

We must now divide both sides by �2. Because the divisor is negative, we reverse the senseof the inequality.

x �4

The solution set consists of all numbers that are less than or equal to �4 and is graphedbelow.

�x �x �4�

�6 �5 �4 �3 �2 �1 210

8

�2

�2x

�2

Example 2

Because inequalities (1), (2), and (3) are all equivalent, the solution set for the original in-equality consists of all numbers that are less than 2. That set can be written

�x �x � 2�

The graph of the solution set is

�2 �1 0 1 2 3 4

NOTE Here we use the bracketto indicate that �4 is includedin the solution set.


Solve and graph.

4x � 3 7x � 12

In working with more complicated inequalities, as was the case with equations, anysigns of grouping must be removed, and like terms combined, before the properties of in-equalities are applied to isolate the variable.


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NOTE Multiply both sides by 6,the LCM of 6 and 3.

NOTE Apply the distributiveproperty on the left.


Solve and graph

5 � 3(x � 2) 1 � x

First, remove the parentheses on the left and combine like terms.

5 � 3x � 6 1 � x

�3x � 11 1 � x

We now proceed as before.

�3x � 11 � 11 1 � 11 � x Subtract 11.

�3x �10 � x

�3x � x �10 � x � x Add x.

�2x �10

�

x � 5

The solution set �x � x � 5� is graphed below.

2 3 4 5 6 7 8

Divide by �2, reversing the senseof the inequality.

�10

�2

�2x

�2

Example 3



� 1 �

6 � 6

6 � 6(1) � 6

3x � 2 � 6 � 2x

�x

3��3x � 2

6 �

�x

3��3x � 2

6� 1�

x

3

3x � 2

6

Example 4


Solve and graph.

4 � 2(x � 5) � �9 � 4x

If fractions are involved in an inequality, you should apply the multiplication property toclear the inequality of fractions as your first step.


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The inequality is now cleared of fractions, and we proceed as before:

3x � 4 � 2x

3x � 2x � 4

x � 4

The solution set is graphed below.

�x � x � 4�

�1 0 1 2 3 4 5 6

The following algorithm summarizes our work thus far in this section in solving linearinequalities.



5x � 1

4� 2

x

2

Step 1 Clear the inequality statement of any fractions by using themultiplication property.

Step 2 Remove any grouping symbols, and combine like terms on either sideof the inequality.

Step 3 Apply the addition property to write an equivalent inequality with thevariable term on one side of the inequality and the constant term onthe other.

Step 4 Apply the multiplication property to write an equivalent inequalitywith the variable isolated on one side of the inequality. Be sure toreverse the sense of the inequality if you multiply or divide by anegative number.

Step 5 Graph the solution set of the original inequality.

Step by Step: Solving Linear Inequalities in one Variable

NOTE In the compoundinequality the word “and” isunderstood.

Let’s now consider two types of inequality statements that arise frequently in mathe-matics. Consider a statement such as

�2 � x � 5

It is called a double or compound inequality because it combines the two inequalities

�2 � x and x � 5

To solve a compound inequality means to isolate the variable in the middle term, as thefollowing example illustrates.

Solving a Compound Inequality

Solve and graph the compound inequality

�3 2x � 1 7

Example 5


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First, we subtract 1 from each of the three members of the compound inequality.

�3 � 1 2x � 1 � 1 7 � 1

or

�4 2x 6

We now divide by 2 to isolate the variable x.

�2 x 3

The solution set consists of all numbers between �2 and 3, including �2 and 3, and iswritten

�x ��2 x 3�

That set is graphed below.

Note: Our solution set is equivalent to

�x �x � �2 and x 3�

Look at the individual graphs.

�x �x � �2�

�x �x 3�

�x �x � �2 and x 3�

Because the connecting word is “and,” we want the intersection of the sets, that is, thosenumbers common to both sets.

�2 30

30

�2 0

�3 �2 �1 0 1 2 3 4

6

2

2x

2

�4

2

NOTE We are really applyingthe addition property to eachof the two inequalities thatmake up the compoundinequality statement.

NOTE Because the points atboth ends of the interval areincluded, we sometimes call thisa closed interval.

NOTE Using set notation, wecould write this as

�x �x � �2� � �x �x 3�

The � represents theintersection of the two sets.

When the coefficient of the variable is a negative number, care must be taken in isolatingthe variable.



�5 2x � 3 3


Solve and graph the compound inequality

�3 � 4 � 3x � 13

Subtract 4 from each member of the inequality.

�7 � �3x � 9

Example 6


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Solve and graph the double inequality.

�5 � 3 � 2x � 5

Now we must divide by �3. The sense of the inequality is reversed whenever we divide bya negative number.

x �3

In the standard smallest-to-largest format, we have

�3 � x �

The solution consists of all numbers between �3 and and is written

�x ��3 � x � That set is graphed below.

�3 0 73

7

3

7

3

7

3

7

3

9

�3

�3x

�3

�7

�3

NOTE This is sometimes calledan open interval.

A compound inequality may also consist of two inequality statements connected bythe word “or.” The following example illustrates the solution of that type of compoundinequality.



2x � 3 � �5 or 2x � 3 5

In this case we must work with each of the inequalities separately.

2x � 3 � �5 or 2x � 3 5

2x � �2 2x 8

x � �1 x 4

The graph of the solution set is shown.

�x �x � �1 or x 4�

Note that because the connecting word is “or” in this case, the solution set of the originalinequality is the union of the two sets, that is, those numbers that belong to either or bothof the sets.

�3 10�1 2 3 4 5 6�2

Example 7

NOTE Add 3.

NOTE Divide by 2.

NOTE In set notation we writethe union as

�x �x � �1� � �x �x 4�


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The following chart summarizes our discussion of solving linear inequalities and the natureof the solution sets of the types of inequalities we have considered in this section.



3x � 4 �7 or 3x � 4 � 7

C H E C K Y O U R S E L F A N S W E R S

1. �x �x 3�

2. �x �x � 5�

3. �x �x � 4. �x �x 3�

5. �x ��1 x 3�

6. �x ��1 � x � 4�

7. �x �x �1 or x � 10 2 3 4 5�3�2�1

11

3

10 2 3 4 5�2 �1

10 2 3�1

1 2 3 4 5

0 1 2�2 �1

32�

�3

2

3 4 5 6 7

1 2 3 4 5

Type of Inequality Graph of Solution Set

If a 0:ax � b � c

If a � 0:

�c � ax � b � c

ax � b � �c or ax � b cr s

r s

r

r

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Exercises

Solve each of the following inequalities. Then graph the solution set.

1. x � 2 � 5 2. x � 3 �4

3. x � 5 � 3 4. x � 4 �2

5. 5x 25 6. 4x � �12

7. �3x �15 8. �7x 21

9. 2x � 3 � 10 10. 5x � 3 17

11. �2x � 7 � 5 12. �3x � 4 � �4

13. 5 � 3x � 14 14. 2 � 5x � 22

15. 3x � 4 2x � 5 16. 4x � 3 3x � 11

17. 8x � 2 2x � 10 18. 5x � 1 x � 9

19. 7x � 3 2x � 13 20. 9x � 2 � 2x � 19

21. 4x � 3 6x � 5 22. 7x � 1 10x � 6

23. 5 � 3x 2x � 3 24. 7 � 5x 3x � 9

Simplify and then solve each of the following inequalities.

25. 5(2x � 1) 25 26. 3(3x � 1) �15

27. 4(5x � 1) 3(3x � 5) 28. 3(2x � 4) 5(3x � 3)

29. 3(x � 1) � 4 � 2(3x � 1) 30. 3(3x � 1) � 4(x � 3) 15

31. 3(x � 7) � 11 2(3x � 5) � x 32. 3(2x � 7) � 5 4(x � 1) � x

2.4

Name

Section Date

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12. 13.

14.

15. 16.

17. 18.

19.

20.

21.

22. 23.

24. 25.

26. 27.

28. 29.

30. 31.

32.

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Clear of fractions and then solve each of the following inequalities.

33. 34.

35. 36.

37. 38.

39. 40.

41. 42.

Solve each of the following compound inequalities. Then graph the solution set.

43. 3 x � 1 5 44. �2 � x � 3 � 3

45. �8 � 2x � 4 46. �6 3x 9

47. 1 2x � 3 6 48. �2 � 3x � 5 � 4

49. �1 � 5 � 3x � 8 50. �7 3 � 2x 8

Solve each of the following compound inequalities. Then graph the solution set.

51. x � 1 � �3 or x � 1 3 52. x � 2 � �5 or x � 2 5

53. 2x � 1 � �7 or 2x � 1 7 54. 2x � 3 � �3 or 2x � 3 3

55. 3x � 1 � �7 or 3x � 1 7 56. 4x � 3 � �5 or 4x � 3 5

Suppose that the revenue a company will receive from producing and selling x items isgiven by R and the cost of those items by C. The company will make a profit only if therevenue is greater than the cost, that is, when R C. Use this information to find thenumber of items that must be produced and sold for the company to make aprofit.

57. R � 50x, C � 1000 � 30x 58. R � 800x, C � 24,000 � 500x

x � 5

4�

x � 1

3

2

3

x � 3

2�

x � 5

5

1

2

x

4�

4x � 3

20�

1

5

x

5�

x � 7

3�

1

3

x

4� 2 �

x

5

x

2�

x

3� 2

x � 2

�4 �6

x � 2

�3 3

x � 5

2� �3

x � 4

3� 5

ANSWERS

33.

34.

35.

36.

37.

38.

39.

40.

41.

42.

43.

44.

45.

46.

47.

48.

49.

50.

51.

52.

53.

54.

55.

56.

57.

58.

88

Recall that the average of a group of test scores is the sum of those test scores divided bythe number of scores. Use this information to solve the following problems.

59. Suppose that Kim has scores of 83, 94, and 91 on three 100-point tests in herchemistry class thus far. Describe the set of scores on the 100-point final test that willgive her an average of 90 or above, so that she will receive an A for the course.

Hint: If x represents her final score, then

83 � 94 � 91 � x

will give her total score for the four tests.

60. Robert has scores of 78, 85, 70, and 83 on four tests. Describe the set of scores hemust have on the 100-point final to average 80 or above for the course.

Solve each of the following problems.

61. A college must decide how many sections of intermediate algebra to offer during thefall quarter. Each section should contain a maximum of 35 students, and the collegeanticipates that a total of 400 students will enroll for the sections. How many sectionsshould be offered?

Hint: If x represents the number of sections, then will give the number of

students per section. Establish an inequality from the given information. Note thatyou can clear the inequality of fractions by multiplying by x because x is the numberof sections and must be a positive number. Also keep in mind that x must be a wholenumber.

62. A student-activities director must order buses for a football game. He anticipates that300 students will sign up, and the capacity of each bus is 40 people. How many busesshould he have available?

63. The mileage markers on a freeway begin at marker 0 at the southern border of a stateand continue to increase toward the northern border. The legal maximum speed onthe freeway is 65 mi/h, and the legal minimum speed is 45 mi/h. If you enter thefreeway at marker 100 and travel north for 2 h, what is the possible range of valuesfor the nearest marker you could legally reach?

Hint: Because distance � rate � time, the minimum distance can be calculated as(45)(2) and the maximum distance as (65)(2). Let m be the marker you could legallyreach, and establish a double-inequality statement for the solution.

64. You enter the freeway at marker 240 and now travel south for 3 h. What is the possiblerange of values for the nearest marker you could legally reach?

65. A new landfill must last at least 30 years for it to receive an operating permit from thelocal community. The proposed site is capable of receiving 570 � 106 metric tons (t)of refuse over its lifespan. How much refuse can the landfill accept each year and stillmeet the conditions of its permit?

400

x

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sANSWERS

59.

60.

61.

62.

63.

64.

65.

89

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66. A garbage burner must receive at least 1350 tons of trash per day to be economicalenough for a community to build it. Local laws restrict truck weight to a 15-ton limit.How many truck deliveries per day will be necessary to supply the burner with itsdaily requirement of trash?

Answers1. �x � x � 7� 3. �x � x � �2�

5. �x � x 5� 7. �x � x � 5�

9. �x � x � 11. �x � x �6�

13. �x � x �3� 15. �x � x 9�

17. �x � x 19. �x � x �2�

21. �x � x � �4� 23. �x � x � 25. �x � x 3�

27. �x � x 1� 29. �x � x �3� 31. �x � x � 5� 33. �x � x � 19�35. �x � x � �11� 37. �x � x � 12� 39. �x � x 15� 41. �x � x 10�

43. �x � 2 x 4� 45. �x � �4 � x � 2� 47. x � 2 x

49. �x � �1 � x � 2� 51. �x � x � �2 or x 4�

53. �x � x � �3 or x 4� 55. x � x � �2 or x

57. x 50 59. x � 92 61. x � 12 63. 190 m 23065. x 19 � 106 t

�2 0 83

�3 0 4

8

3��2 0 420�1

92

0 220�40 42

9

2�

25

00�4

2

5

0�243

0

4

3

0 90�3

0�672

0

7

2

0 50 5

0�20 7

ANSWERS

66.

90

2.3 Problem Solving and Applications of Linear Equations€¦ · Problem Solving and Applications...

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Transcript of 2.3 Problem Solving and Applications of Linear Equations€¦ · Problem Solving and Applications...