Slide 1.2- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Slide 1.2- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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

OBJECTIVES

Applications of Linear Equations

Learn procedures for solving applied problems.

Learn to solve finance problems.

Learn to solve uniform-motion problems.

Learn to solve work-rate problems.

Learn to solve mixture problems

SECTION 1.2

1

2

3

4

5


Procedure for Solving Applied ProblemsStep 1 Read the problem as many times as needed

to understand it thoroughly. Pay close attention to the questions asked to help identify the quantity the variable should represent.

Step 2 Assign a variable to represent the quantity you are looking for, and, when necessary, express all other unknown quantities in terms of this variable. Frequently, it is helpful to draw a diagram to illustrate the problem or to set up a table to organize the information.


Step 3 Write an equation that describes the situation.

Step 4 Solve the equation.

Step 5 Answer the question asked in the problem.

Step 6 Check the answer against the description of the original problem (not just the equation solved in step 4).

Procedure for Solving Applied Problems


EXAMPLE 1 Analyzing Investments

Tyrick invests $15,000, some in stocks and the rest in bonds. If he invests twice as much in stocks as he does in bonds, how much does he invest in each?

Solution

Step 2 Let x = the amount invested in stocks. The rest of the $15,000 investment ($15,000 – x) is invested in bonds. We have one more important piece of information to use:



Solution continued

Step 2 cont.

Amount invested in stocks, x =

Twice the amount invested in bonds, 15,000 – x

x 2 15, 000 x Step 3 Replace the verbal description with algebraic expressions.

x 30,000 2xStep 4 Distributive property



Solution continued

Step 5 Tyrick invests $10,000 in stocks and $15,000 – $10,000 = $5000 in bonds.

x 30,000 2xStep 4 Distributive property3x 30,000 Add 2x to both sides.

x 10,000 Divide both sides by 3.

Step 6 Tyrick’s total investment is $10,000 + $5000 = $15,000, and $10,000 (stocks) is twice $5,000 (bonds).


If a principal of P dollars is borrowed for a period of t years with interest rate r (expressed as a decimal) computed yearly, then the total interest paid at the end of years is

Interest computed with this formula is called simple interest. When interest is computed yearly, the rate r is called an annual interest rate (or per annum interest rate).

SIMPLE INTEREST

I Prt.


EXAMPLE 2 Solving Problems Involving Simple Interest

Ms. Sharpy invests a total of $10,000 in blue-chip and technology stocks. At the end of a year, the blue-chips returned 12% and the technology stocks returned 8% on the original investments. How much was invested in each type of stock if the total interest earned was $1060?

Solution

We are asked to find two amounts: that invested in blue-chip stocks and that invested in technology stocks.

Step 1



Solution continued

If we know how much was invested in blue-chip stocks, then we know that the rest of the $10,000 was invested in technology stocks.

Let x = amount invested in blue-chip stocks. Then10,000 – x = amount invested in technology stocks.

Step 2



Solution continued

Invest P r t I = Prt

Blue x 0.12 1 0.12x

Tech 10,000 – x 0.08 1 0.08(10,000 – x)

Interest fromBlue-chip

Interest fromtechnology =

TotalInterest

+



Solution continued

Step 3

$ in blue-chip stocks

0.12x 0.08 10,000 x 1060

12x 8 10, 000 x 106, 000

Step 4 12x 80,000 8x 106,000

4x 26,000

x 6500

10, 000 x 10, 000 6500

3500$ in technology stocks



Solution continued

Step 5 Ms. Sharpy invests $3500 in technology stocks and $6500 in blue- chip stocks.

Step 6 $3500 $6500 $10,000

12% of $6500 $780

8% of $3500 $280

Total interest earned = $1060.


If an object moves at a rate (average speed) r, then the distance traveled d in time t is

UNIFORM MOTION

d rt.


EXAMPLE 3 Solving a Uniform-Motion Problem

A motorcycle policeman is chasing a car that is speeding at 70 miles per hour. The policeman is 3 miles behind the car and is traveling 80 miles per hour. How long will it be before the policeman overtakes the car?

Solution

We are asked to find the amount of time before the policeman overtakes the car. Draw a sketch to help visualize the problem.

Step 2



Solution continued

Let x = distance of car before being overtaken

x + 3 = distance of motorcycle before overtaking the car

Step 2



Solution continued

The time from the start of the chase to the interception point is the same for both the car and the motorcycle.

Objectd

milesr

mph hours

Car x 70

Motorcycle x + 3 80

t d

rx

70

x 3

80



Solution continued

The time required to overtake the car isStep 5

Step 3 x

70

x 3

80

Step 4 8x 7 x 3 8x 7x 21

x 21

t x

70

21

70

3

10.



Solution continued

In of an hour, or 18 minutes, the

policeman overtakes the car.

3

10

Step 6 Policeman travels3

10g80 24 miles.

Car travels3

10g70 21 miles.

Policeman started 3 miles behind the car so he does indeed overtake the car.


EXAMPLE 4 Dealing with a Bomb Threat on the QE II

The Queen Elizabeth II was 1000 miles from Britain and traveling toward Britain at 32 miles per hour. A Hercules aircraft was flying from Britain directly toward the ship and averaging 300 miles per hour. How long would the passengers have to wait for the aircraft to meet the ship?

Solution

Let t = time elapsed before meet.32t = distance ship travels.300t = distance aircraft travels.


EXAMPLE 4 Dealing with a Bomb Threat on the QE II

Solution continued

The aircraft and ship meet after about 3 hours.

Distance ship traveled

Distance aircraft traveled = 1000 miles+

32t 300t 1000

332t 1000

t 1000

3323.01


The portion of a job completed per unit of time is called the rate of work.

WORK RATE

If a job can be completed in x units of time (seconds, hours, days, etc.), then the portion of the job completed in one unit of time (1 second, 1 hour, 1 day, etc.) is 1/x. The portion of the job completed in t units of time (t seconds, t hours, t days, etc.) is t • 1/x . When the portion of the job completed is 1, the job is done.


EXAMPLE 5 Solving a Work-Rate Problem

One copy machine copies twice as fast as another. If both copiers work together, they can finish a particular job in 2 hours. How long would it take each copier, working alone, to do the job?

Solution

Step 1 The speed of one copier is twice the speed of the other. We find out how long it takes the faster copier to do the job working alone. The slower copier will take twice as long.



Solution continued

Step 2 x = number of hours for the faster copier to complete the job alone.

2x = number of hours for the faster copier to complete the job alone.

= portion of the job the faster copier does in 1 hour.

1

x

= portion of the job the faster copier does in 1 hour.

1

2x



Solution continued

Portion done in 1 hr

Time work together

Potion done by each

Faster 2

Slower 2

1

x

1

2x2

1

2x

1

x

21

x

2

x

Slower copier portion

Faster copier portion

= 1+



Solution continued

Step 3

2 1 x

3 x

Step 4

2

x

1

x1

Faster copier takes 3 hr, slower 6 hrStep 5

Together = 1 job

Faster copierStep 62g

1

3

2

3 job

Slower copier2g

1

6

1

3 job


EXAMPLE 6 Solving a Mixture Problem

A full 6-quart radiator contains 75% water and 25% pure antifreeze. How much of this mixture should be drained and replaced by pure antifreeze so that the resulting 6-quart mixture is 50% pure antifreeze?

Solution

Step 1 Find the quantity of the radiator mixture that should be drained (25% pure antifreeze) and replaced by pure (100%) antifreeze.



Solution continued

Let x = quarts 25% antifreeze drained. x = quarts pure antifreeze added.0.25x = quarts pure antifreeze drained.

Step 2

Pure antifreeze final mix

Pure antifreeze original mix

Pure antifreeze drained

Pure antifreeze added

= – +

(50% of 6) = (25% of 6) – (25% of x) + x

0.5 6 0.25 6 0.25x xStep 3



Solution continued

0.5 6 0.25 6 0.25x xStep 3

3 1.5 0.75x

1.5 0.75x

2 x

Step 4

Step 5 Drain 2 quarts of mixture from the radiator.

Step 6 Drain 2 qt from the 6, leaves 4 qt: 1 qt pure antifreeze, 3 qt water. Add 2 qt antifreeze, now have 3 qt antifreeze, 3 qt water, so it’s 50% pure antifreeze solution.

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Slide 1.2- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley