8.5 Add & Subtract Rational
Expressions
Algebra II
Ex. 1) Remember: When adding or
subtracting fractions, you need a
common denominator!
3 1 4
a.  
5 5 5
c.
1
2
3
4
4 3
2 1
1
b.   

6 6
3 2
6
1 4
2
4
 * 

2 3
3
6
Ex. 2)
4
2
3
7


a.

2x
x
2x 2x
3x
6
3x  6
b.


x4 x4
x4
3( x  2)
or
x4
Example 3:
4
x
 3
3
2
3x 6 x  3x
4
x
 3 2
3x 3x (2 x  1)
** Needs a common denominator 1st!
Sometimes it helps to factor the
denominators to make it easier to find
your LCD.
LCD: 3x3(2x+1)
4(2 x  1)
x
 x
 3
 2
 
3 x (2 x  1) 3 x (2 x  1)  x 
4(2 x  1)  x

3x 3 (2 x  1)
2
x 2  8x  4
 3
3x (2 x  1)
Example 4:
x 1
1
 2
2
x 6 x  9 x  9
x 1
1


( x  3)( x  3) ( x  3)( x  3)
LCD: (x+3)2(x-3)
( x  1)( x  3)
1
 x 3




2
( x  3) ( x  3) ( x  3) ( x  3)  x  3 
( x  1)( x  3)  ( x  3)

2
( x  3) ( x  3)
x 2  3x  x  3  x  3

( x  3) 2 ( x  3)
x 2  3x  6

( x  3) 2 ( x  3)
Complex Fraction – a fraction with
a fraction in the numerator and/or
denominator.
1
Such as:
2
3
5
How would you simplify this complex fraction?
Multiply the top by the reciprocal of the bottom!
1 5 5
* 
3 2 6
Steps to make complex fractions easier.
1. Condense the numerator and
denominator into one fraction each. (if
necessary)
2. Multiply the numerator by the reciprocal
of the denominator.
3. Simplify the remaining fraction.
Example 1:
3
x4

3
( x  4)
( x  1)
3( x  4)
1
3


( x  4)( x  1) ( x  4)( x  1)
x  4 x 1
3
3
( x  4)
( x  4)
3
( x  4)( x  1)



*
x  1  3 x  12
4 x  11
( x  4)
4 x  11
( x  4)( x  1)
( x  4)( x  1)
3( x  4)( x  1)

( x  4)( 4 x  11)
3( x  1)
3x  3

or
4 x  11 4 x  11
Example 2:
2
x 1
4
1

x 1 x
2
 x 1
5x  1
x( x  1)
2
x 1

4x
( x  1)

x( x  1) x( x  1)
2
x( x  1)

*
x  1 5x  1
2
 x 1
4x  x 1
x( x  1)
2 x( x  1)

( x  1)(5 x  1)
2x

5x 1
Example 3:
4
2

2
x 9 x 3
1
1

x 3 x 3
4
2

( x  3)( x  3) ( x  3)

1
1

( x  3) ( x  3)
4
2( x  3)
4  2( x  3)
2 x  10

( x  3)( x  3) ( x  3)( x  3)
( x  3)( x  3)
( x  3)( x  3)



( x  3)
( x  3)
(
x

3
)

(
x

3
)
2x

( x  3)( x  3) ( x  3)( x  3)
( x  3)( x  3)
( x  3)( x  3)
2 x  10
( x  3)( x  3) (2 x  10)( x  3)( x  3)

*

( x  3)( x  3)
2x
( x  3)( x  3)2 x
2 x  10

2x
2( x  5)

2x
x5

x
Assignment