Haberman / Kling
MTH 95
Section III: Rational Expressions, Equations, and Functions
Module 3: Adding and Subtracting Rational Expressions
Adding and subtracting rational expressions works the same way as adding and subtracting
fractions.
EXAMPLE:
ADDING FRACTIONS :
5 2 5 3
  
6 9 6 3
15


18
15 

18
19

18
SUBTRACTING FRACTIONS :
2 2
4
11
4 2 11 5
  (create common denominators) 



 

9 2
15
6
15 2
6 5
4
8
55


18
30 30
4
8  55
 (add/subtract numerators) 


30
47

30

The key step in adding or subtracting fractions involves creating common
denominators. This is also the key step in adding or subtracting rational
expressions.
To add or subtract rational expressions with like denominators, add or
subtract the numerators and keep the same denominator:
AB
A
B


C
C
C
AND
AB
A
B


C
C
C
It is often useful to factor denominators so that it is easier to create common denominators.
EXAMPLE: Perform the indicated operations; simplify your result completely.
a.
x
4
 2
x  9 x  20
x  7 x  12
b.
2
4

p
3
p 9
c.
t2  2
1
t

 2
t 1 t  2 t t  2
d.
3 y2
12  12 y

y2
2 y
2
2
2
SOLUTIONS:
a.
x
4
x
4
 2


(factor denominators)
x  9 x  20
x  7 x  12 ( x  5)( x  4) ( x  4)( x  3)
x
x3
4
x5




( x  5)( x  4) x  3 ( x  4)( x  3) x  5
x( x  3)
4( x  5)


( x  5)( x  4)( x  3) ( x  4)( x  3)( x  5)
x( x  3)  4( x  5)

(subtract numerators)
( x  5)( x  4)( x  3)
2



b.
x 2  x  20
( x  5)( x  4)( x  3)
( x  5) ( x  4)
( x  5) ( x  4) ( x  3)
x5
, for x  4
( x  5)( x  3)
2
4
2
4



(factor denominators)
p  3 ( p  3)( p  3)
p3
p 9
p3
2
4



(create common denominators)
( p  3)( p  3)
p3 p3
2  4  ( p  3)

(add numerators)
( p  3)( p  3)
2  4 p  12

( p  3)( p  3)
4 p  10

( p  3)( p  3)
2(2 p  5)

( p  3)( p  3)
2
t2  2
t2  2
1
t
1
t
c.





t  1 t  2 t 2  t  2 t  1 t  2 (t  1)(t  2)
(factor denominators)
t 1
t2  2
1 t2
t





t  1 t  2 t  2 t  1 (t  1)(t  2)

t  2  t (t  1)  t 2  2
(t  1)(t  2)
t  2  t2  t  t2  2
(t  1)(t  2)
0

(t  1)(t  2)
 0, for t  1, t  2

(add / subtract numerators)
3
d. For this example it is helpful to recognize the following fact: –(a – b) = b – a
3y2
12  12 y
3y2
 1  12  12 y


  
y2
2 y
y  2  1  2  y

3y2
1  (12  12 y )

y2
y2

3 y 2  (1)(12  12 y )
y2

3 y 2  12  12 y
y2

3( y 2  4 y  4)
y2

3 ( y  2) ( y  2)
y2
 3( y  2), for y  2
Try these yourself and check your answers.
Perform the indicated operations; simplify your result completely.
a.
1
x

2x  3 x  4
b.
b2
4
 2
b  3 b  2b  3
SOLUTIONS:
a.
x4
2x  3
1
x
1
x





2x  3
x  4 2x  3 x  4
x  4 2x  3

x  4  x(2 x  3)
(2 x  3)( x  4)

x  4  2 x 2  3x
(2 x  3)( x  4)

2 x2  2 x  4
(2 x  3)( x  4)

2( x 2  x  2)
(2 x  3)( x  4)

2( x  1)( x  2)
(2 x  3)( x  4)
(create common denominators)
4
b.
b2
b2
4
4
 2


b  3 b  2b  3 b  3 (b  3)(b  1)
(factor denominators)
b 1
b2
4


b  3 b  1 (b  3)(b  1)
4(b  1)  (b  2)

(b  3)(b  1)


4b  4  b  2
(b  3)(b  1)

3b  2
(b  3)(b  1)
(create common denominators)
EXAMPLE: If f ( x)   3x  5 , find and simplify the expression
f ( x  h)  f ( x )
.
h
SOLUTION:
f ( x  h)  f ( x) 3( x  h)  5  (3 x  5)

h
h
3x  3h  5  3x  5)

h
3 h

h
  3, for h  0
Try this one yourself and check your answer.
If g ( x)  7 x  10 , find and simplify the expression
g ( x + h) - g ( x )
.
h
SOLUTION:
g ( x  h)  g ( x) 7( x  h)  10  (7 x  10)

h
h
7 x  7h  10  7 x  10

h
7h

h
 7, for h  0
5
EXAMPLE
If g ( x) 
4
, find and simplify the expression g (a  2)  g (a ) .
3x
SOLUTION:
g (a  2)  g (a) 

4
4

3(a  2) 3a
4
a
4 a2
 

3(a  2) a 3a a  2
(create common denominators)
4(a  2)
4a

3a(a  2) 3a(a  2)
4a  4(a  2)

3a(a  2)
8

3a(a  2)


8
3a(a  2)
Try this one yourself and check your answer.
If f ( x) 
1
, find and simplify the expression f ( w  1)  f ( w  1) .
2x
SOLUTION:
f ( w  1)  f ( w  1) 

1
1

2( w  1) 2( w  1)
w 1
w 1
1
1



2( w  1) w  1 2( w  1) w  1
w  1  ( w  1)
2( w  1)( w  1)
w 1 w1

2( w  1)( w  1)
2

2( w  1)( w  1)


1
( w  1)( w  1)
(obtain a common denominator)