Download Chapter 5, Section 3 - Palm Beach State College

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Sec 5.3 - 1
Chapter 5
Systems of Linear Equations
Copyright © 2010 Pearson Education, Inc. All rights reserved
Sec 5.3 - 2
5.3
Applications of Systems
of Linear Equations
Copyright © 2010 Pearson Education, Inc. All rights reserved
Sec 5.3 - 3
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.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
Sec 5.3 - 4
5.3 Applications of Systems of Linear Equations
Solving 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.
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Sec 5.3 - 5
5.3 Applications of Systems of Linear Equations
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
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Sec 5.3 - 6
5.3 Applications of Systems of Linear Equations
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 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)
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Sec 5.3 - 7
5.3 Applications of Systems of Linear Equations
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 4 Solve the system of equations. We substitute 3W + 20 for L, in
equation (1), and solve for W.
2L + 2W = 800
2(3W + 20) + 2W = 800
6W + 40 + 2W = 800
8W + 40 = 800
(1)
Let L = 3W + 20.
Distributive property
Combine terms.
8W = 760
Subtract 40.
W = 95
Divide by 8.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
Sec 5.3 - 8
5.3 Applications of Systems of Linear Equations
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 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
Copyright © 2010 Pearson Education, Inc. All rights reserved.
Sec 5.3 - 9
5.3 Applications of Systems of Linear Equations
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 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.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
Sec 5.3 - 10
5.3 Applications of Systems of Linear Equations
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 5EXAMPLE
soft drinks and 2
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.
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Sec 5.3 - 11
5.3 Applications of Systems of Linear Equations
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 5EXAMPLE
soft drinks and 2
4 hamburgers is $32.25. Find the price of a
single hamburger and a soft drink.
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
s + 2h = 12.15.
By similar reasoning, the second equation is
5s + 4h = 32.25.
Therefore, the system is
s + 2h = 12.15
5s + 4h = 32.25.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
(1)
(2)
Sec 5.3 - 12
5.3 Applications of Systems of Linear Equations
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 5EXAMPLE
soft drinks and 2
4 hamburgers is $32.25. Find the price of a
single hamburger and a soft drink.
Step 4 Solve the system of equations.
s + 2h = 12.15
5s + 4h = 32.25
–5s – 10h = –60.75
5s +
4h = 32.25
–6h = –28.50
h = 4.75
Copyright © 2010 Pearson Education, Inc. All rights reserved.
(1)
(2)
Multiply each side of (1) by –5.
(2)
Add.
Divide by –6.
Sec 5.3 - 13
5.3 Applications of Systems of Linear Equations
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 5EXAMPLE
soft drinks and 2
4 hamburgers is $32.25. Find the price of a
single hamburger and a soft drink.
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.
s + 2h = 12.15
s + 2(4.75) = 12.15
s + 9.50 = 12.15
s = 2.65
Copyright © 2010 Pearson Education, Inc. All rights reserved.
(1)
Let h = 4.75.
Multiply.
Subtract 9.50.
Sec 5.3 - 14
5.3 Applications of Systems of Linear Equations
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 5EXAMPLE
soft drinks and 2
4 hamburgers is $32.25. Find the price of a
single hamburger and a soft drink.
Step 5 State the answer. The price of a single soft drink is $2.65 and the
price of a hamburger is $4.75.
Step 6 Check that these values satisfy the conditions stated in the problem.
5(2.65)
1(2.65)
+
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4(4.75)
2(4.75)
= $32.25
$12.15
Sec 5.3 - 15
5.3 Applications of Systems of Linear Equations
EXAMPLE 3
Solving a Mixture Problem
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 1 Read the problem. Two solutions of different strengths are being
mixed together to get a specific amount of a solution with an “inbetween” strength.
Step 2 Assign a variable. Let x = the number of ounces of 10% solution and
y = the number of ounces of 25% solution.
10%
x oz
+
25%
y oz
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=
25%
22%
10%
40 oz
Sec 5.3 - 16
5.3 Applications of Systems of Linear Equations
EXAMPLE 3
Solving a Mixture Problem
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.
Percent (as a decimal)
Number of Ounces
Ounces of Pure Acid
10% = 0.10
x
0.10x
25% = 0.25
y
0.25y
22% = 0.22
40
0.22(40)
Step 3 Write a system of equations.
10%
x oz
+
x
y ==
+
25%
.10x + .25y =
y oz
Copyright © 2010 Pearson Education, Inc. All rights reserved.
40
22%
8.8
40 oz
(1)
(2)
Sec 5.3 - 17
5.3 Applications of Systems of Linear Equations
EXAMPLE 3
Solving a Mixture Problem
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 4 Solve the system.
–10x
10x
x
=
40
(1)
0.10x + 0.25y =
8.8
(2)
+ –10y =
+
y
+
–400
Multiply each side of (1) by –10.
25y =
880
Multiply each side of (2) by 100.
15y =
480
Add.
y =
32
Divide by 15.
Because y = 32 and x + y = 40, x = 8.
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Sec 5.3 - 18
5.3 Applications of Systems of Linear Equations
EXAMPLE 3
Solving a Mixture Problem
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 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)
Number of Ounces
Ounces of Pure Acid
0.10
8
0.8
0.25
32
8
0.22
40
8.8
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Sec 5.3 - 19
5.3 Applications of Systems of Linear Equations
EXAMPLE 4
Solving a Motion Problem
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 1 Read the problem again. Given the distances traveled, we need to
find the speed of each vehicle.
Step 2 Assign variables.
Let x = the speed of the car
and y = the speed of the bus.
Since d = rt,
t = dr .
distance (d)
rate (r)
time (t)
Car
310
x
310
x
Bus
290
y
290
y
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Sec 5.3 - 20
5.3 Applications of Systems of Linear Equations
EXAMPLE 4
Solving a Motion Problem
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 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 table
310 = 290 .
Multiplying both sides by xy gives 310y = 290x.
x
y
distance (d)
rate (r)
time (t)
Car
310
x
310
x
Bus
290
y
290
y
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Sec 5.3 - 21
5.3 Applications of Systems of Linear Equations
EXAMPLE 4
Solving a Motion Problem
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 4 Solve the system of equations using substitution.
x = y + 4
310y = 290x
(1)
(2)
310y = 290x
(2)
310y = 290( y + 4)
Let x = y + 4.
310y = 290y + 1160
Distributive property
20y = 1160
y = 58
Subtract 290y.
Divide by 20.
Because x = y + 4, the value of x is 58 + 4 = 62.
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Sec 5.3 - 22
5.3 Applications of Systems of Linear Equations
EXAMPLE 4
Solving a Motion Problem
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 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:
310
t = dr =
= 5
62
Times are equal.
290
d
= 5
Bus: t = r =
58
Since 62 – 58 = 4, the conditions of the problem are satisfied.
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Sec 5.3 - 23
5.3 Applications of Systems of Linear Equations
EXAMPLE 5
Solving a Problem Involving Prices
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 1 Read the problem again. There are three unknowns.
Step 2 Assign variables to represent the three unknowns.
x = number of loaves of honey wheat,
y = number of loaves of sunflower,
and z = number of loaves of French bread.
Let
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Sec 5.3 - 24
5.3 Applications of Systems of Linear Equations
EXAMPLE 5
Solving a Problem Involving Prices
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 3 Write a system of equations.
x = loaves of honey wheat, y = loaves of sunflower, z = loaves of French bread
y = 2x,
or
y – 2x = 0
(1)
z = y – 3,
or
z – y = –2
(2)
2.69x + 2.89y + 3.39z = 96.58
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(3)
Sec 5.3 - 25
5.3 Applications of Systems of Linear Equations
EXAMPLE 5
Solving a Problem Involving Prices
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 4 Solve the system of three equations using substitution.
y = 2x
(1)
z = y – 3
(2)
z = 2x – 3
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Let y = 2x.
(4)
Sec 5.3 - 26
5.3 Applications of Systems of Linear Equations
EXAMPLE 5
Solving a Problem Involving Prices
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 4 Solve the system of three equations using substitution.
y = 2x
(1)
z = 2x – 3
(4)
269x + 289y + 339z = 9658
Multiply each side (3) by 100.
269x + 289(2x) + 339(2x – 3) = 9658
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Substitute in (3).
Sec 5.3 - 27
5.3 Applications of Systems of Linear Equations
EXAMPLE 5
Solving a Problem Involving Prices
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 4 Solve the system of three equations using substitution.
269x + 578x + 678x – 1017 = 9658
1525x – 1017 = 9658
1525x = 10,675
x = 7
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Multiply & distribute.
Combine terms.
Add 1017.
Divide by 1525.
Sec 5.3 - 28
5.3 Applications of Systems of Linear Equations
EXAMPLE 5
Solving a Problem Involving Prices
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 4 Solve the system of three equations using substitution.
Because x = 7 and y = 2x, y = 14.
Also, because y = 14 and z = y – 3, z = 11.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
Sec 5.3 - 29
5.3 Applications of Systems of Linear Equations
EXAMPLE 5
Solving a Problem Involving Prices
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 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.
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Sec 5.3 - 30
5.3 Applications of Systems of Linear Equations
EXAMPLE 5
Solving a Problem Involving Prices
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 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 - 31