copyright © 2010 pearson education, inc. all rights reserved sec 2.4 - 1

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Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.4 - 1

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Page 1: Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.4 - 1

Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.4 - 1

Page 2: Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.4 - 1

Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.4 - 2

Linear Equations and Applications

Chapter 2

Page 3: Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.4 - 1

Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.4 - 3

2.4

Further Applications

of Linear Equations

Page 4: Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.4 - 1

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 4

2.4 Further Applications of Linear Equations

Objectives

1. Solve problems about different

denominations of money.

2. Solve problems about uniform

motion.

3. Solve problems about angles.

Page 5: Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.4 - 1

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 5

2.4 Further Applications of Linear Equations

Problems About Different Denominations of Money

Problem-Solving Hint

In problems involving money, use the fact that

For example, 67 nickels have a monetary value of $.05(67) = $3.35. Forty-two five dollar bills have a value of $5(42) = $210.

number of monetaryunits of the same kind

X denomination =total monetary

value.

Page 6: Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.4 - 1

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 6

2.4 Further Applications of Linear Equations

Problems About Different Denominations of Money

Elise has been saving dimes and quarters in a toy bank. Every time she saves a coin, she pulls a small lever, and the bank records the number of coins that have been deposited as well as the total amount in the bank.

$29.95

202 coins

The bank contains $29.95, consisting of 202 coins. How many of each type coin does the bank contain?

Page 7: Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.4 - 1

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 7

2.4 Further Applications of Linear Equations

Problems About Different Denominations of Money

Continued.

Step 1

$29.95

202 coins

Read the problem.

The problem asks that we find the number of dimes and quarters that have been saved in the bank.

Page 8: Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.4 - 1

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 8

2.4 Further Applications of Linear Equations

Problems About Different Denominations of Money

Continued.

Step 2

Assign a variable. Let x represent the number of dimes;

then 202 – x represents the number of quarters.

Denomination Number of Coins Total Value

$0.10 x 0.10x

$0.25 202 – x 0.25(202 – x)

202 $29.95

Page 9: Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.4 - 1

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 9

2.4 Further Applications of Linear Equations

Problems About Different Denominations of Money

Continued.

Step 3

Write an equation.

Denomination Number of Coins Total Value

$0.10 x 0.10x

$0.25 202 – x 0.25(202 – x)

202 $29.95

TotalsFrom the last column of the table,

Page 10: Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.4 - 1

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 10

2.4 Further Applications of Linear Equations

Problems About Different Denominations of Money

Continued.

Step 4

Solve.

Multiply by 100.

Distributive prop.

Subtract 5,050.

Divide by –15.

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Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 11

2.4 Further Applications of Linear Equations

Problems About Different Denominations of Money

Step 5

Elise has 137 dimes and 202 – x = 202 – 137 = 65 quarters in the bank.

Step 6

The bank has 137 + 65 = 202 coins, and the value of the coins is $.10(137) + $.25(65) = $29.95.

State the answer.

Check.

Caution Always be sure your answer is reasonable!

Continued.

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Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 12

2.4 Further Applications of Linear Equations

Solving a Motion Problem (Opposite Directions)

Two snowmobiles leave the same place, one going east and one going west. The eastbound snowmobile averages 24 mph, and the westbound snowmobile averages 32 mph. How long will it take them to be 245 miles apart?

W E

Page 13: Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.4 - 1

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 13

2.4 Further Applications of Linear Equations

Solving a Motion Problem (Opposite Directions)

Step 1

Read the problem.

We must find the time it takes for the two snowmobiles to be 245 miles apart.

Caution The sum of their distances must be 245 mi. Each does not travel 245 mi.

Continued.

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2.4 Further Applications of Linear Equations

Solving a Motion Problem (Opposite Directions)

Step 2

Assign a variable.The sketch shows what is happening in the problem. Let x represent the time traveled by each snowmobile.

32 mph 24 mph

Starting point

Total distance = 245 mph

Rate Time Distance

Eastbound 24 x 24x

Westbound 32 x 32x

245

Continued.

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Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 15

2.4 Further Applications of Linear Equations

Solving a Motion Problem (Opposite Directions)

Step 3 Write an equation.

Rate Time Distance

Eastbound 24 x 24x

Westbound 32 x 32x

245

Continued.

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Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 16

2.4 Further Applications of Linear Equations

Solving a Motion Problem (Opposite Directions)

Step 4 Solve.

Combine like terms.

Divide by 56; lowest terms

Continued.

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Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 17

2.4 Further Applications of Linear Equations

Solving a Motion Problem (Opposite Directions)

Step 4 State the answer.

The snowmobiles travel hr, or 4 hr and 22½ min.

Step 5 Check.

Distance traveled by eastbound

Distance traveled by westbound

Total distance traveled OK

Continued.

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Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 18

2.4 Further Applications of Linear Equations

Solving a Motion Problem (Same Direction)

Brandon works 360 miles away from his home and returns on weekends. For the trip home, he travels 6 hours on interstate highways and 1 hour on two-lane roads. If he drives 25 mph faster on the interstate highways than he does on the two-lane roads, determine how fast he travels on each part of the trip home.

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2.4 Further Applications of Linear Equations

Solving a Motion Problem (Same Direction)

Step 2

Read the problem.

We are asked to find the speed Brandon drives on the interstate highways and the speed he drives on the two-lane roads.

Continued.

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Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 20

2.4 Further Applications of Linear Equations

Solving a Motion Problem (Same Direction)

Step 3 Assign a variable.

The problem asks for two speeds. We can let Brandon’s speed on the two-lane highways be x. Then

the speed on the interstate highways must be x + 25.

For the interstate highways,

and for the two-lane roads

.

Continued.

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2.4 Further Applications of Linear Equations

Solving a Motion Problem (Same Direction)

Step 2 Assign a variable.

Summarizing this information in a table, we have:

Rate Time Distance

Interstate x + 25 6 6(x + 25)

Two-lane x 1 x

360 total

Continued.

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Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 22

2.4 Further Applications of Linear Equations

Solving a Motion Problem (Same Direction)

Step 3 Write an equation.

Rate Time Distance

Interstate x + 25 6 6(x + 25)

Two-lane x 1 x

360

Continued.

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Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 23

2.4 Further Applications of Linear Equations

Solving a Motion Problem (Same Direction)

Step 3 Solve.

Distributive prop.

Collect like terms.

Subtract 150.

Inverse prop.

Divide by 7.

Continued.

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Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 24

2.4 Further Applications of Linear Equations

Solving a Motion Problem (Same Directions)

Step 5 State the answer.Brandon drives the two-lane roads at a speed of 30 mph; he drives the interstate highways at x + 25 = 30 + 25 = 55 mph.

Step 6 Check by finding the distances using:

Distance traveled on interstate highways

Distance traveled on two-lane roads

Total distance traveled 360 mi OK

Continued.

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Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 25

2.4 Further Applications of Linear Equations

Solving Problems Involving Angles of a Triangle

From Euclidean Geometry

The sum of the angle measures of any triangle equal 180º.

A B

C

D

E

F

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2.4 Further Applications of Linear Equations

Finding Angle Measures

Find the value of x and determine the measure of each angle in the figure.

Page 27: Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.4 - 1

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 27

2.4 Further Applications of Linear Equations

Finding Angle Measures

Step 1 Read the problem.

We are asked to find the measure of each angle in the triangle.

Step 2 Assign a variable.

Let x represent the measure of one angle.

Continued.

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Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 28

2.4 Further Applications of Linear Equations

Finding Angle Measures

Step 3

Write an equation.

The sum of the three measures shown in the figure must equal 180º.

Continued.

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2.4 Further Applications of Linear Equations

Finding Angle Measures

Step 4

Solve the equation.

Collect like terms.

Subtract 110.

Divide by 2.

Continued.

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Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.4 - 30

2.4 Further Applications of Linear Equations

Finding Angle Measures

Step 5 State the answer.

One angle measures 35º, another measures 2x + 25 =

2(35) + 25 = 95º, and the third angle measures 85 – x = 85 – 35 = 50º.

Step 6 Check.

Since 35º + 95º + 50º = 180º, the answer is correct.

Continued.