When adding and subtracting fractions, it is necessary to find a
common denominator – preferably the lowest common denominator
(LCD). This is what makes addition and subtraction of fractions
more tedious than multiplication and division. A common denominator
is NOT required for multiplication or division.
The process of determining the LCD is critical when adding or
subtracting fractions but before dealing with examples where the
denominators are different, it might be useful to work with some
where the denominators are the same.
In these cases, it will not be necessary to determine the LCD
because the denominators will start out that way.
When the denominators of the fractions being added or subtracted
start out the same, it is necessary to combine the numerators over
that same denominator.
It may be possible to factor the denominator and the newly combined
numerator in a later step.
5x 7 x

6
6
12x

6
 2x
5x 7 x

4
4
12x

4
 3x
x2 3x

5
5
x 2  3x

5
x(x  3)

5
3x
15

x 2  5x x 2  5x

3x
15

x(x  5) x(x  5)
3x  15

x(x  5)
3(x  5)

x(x  5)

3
x
2x2  5x
20x  8

x2  6x  16 x2  6x  16
2x2  5x
20x  8


(x  8)(x  2) (x  8)(x  2)
2x2  5x  20x  8

(x  8)(x  2)
2x2  15x  8

(x  8)(x  2)
(2x  1)(x  8)

(x  8)(x  2)
(2x  1)

(x  2)
3x2 - 10x – 8
3x2 + 2x - 12x – 8
x(3x + 2) - 4(3x + 2)
(3x + 2)(x - 4)
2x2 + 15x – 8
2×-8
2x2 – 1x + 16x – 8
= -16
x(2x – 1) + 8(2x – 1)
-1
16
(2x – 1)(x + 8)
-2
8
-4
4
x2
3×-8
= -24
1
-24
2
-12
3
-8
4
-6
16
3x 2  10x  8 3x 2  10x  8
x2
16


(3x  2)(x  4) (3x  2)(x  4)
x2  16

(3x  2)(x  4)



(x  4)(x  4)
(3x  2)(x  4)
( x  4)
(3x  2)
When adding and subtracting fractions where the denominators are
different, it is necessary to find the LCD. This is what makes addition
and subtraction of fractions more tedious than multiplication and
division. A common denominator is NOT required for multiplication or
division.
The LCD is basically the simplest expression that all of the starting
denominators can divide into. If the denominators are just numerical,
then the LCD is actually the lowest common multiple of each of the
original denominators.
3 5

5 9
7
5

18 12
Multiples of 5
Multiples of 18
= 5, 10, 15, 20, 25, 30, 35, 45, 50 ,55, …
= 18, 36, 54, 72, 90, 108, …
Multiples of 9
Multiples of 12
= 9, 18, 27, 36, 45, 54 ,63, …
= 12, 24, 36, 48, 60, 72, 84, …
 LCM = 45
 LCM = 36
The steps involved in the addition of algebraic fractions can be
appreciated by analyzing the steps of adding of a simpler pair of
numeric fractions.
3 5

8 6
To add these 2 fractions, we must find the
lowest multiple of denominators 6 and 8 – in
other words ‘the lowest common
denominator’.
LCD = 24
The LCD for these 2 fractions is 24 because 24 is the lowest
number that both 6 and 8 can divide into evenly. 24  8 = 3 and
24  6 = 4
After choosing the LCD, it is now necessary to rewrite each of the
fractions with the LCD instead of the original denominators. NOTE:
If the denominator is changed then the numerator must be changed
by the same factor to preserve the value of the original fractions.
3 3
9


8 3
24
5 4
20


6 4
24
3 5

8 6
9 20


24 24
29

24
Once the fractions have been written with a common
denominator, the operation can be completed in the
numerator.
This whole process basically has 3 critical steps:
1. Choosing the LCD.
- Technically any common denominator will yield the correct answer (not
just the lowest), however if a greater denominator is used it causes all
the numbers thereafter to be bigger and therefore more difficult to
work with. When working with algebraic fractions a common
denominator that is NOT as small as possible will often be crippling and
make the expressions too complicated to work with.
2. Adjusting the numerator of each fraction.
- When the denominator of a fraction is changed to the LCD, the
numerator must be changed by the same factor.
3. Performing operations in numerator.
- Only the numerators are added/subtracted. The denominator
remains whatever the LCD is.
When adding or subtracting fractions with different denominators, we will
deal with them first by working with fractions that have monomial (one-term)
denominators.
5x 7 x
LCD = 6

7 x 2x

12
3
7 x 8x


12 12
7 x  8x

12
15x

12
5x

4
2
3
15x 14x


6
6
15x  14x

6
29x

6
LCD = 12
5x 3

 15x
3
2
6
7x 2

 14x
3 2
6
7x 1
  7x
12 1
12
2x 4
  8x
4
3
12
To determine the LCD for monomial expressions involving variables,
it is helpful to visualize the basic or prime factors of each of the
denominators of the fractions being added or subtracted.
1
5

2 3
6x y z 9xy 3z 2
6x2y3z = 2  3  x  x  y  y  y  z
9xy3z2 = 3  3  x  y  y  y  z  z
 8 basic factors
 8 basic factors
The LCD must contain all of those basic factors. If a factor exists
in one of the original denominators more than once, then it must
also be in the LCD more than once. However, if a factor is in both
denominators only once, then it only needs to be in the LCD once.
6x2y3z = 2  3  x  x  y  y  y  z
9xy3z2 = 3  3  x  y  y  y  z  z
 8 basic factors
 8 basic factors
Referring to the example, the LCD must contain a factor of 2 and 3
because these are factors of 6x2y3z.
LCD = 2  3  3  x  x  y  y  y  z  z = 18x2y3z2
It must contain a second factor of 3 because there are two factors
of 3 in 9xy3z2.
It must contain two factors of x because there are two factors of x
in 6x2y3z. 9xy3z2 has only 1 factor of x and it is already there once.
It must contain a three factors of y because there are 3 factors of y
in 6x2y3z. 9xy3z2 also has 3 factors of y but it is already part of the
LCD 3 times.
It must contain a factor of z because there is a factor of z in 6x2y3z.
9xy3z2 has 2 factors of z. The LCD already has been given one but it
must be given a second.
The LCD has 10 basic factors.
1
5

2 3
6x y z 9xy 3z 2
Another way to determine the LCD is to determine the numeric part
of the LCD by choosing the lowest common multiple and then
determine the variable part by choosing the greatest exponent
from each of the variables of the original denominators.
Multiples of 6 = 6, 12, 18, 24, 30, 36, 42, …
Multiples of 9 = 9, 18, 27, 36, 45, 54, …
6x2y3z
LCM = 18
9xy3z2
The LCD for the x variable will be x2 because 2 is the greatest
exponent for the x variable.
The LCD for the y variable will be y3 because 3 is the greatest
exponent for the y variable.
The LCD for the z variable will be z2 because 2 is the greatest
exponent for the z variable.
LCM = 18x2y3z2
5 7

x y
5y 7x


xy xy
LCD = xy
7 x
7x
 
y x
xy
5y  7x

xy
7x  5y

xy
5 3

x x2
5x 3
 2  2
x
x
5x  3

x2
y
5
5y
 
y
x
xy
LCD = x2
5x
5 x
  2
x x
x
3
3 1
 2
2 
x
x
1
5
3

2 3
6x y
10x 3 y
LCD = 30x3y3
9y2
25x


3 3
30x y
30x 3 y 3
25x  9 y2

30x 3 y 3
a
c
LCD = ab2c2d2
 2 2
2
bc d ab d
a 3bd
c3
 2 2 2  2 2 2
ab c d
ab c d
a 3bd  c3

ab2c2d2
2
25x
5x
5


2 3
5x 30x 3 y 3
6x y
9y2
3
3y 2

3 
2
10x y 3y
30x 3 y 3
a3bd
a2
abd


2
bc d abd
ab2c2d2
c
c2
 2
2 2
ab d
c
c3

ab2c2d2
To determine the LCD for denominators having expressions involving 2
or more terms, it is again helpful to visualize the basic or prime factors
of each of the denominators of the fractions being added or
subtracted.
For this reason, it is necessary to factor the denominators to their
most basic factors at the very beginning.
1
5
 2
2
x  6x  8 x  4
1
5


(x  2)(x  4) (x  2)(x  2)
There are 2 factors in each denominator. The LCD must contain those
factors.
The first denominator has the factors (x + 2) and (x + 4) so they must
be a part of the LCD.
The second denominator has the factors (x + 2) and (x - 2) so they
must be a part of the LCD. Since (x + 2) already is there, it doesn’t
need to be placed there a second time.
LCD = (x + 2)(x + 4)(x – 2)
7
14

4
3
2
6x  36x  48x
9x 3  36x2  36x
7
14


2
6x (x  2)(x  4) 9x(x  2)(x  2)
7
14


2  3  x  x(x  2)(x  4) 3  3  x(x  2)(x  2)
There are 6 factors in the first denominator and 5 factors in the second.
The LCD must contain those factors.
The first denominator has the factors 2, 3, x, x a second time, (x + 2)
and (x + 4) so they must be a part of the LCD.
The second denominator has the factors 3, 3 a second time, x, (x + 2)
and (x + 2) a second time so they must be a part of the LCD. Since 3,
x, and (x + 2) are already there one time, they don’t need to be placed
there again. However because 3 and (x + 2) are in the second
denominator twice, they must be a part of the LCD twice.
LCD = 23xx(x + 2)(x + 4)3(x + 2)
= 233xx (x + 2)(x + 2)(x + 4)
= 18x2(x + 2)2(x + 4)
7
14

6x 4  36x 3  48x2 9x 3  36x2  36x
7
14


6x2 (x  2)(x  4) 9x(x  2)(x  2)
21(x  2)  28x(x  4)

18x2 (x  2)2 (x  4)
21x  42  28x2  112x

18x2 (x  2)2 (x  4)
28x2  133x  42

18x2 (x  2)2 (x  4)
7( 4x2  19x  6)

18x2 (x  2)2 (x  4)
LCD = 18x2(x + 2)2(x + 4)
3(x  2)
7

 21(x + 2)
2
6x (x  2)(x  4) 3(x  2)
LCD
2x(x  4) 28x(x + 4)
14


9x(x  2)(x  2) 2x(x  4)
LCD
7
14

6x 4  36x 3  48x2 9x 3  36x2  36x
7
14


6x2 (x  2)(x  4) 9x(x  2)(x  2)
21(x  2)  28x(x  4)

18x2 (x  2)2 (x  4)
21x  42  28x2  112x

18x2 (x  2)2 (x  4)
 28x2  91x  42

18x2 (x  2)2 (x  4)
 7( 4x2  13x  6)

18x2 (x  2)2 (x  4)
Be careful with subtraction
of fractions.
LCD = 18x2(x + 2)2(x + 4)
3(x  2)
7

 21(x + 2)
2
6x (x  2)(x  4) 3(x  2)
LCD
2x(x  4) 28x(x + 4)
14


9x(x  2)(x  2) 2x(x  4)
LCD
1
5
 2
2
x  7 x  12 x  x  6
1
5


(x  3)(x  4) (x  3)(x  2)
1(x  2)  5(x  4)

(x  3)(x  4)(x  2)
x  2  5x  20

(x  3)(x  4)(x  2)
6x  18

(x  3)(x  4)(x  2)

6(x  3)
(x  3)(x  4)(x  2)
6

(x  4)(x  2)
LCD = (x + 3)(x + 4)(x – 2)
1
(x  2)

(x  3)(x  4)
(x  2)
5
( x  4)

(x  3)(x  2)
( x  4)


1(x – 2)
(x  3)(x  4)(x  2)
5(x + 4)
(x  3)(x  4)(x  2)