2.7Applications of Linear Equations

Use percent in solving problems involving rates.Recall that percent means “per hundred.” Thus,

percents are ratios in which the second number is always 100. For example, 50% represents the ratio 50 to 100 and 27% represents the ratio 27 to 100.

PROBLEM-SOLVING HINTPercents are often used in problems involving mixing different

concentrations of a substance or different interest rates. In each case, to get the amount of pure substance or the interest, we multiply.

In an equation, percent is always written as a decimal. For example, 35% is written 0.35, not 35, and 7 % is written 0.07 not 7.

Interest Problems (annual) principle × rate (%) = interest

p × r = I

Mixture Problems base × rate (%) = percentage

b × r = p

EXAMPLE 1

What is the amount of pure acid in 40 L of a 16% acid solution?

Find the annual interest if $5000 is invested at 4%.

Using Percents to Find Percentages

Solution:

Solution:

40 0.16 6.4L L

$5000 0.04 $200

PROBLEM-SOLVING HINTIn the examples that follow, we use tables to organize the

information in the problems. A table enables us to more easily set up an equation, which is usually the most difficult step.

EXAMPLE 2

Kg of Percentage Kg of

Metal (as a decimal) Copper

x 0.4 0.4x

80 0.7 80(0.7)=56

x+80 0.5 0.5(x + 80)

Solution: Let x = kg of 40% copper metal.

Solving a Mixture Problem

160 kg of the 40% copper metal is needed.

A certain metal is 40% copper. How many kilograms of this metal must be mixed with 80 kg of a metal that is 70% copper to get a metal that is 50% copper?

4 15 55 60x xx x

1 161 1

0x

160x

0.4 5610 10 10 0.5 80x x

5604 560 5 400 560x x

Then, 0.4 56 .5 80 .x x

€

4x + 560 = 5x + 400

€

4x = 5x −160

EXAMPLE 3

Amount Invested Rate of Interest for

in Dollars Interest One Year

x 0.05 0.05x

2x + 3000 0.08 0.08(2x+3000)

Solution: Let x = amount invested at 5%.

$7000 was invested at 5% interest.

With income earned by selling a patent, an engineer invests some money at 5% and $3000 more than twice as much at 8%. The total annual income from the investment is $1710. Find the amount invested at 5%.

21 147021 21

00x 7000x

100 1000.08 2 3000 0.05 0 171010x x

2400016 24000 5 1710 2400000x x

€

Then, 0.08(2x + 3000) + 0.05x =1710

€

8(2x + 3000) + 5x =171000

€

16x + 24000 + 5x =171000

€

21x =147000

Textbook Homework2.1

1-71 ODD2.2

1-75 ODD

2.3 5 - 53 ODD

2.4 5 – 65 ODD

2.55 – 85 EOO

2.63 – 67 EOO

2.77 – 53 EOO

OR MYMATHLAB!

Handout!