Transcript Ch4-Sec 4.4
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4.4 – Slide 1
Chapter 4
Systems of Linear Equations
and Inequalities
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4.4 – Slide 2
4.4
Applications of Linear Systems
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4.4 – Slide 3
4.4 Applications of Linear Systems
Objectives
1.
2.
3.
4.
Solve problems about unknown numbers.
Solve problems about quantities and their costs.
Solve problems about mixtures.
Solve problems about distance, rate (or speed),
and time.
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4.4 – Slide 4
4.4 Applications of Linear Systems
Solving an Applied Problem with Two Variables
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 two equations using both variables.
Step 4 Solve the system of two equations.
Step 5 State the answer to the problem. Is the answer
reasonable?
Step 6 Check the answer in the words of the original
problem.
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4.4 – Slide 5
4.4 Applications of Linear Systems
Solving Problems about Unknown Numbers
Example 1
Amy and Bryan just celebrated a birthday together. Amy is 3
years older than Bryan. However, if Bryan doubled his age and
subtracted 5, the result would be 4 years older than Amy. How
old are Amy and Bryan?
Step 1 Read the problem carefully. We are looking for Amy
and Bryan’s ages.
Step 2 Assign variables. Let x = Amy’s age, and y = Bryan’s
age.
Step 3 Write two equations. Since Amy is 3 years older than
Bryan, x = 3 + y. And, since Bryan’s age doubled minus
5 is 4 years more than Amy’s age, 2y – 5 = 4 + x.
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4.4 – Slide 6
4.4 Applications of Linear Systems
Solving Problems about Unknown Numbers
Example 1 (continued)
Step 4 Solve.
x=3+y
2y – 5 = 4 + x
x – y=3
+ –x + 2y = 9
y = 12
Step 5 State the answer. We will substitute y = 12 into the
first equation to solve for x.
x–y=3
x – 12 = 3
+12 +12
x = 15
Amy is 15 years old, and Bryan is 12 years old.
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4.4 – Slide 7
4.4 Applications of Linear Systems
Solving Problems about Unknown Numbers
Example 1 (concluded)
Step 6 Check.
x–y=3
2y – 5 = 4 + x
15 – 12 = 3 ?
3=3
2(12) – 5 = 4 + 15 ?
24 – 5 = 19
?
19 = 19
CAUTION
If an applied problem asks for two values, be sure to
give both of them in your answer.
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4.4 – Slide 8
4.4 Applications of Linear Systems
Solving Problems about Quantities and Their Costs
Example 2
Carrie and Diego are headed to the vending machine and have
decided to pool their money. Carrie has only quarters, Diego
has only dimes, and together they have ten coins worth $2.05.
How many quarters does Carrie have, and how many dimes
does Diego have?
Step 1 Read. We are looking for the number of quarters for
Carrie and the number of dimes for Diego.
Step 2 Assign
variables.
# of Coins
Coin Value
Total Value
Carrie
x
0.25
0.25x
Diego
y
0.10
0.10y
Total
x+y
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0.25x + 0.10y
4.4 – Slide 9
4.4 Applications of Linear Systems
Solving Problems about Quantities and Their Costs
Example 2 (continued)
Step 3 Write two equations. Since they had 10 coins together,
x + y = 10. And, since their coins were worth $2.05,
0.25x
+ 0.10y = 2.05.
Step 5 State the Answer.
Step 4 Solve.
We will substitute
–0.10(x + y)= (10)·–0.10
x = 7 into the first
0.25x + 0.10y = 2.05
equation to solve for y.
–0.10x – 0.10y = –1
+ 0.25x + 0.10y = 2.05
0.15x = 1.05
0.15 0.15
x=7
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7 + y = 10
–7
–7
y=3
Carrie had 7 quarters
and Diego had 3 dimes.
4.4 – Slide 10
4.4 Applications of Linear Systems
Solving Problems about Quantities and Their Costs
Example 2 (concluded)
Step 6 Check.
x + y = 10
0.25x + 0.10y = 2.05
7 + 3 = 10 ?
10 = 10
0.25(7) + 0.10(3) = 2.05 ?
1.75 + 0.30 = 2.05 ?
2.05 = 2.05
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4.4 – Slide 11
4.4 Applications of Linear Systems
Solving Problems about Mixtures
Example 3
A pharmacist needs 100L of 50% alcohol solution. She has on
hand 30% alcohol solution and 80% alcohol solution, which she
can mix. How many liters of each will be required to make the
100L of 50% alcohol solution?
Step 1 Read. We are looking for the number of liters of 30%
and 80% alcohol solutions to get a 50% alcohol solution.
Step 2 Assign
Variables.
Percent
# of Liters
Liters of Alcohol
0.30
x
0.30x
0.80
y
0.80y
0.50
100
0.50(100)
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4.4 – Slide 12
4.4 Applications of Linear Systems
Solving Problems about Mixtures
Example 3 (continued)
Step 3 Write two equations. Since we need 100L of
solution, x + y = 100. Since it must be 50% alcohol,
0.30x + 0.80y = 0.50(100).
Step 5 State the answer.
Step 4 Solve.
We substitute y = 40
into the first equation
–0.30(x + y)= (100)·–0.30
to solve for x.
0.30x + 0.80y = 50
x + 40 = 100
–40 –40
–0.30x – 0.30y = –30
x = 60
+ 0.30x + 0.80y = 50
The pharmacist needs 60L of
0.50y = 20
30% alcohol solution and 40L
0.50 0.50
of 80% alcohol solution.
y = 40
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4.4 – Slide 13
4.4 Applications of Linear Systems
Solving Problems about Mixtures
Example 3 (concluded)
Step 6 Check.
x + y = 100
0.30x + 0.80y = 50
60 + 40 = 100 ?
100 = 100
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0.30(60) + 0.80(40) = 50 ?
18 + 32 = 50 ?
50 = 50
4.4 – Slide 14
4.4 Applications of Linear Systems
Solving Problems about Distance, Rate (or Speed), and Time
Example 4
A plane flies 560 mi in 1.75 hr traveling with the wind. The
return trip against the same wind takes the plane 2hr. Find the
speed of the plane and the speed of the wind.
Step 1 Read. We are looking for the speed of the plane and the
speed of the wind.
Step 2 Assign variables.
Let x = speed of the plane
and y = speed of the wind.
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r
t
d
With
Wind
x+y
1.75
560
Against
Wind
x–y
2
560
4.4 – Slide 15
4.4 Applications of Linear Systems
Solving Problems about Distance, Rate (or Speed), and Time
Example 4 (continued)
Step 3 Write two equations. The trip with the wind is
(x + y)1.75 = 560. The trip against the wind is (x – y)2 = 560.
Step 5 State the answer.
Step 4 Solve.
We will substitute
2(1.75x + 1.75y)= (560)2
x = 300 into the second
equation to solve for y.
1.75(2x – 2y)= (560)1.75
2(300) – 2y = 560
600 – 2y = 560
3.50x + 3.50y = 1120
–600
–600
+ 3.50x – 3.50y = 980
–2y = –40
7x = 2100 The speed of the
2
2
plane was 300 mph
7
7
y = 20
x = 300 and the speed of the
wind was 20 mph.
4.4 – Slide 16
Copyright © 2010 Pearson Education, Inc. All rights reserved.
4.4 Applications of Linear Systems
Solving Problems about Distance, Rate (or Speed), and Time
Example 4 (concluded)
Step 6 Check.
1.75x + 1.75y = 560
2x – 2y = 560
1.75(300) + 1.75(20) = 560 ?
2(300) – 2(20) = 560 ?
525 + 35 = 560 ?
600 – 40 = 560 ?
560 = 560
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560 = 560
4.4 – Slide 17