copyright © 2010 pearson education, inc. all rights reserved sec 5.3 - 1


Systems of Linear Equations

Chapter 5


5.3

Applications of Systems

of Linear Equations

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 4

5.3 Applications of Systems of Linear Equations

Objectives

1. Solve geometry problems using two variables.

2. Solve money problems using two variables.

3. Solve mixture problems using two variables.

4. Solve distance-rate-time problems using two variables.

5. Solve problems with three variables using a system of three equations.


5.3 Applications of Systems of Linear EquationsSolving an Applied Problem

by Writing a System of Equations

Step 1 Read the problem, several times if necessary, until you understand

what is given and what is to be found.

Step 2 Assign variables to represent the unknown values, using diagrams

or tables as needed. Write down what each variable represents.

Step 3 Write a system of equations that relates the unknowns.

Step 4 Solve the system of equations.

Step 5 State the answer to the problem. Does it seem reasonable?

Step 6 Check the answer in the words of the original problem.



EXAMPLE 1 Finding the Dimensions of a Parking Lot

The length of a rectangular parking lot is 20 ft more than three times its width.

The perimeter of the parking lot is 800 ft. What are the dimensions of the

parking lot?

Step 1 Read the problem again. We must find the dimensions of the parking

lot.

Step 2 Assign variables. Let L = the length and W = the width.

L

W






parking lot?

Step 3 Write a system of equations. Because the perimeter is 800 ft, we

find one equation by using the perimeter formula:

2L + 2W = 800.

Because the length is 20 ft more than three times its width, we have

L = 3W + 20.

The system is, therefore,

2L + 2W = 800 (1)

L = 3W + 20. (2)






parking lot?

Step 4 Solve the system of equations. We substitute 3W + 20 for L, in

equation (1), and solve for W.

2L + 2W = 800 (1)

2(3W + 20) + 2W = 800 Let L = 3W + 20.

6W + 40 + 2W = 800 Distributive property

8W + 40 = 800 Combine terms.

8W = 760 Subtract 40.

W = 95 Divide by 8.






parking lot?

Step 4 Solve the system of equations. We just solved the equation

2L + 2W = 800 and found W = 95.

Now, let W = 95 in the equation L = 3W + 20 to find L.

L = 3W + 20

L = 3(95) + 20 Let W = 95.

L = 305






parking lot?

Step 5 State the answer. The length is 305 ft and the width is 95 ft.

Step 6 Check. The perimeter is 2(305) + 2(95) = 800 ft, and the length, 305 ft,

is 20 ft more than three times the width, since 3(95) + 20 = 305. The

answer is correct.


EXAMPLE 2

Solving a Problem about Prices

At a local restaurant, the price of 1 soft drink and 2 hamburgers is $12.15 and

the price of 5 soft drinks and 4 hamburgers is $32.25. Find the price of a

single hamburger and a soft drink.

Step 1 Read the problem again. There are two unknowns.

Step 2 Assign variables. Let s represent the price of one soft drink and h

represent the price of one hamburger.

EXAMPLE 2



EXAMPLE 2





Step 3 Write a system of equations. Because one soft drink and 2

hamburgers cost a total of $12.15, one equation for the system is

EXAMPLE 2


s + 2h = 12.15.

By similar reasoning, the second equation is

5s + 4h = 32.25.

Therefore, the system is

s + 2h = 12.15 (1)

5s + 4h = 32.25. (2)


EXAMPLE 2





Step 4 Solve the system of equations.

EXAMPLE 2


–5s – 10h = –60.75 Multiply each side of (1) by –5.

5s + 4h = 32.25 (2)

–6h = –28.50 Add.

h = 4.75 Divide by –6.

s + 2h = 12.15 (1)

5s + 4h = 32.25 (2)


EXAMPLE 2





Step 4 Solve the system of equations. We just found h = 4.75. Now, let

h = 4.75 in the equation s + 2h = 12.15 to find s.

EXAMPLE 2


s + 2h = 12.15 (1)

s + 2(4.75) = 12.15 Let h = 4.75.

s + 9.50 = 12.15 Multiply.

s = 2.65 Subtract 9.50.



EXAMPLE 2





Step 5 State the answer. The price of a single soft drink is $2.65 and the

price of a hamburger is $4.75.

EXAMPLE 2

Step 6 Check that these values satisfy the conditions stated in the problem.

5(2.65) 4(4.75) = $32.25+1(2.65) 2(4.75) = $12.15+



EXAMPLE 3 Solving a Mixture Problem

Step 1 Read the problem. Two solutions of different strengths are being

mixed together to get a specific amount of a solution with an “in-

between” strength.

How many ounces each of 10% hydrochloric acid and 25% hydrochloric acid

must be combined to get 40 oz of solution that is 22% hydrochloric acid?

Step 2 Assign a variable. Let x = the number of ounces of 10% solution and

y = the number of ounces of 25% solution.

+ =10% 25% 10%

25%

x oz y oz 40 oz

22%


Step 2 Assign a variable. Let x = the number of ounces of 10% solution and

y = the number of ounces of 25% solution.





Percent (as a decimal) Ounces of Pure Acid

0.10x

Number of Ounces

0.25y

0.22(40)

x y 40

Step 3 Write a system of equations.

+ =+ =10% 25% 22%

x oz y oz 40 oz.10x .25y 8.8+ =

(1)

(2)

x

y

40

10% = 0.10

25% = 0.25

22% = 0.22


Step 4 Solve the system.





x y 40+ =

0.10x 0.25y 8.8+ =

(1)

(2)

–10x –10y –400+ =

10x 25y 880+ =

Multiply each side of (1) by –10.

Multiply each side of (2) by 100.

15y 480= Add.

y 32= Divide by 15.

Because y = 32 and x + y = 40, x = 8.


Step 5 State the answer. The desired mixture will require 8 oz of the 10%

solution and 32 oz of the 25% solution.





Step 6 Check that these values satisfy both equations of the system.

Percent (as a decimal) Ounces of Pure AcidNumber of Ounces

0.10

0.25

8

32

0.22

0.8

8

40 8.8



EXAMPLE 4 Solving a Motion Problem

Step 1 Read the problem again. Given the distances traveled, we need to

find the speed of each vehicle.

A car travels at 310 miles in the same time that a bus travels 290 miles. If the

speed of the car is 4 mph faster than the speed of the bus, find both speeds.

Step 2 Assign variables.

Let x = the speed of the car

and y = the speed of the bus.

distance (d) time (t)rate (r)

Car

Bus

x

y

310

290

Since d = rt,

t = .dr

310x

290y




Step 3 Write a system of equations. The problem states that the car travels

4 mph faster than the bus. Since the two speeds are x and y,



x = y + 4.

Both vehicles travel for the same time, so from the table310

x290

y= .

distance (d) time (t)rate (r)

Car

Bus

x

y

310

290

310x

290y

Multiplying both sides by xy gives 310y = 290x.




Step 4 Solve the system of equations using substitution.



x = y + 4 (1)

310y = 290x (2)

310y = 290x (2)

310y = 290( y + 4) Let x = y + 4.

310y = 290y + 1160 Distributive property

20y = 1160 Subtract 290y.

y = 58 Divide by 20.

Because x = y + 4, the value of x is 58 + 4 = 62.




Step 5 State the answer. The car’s speed is 62 mph, and the speed of the

bus is 58 mph.



Step 6 Check. This is especially important since one of the equations had

variable denominators.

Car: t = dr = 310

62= 5

Bus: t = dr = 290

58= 5

Since 62 – 58 = 4, the conditions of the problem are satisfied.

Times are equal.



EXAMPLE 5 Solving a Problem Involving Prices

Step 1 Read the problem again. There are three unknowns.

At a local bakery, a loaf of honey wheat bread costs $2.69, a loaf of sunflower

bread costs $2.89, and a loaf of French bread costs $3.39. On a recent day,

twice as many loaves of sunflower were sold as honey wheat. The number of

loaves of French bread sold was three less than the number of loaves of

sunflower. Total receipts for these breads were $96.58. How many loaves of

each type of bread were sold?

Step 2 Assign variables to represent the three unknowns.

Let x = number of loaves of honey wheat,

y = number of loaves of sunflower,

and z = number of loaves of French bread.




Step 3 Write a system of equations.







y = 2x, or y – 2x = 0 (1)

x = loaves of honey wheat, y = loaves of sunflower, z = loaves of French bread

z = y – 3, or z – y = –2 (2)

2.69x + 2.89y + 3.39z = 96.58 (3)




Step 4 Solve the system of three equations using substitution.







y = 2x (1)

z = y – 3 (2)

z = 2x – 3 Let y = 2x. (4)











y = 2x (1)

z = 2x – 3 (4)

269x + 289y + 339z = 9658 Multiply each side (3) by 100.

269x + 289(2x) + 339(2x – 3) = 9658 Substitute in (3).











269x + 578x + 678x – 1017 = 9658 Multiply & distribute.

1525x – 1017 = 9658 Combine terms.

1525x = 10,675 Add 1017.

x = 7 Divide by 1525.











Because x = 7 and y = 2x, y = 14.

Also, because y = 14 and z = y – 3, z = 11.




Step 5 State the answer. The solution set is { (7, 14, 11) }, meaning that

7 loaves of honey wheat, 14 loaves of sunflower, and 11 loaves

of French bread were sold.
















Step 6 Check. Since 14 = 2(7), the number of loaves of sunflower sold is

twice the number of loaves as honey wheat. Also, 14 – 3 = 11, so the

number of loaves of French bread is three less than the number of

loaves of sunflower. The total from the receipts is $96.58 as stated.

2.69(7) + 2.89(14) + 3.39(11) = 96.58.

copyright © 2010 pearson education, inc. all rights reserved sec 5.3 - 1

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