Section 6.5 Addition and Subtraction with Unlike Denominators
To add or subtract expressions that have different denominators:
1) Find the LCD
2) For each rational expression, find an equivalent rational expression using the LCD as the
denominator. To do this, multiply each rational expression by an expression equivalent to 1 using
the factors of the LCD that are missing from that expressions denominator.
3) Add or subtract the numerators and write the sum or difference over the LCD. If subtracting, make
sure to distribute the “minus” sign thru any grouping symbols.
4) Remove all grouping symbols in the numerator and combine like terms, if possible.
5) Completely factor the numerator and simplify if possible.
#5
Sec. 6.5
LCD  x 2 y 2
4
6

xy 2 x 2 y
Find a fraction equivalent to
4
xy 2
which has the LCD as its denominator. If we compare the denom. of
4
to x 2 y 2 , the denom. is missing a factor of x.
2
xy
6
y
6y
Similarly,
.


2
2 2
x y y x y
Hence, by step 2, we have
4
6
4x
6y
4 x  6 y 22 x  3 y 





2
2
2 2
2 2
2 2
xy
x y x y
x y
x y
x2 y 2
4
x
4x
.


2
2 2
x x y
xy
Section 6.5 Addition and Subtraction with Unlike Denominators (continued) P. 2
#32
p. 292
t
5

t  3 4t  3
By step 2a of How to Find the LCD, LCD
 4t  3
t
4
4t
 
t  3 4 4t  3
t
5
4t
5



t  3 4t  3 4t  3 4t  3

4t  5
4t  3
Note: Do not cancel the 4. See the box on p. 269 of your textbook.
Section 6.5 Addition and Subtraction with Unlike Denominators (continued) P. 3
#54,
p. 292
3  x   x  3   x  3
so, replace 3  x with   x  3

5x
4

x2  9 3  x

5x
4
5x
4



x  3x  3  x  3 x  3x  3 x  3

5x
4

x  3x  3 x  3

x  3
5x
4


x  3x  3 x  3 x  3

5 x  4 x  3
x  3x  3

9 x  12
3(3x  4)
5 x  4 x  12


x  3x  3 x  3x  3 x  3x  3
LCD  x  3x  3
Section 6.5 Addition and Subtraction with Unlike Denominators (continued) P. 4
#24,
p. 292



missing a factor of
z  1.
LCD  z  1 z  1
The denominator of the first RE is missing a factor
of z  1, while the denominator of the 2nd RE is
2z
3z

z 1
z 1
2z
z 1
3z
z 1



z 1 z 1 z 1 z 1
2 z  z  1
3z  z  1
2 z  z  1  3z  z  1


z  1z  1 z  1z  1
z  1z  1
 z ( z  5)
2 z 2  2 z  3z 2  3z
 z 2  5z



z  1z  1
z  1z  1 ( z  1)( z  1)