Assignment: Add and Subtract Rational Expressions
Choose any four (4) of the following five problems to solve. Show all work leading to your
answer.
1. Randy drives his car to work each day. Suppose the expression
𝟔
+
𝟔
represents the
𝒙
𝟑𝒙
total time it takes Randy to drive to work in a day, where x represents the average speed of
the car in miles per hour, during the first part of the drive.
a. Use addition to simplify the expression
𝟔
+
𝟔
. Show all the steps necessary to write the
𝒙
𝟑𝒙
expression with a common denominator and to simplify the expression.

When the same factor appears in the numerator and denominator of either fraction, we
can “cancel” these common factors by dividing both terms by the factor.
6
6
+ 3𝑥
𝑥

6
2.3
𝑥
3𝑥
= +
6
2
𝑥
𝑥
= +
After both fractions have the same denominator x, we can find the sum by adding the
numerators and keeping the denominator the same.
6
𝑥
2
6+2
𝑥
𝑥
+ =
 So,
𝟔
𝒙
+
𝟔
𝟑𝒙
=
=
𝟖
𝒙
𝟖
𝒙
b. If the speed of the car was 24 mph during the first part of the drive, find the total
amount of time it took Randy to drive to work that day.

The sum
𝟖
𝒙
represents the total time it takes Randy to drive to work, where x represents
the average speed of the car in miles per hour So if Randy drives his car to work at a
speed of 24 mph, for example, we could find the total time it takes him to drive to work
by substituting 24 for x in the factor
𝟖
𝟐𝟒
=
𝟏
𝟑
𝟖
𝒙
as below:
 In other words, it would take Randy one – third of an hour, or 20 minutes, to
drive work that day.
2. Suppose the expression
𝟑𝟐𝟓
𝒙
𝟑𝟐𝟓
− 𝒙+𝟏𝟓
represents the difference between the times,
measured in hours, it took two moving trucks to drive 325 miles from one town to another,
where x represents the average speed of the slower truck during the drive.
a. Use subtraction to simplify the expression
𝟑𝟐𝟓
𝟑𝟐𝟓
𝒙 − 𝒙+𝟏𝟓 . Show all the steps necessary
to write the expression with a common denominator and to simplify the expression.

First, find the lowest common denominator (LCD). The LCD of x and is 𝒙 + 𝟏𝟓 is
x(x+15). Then rewrite the sum with the LCD.
325 (𝑥+15)
𝑥

.
(𝑥+15)
-
325
𝑥+15
.
𝑥
𝑥
=
325(𝑥+15)
𝑥(𝑥+15)
-
325𝑥
𝑥(𝑥+15)
After both fractions have the same denominator x(x+15), we can find the difference by
adding the numerators and keeping the denominator the same.
325(𝑥+15)
𝑥(𝑥+15)
 So,
325𝑥
- 𝑥(𝑥+15) =
𝟑𝟐𝟓
𝒙
−
𝟑𝟐𝟓
𝒙+𝟏𝟓
=
325(𝑥+15)−325𝑥
𝑥(𝑥+15)
=
𝟑𝟐𝟓(𝟐𝒙+𝟏𝟓)
𝒙(𝒙+𝟏𝟓)
𝟑𝟐𝟓(𝟐𝒙+𝟏𝟓)
𝒙(𝒙+𝟏𝟓)
b. If the slower truck drove at an average speed of 50 mph, find the difference between
the driving times.

This difference
325(2𝑥+15)
𝑥(𝑥+15)
represents the different between the time two moving
trucks to drive from one town to another , where x represents the average speed of the of
the slower truck during the drive .So if the slower truck drove at an average speed of 50
mph, for example, we could find the difference between the driving times by substituting
50 for x in the factor
325(2𝑥+15)
as below:
𝑥(𝑥+15)
325(2.𝟓𝟎+15)
50(50+15)
=
37375
3250
𝟏
= 11 𝟐
 In other words, the difference between the driving times of the two moving
𝟏
trucks is 11 hours , or 11hours and 30 minutes.
𝟐
3. Two cement mixing trucks are scheduled to pour cement for the foundation of a new
apartment building. Suppose the expression
𝟏𝟎𝟎
𝒙
𝟏𝟎𝟎
+ 𝒙+𝟏represents the total time expected
for both trucks to complete the job, where x represents the rate in gallons per minute, at
which the slower truck can pour concrete.
a. Use addition to simplify the expression
𝟏𝟎𝟎
+
𝒙
𝟏𝟎𝟎
𝒙+𝟏
.Show all the steps necessary to
write the expression with a common denominator and to simplify the expression.

First, find the lowest common denominator (LCD). The LCD of x and is 𝒙 + 𝟏 is
x(x+1). Then rewrite the sum with the LCD.
100
𝑥

(𝒙+𝟏)
100 𝑥
. (𝒙+𝟏)+ 𝑥+1.𝑥 =
100(𝑥+1)
𝑥(𝑥+1)
100𝑥
+ 𝑥(𝑥+1)
After both fractions have the same denominator x(x+1), we can find the difference by
adding the numerators and keeping the denominator the same.
100(𝑥+1)
𝑥(𝑥+1)
 So,
𝟏𝟎𝟎
𝒙
100𝑥
+ 𝑥(𝑥+1) =
𝟏𝟎𝟎
+ 𝒙+𝟏 =
100(𝑥+1)+100𝑥
𝑥(𝑥+1)
=
𝟏𝟎𝟎(𝟐𝒙+𝟏)
𝒙(𝒙+𝟏)
𝟏𝟎𝟎(𝟐𝒙+𝟏)
𝒙(𝒙+𝟏)
c. If the slower truck can pour concrete at a rate of 24 gallons per minute, find the total
amount of time it would take both cement mixing trucks to complete the job.

The sum
𝟏𝟎𝟎(𝟐𝒙+𝟏)
𝒙(𝒙+𝟏)
represents the total time expected for both trucks to complete the
job, where x represents the rate in gallons per minute, at which the slower truck can pour
concrete. So if the slower truck can pour concrete at a rate of 24 gallons per minute , for
example, we could find the total time it would take both cement mixing trucks to
complete the job by substituting 24 for x in the sum
100(2.𝟐𝟒+1)
4900
𝟏𝟎𝟎(𝟐𝒙+𝟏)
𝒙(𝒙+𝟏)
as below:
𝟏
= 600 = 8 𝟔
𝟐𝟒(24+1)
 In other words, it would take both cement mixing trucks
8
𝟏
𝟔
hours, or 8
hours and 10 minutes , to complete the job .
5. A school of salmon swims 100 feet against the current in a river and then swims the
𝟏𝟎𝟎
same distance through a stretch of still water. Suppose the expression
−
𝒙−𝟓
𝟏𝟎𝟎
represents the difference between the times it takes the salmon to swim against
𝒙
the current versus in the still water, where x represents the average speed of the salmon
in still water.
𝟏𝟎𝟎
𝟏𝟎𝟎
a. Use subtraction to simplify the expression 𝒙−𝟓 − 𝒙 . Show all the steps necessary
to write the expression with a common denominator and to simplify the expression.

First, find the lowest common denominator (LCD). The LCD of x and is 𝑥 − 𝟓 is
x( x- 5). Then rewrite the sum with the LCD.
100 𝒙
100
𝑥−5 𝒙
𝑥
. −

.
(𝒙−𝟓)
=
(𝒙−𝟓)
𝟏𝟎𝟎𝒙
𝒙(𝒙−𝟓)
–
𝟏𝟎𝟎(𝒙−𝟓)
𝒙(𝒙−𝟓)
After both fractions have the same denominator x(x - 5), we can find the difference by
adding the numerators and keeping the denominator the same.
100𝑥
𝑥(𝑥−5)
–

100(𝑥−5)
𝑥(𝑥−5)
So,
𝟏𝟎𝟎
𝒙−𝟓
−
=
100𝑥−100(𝑥−5)
𝟏𝟎𝟎
𝒙
𝑥(𝑥−5)
=
𝟓𝟎𝟎
𝒙(𝒙−𝟓)
=
100𝑥−100𝑥∓500
𝑥(𝑥−5)
=
𝟓𝟎𝟎
𝒙(𝒙−𝟓)
b. Find the difference between the times if the average speed of the salmon is 20 feet per
minute

The difference
𝟓𝟎𝟎
𝒙(𝒙−𝟓)
represents the different between the time between the times
it takes the salmon to swim against the current versus in the still water, where x
represents the average speed of the salmon in still water. So if the average speed of
the salmon is 20 feet per minute, for example, we could find the difference between
the times by substituting 20 for x in the difference
𝟓𝟎𝟎
𝟐𝟎(𝟐𝟎−𝟓)
=
𝟓𝟎𝟎
𝟑𝟎𝟎
=
𝟓
𝟑
=1
𝟓𝟎𝟎
as below:
𝒙(𝒙−𝟓)
𝟐
𝟑
 In other words, the difference between the times the salmon to swim against
𝟐
the current versus in the still water 1 𝟑 minutes, or 1minutes and 40 seconds.