Intermediate Algebra (11th edition)

(Marvins-Underground-K-12) #1

SECTION 7.5 Applications of Rational Expressions 397


OBJECTIVE 3 Solve applications by using proportions.A ratiois a compar-


ison of two quantities. The ratio of ato bmay be written in any of the following ways.


ato b,orRatio of ato b


Ratios are usually written as quotients in algebra. A proportionis a statement that


two ratios are equal.


Proportion

a


b


=


c


d


a


b


a:b,


Solving a Proportion

In 2008, about 15 of every 100 Americans had no health insurance. The population


at that time was about 302 million. How many million Americans had no health


insurance? (Source:U.S. Census Bureau.)


Step 1 Readthe problem.


Step 2 Assign a variable.Let the number of Americans (in millions) who had


no health insurance.


Step 3 Write an equation.The ratio 15 to 100 should equal the ratio xto 302.


Write a proportion.

Step 4 Solve.


Simplify.

x=45.3 Divide by 100.


4530 = 100 x


30,200a


15


100


b =30,200a


x


302


b


15


100


=


x


302


x=


EXAMPLE 4


Multiply by a common denom-
inator,. 100 # 302 =30,200

Step 5 State the answer.There were 45.3 million Americans with no health


insurance in 2008.


Step 6 Checkthat the ratio of 45.3 million to 302 million equals 10015. NOW TRY


NOW TRY
EXERCISE 3
Solve for N.


NR
N-n

=t

NOW TRY
EXERCISE 4
In 2008, about 13 of every 100
Americans lived in poverty.
The population at that time
was about 302 million. How
many Americans lived in
poverty in 2008? (Source:
U.S. Census Bureau.)


NOW TRY ANSWERS
3.
4.39.26 million


N=tnt-R

Solving a Formula for a Specified Variable

Solve for n.


Multiply by

Distributive property on the left
Subtract nrI.
Factor out n.

,orn= Divide by E-rI. NOW TRY


RI


E-rI


RI


E- rI


=n


RI=n 1 E- rI 2


RI=nE-nrI


RI+ nrI=nE


1 R+ nr 2 I= 1 R+nr 2 R+nr.


nE


R+nr


I=


nE


R+nr


EXAMPLE 3


CAUTION Refer to the steps in Examples 2 and 3that factor out the desired


variable. This variable mustbe a factor on only one side of the equation, so that each


side can be divided by the remaining factor in the last step.

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