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Addition and Subtraction of Fractions with Like Denominators

Module by: Wade Ellis, Denny Burzynski. E-mail the authorsEdited By: Math Editors

Summary: This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses how to add and subtract fractions with like denominators. By the end of the module students should be able to add and subtract fractions with like denominators.

Section Overview

  • Addition of Fraction With Like Denominators
  • Subtraction of Fractions With Like Denominators

Addition of Fraction With Like Denominators

Let's examine the following diagram.

A rectangle divided in five parts. Each part is labeled one-fifth. Two of the parts are shaded, and labeled two-fifths. A third part is shaded, and is labeled one-fifth.
 2 one-fifths and 1 one fifth is shaded.

It is shown in the shaded regions of the diagram that

(2 one-fifths) + (1 one-fifth) = (3 one-fifths)

That is,

2 5 + 1 5 = 3 5 2 5 + 1 5 = 3 5 size 12{ { {2} over {5} } + { {1} over {5} } = { {3} over {5} } } {}

From this observation, we can suggest the following rule.

Method of Adding Fractions Having Like Denominators

To add two or more fractions that have the same denominators, add the numer­ators and place the resulting sum over the common denominator. Reduce, if necessary.

Sample Set A

Find the following sums.

Example 1

37+2737+27 size 12{ { {3} over {7} } + { {2} over {7} } } {}. The denominators are the same. Add the numerators and place that sum over 7.

3 7 + 2 7 = 3 + 2 7 = 5 7 3 7 + 2 7 = 3 + 2 7 = 5 7 size 12{ { {3} over {7} } + { {2} over {7} } = { {3+2} over {7} } = { {5} over {7} } } {}

Example 2

18+3818+38 size 12{ { {1} over {8} } + { {3} over {8} } } {}. The denominators are the same. Add the numerators and place the sum over 8. Reduce.

1 8 + 3 8 = 1 + 3 8 = 4 8 = 1 2 1 8 + 3 8 = 1 + 3 8 = 4 8 = 1 2 size 12{ { {1} over {8} } + { {3} over {8} } = { {1+3} over {8} } = { {4} over {8} } = { {1} over {2} } } {}

Example 3

49+5949+59 size 12{ { {4} over {9} } + { {5} over {9} } } {}. The denominators are the same. Add the numerators and place the sum over 9.

4 9 + 5 9 = 4 + 5 9 = 9 9 = 1 4 9 + 5 9 = 4 + 5 9 = 9 9 = 1 size 12{ { {4} over {9} } + { {5} over {9} } = { {4+5} over {9} } = { {9} over {9} } =1} {}

Example 4

78+5878+58 size 12{ { {7} over {8} } + { {5} over {8} } } {}. The denominators are the same. Add the numerators and place the sum over 8.

7 8 + 5 8 = 7 + 5 8 = 12 8 = 3 2 7 8 + 5 8 = 7 + 5 8 = 12 8 = 3 2 size 12{ { {7} over {8} } + { {5} over {8} } = { {7+5} over {8} } = { {"12"} over {8} } = { {3} over {2} } } {}

Example 5

To see what happens if we mistakenly add the denominators as well as the numerators, let's add

1 2 + 1 2 1 2 + 1 2 size 12{ { {1} over {2} } + { {1} over {2} } } {}

Adding the numerators and mistakenly adding the denominators produces

1 2 + 1 2 = 1 + 1 2 + 2 = 2 4 = 1 2 1 2 + 1 2 = 1 + 1 2 + 2 = 2 4 = 1 2 size 12{ { {1} over {2} } + { {1} over {2} } = { {1+1} over {2+2} } = { {2} over {4} } = { {1} over {2} } } {}

This means that two 1212 size 12{ { {1} over {2} } } {}'s is the same as one 1212 size 12{ { {1} over {2} } } {}. Preposterous! We do not add denominators.

Practice Set A

Find the following sums.

Exercise 1

110+310110+310 size 12{ { {1} over {"10"} } + { {3} over {"10"} } } {}

Solution

2525 size 12{ { {2} over {5} } } {}

Exercise 2

14+1414+14 size 12{ { {1} over {4} } + { {1} over {4} } } {}

Solution

1212 size 12{ { {1} over {2} } } {}

Exercise 3

711+411711+411 size 12{ { {7} over {"11"} } + { {4} over {"11"} } } {}

Solution

1

Exercise 4

35+1535+15 size 12{ { {3} over {5} } + { {1} over {5} } } {}

Solution

4545 size 12{ { {4} over {5} } } {}

Exercise 5

Show why adding both the numerators and denominators is preposterous by adding 3434 size 12{ { {3} over {4} } } {} and 3434 size 12{ { {3} over {4} } } {} and examining the result.

Solution

34+34=3+34+4=68=3434+34=3+34+4=68=34 size 12{ { {3} over {4} } + { {3} over {4} } = { {3+3} over {4+4} } = { {6} over {8} } = { {3} over {4} } } {}, so two 3434 size 12{ { {3} over {4} } } {}’s= one 3434 size 12{ { {3} over {4} } } {} which is preposterous.

Subtraction of Fractions With Like Denominators

We can picture the concept of subtraction of fractions in much the same way we pictured addition.

A visualization of a subtraction problem. There are three rows displayed, and each row has an element that corresponds with it. In the first row are three rectangles, each divided into five parts. Each part in each fraction is labeled one-fifth. The first rectangle has three shaded parts. Next to this is the statement, take away. Next to this is the second rectangle, with one part shaded. Next to this is an equals sign. Finally, the third rectangle has two shaded parts. The second row reads three-fifths minus one-fifth equals two-fifths. The third row shows the same equation written in words.

From this observation, we can suggest the following rule for subtracting fractions having like denominators:

Subtraction of Fractions with Like Denominators

To subtract two fractions that have like denominators, subtract the numerators and place the resulting difference over the common denominator. Reduce, if possible.

Sample Set B

Find the following differences.

Example 6

35−1535−15 size 12{ { {3} over {5} } - { {1} over {5} } } {}. The denominators are the same. Subtract the numerators. Place the difference over 5.

3 5 − 1 5 = 3 − 1 5 = 2 5 3 5 − 1 5 = 3 − 1 5 = 2 5 size 12{ { {3} over {5} } - { {1} over {5} } = { {3 - 1} over {5} } = { {2} over {5} } } {}

Example 7

86−2686−26 size 12{ { {8} over {6} } - { {2} over {6} } } {}. The denominators are the same. Subtract the numerators. Place the difference over 6.

8 6 − 2 6 = 8 − 2 6 = 6 6 = 1 8 6 − 2 6 = 8 − 2 6 = 6 6 = 1 size 12{ { {8} over {6} } - { {2} over {6} } = { {8 - 2} over {6} } = { {6} over {6} } =1} {}

Example 8

169−29169−29 size 12{ { {"16"} over {9} } - { {2} over {9} } } {}. The denominators are the same. Subtract numerators and place the difference over 9.

16 9 − 2 9 = 16 − 2 9 = 14 9 16 9 − 2 9 = 16 − 2 9 = 14 9 size 12{ { {"16"} over {9} } - { {2} over {9} } = { {"16" - 2} over {9} } = { {"14"} over {9} } } {}

Example 9

To see what happens if we mistakenly subtract the denominators, let's consider

7 15 − 4 15 = 7 − 4 15 − 15 = 3 0 7 15 − 4 15 = 7 − 4 15 − 15 = 3 0 size 12{ { {7} over {"15"} } - { {4} over {"15"} } = { {7 - 4} over {"15" - "15"} } = { {3} over {0} } } {}

We get division by zero, which is undefined. We do not subtract denominators.

Practice Set B

Find the following differences.

Exercise 6

1013−8131013−813 size 12{ { {"10"} over {"13"} } - { {8} over {"13"} } } {}

Solution

213213 size 12{ { {2} over {"13"} } } {}

Exercise 7

512−112512−112 size 12{ { {5} over {"12"} } - { {1} over {"12"} } } {}

Solution

1313 size 12{ { {1} over {3} } } {}

Exercise 8

12−1212−12 size 12{ { {1} over {2} } - { {1} over {2} } } {}

Solution

0

Exercise 9

2610−14102610−1410 size 12{ { {"26"} over {"10"} } - { {"14"} over {"10"} } } {}

Solution

6565 size 12{ { {6} over {5} } } {}

Exercise 10

Show why subtracting both the numerators and the denominators is in error by performing the subtraction 59−2959−29 size 12{ { {5} over {9} } - { {2} over {9} } } {}.

Solution

59−29=5−29−9=3059−29=5−29−9=30 size 12{ { {5} over {9} } - { {2} over {9} } = { {5 - 2} over {9 - 9} } = { {3} over {0} } } {}, which is undefined

Exercises

For the following problems, find the sums and differences. Be sure to reduce.

Exercise 11

38+2838+28 size 12{ { {3} over {8} } + { {2} over {8} } } {}

Solution

5858 size 12{ { {5} over {8} } } {}

Exercise 12

16+2616+26 size 12{ { {1} over {6} } + { {2} over {6} } } {}

Exercise 13

910+110910+110 size 12{ { {9} over {"10"} } + { {1} over {"10"} } } {}

Solution

1

Exercise 14

311+411311+411 size 12{ { {3} over {"11"} } + { {4} over {"11"} } } {}

Exercise 15

915+415915+415 size 12{ { {9} over {"15"} } + { {4} over {"15"} } } {}

Solution

13151315 size 12{ { {"13"} over {"15"} } } {}

Exercise 16

310+210310+210 size 12{ { {3} over {"10"} } + { {2} over {"10"} } } {}

Exercise 17

512+712512+712 size 12{ { {5} over {"12"} } + { {7} over {"12"} } } {}

Solution

1

Exercise 18

1116−2161116−216 size 12{ { {"11"} over {"16"} } - { {2} over {"16"} } } {}

Exercise 19

316−316316−316 size 12{ { {3} over {"16"} } - { {3} over {"16"} } } {}

Solution

0

Exercise 20

1523−2231523−223 size 12{ { {"15"} over {"23"} } - { {2} over {"23"} } } {}

Exercise 21

16−1616−16 size 12{ { {1} over {6} } - { {1} over {6} } } {}

Solution

0

Exercise 22

14+14+1414+14+14 size 12{ { {1} over {4} } + { {1} over {4} } + { {1} over {4} } } {}

Exercise 23

311+111+511311+111+511 size 12{ { {3} over {"11"} } + { {1} over {"11"} } + { {5} over {"11"} } } {}

Solution

911911 size 12{ { {9} over {"11"} } } {}

Exercise 24

1620+120+2201620+120+220 size 12{ { {"16"} over {"20"} } + { {1} over {"20"} } + { {2} over {"20"} } } {}

Exercise 25

128+28+18128+28+18 size 12{ { {"12"} over {8} } + { {2} over {8} } + { {1} over {8} } } {}

Solution

158158 size 12{ { {"15"} over {8} } } {}

Exercise 26

115+815+615115+815+615 size 12{ { {1} over {"15"} } + { {8} over {"15"} } + { {6} over {"15"} } } {}

Exercise 27

38+28−1838+28−18 size 12{ { {3} over {8} } + { {2} over {"8"} } - { {1} over {"8"} } } {}

Solution

1212 size 12{ { {1} over {2} } } {}

Exercise 28

1116+916−5161116+916−516 size 12{ { {"11"} over {"16"} } + { {9} over {"16"} } - { {5} over {"16"} } } {}

Exercise 29

420−120+920420−120+920 size 12{ { {4} over {"20"} } - { {1} over {"20"} } + { {9} over {"20"} } } {}

Solution

3535 size 12{ { {3} over {5} } } {}

Exercise 30

710−310+1110710−310+1110 size 12{ { {7} over {"10"} } - { {3} over {"10"} } + { {"11"} over {"10"} } } {}

Exercise 31

165−15−25165−15−25 size 12{ { {"16"} over {5} } - { {1} over {5} } - { {2} over {5} } } {}

Solution

135135 size 12{ { {"13"} over {5} } } {}

Exercise 32

2135−1735+31352135−1735+3135 size 12{ { {"21"} over {"35"} } - { {"17"} over {"35"} } + { {"31"} over {"35"} } } {}

Exercise 33

52+162−1252+162−12 size 12{ { {5} over {2} } + { {"16"} over {2} } - { {1} over {2} } } {}

Solution

10

Exercise 34

118+318+118+418−518118+318+118+418−518 size 12{ { {1} over {"18"} } + { {3} over {"18"} } + { {1} over {"18"} } + { {4} over {"18"} } - { {5} over {"18"} } } {}

Exercise 35

622−222+422−122+1122622−222+422−122+1122 size 12{ { {6} over {"22"} } - { {2} over {"22"} } + { {4} over {"22"} } - { {1} over {"22"} } + { {"11"} over {"22"} } } {}

Solution

911911 size 12{ { {9} over {"11"} } } {}

The following rule for addition and subtraction of two fractions is preposterous. Show why by performing the operations using the rule for the following two problems.

Preposterous Rule

To add or subtract two fractions, simply add or subtract the numerators and place this result over the sum or difference of the denominators.

Exercise 36

310−310310−310 size 12{ { {3} over {"10"} } - { {3} over {"10"} } } {}

Exercise 37

815+815815+815 size 12{ { {8} over {"15"} } + { {8} over {"15"} } } {}

Solution

1630=851630=85 size 12{ { {"16"} over {"30"} } = { {8} over {5} } } {}(using the preposterous rule)

Exercise 38

Find the total length of the screw.
A screw. The head of the screw is three thirty-seconds of an inch. The shaft of the screw is sixteen thirty-seconds of an inch.

Exercise 39

Two months ago, a woman paid off 324324 size 12{ { {3} over {"24"} } } {} of a loan. One month ago, she paid off 524524 size 12{ { {5} over {"24"} } } {} of the total loan. This month she will again pay off 524524 size 12{ { {5} over {"24"} } } {} of the total loan. At the end of the month, how much of her total loan will she have paid off?

Solution

13241324 size 12{ { {"13"} over {"24"} } } {}

Exercise 40

Find the inside diameter of the pipe.
A pipe with a thickness of two-sixteenths, and a total diameter of eleven-sixteenths.

Exercises for Review

Exercise 41

((Reference)) Round 2,650 to the nearest hundred.

Solution

2700

Exercise 42

((Reference)) Use the numbers 2, 4, and 8 to illustrate the associative property of addition.

Exercise 43

((Reference)) Find the prime factors of 495.

Solution

32⋅5⋅1132⋅5⋅11 size 12{3 rSup { size 8{2} } cdot 5 cdot "11"} {}

Exercise 44

((Reference)) Find the value of 34⋅1625⋅5934⋅1625⋅59 size 12{ { {3} over {4} } cdot { {"16"} over {"25"} } cdot { {5} over {9} } } {}.

Exercise 45

((Reference)) 8383 size 12{ { {8} over {3} } } {} of what number is 179179 size 12{1 { {7} over {9} } } {}?

Solution

2323 size 12{ { {2} over {3} } } {}

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