Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.3 - 1.


Linear Equations and Applications

Chapter 2


2.3

Applications of Linear Equations

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 4

2.3 Applications of Linear Equations

Objectives

1. Translate from words to mathematical expressions.

2. Write expressions from given information.

3. Distinguish between expressions and equations.

4. Use the six steps in solving an applied problem.

5. Solve percent problems.

6. Solve investment problems.

7. Solve mixture problems.



Problem-Solving Hint

PROBLEM-SOLVING HINT

Usually there are key words and phrases in a verbal problem that translate

into mathematical expressions involving addition, subtraction, multiplication,

and division. Translations of some commonly used expressions follow.



Translating from Words to Mathematical Expressions

Verbal Expression

The sum of a number and 2

Mathematical Expression

(where x and y are numbers)

Addition

3 more than a number

7 plus a number

16 added to a number

A number increased by 9

The sum of two numbers

x + 2

x + 3

7 + x

x + 16

x + 9

x + y




Verbal Expression

4 less than a number



Subtraction

10 minus a number

A number decreased by 5

A number subtracted from 12

The difference between two

numbers

x – 4

10 – x

x – 5

12 – x

x – y




Verbal Expression

14 times a number



Multiplication

A number multiplied by 8

Triple (three times) a number

The product of two numbers

14x

8x

3x

xy

of a number (used with

fractions and percent)

34 x3

4




Verbal Expression

The quotient of 6 and a number



Division

A number divided by 15

The ratio of two numbers

or the quotient of two numbers

(x ≠ 0)6x

(y ≠ 0)xy

x15


CAUTION

Because subtraction and division are not commutative operations, be carefulto correctly translate expressions involving them. For example, “5 less than anumber” is translated as x – 5, not 5 – x. “A number subtracted from 12” isexpressed as 12 – x, not x – 12. For division, the number by which we are dividing is the denominator, andthe number into which we are dividing is the numerator. For example, “a number divided by 15” and “15 divided into x” both translate as . Similarly,“the quotient of x and y” is translated as .


Caution

x15x

y



Indicator Words for Equality

Equality

The symbol for equality, =, is often indicated by the word is. In fact, any

words that indicate the idea of “sameness” translate to =.



Translating Words into Equations

Verbal Sentence Equation

16x – 25 = 87If the product of a number and 16 is decreased

by 25, the result is 87.

= 48The quotient of a number and the number plus

6 is 48.x + 6

x

+ x = 54The quotient of a number and 8, plus the

number, is 54.8x

Twice a number, decreased by 4, is 32. 2x – 4 = 32



Distinguishing between Expressions

and Equations

(a) 4(6 – x) + 2x – 1

(b) 4(6 – x) + 2x – 1 = –15

There is no equals sign, so this is an expression.

Because of the equals sign, this is an equation.

Decide whether each is an expression or an equation.

Note that the expression in part (a) simplifies to the expression –2x + 23

and the equation in part (b) has solution 19.


Solving an Applied Problem

Step 1 Read the problem, several times if necessary, until you understandwhat is given and what is to be found.

Step 2 Assign a variable to represent the unknown value, using diagrams or tables as needed. Write down what the variable represents. Express any other unknown values in terms of the variable.

Step 3 Write an equation using the variable expression(s).

Step 4 Solve the equation.

Step 5 State the answer to the problem. Does it seem reasonable?

Step 6 Check the answer in the words of the original problem.


Six Steps to Solving Application Problems



Solving a Geometry Problem

Step 1 Read the problem. We must find the length and width of the rectangle.

The length is 2 ft more than three times the width and the perimeter is

124 ft.

The length of a rectangle is 2 ft more than three times the width. The perimeter

of the rectangle is 124 ft. Find the length and the width of the rectangle.

Step 2 Assign a variable. Let W = the width; then 2 + 3W = length.

Make a sketch.

W

2 + 3W

Step 3 Write an equation. The perimeter of a rectangle is given by the

formula P = 2L + 2W.

124 = 2(2 + 3W) + 2W Let L = 2 + 3W and P = 124.




Step 4 Solve the equation obtained in Step 3.



124 = 2(2 + 3W) + 2W

124 = 4 + 6W + 2W

124 – 4 = 4 + 8W – 4

120 8W8 8

15 = W

Distributive property

124 = 4 + 8W Combine like terms.

Subtract 4.

Divide by 8.

120 = 8W

=




Step 5 State the answer. The width of the rectangle is 15 ft and the length is

2 + 3(15) = 47 ft.



Step 6 Check the answer by substituting these dimensions into the words of

the original problem.



Finding Unknown Numerical Quantities

Step 1 Read the problem. We are asked to find the total number of each type

of cookie baked.

A local grocery store baked a combined total of 912 chocolate chip cookies

and sugar cookies. If they baked 336 more chocolate chip cookies than sugar

cookies, how many of each did the store bake?

Step 2 Assign a variable. Let s = the total number of sugar cookies baked.

Since there were 336 more chocolate chip cookies baked,

s + 336 = the total number of chocolate chip cookies baked.

Step 3 Write an equation. The sum of the numbers of the cookies baked is

912, sosugar chocolate chip = total+

s (s + 336) = 912+








s + (s + 336) = 912

2s + 336 = 912 Combine like terms.

Subtract 336.

Divide by 2.

2s + 336 – 336 = 912 – 336

2s = 576

2 2

s = 288

2s 576=




Step 5 State the answer. We let s represent the number of sugar cookies, so

there were 288 sugar cookies baked. Also,




(s + 336) = 624 is the number of chocolate chip cookies baked.

Step 6 Check. 624 is 336 more than 288, and the sum of 624 and 288 is 912.

The conditions of the problem are satisfied, and our answer checks.


CAUTION

A common error in solving applied problems is forgetting to answer all thequestions asked in the problem. In the previous example, we were asked tofind the number of each type of cookie baked, so there was an extra step at the end to find the number of chocolate chip cookies baked.


Caution



Percent Reminder

Percent

Recall from Section 2.2 that percent means “per one hundred,” so 3%

means 0.03, 12% means 0.12, and so on.



Solving a Percent Problem

Step 1 Read the problem. We are given that the number of raffle tickets sold

increased by 350% from the first to the second day, and there were

1440 raffle tickets sold altogether. We must find the number of raffle

tickets sold the first day.

During a 2-day fundraiser, a local school sold 1440 raffle tickets. If they sold

350% more raffle tickets on the second day than the first day, how many raffle

tickets did they sell on the first day?

Step 2 Assign a variable. Let x = the total number raffle tickets sold on the

first day of the fundraiser.

350% = 350(.01) = 3.5,

so 3.5x represents the number of raffle tickets sold on the second day

of the fundraiser.







Step 3 Write an equation from the information given.

tickets sold the first day tickets sold the first day = total tickets sold+

x (3.5x) = 1440+


1x + 3.5x = 1440 Identity property

4.5x = 1440 Combine like terms.

x = 320 Divide by 4.5.







Step 5 State the answer. There were 320 raffle tickets sold on the first day

of the 2-day fundraiser.

Step 6 Check that the number of raffle tickets sold on the second day,

1440 – 320 = 1120, is 350% of 320.


CAUTION

Avoid two common errors that occur in solving problems like the one in theprevious example.

1. Do not try to find 350% of 1440 and subtract that amount from 1440. The 350% should be applied to the number of tickets sold on the first day, not the total number of tickets sold.

2. Do not write the equation as

The percent must be multiplied by some amount; in this case, the amount is the number of tickets sold on the first day, giving 3.5x.


Caution

x + 3.5 = 1440. Incorrect



Solving an Investment Problem

Step 1 Read the problem. We must find the amount invested in each account.

A local company has $50,000 to invest. It will put part of the money in an

account paying 3% interest and the remainder into stocks paying 5%. If the

total annual income from these investments will be $2180, how much will be

invested in each account?

Step 2 Assign a variable.

Let x = the amount to invest at 3%;

50,000 – x = the amount to invest at 5%.

Rate (as a decimal) Interest

.03 0.03x

Principle

.05

x

50,000 – x

The formula for interest is I = p r t.

Time

1

1 .05(50,000 – x)

50,000 2180 Totals








Rate (as a decimal) Interest

0.03 0.03x

Principle

0.05

x

50,000 – x

Time

1

1 0.05(50,000 – x)

50,000 2180 Totals

Step 3 Write an equation. The last column of the table gives the equation.

interest at 3% interest at 5% = total interest+

0.03x 0.05(50,000 – x) = 2180+








Step 4 Solve the equation. We do so without clearing decimals.

0.03x + 0.05(50,000) – 0.05x = 2180 Distributive property

0.03x 0.05(50,000 – x) = 2180+

0.03x + 2500 – 0.05x = 2180 Multiply.

–0.02x + 2500 = 2180 Combine like terms.

–0.02x = –320 Subtract 2500

x = 16,000 Divide by –.02.








Step 5 State the answer. The company will invest $16,000 at 3%. At 5%, the

company will invest $50,000 – $16,000 = $34,000.

and

Step 6 Check by finding the annual interest at each rate; they should total

$2180.

0.03($16,000) = $480 0.05($34,000) = $1700

$480 + $1700 = $2180, as required.





In the previous example, we chose to let the variable represent the amount

invested at 3%. Students often ask, “Can I let the variable represent the

other unknown?” The answer is yes. The equation will be different, but in

the end the two answers will be the same.



Solving a Mixture Problem

Step 1 Read the problem. The problem asks for the amount of 80% solution

to be used.

A chemist must mix 12 L of a 30% acid solution with some 80% solution to get

a 60% solution. How much of the 80% solution should be used?

Step 2 Assign a variable. Let x = the number of liters of 80% solution to be

used.

+ =30% 80% 30%

80%

12 L Unknown

number of liters, x

(12 + x)L


Step 2 Assign a variable. Let x = the number of liters of 80% solution





Percent (as a decimal) Liters of Pure Acid0.30 0.30(12) = 3.6

Number of Liters

0.80

12

x 0.08x12 + x 0.60(12 + x)0.60

Sum must be equal

3.6 .08x .60(12 + x)Step 3 Write an equation. + =

+

=30% 80% 30%

80%

12 L x L (12 + x)L


Step 4 Solve.





3.6 0.08x 0.60(12 + x)+ =

3.6 + 0.80x = 7.2 + 0.60x Distributive property

0.20x = 3.6

Subtract 3.6 and 0.60x.


Step 5 State the answer. The chemist should use 18 L of 80% solution.


Step 6 Check. 12 L of 30% solution plus 18 L of 80% solution is





12(0.30) + 18(0.8) = 18 L

of acid. Similarly, 12 + 18 or 30 L of 60% solution has

30(0.6) = 18 L

Percent (as a decimal) Liters of Pure AcidNumber of Liters

0.30

0.80

12

18

0.60

3.6

14.4

30 18

of acid in the mixture. The total amount of pure acid is 18 L both

before and after mixing, so the answer checks.





When pure water is added to a solution, remember that water is 0% of the

chemical (acid, alcohol, etc.). Similarly, pure chemical is 100% chemical.




Step 1 Read the problem. The problem asks for the amount of pure alcohol

to be used.

A chemist must mix 8 L of a 10% alcohol solution with pure alcohol to get a

a 40% solution. How much of the pure alcohol solution should be used?

Step 2 Assign a variable. Let x = the number of liters of 100% solution to be

used.

+ =10% 100% 10%

100%

8 L Unknown

number of liters, x

(8 + x)L


Step 2 Assign a variable. Let x = the number of liters of 100% solution





Percent (as a decimal) Pure Alcohol (L)

0.10 0.10(8) = .8Number of Liters

1.00

8

x 1x

8 + x 0.40(8 + x)0.40

Sum must be equal

.8 1x .40(8 + x)Step 3 Write an equation. + =

+ =

10% 100%10%

100%

8 L x L (8 + x)L


Step 4 Solve.





0.8 1x 0.40(8 + x)+ =

0.8 + 1x = 3.2 + 0.40x Distributive property

0.60x = 2.4

Subtract 0.8 and 0.40x.


Step 5 State the answer. The chemist should use 4 L of pure alcohol.


Step 6 Check. 8 L of 10% solution plus 4 L pure alcohol is





8(0.10) + 4(1) = 4.8 L

of acid. Similarly, 8 + 4 or 12 L of pure alcohol has

12(0.40) = 4.8 L

Percent (as a decimal) Pure Alcohol (L)Number of Liters

0.10

1.00

8

4

0.40

0.8

4

12 4.8

of alcohol in the mixture. The total amount of pure acid is 4.8 L both

before and after mixing, so the answer checks.

Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.3 - 1.

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