HFCC Math Lab
Arithmetic - 2
Addition and Subtraction of Fractions
Part 1: Addition and Subtraction of Fractions with the Same Denominators
As you perhaps already know, it is very easy to add and subtract fractions with the same
denominators. The following procedure is used.
A)
To add two or more fractions with the same denominators, add all the numerators
and place this sum over the common denominator. Reduce the resulting fraction to
the lowest terms.
B)
To subtract two fractions with the same denominators, write the first numerator
minus the second numerator and place this difference over the common
denominator. Reduce the resulting fraction to lowest terms.
Note: Never add or subtract the denominators.
Note: Always reduce your answer to lowest terms.
Consider the following examples:
2 4

9 9
Ex 1: Find
2 4 2  4 6 6 3 2
 
 

9 9
9
9 9 3 3
Ex 2: Find
5 4

15 6
Before adding or subtracting fractions, the first step is to make sure that each fraction in
the problem is reduced to lowest terms.
We have
5
55 1


15 15  5 3
Therefore,
5 4 1 2 1 2 3
   
 1
15 6 3 3
3
3
Revised 02 /10
4 42 2


6 62 3
and
1
4 3 14
 
20 5 35
Ex 3: Find
4

20
44


20  4
1


5
Reduce each fraction to lowest terms, then add.
3
14

5
35
3
14  7

5
35  7
3
2
1 3  2 6
1


  1
5
5
5
5
5
13 5

24 24
Ex 4: Find
13 5 13  5 8
8 8 1
 =



24 24
24
24 24  8 3
29 11

72 72
Ex 5: Find
29 11 29  11 18 18  9 2 2  2 1




 

72 72
72
72 72  9 8 8  2 4
Let us now consider some examples in which addition and subtraction appear in the same
problem.
11 3 15
 
16 16 16
Ex 6: Find
Remembering the Order of Operations, we add and subtract,
working from left to right.
11 3 15 11  3  15 8  15 23
7
  


1
16 16 16
16
16
16
16
10 9 8
 
14 21 28
Ex 7: Find
First, reduce each fraction to lowest term.
10
9
8


14
21
28
10  2
93
84



14  2
21  3
28  4
5
3
2



7
7
7
53 2
82
6



Remember to add and subtract from left to right.
7
7
7
Revised 02 /10
2
Part 2: Addition and Subtraction of Fractions with Different Denominators.
Recall that we can add and subtract fractions only when their denominators are the same.
Therefore, the first step in adding and subtracting fractions with different denominators is to find
the least common denominator (LCD) of the fractions.
The Least Common Denominator (LCD) of given fractions is the Least Common Multiple
(LCM) of their Denominators.
The least common multiple (LCM) of two or more numbers is the least (smallest) number which
is divisible by each of the given numbers. For example, the least common multiple of 3, 6, and 9
is 18 because no number smaller than 18 is divisible by 3, 6, and 9. Frequently, we are able to
find LCM of two or more numbers by inspection. But in those cases where we are not able to
find the LCM by inspection, we use the following procedure.
Step 1.
Express each of the numbers as the product of two or more prime factors.
(Recall that the first few prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 etc).
If a prime factor appears more than once, write the product using
exponential notation.
Step 2.
List all the different prime factors used in Step 1.
Step 3.
Find the greatest exponent of each prime factor used in Step 1. (Remember
that if there is no exponent, it is understood to be 1).
Step 4.
Multiply all the prime factors with the exponents determined in Step 3. This
product is the LCM.
Ex8:
Find the LCM of 24 and 36.
Step 1:
Write 24 and 36 as the product of two or more prime factors as shown below.
24
2
4


2  2
36
6

3
2
6

6

3  2

Thus 24  2  2  2  3  23  3
36  2  3  2  3  22  32
Step 2:
Revised 02 /10
2 and 3 are the different prime factors used in Step 1.
3
3
Step 3:
The greatest exponent of 2 is 3 and the greatest exponent of 3 is 2.
Step 4:
The LCM of 24 and 36 is 23  32  8  9  72
Procedure for Adding and Subtracting Fractions with Different Denominators
Step 1.
Find the LCD of the given fractions. That is, find the LCM of their
denominators.
Step 2.
Use the fundamental principle of fractions to write each fraction as an
equivalent fraction with the LCD as the new denominator. (Recall that the
fundamental principle of fractions allows us to multiply or divide both the
numerator and the denominator of any fraction by the same non-zero
number to get an equivalent fraction.)
Step 3.
Add or subtract the fractions with the same denominators obtained in Step 2.
Step 4.
Reduce the fraction obtained in Step 3 to lowest terms.
Ex 9: Add
5 7

6 8
Step 1: By inspection, the LCM of 6 and 8 is 24. So, the LCD of the fractions is 24.
Step 2: Write
5
7
as equivalent fractions with LCD 24 as the new denominator.
and
6
8
Since 24  6  4 , then
Since 24   3 , then
5 5  4 20


6 6  4 24
7 7  3 21


8 8  3 24
Step 3:
5 7 20 21 20  21 41
 



6 8 24 24
24
24
Step 4:
41 17
1
24
24
Therefore,
Revised 02 /10
5 7
17
 1
6 8
24
4
Ex.10: Add
11 19

24 40
Step 1:
The LCM of 24 and 40 is not obvious. So, we write 24 and 40 as the product of
two or more prime factors.
24
2
40
4


2  2
Thus
6

3
2
4

10

2  2

5
24  23  3
40  23  5
Therefore the LCM of 24 and 40 is 23  3  5  8  3  5  24  5  120
So, the LCD of the fractions is 120.
11
19
as equivalent fractions with LCD 120 as the new
and
24
40
denominator.
Step 2: Write
Step 3:
Since 120  24  5 , then
11 11 5 55


24 24  5 120
Since 120    3 , then
19 19  3 57


40 40  3 120
11 19 55 57 55  57 112





24 40 120 120
120
120
Step 4: Reduce
112
to the lowest terms.
120
112 112  2 56 56  2 28 28  2 14






120 120  2 60 60  2 30 30  2 15
Therefore,
Revised 02 /10
11 19 14


24 40 15
5
Ex 11: Find
3 5 9
 
14 21 28
Step 1: The LCM of 14, 21 and 28 is 84.
Therefore, the LCD of the fractions is 84. Use the method described on page 3
to verify.
3
3  6 18


14 14  6 84
5
5  4 20
Since 84  21  4, then 

21 21 4 84
Step 2: Since 84 14  6, then
Since 84  28  3, then
9
9  3 27


28 28  3 84
Step 3:
3 5 9 18 20 27 18  20  27 65
 
  


14 21 28 84 84 84
84
84
Step 4:
65
is already in lowest terms.
84
3 5 9 65
 

14 21 28 84
Therefore,
Ex 12: Find
11 2

12 3
Step 1:
The LCM of 12 and 3 is 12.
Therefore, the LCD of the fractions is 12.
Step 2:
11 11

12 12
(The denominator of
2 2 4 8


3 3  4 12
Step 3:
11 2 11 8 11  8 3
   

12 3 12 12
12
12
Step 4:
3
33 1


12 12  3 4
Therefore,
Revised 02 /10
11 2 1
 
12 3 4
6
11
is already the same as the LCD, 12.)
12
Ex 13: Find
23 10
10 10  2 5
. First, reduce the second fraction to lowest terms.



36 48
48 48  2 24
23 10 23 5



36 48 36 24
Step 1: The LCM of 36 and 24 is 72. So, the LCD of the fractions is 72. Use the
method described on page 3 to verify.
Step 2: Since 72  36  2, then
Since 72  24  3, then
23 23  2 46


36 36  2 72
5
5  3 15


24 24  3 72
Step 3:
23 5 46 15 46  15 31





36 24 72 72
72
72
Step 4:
31
is already in lowest terms.
72
Therefore,
23 10 31


36 48 72
Ex14: Perform the indicated operations and simplify.
7 3 23
 
10 20 40
Step 1:
The LCM of 10, 20, and 40 is 40. So, the LCD of the fractions is 40.
Step 2:
7
7  4 28


10 10  4 40
3
3 2
6


20 20  2 40
23 23

40 40
Step3:
7 3 23 28 6 23 28  6  23 22  23 45
 

 



10 20 40 40 40 40
40
40
40
Step4:
45 45  5 9
1

 1
40 40  5 8
8
Therefore,
Revised 02 /10
7 3 23
1
 
1
10 20 40
8
7
Exercises: Perform the indicated operations.
1.
5 7 2
 
9 9 9
13.
7 2

10 15
2.
10 15 1
 
12 18 6
14.
12 14

18 21
3.
7 16

15 25
15.
9
3

27 18
4.
9 13

20 50
16.
17 8

20 25
5.
15 9

28 20
17.
3
3

40 50
6.
5 3 7
 
6 8 12
18.
7 3

18 16
7.
3 4 7
 
10 15 20
19.
31 20

33 33
8.
2 3 5
 
3 5 6
20.
53 17

54 54
9.
18 5 7
 
24 20 14
21.
2 5 4
 
3 6 9
10.
6 7 9
 
18 21 27
22.
8 2 5
 
9 18 10
11.
1 1 1
 
3 5 15
23.
23 1 1
 
24 2 3
12.
1 1 1
 
6 7 42
24.
7 5 3
 
12 18 4
Solutions to odd-numbered problems and answers to even-numbered problems:
1.
5 7 2 5  7  2 14 5
  
 1
9 9 9
9
9
9
Revised 02 /10
2.
8
1
5
6
3.
7 16

15 25
The LCD is 75.
4.
71
100
7 16 7  5 16  3 35 48 35  48 83
8







=1
15 25 15  5 25  3 75 75
75
75
75
5.
15 9

28 20
6. 1
The LCD is 140.
19
24
15 9 15  5 9  7
75 63 138 138  2 69








28 20 28  5 20  7 140 140 140 140  2 70
7.
3 4 7
The LCD is 60.
 
10 15 20
8. 2
1
10
3 4 7
3  6 4  4 7  3 18 16 21
 




 
10 15 20 10  6 15  4 20  3 60 60 60
18  16  21 55 55  5 11

 

60
60 60  5 12
9.
18 5 7
 
24 20 14
18 5 7 3 1 1
    
24 20 14 4 4 2

11.
10. 1
First reduce each fraction to lowest terms.
The LCD is 4.
3 1 2 3 1  2 2  2 4
  

 1
4 4 4
4
4
4
1 1 1
 
3 5 15
The LCD is 15.
12.
1
3
1 1 1 1 5 1  3 1 5 3 1
  

   
3 5 15 3  5 5  3 15 15 15 15

13.
5  3 1 9
9 3 3
 

15
15 15  3 5
7 2

10 15
14. 0
The LCD is 30.
7 2
7  3 2  2 21 4 21  4 17
 

  

10 15 10  3 15  2 30 30
30
30
Revised 02 /10
9
15.
9
3

27 18
First reduce each fraction to the lowest terms.
9 3 1 1
  
27 18 3 6
16.
53
100
18.
29
144
The LCD is 6.
9 3 1 1 2 1 2 1 1
     

27 18 3 6 6 6
6
6
17.
3
3

40 50
The LCD is 200.
3
3
35
3 4
15
12 15  12
3
 





40 50 40  5 50  4 200 200
200
200
19.
31 20 31  20 11 11 11 1





33 33
33
33 33 11 3
20.
2
3
21.
2 5 4
 
3 6 9
22.
1
2
The LCD is 18
2 5 4 2  6 5  3 4  2 12 15 8
  


  
3 6 9 3  6 6  3 9  2 18 18 18

23.
12  15  8 27  8 19
1

 1
18
18
18 18
23 1 1
 
24 2 3
The LCD is 24.
24. 1
1
18
23 1 1 23 112 1 8 23 12 8
  



 
24 2 3 24 2 12 3  8 24 24 24

23  12  8 11  8 3
33 1




24
24
24 24  3 8
Note: For further explanation and more practice go to
http://themathpage.com. Click on Skill in Arithmetic, Lesson 21, 22, 24.
Revised 02 /10
10