§ 6.7
Formulas and Applications of Rational
Equations
Applications of Rational Equations
EXAMPLE
Solve the formula for R: I 
E
.
Rr
SOLUTION
E
I
Rr
R  r   I  R  r  
I  R  r   E
I
E
Rr
This is the original equation.
E
Rr
Multiply both sides by the
LCD, R + r.
Simplify.
Divide both sides by R + r.
Blitzer, Intermediate Algebra, 4e – Slide #104
Applications of Rational Equations
EXAMPLE
A company is planning to manufacture small canoes. Fixed
monthly cost will be $20,000 and it will cost $20 to produce each
canoe.
(a)Write the cost function, C, of producing x canoes.
(b) Write the average cost function, C, of producing x canoes.
(c) How many canoes must be produced each month for the
company to have an average cost of $40 per canoe?
Blitzer, Intermediate Algebra, 4e – Slide #105
Applications of Rational Equations
CONTINUED
SOLUTION
(a) The cost function, C, is the sum of the fixed cost and the variable
costs.
Cx  20,000 20x
Fixed cost is
$20,000.
Variable cost: $20
for each canoe
produced.
(b) The average cost function, C , is the sum of fixed and variable
costs divided by the number of canoes produced.
20,000  20 x
C x  
x
Blitzer, Intermediate Algebra, 4e – Slide #106
Applications of Rational Equations
CONTINUED
(c) We are interested in the company’s production level that results
in an average cost of $40 per canoe. Substitute 40, the average cost,
for C x  and solve the resulting rational equation for x.
40 
20,000  20 x
x
Substitute 40 for Cx.
40x  20,000 20x
Multiply both sides by the LCD, x.
20x  20,000
Subtract 20x from both sides.
x  1,000
Divide both sides by 20.
The company must produce 1,000 canoes each month for an average
cost of $40 per canoe.
Blitzer, Intermediate Algebra, 4e – Slide #107
Applications of Rational Equations
Time in Motion
d
t
r
Distance traveled
Time traveled 
Rate of travel
Blitzer, Intermediate Algebra, 4e – Slide #108
Applications of Rational Equations
EXAMPLE
An engine pulls a train 140 miles. Then a second engine, whose
average rate is 5 miles per hour faster than the first engine, takes
over and pulls the train 200 miles. The total time required for both
engines is 9 hours. Find the average rate of the first engine.
SOLUTION
1) Let x represent one of the quantities. Let
x = the rate of the first engine.
2) Represent other quantities in terms of x. Because the average
rate of the second engine is 5 miles per hour faster than the
average rate of the first engine, let
x + 5 = the rate of the second engine.
Blitzer, Intermediate Algebra, 4e – Slide #109
Applications of Rational Equations
CONTINUED
3) Write an equation that describes the conditions. By reading
the problem again, we discover that the crucial idea is that the time
for both engines’ trips is 9 hours. Thus, the time of the first engine
plus the time of the second engine is 9 hours.
Train 1
Train 2
Distance
Rate
Time
140
x
140
x
200
x5
200
x+5
The sum of the
two times is 9
hours.
We are now ready to write an equation that describes the problems’
conditions.
Blitzer, Intermediate Algebra, 4e – Slide #110
Applications of Rational Equations
CONTINUED
plus
time of the
second train
Time of the
first train
equals
140 200

9
x
x5
9 hours.
4) Solve the equation and answer the question.
140 200

9
x
x5
 140 200 
x  x  5  

  x  x  5  9
x5
 x
x  x  5 
140
200
 x  x  5 
 x  x  5  9
x
x5
This is the equation for the
problems’ conditions.
Multiply both sides by the
LCD, x(x + 5).
Use the distributive
property on both sides.
Blitzer, Intermediate Algebra, 4e – Slide #111
Applications of Rational Equations
CONTINUED
140x  5  200x  9 xx  5
140x  700 200x  9 x 2  45x
340x  700  9 x 2  45x
0  9 x 2  295x  700
0  x  359 x  20
0  x  35
35  x
0  9 x  20
 20
x
9
Simplify.
Use the distributive property.
Combine like terms.
Subtract 340x + 700 from both
sides.
Factor the right side.
Set each variable factor equal
to zero.
Solve for x.
Blitzer, Intermediate Algebra, 4e – Slide #112
Applications of Rational Equations
CONTINUED
Because x represents the average rate of the first engine, we reject
the negative value, -20/9. The rate of the first engine is 35 miles
per hour.
5) Check the proposed solution in the original wording of the
problem. Do the two engines’ trips take a combined 9 hours?
Because the rate of the second engine is 5 miles per hour faster
than the rate of the first engine, the rate of the second engine is 35
+ 5 = 40 miles per hour.
Distance 140
Time of the first engine 

 4 hours
Rate
35
Distance 200
Time of the second engine 

 5 hours
Rate
40
The total time is 4 + 5 = 9 hours. This checks correctly.
Blitzer, Intermediate Algebra, 4e – Slide #113
Applications of Rational Equations
Work Problems
In work problems, the number 1 represents one whole job
completed. Equations in work problems are often based
on the following condition:
Fractional part of
the job done by
the first person
+
fractional part of
the job done by
the second person
=
1 (one whole job
completed).
Blitzer, Intermediate Algebra, 4e – Slide #114
Applications of Rational Equations
EXAMPLE
A hurricane strikes and a rural area is without food or water. Three
crews arrive. One can dispense needed supplies in 10 hours, a
second in 15 hours, and a third in 20 hours. How long will it take
all three crews working together to dispense food and water?
SOLUTION
1) Let x represent one of the quantities. Let
x = the time, in hours, for all three crews to do the job
working together.
2) Represent other quantities in terms of x. There are no other
unknown quantities.
Blitzer, Intermediate Algebra, 4e – Slide #115
Applications of Rational Equations
CONTINUED
3) Write an equation that describes the conditions. Working
together, the three crews can dispense the supplies in x hours. We
construct a table to find the fractional part of the task completed by
the three crews in x hours.
Fractional
part of job
completed
in 1 hour
Time
working
together
Fractional
part of job
completed
in x hours
First Crew
1/10
x
x/10
Second Crew
1/15
x
x/15
Third Crew
1/20
x
x/20
Blitzer, Intermediate Algebra, 4e – Slide #116
Applications of Rational Equations
CONTINUED
Because all three teams working together can complete the job in x
hours,
x
x
x
 
 1.
10 15 20
4) Solve the equation and answer the question.
x
x
x
 
1
10 15 20
x 
 x x
60      601
 10 15 20 
6
4 x
3
x
x
60   60   60 
 60
10
15
20
This is the equation for the
problem’s conditions.
Multiply both sides by 60, the
LCD.
Use the distributive property on
each side.
Blitzer, Intermediate Algebra, 4e – Slide #117
Applications of Rational Equations
CONTINUED
6 x  4 x  3x  60
13 x  60
x  4 .6
Simplify.
Combine like terms.
Divide both sides by 13.
Because x represents the time that it would take all three crews to
get the job done working together, the three crews can get the job
done in about 4.6 hours.
5) Check the proposed solution in the original wording of the
problem. Will the three crews complete the job in 4.6 hours?
Because the first crew can complete the job in 10 hours, in 4.6
hours, they can complete 4.6/10, or 0.46, of the job.
Blitzer, Intermediate Algebra, 4e – Slide #118
Applications of Rational Equations
CONTINUED
Because the second crew can complete the job in 15 hours, in 4.6
hours, they can complete 4.6/15, or 0.31, of the job. Because the
third crew can complete the job in 20 hours, in 4.6 hours, they can
complete 4.6/20, or 0.23, of the job. Notice that 0.46 + 0.31 + 0.23
= 1, which represents the completion of the entire job, or one whole
job.
Blitzer, Intermediate Algebra, 4e – Slide #119