Copyright © 2013 Pearson Education, Inc.

Section 2.3

Introduction to Problem Solving

Steps for Solving a ProblemStep 1: Read the problem carefully and be sure that you

understand it. (You may need to read the problem more than once.) Assign a variable to what you are being asked to find. If necessary, write other quantities in terms of this variable.

Step 2: Write an equation that relates the quantities described in the problem. You may need to sketch a diagram or refer to known formulas.

Step 3: Solve the equation. Use the solution to determine the solution(s) to the original problem. Include any necessary units.

Step 4: Check your solution in the original problem. Does it seem reasonable?

Page 111

Example

Translate the sentence into an equation using the variable x. Then solve the resulting equation.a.Six times a number plus 7 is equal to 25.b.The sum of one-third of a number and 9 is 18.c.Twenty is 8 less than twice a number. Solutiona. 6x + 7 = 25

6x = 186 18

6 6

x

3x

1b. 9 18

3 x

27x

19

3x

c. 20 = 2x − 8

28 = 2x

14 = x

28 2

2 2

x

Page 112

Consecutive Integers

English Phrase Algebraic Expression

Example

Two consecutive integers x, x + 1 5, 6

Three consecutive integers x, x + 1, x + 2 8, 9, 10

Two consecutive even integers x, x + 2 6, 8

Two consecutive odd integers x, x + 2 -5, -3

Three consecutive even integers x, x + 2, x + 4 2, 4, 6

Three consecutive odd integers x, x + 2, x + 4 3, 5, 7

Algebraic Translations of English Phrases

Page 111

Numbers Example

The sum of three consecutive integers is 126. Find the three numbers.SolutionStep 1: Assign a variable to an unknown quantity.

n: smallest of the three integersn + 1: next integern + 2: largest integer

Step 2: Write an equation that relates these unknown quantities.

n + (n + 1) + (n + 2) = 126

Page 113

Numbers Example (cont)

Step 3: Solve the equation in Step 2.n + (n + 1) + (n + 2) = 126(n + n + n) + (1 + 2) = 126

3n + 3 = 1263n = 123

n = 41So the numbers are 41, 42, and 43.

Step 4: Check your answer. The sum of these integers is 41 + 42 + 43 = 126. The answer checks.

Page 113

Numbers Example 2

The sum of three odd integers is 129. Find the three numbers.SolutionStep 1: Assign a variable to an unknown quantity.

n: smallest of the three integersn + 2: next integern + 4: largest integer

Step 2: Write an equation that relates these unknown quantities.

n + (n + 2) + (n + 4) = 129

Page 113

Numbers Example 2 (cont)

Step 3: Solve the equation in Step 2.n + (n + 2) + (n + 4) = 129(n + n + n) + (2 + 4) = 129

3n + 6 = 1293n = 123

n = 41So the numbers are 41, 43, and 45.

Step 4: Check your answer. The sum of these integers is 41 + 43 + 45 = 129. The answer checks.

Page 113

Other types

Four subtracted from six times a number is 68. Find the number?

6846 x Add 4

726 x Divide 6

12x

A taxi charges $2.00 to turn on the meter plus $0.25 for each eighth of a mile. If you have $10.00, how many eighths of a mile can you go? How many miles is that?

1025.2 x

825. x

32x

Sub 2

Divide .25

32 eighths4 miles

Note: To write x% as a decimal number, move the decimal point in the number x two places to the left and then remove the % symbol.

Slide 10

Page 114

Percent Example

Convert each percentage to fraction and decimal notation.a.47% b. 9.8% c. 0.9%Solution

Fraction Notation Decimal Notationa.

b.

c.

4747%

100 47% 0.47

9.8 98 49 2 499.8%

100 1000 500 2 500

9.8% 0.098

0.9 90.9%

100 1000 0.9% 0.009

Page 114

Percent Example

Convert each real number to a percentage.a. 0.761 b. c. 6.3

Solutiona. Move the decimal point two places to the right and

then insert the % symbol to obtain 0.761 = 76.1%

b.

c. Move the decimal point two places to the right and then insert the % symbol to obtain 6.3 = 630%. Note that percentages can be greater than 100%.

2

5

2 20.40, so 40%.

5 5

Page 114

Percent Example

The price of an oil change for an automobile increased from $15 to $24. Calculate the percent increase.

Solutionold value

ol

-

d

new

valu

e

value100 15

1

24

5

- 100 60%

15)1524(

p

BpAA = p times B

Base = 15amount = 24-15=9 159 p %606.

15

9p

Page 115

15

15

15

9 p

a.

b.

c.

Using the Percent Formula

50%9 a

b %609

9 is 60% of what?

Amount = Percent times Base or a=pb

5009. a

5.4amount

What is 9% of 50?

5018 p

18 is what percent of 50?

b 6.9

15base

5018 p

%3636. percent

6.

6.

6.

9 b

50

50

50

18 p

d.

Using the Percent Formula

940329 p

Amount = Percent times Base or a=pb

A television regularly sells for $940. The sale price is $611. Find the percent decrease in the television’s price?

Base = 940amount = 940-611=329

Decrease = 940-611=329

940

940

940

329 p

%3535.940

329p

Percent Example

Number 72 on page 122, Tuition Increase:Tuition is currently $125 per credit. There are plans to raise tuition by 8% for next year. What will the new tuition be per credit?

Solution

125%108

A

BpAA = p times B

Base = $125percent = 108%

135$

12508.1

A

A

Page 115

Base = $125percent = 8% 125%8

A

BpA

$135 $10 plus $125 is tuition New

10$

12508.0

A

A

New tuition.

Tuition increase.

Percent Example

Number 69 on page 122, Voter Turnout:In the 1980 election for president there were 86.5 million voters, whereas in 2008 there were 132.6 million voters. Fine the percent change in the number of voters.

Solution

5.861.46 p

BpAA = p times B

Base = 86.5Amount = 132.6 – 86.5 = 46.1

%3.53533.5.86

1.46

5.861.46

p

p

Page 115

Percent Example

A car salesman sells a total of 85 cars in the first and second quarter of the year. In the second quarter, he had an increase of 240% over the previous quarter. How many cars did the salesman sell in the first quarter? SolutionStep 1: Assign a variable.

x: the amount sold in the first quarter.

Step 2: Write an equation. x + 2.4x = 85 (Note: A=pB which is A=240%*B)

Page 115

Percent Example (cont)

Step 3: Solve the equation in Step 2. x + 2.4x = 85

3.4x = 85 x = 25

In the first quarter the salesman sold 25 cars.

Step 4: Check your answer. An increase of 240% of 25 is 2.4 × 25 = 60.

Thus the amount of cars sold in the second quarter would be 25 + 60 = 85.

Page 115

Percent Example

5644. xx

After a 40% price reduction, an exercise machine sold for $564. What was the exercise machine’s price before this reduction?

Divide .6

x

bpa

or

6.564

5646. x

940x

Motion Formula

d = r · td = distancer = ratet = time Distance equals rate times time.

A Formula for Motion

Motion Example

A truck driver travels for 4 hours and 30 minutes at a constant speed and travels 252 miles. Find the speed of the truck in miles per hour.SolutionStep 1: Let r represent the truck’s rate, or speed, in miles.

Step 2: The rate is to be given in miles per hour, so change the 4 hours and 30 minutes to 4.5 or 9/2 hours.

d = rtdistance = 252time = 4 hours 30 minutes

9252

2r

Page 117

Motion Example (cont)

Step 3: Solve the equation.

Step 4: d = rt

9252

2r

2 2

9 9

9252

2r

56 r

The speed of the truck is 56 miles per hour.

956 252 miles

2

The answer checks.

Page 117

Motion Example Page 117

Rate Time Distance

Car one

Car two

171 6)1.5(r 1.5r

Number 90 on page 123, Finding Speeds:Two cars pass on a straight highway while traveling in opposite directions. One car is traveling 6 miles per hour faster than the other car. After 1.5 hours the two cars are 171 miles apart. Find the speed of each car.

r

r+6

1.5

1.5

1.5r

1.5(r+6)

171

Motion Example, cont Page 117

Number 90 on page 123, Finding Speeds:Two cars pass on a straight highway while traveling in opposite directions. One car is traveling 6 miles per hour faster than the other car. After 1.5 hours the two cars are 171 miles apart. Find the speed of each car.

Solve the equation and answer the question.1.5r + 1.5(r+6) = 1711.5r + 1.5r + 9 = 171 Distributive property3r + 9 = 171 Combine like terms.3r = 162 Subtract 9 r = 54 Divide both sides by 3

The first car will be 54mph, second car will be 60mph.

Rate Time Distance

Faster

Slower

420 50x 55x

Motion Problem

Motion Problem: Two cars leave from the same place, traveling in opposite directions. The rate of the faster car is 55 miles per hour. The rate of the slower car is 50 miles per hour. In how many hours will the cars be 420 miles apart?

420105x

55 mph

50 mph171

X

X

55x

50x

hours 4x

105

420

105

105x

Mixture Example

A chemist mixes 100 mL of a 28% solution of alcohol with another sample of 40% alcohol solution to obtain a sample containing 36% alcohol. How much of the 40% alcohol was used?Solution

Step 1: Assign a variable.

x: milliliters of 40%x + 100: milliliters of 36%

.

Concentration Solution Amount

(milliliters)

Pure alcohol

28%=0.28 100

40%=0.40

36%=0.36

Page 118

28

.4x

0.36x+36

x x+100

Mixture Example (cont)

A chemist mixes 100 mL of a 28% solution of alcohol with another sample of 40% alcohol solution to obtain a sample containing 36% alcohol. How much of the 40% alcohol was used?

Step 2: Write an equation. 0.28(100) + 0.4x = 0.36(x + 100)

Concentration Solution Amount (milliliters)

Pure alcohol

0.28 100 28

0.40 x 0.4x

0.36 x + 100 0.36x + 36

Column three.

Mixture Example (cont)

Step 3: Solve the equation in Step 2.0.28(100) + 0.4x = 0.36(x + 100) 28(100) + 40x = 36(x + 100)

2800 + 40x = 36x + 3600 2800 + 4x = 3600

4x = 800x = 200

200 mL of 40% alcohol solution was added to the 100 mL of the 28% solution.

Mixture Example (cont)

Step 4: Check your answer. If 200 mL of 40% solution are added to the 100 mL of 28% solution, there will be 300 mL of solution.

200(0.4) + 100(0.28) = 80 + 28 = 108 of pure alcohol.

The concentration is or 36%.

1080.36,

300

Solving a Mixture Problem

Mixture Problem: How many ounces of a 50% alcohol solution must mixed with a 20% alcohol solution to make 240 ounces of 40% alcohol solution?.

962402.5. xx

Quantity Percent Solution

50 percent sol 50%

20 percent sol 20%

New sol 240 40%

x 240-x

.5x

.2(240-x)

96

Solving a Mixture Problem

96483.

96482.5.

962.485.

x

xx

xx

483. x

ounces 160x

Answer: 160 ounces of 50% and 80 ounces of 20%

Mixture Problem continued: How many ounces of a 50% alcohol solution must mixed with a 20% alcohol solution to make 240 ounces of 40% alcohol solution?.

962402.5. xx

Quantity Percent Solution

50 percent sol x 50% .5X

20 percent sol 240-x 20% .2(240-x)

New sol 240 40% 96

Distributive propertyCombine like terms.

Subtract 48

Divide both sides by .3

Suppose you invested $25,000, part at 9% simple interest and the remainder at 12%. If the total yearly interest from these investments was $2550, find the amount invested at each rate.

Problems Involving Simple Interest

Principal Rate Time Interest

One 9% 1

Two 12% 1

total $25000 $2550

x = the amount in the first account.

25000 – x = amount in the second account.

.09x +.12(25000 – x) = 2550

x 25000-x

.09x

.12(25000-x)

(continued) Suppose you invested $25,000, part at 9% simple interest and the remainder at 12%. If the total yearly interest from these investments was $2550, find the amount invested at each rate.

Problems Involving Simple Interest

Principal Rate Time Interest

One X 9% 1 .09X

Two 25000-X 12% 1 .12(25000-X)

total $25000 $2550

4. Solve the equation and answer the question

.09x + .12(25000 – x) = 2550

.09x + .12(25000)-.12x = 2550 distributive property

-.03x + 3000 = 2550 combine like terms .09x-.12x

-3000 -3000 subtract 3000

-.03x = -450 divide by -.03

x = 15000

Answer is: $15,000 at 9% and $10,000 at 12%

Teaching Example 12 on page 119, College Loans:A student borrows two amounts of money. The interest rate on the first amount is 4% and the interest rate on the second amount is 6%. If he borrowed $5000 more at 4% than at 6%, then the total interest for one year is $950. How much does the student owe at each rate?

Problems Involving Simple Interest

Principal Rate Time Interest

One 4% 1

Two 6% 1

total $950

x = the amount in the first loan.

x-5000 = amount in the second loan.

Calculate interest

x x-5000

.04x

.06(x-5000)

.04x +.06(x-5000) = 950

.04x +.06(x-5000) = 950

.04x +.06x-300 = 950 Distribute.

.10x -300 = 950 Combine like terms.

.10x = 1250 Add 300 from both sides.x = 12500 Divide both sides by .1.

Invest $12500 in the 4% account and $7500 in the 6% account.

Problems Involving Simple Interest

Teaching Example 12 on page 119, College Loans (continued):

A student borrows two amounts of money. The interest rate on the first amount is 4% and the interest rate on the second amount is 6%. If he borrowed $5000 more at 4% than at 6%, then the total interest for one year is $950. How much does the student owe at each rate?

P R T Interest

One 4% 1

Two 6% 1

total $950

x x-5000

.04x

.06(x-5000)

Solving a Mixture Problem

Mixture Problem: How many pounds of cashews selling for $8.96 per pound must be mixed with 12 pounds of chocolates selling for $4.48 per pound to create a mixture that sells for $7.28 per pound?.

1228.71248.496.8 xx

Quantity Dollars or price

Total price

Cashews $8.96

Chocolates 12 $4.48

Mix $7.28

x

x+12

8.96x

4.48(12)

7.28(x+12)

Solving a Mixture Problem (cont)

36.8728.776.5396.8 xx

60.3368.1 x

20x

Answer: 20 pounds

Mixture Problem (cont): How many pounds of cashews selling for $8.96 per pound must be mixed with 12 pounds of chocolates selling for $4.48 per pound to create a mixture that sells for $7.28 per pound?

Quantity Dollars or price

Total price

Cashews X $8.96 8.96X

Chocolates 12 $4.48 4.48(12)

Mix X+12 $7.28 7.28(x+12)

1228.71248.496.8 xx

DONE

Objectives

• Steps for Solving a Problem

• Percent Problems

• Distance Problems

• Other Types of Problems