§ 6.7 Formulas and Applications of Rational Equations

§ 6.7

Formulas and Applications of Rational Equations

Blitzer, Intermediate Algebra, 5e – Slide #2 Section 6.7

Applications of Rational Equations

Formulas and mathematical models often contain rational expressions. You can often solve for a different variable by using the procedure for solving rational equations. The goal when solving for a specified variable is to get that variable alone on one side of the equation. To do so, collect all terms with this variable on one side and all other terms on the other side. Sometimes it is necessary to factor out the variable you are solving for.



EXAMPLEEXAMPLE

Solve the formula for R:

SOLUTIONSOLUTION

.rR

EI

rR

EI

This is the original equation.

rR

ErRIrR

Multiply both sides by the

LCD, R + r.

ErRI Simplify.

rI

ER

I

ErR

Divide both sides by I and subtract r.



Check Point 1Check Point 1

Solve the formula for x: LCD= xyzzyx

111

Multiply both sides by the LCD, xyz.

Subtract xz from both sides.

Factor x and divide both sides by (y-z).

zxyz

yxyz

xxyz

111

xyxzyz

xzxyyz

)( zyxyz

zy

yzx



EXAMPLEEXAMPLE

A company is planning to manufacture small canoes. Fixed monthly cost will be $20,000 and it will cost $20 to produce each canoe.

(a)Write the cost function, C, of producing x canoes.

(b) Write the average cost function, , of producing x canoes.

(c) How many canoes must be produced each month for the company to have an average cost of $40 per canoe?

C



SOLUTIONSOLUTION

C

(a) The cost function, C, is the sum of the fixed cost and the variable costs.

CONTINUECONTINUEDD

xxC 20000,20

Variable cost: $20 for each canoe produced.

Fixed cost is $20,000.

(b) The average cost function, , is the sum of fixed and variable costs divided by the number of canoes produced.

x

xxC

20000,20



. xC

(c) We are interested in the company’s production level that results in an average cost of $40 per canoe. Substitute 40, the average cost, for and solve the resulting rational equation for x.

CONTINUECONTINUEDD

x

x20000,2040

xC

Substitute 40 for

xx 20000,2040 Multiply both sides by the LCD, x.

000,2020 x Subtract 20x from both sides.

000,1x Divide both sides by 20.

The company must produce 1,000 canoes each month for an average cost of $40 per canoe.



Check Point 2 on page Check Point 2 on page 454454

A company is planning to manufacture wheelchairs that are light, fast, and beautiful. Fixed monthly cost will be $500,000 and it will cost $400 to produce each radically innovative chair.

(a)Write the cost function, C, of producing x wheelchairs.

(b) Write the average cost function, , of producing x wheelchairs.

(c) How many wheelchairs must be produced each month for the company to have an average cost of $450 per wheelchairs?

C



SOLUTIONSOLUTION

C

(a) The cost function, C, is the sum of the fixed cost and the variable costs.

CONTINUECONTINUEDD

xxC 400000,500

Variable cost: $400 for each wheelchair produced.

Fixed cost is $500,000.

(b) The average cost function, , is the sum of fixed and variable costs divided by the number of canoes produced.

x

xxC

400000,500



. xC

(c) We are interested in the company’s production level that results in an average cost of $450 per wheelchair. Substitute 450, the average cost, for and solve the resulting rational equation for x.

CONTINUECONTINUEDD

x

x400000,500450

xC

Substitute 450 for

xx 400000,500450 Multiply both sides by the LCD, x.

000,50050 x Subtract 400x from both sides.

000,10x Divide both sides by 50.

The company must produce 10,000 wheelchairs each month for an average cost of $450 per wheelchairs. (Look at questions 15-18 on page 459.)



Time in Motion

r

dt

travelof Rate

traveledDistance traveledTime



EXAMPLEEXAMPLE

An engine pulls a train 140 miles. Then a second engine, whose average rate is 5 miles per hour faster than the first engine, takes over and pulls the train 200 miles. The total time required for both engines is 9 hours. Find the average rate of the first engine.

SOLUTIONSOLUTION

1) Let x represent one of the quantities. Let

x = the rate of the first engine.

2) Represent other quantities in terms of x. Because the average rate of the second engine is 5 miles per hour faster than the average rate of the first engine, let

x + 5 = the rate of the second engine.



3) Write an equation that describes the conditions. By reading the problem again, we discover that the crucial idea is that the time for both engines’ trips is 9 hours. Thus, the time of the first engine plus the time of the second engine is 9 hours.

CONTINUECONTINUEDD

Distance Rate Time

Train 1 140 x

Train 2 200 x + 5

x

140

5

200

x

The sum of the two times is 9 hours.



We are now ready to write an equation that describes the problems’ conditions.



CONTINUECONTINUEDD

95

200140

xx

This is the equation for the problems’ conditions.

Time of the first train

plustime of the second train

equals

9 hours.

4) Solve the equation and answer the question.

95

200140

xx

Multiply both sides by the LCD, x(x + 5).

955

2001405

xx

xxxx

Use the distributive property on both sides.

955

2005

1405

xx

xxx

xxx



CONTINUECONTINUEDD

Simplify. 592005140 xxxx

Use the distributive property.xxxx 459200700140 2 Combine like terms.

xxx 459700340 2 Subtract 340x + 700 from both sides.

70029590 2 xx

Factor the right side. 209350 xx

Set each variable factor equal to zero.

2090 x350 x

Solve for x.x

9

20x35



CONTINUECONTINUEDD

hours 435

140

Rate

Distanceenginefirst theof Time

Because x represents the average rate of the first engine, we reject the negative value, -20/9. The rate of the first engine is 35 miles per hour.5) Check the proposed solution in the original wording of the problem. Do the two engines’ trips take a combined 9 hours? Because the rate of the second engine is 5 miles per hour faster than the rate of the first engine, the rate of the second engine is 35 + 5 = 40 miles per hour.

hours 540

200

Rate

Distanceengine second theof Time

The total time is 4 + 5 = 9 hours. This checks correctly.

EXAMPLE A plane travels 100 mi against the wind in the same time that it takes to travel 120 mi with the wind. The wind speed is 20 mph. Find the speed of the plane in still air.

Let x = the speed of the plane in still air.

Use d = rt, to complete the table


D r t

Against Wind

100 x – 20

With

Wind

120 x + 20

100

20x

100 120

20 20x x

20

120

x

continued

Multiply by the LCD (x – 20)(x + 20).

100 120

20 20x x

100 120

20 20( 20)( 20) ( 20)( 20)

x xx x x x

100( 20) 120( 20)x x

100 2000 120 2400x x

4400 20x

220 x


The speed of the airplane is 220 mph in still air.

Cross multiply?

Check Point 3 on page 455After riding at a steady speed for 40 miles, a bicyclist had a flat tire and walked 5 miles to a repair shop. The cycling rate was 4 times faster than the walking rate. If the time spent cycling and walking was 5 hours, at what rate was the cyclist riding?

Let x = the rate walking.


d r t

Cycling 40 4x

Walking 5 x

x4

40

x

5

continued


EXAMPLEDana Kenly drove 300 mi north from San Antonio, mostly on the freeway. She usually averaged 55 mph, but an accident slowed her speed through Dallas to 15 mph. If her trip took 6 hours, how many miles did she drive at the reduced speed?

Let x = the distance at reduced speed.


d r t

Normal Speed (on freeway)

300 – x 55

Reduced Speed x 15

Total 6

300

55

x

15

x

300

55

x 15

x6

continued

She drove 11 ¼ miles at the reduced speed.

She drove 11 ¼ miles at reduced speed.

3006

55 15165 165

x x

3(300 ) 11 990x x 900 3 11 990x x

8 90x 14

90 or 11

8x


Multiply by the LCD, 165.



Work Problems

In work problems, the number 1 represents one whole job completed. Equations in work problems are often based on the condition that the sum of the separate amount s of the job completed by each person (or machine) working on that job is equal to the whole job, or 1. Suppose there were only two people working on a job and that those two people completed the whole job. Then the following would be true:

Fractional part of the job done by the first person

fractional part of the job done by the second person

1 (one whole job completed).+ =

If it takes a person 10 hours to complete a job working alone, his rate is 1/10. If he ends up working x hours on that job, the fractional part of the job that he gets done is (x)(1/10) or x/10.



EXAMPLEEXAMPLE

A hurricane strikes and a rural area is without food or water. Three crews arrive. One can dispense needed supplies in 10 hours, a second in 15 hours, and a third in 20 hours. How long will it take all three crews working together to dispense food and water?

SOLUTIONSOLUTION

1) Let x represent one of the quantities. Let x = the time, in hours, for all three crews to do the job working together.

2) Represent other quantities in terms of x. There are no other unknown quantities.



3) Write an equation that describes the conditions. Working together, the three crews can dispense the supplies in x hours. We construct a table to find the fractional part of the task completed by the three crews in x hours.

CONTINUECONTINUEDD

Fractional part of job completed

in 1 hour

Time working together

Fractional part of job completed in x

hours

First Crew 1/10 x x/10

Second Crew 1/15 x x/15

Third Crew 1/20 x x/20

Together =1/x =1



Because all three teams working together can complete the job in x hours,

CONTINUECONTINUEDD

.1201510

xxx

4) Solve the equation and answer the question.

1201510

xxx This is the equation for the

problem’s conditions.

160201510

60

xxx Multiply both sides by 60, the LCD.

6020

6015

6010

60 xxx Use the distributive property on

each side.

6 4 3



CONTINUECONTINUEDD

60346 xxx Simplify.

6013 x Combine like terms.

6.4x Divide both sides by 13.

Because x represents the time that it would take all three crews to get the job done working together, the three crews can get the job done in about 4.6 hours.

5) Check the proposed solution in the original wording of the problem. Will the three crews complete the job in 4.6 hours? Because the first crew can complete the job in 10 hours, in 4.6 hours, they can complete 4.6/10, or 0.46, of the job.



CONTINUECONTINUEDD

Because the second crew can complete the job in 15 hours, in 4.6 hours, they can complete 4.6/15, or 0.31, of the job. Because the third crew can complete the job in 20 hours, in 4.6 hours, they can complete 4.6/20, or 0.23, of the job. Notice that 0.46 + 0.31 + 0.23 = 1, which represents the completion of the entire job, or one whole job.

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE

Stan needs 45 minutes to do the dishes, while Bobbie can do them in 30 minutes. How long will it take them if they work together?

Let x = the time it will take them working together.

Rate Time Working Together

Fractional Part of the Job Done

Stan

Bobbie

1

451

30

1

45x

1

30x


x

x

continued

Part done Part done 1 wholeby Stan plus by Bobbie equals job.

Multiply by the LCD, 90.

1

45x 1

30x 1

1 190 90 1

45 30x x

2 3 90x x

5 90x 18x


It will take them 18 minutes working together.



Check Point 4 on page 457A new underwater tunnel is being built using tunnel-boring machines that begin at opposite ends of the tunnel. One tunnel-boring machine can complete the tunnel in 18 months. A faster machine can tunnel to the other side in 9 months. If both machines start at opposite ends and work at the same time, in how many months will the tunnel be finished?

Rate Time Working Together

Fractional Part of the Job Done

Slower machine

Faster machine

18

1

9

1

x

x18

x

9

x

.1918

xx



CONTINUEDCONTINUED

Solve the equation and answer the question.

1918

xx This is the equation

118918

18

xx Multiply both sides by 18, the LCD.

189

1818

18 xx Use the distributive property on

each side.

2

182 xx Simplify.


6x Divide both sides by 3.



Check Point 5An experienced carpenter can panel a room 3 times faster than an apprentice can. Working together, they can panel the room in 6 hours. How long would it take each person working alone to do the job?

Fractional part of job completed

in 1 hour

Time Working Together

Fractional Part of the Job Completed

in 6 hours

Experienced carpenter

Apprenticex

1

x3

1

6

6

x

6

x3

6

13

66

xx



CONTINUEDCONTINUED

Solve the equation and answer the question.

This is the equation

133

663

x

xxx Multiply both sides by 3x, the

LCD.

xx

x

x

x3

3

6363

Use the distributive property on each side.

x3618 Simplify.


8x Divide both sides by 3.

13

66

xx



Do 48 on page 461A pool can be filled by a pipe in 3 hours. It akes 3 times as long for another pipe to empty the pool. How long will it take to fill the pool if both pipes are open?

DONE

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§ 6.7 Formulas and Applications of Rational Equations