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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 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.

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

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

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

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

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

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

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

Practice set a

Find the following sums.

1 10 + 3 10 size 12{ { {1} over {"10"} } + { {3} over {"10"} } } {}

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

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

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

7 11 + 4 11 size 12{ { {7} over {"11"} } + { {4} over {"11"} } } {}

1

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

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

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

3 4 + 3 4 = 3 + 3 4 + 4 = 6 8 = 3 4 size 12{ { {3} over {4} } + { {3} over {4} } = { {3+3} over {4+4} } = { {6} over {8} } = { {3} over {4} } } {} , so two 3 4 size 12{ { {3} over {4} } } {} ’s= one 3 4 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.

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

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

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

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

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.

10 13 8 13 size 12{ { {"10"} over {"13"} } - { {8} over {"13"} } } {}

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

5 12 1 12 size 12{ { {5} over {"12"} } - { {1} over {"12"} } } {}

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

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

0

26 10 14 10 size 12{ { {"26"} over {"10"} } - { {"14"} over {"10"} } } {}

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

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

5 9 2 9 = 5 2 9 9 = 3 0 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.

3 8 + 2 8 size 12{ { {3} over {8} } + { {2} over {8} } } {}

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

1 6 + 2 6 size 12{ { {1} over {6} } + { {2} over {6} } } {}

9 10 + 1 10 size 12{ { {9} over {"10"} } + { {1} over {"10"} } } {}

1

3 11 + 4 11 size 12{ { {3} over {"11"} } + { {4} over {"11"} } } {}

9 15 + 4 15 size 12{ { {9} over {"15"} } + { {4} over {"15"} } } {}

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

3 10 + 2 10 size 12{ { {3} over {"10"} } + { {2} over {"10"} } } {}

5 12 + 7 12 size 12{ { {5} over {"12"} } + { {7} over {"12"} } } {}

1

11 16 2 16 size 12{ { {"11"} over {"16"} } - { {2} over {"16"} } } {}

3 16 3 16 size 12{ { {3} over {"16"} } - { {3} over {"16"} } } {}

0

15 23 2 23 size 12{ { {"15"} over {"23"} } - { {2} over {"23"} } } {}

1 6 1 6 size 12{ { {1} over {6} } - { {1} over {6} } } {}

0

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

3 11 + 1 11 + 5 11 size 12{ { {3} over {"11"} } + { {1} over {"11"} } + { {5} over {"11"} } } {}

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

16 20 + 1 20 + 2 20 size 12{ { {"16"} over {"20"} } + { {1} over {"20"} } + { {2} over {"20"} } } {}

12 8 + 2 8 + 1 8 size 12{ { {"12"} over {8} } + { {2} over {8} } + { {1} over {8} } } {}

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

1 15 + 8 15 + 6 15 size 12{ { {1} over {"15"} } + { {8} over {"15"} } + { {6} over {"15"} } } {}

3 8 + 2 8 1 8 size 12{ { {3} over {8} } + { {2} over {"8"} } - { {1} over {"8"} } } {}

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

11 16 + 9 16 5 16 size 12{ { {"11"} over {"16"} } + { {9} over {"16"} } - { {5} over {"16"} } } {}

4 20 1 20 + 9 20 size 12{ { {4} over {"20"} } - { {1} over {"20"} } + { {9} over {"20"} } } {}

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

7 10 3 10 + 11 10 size 12{ { {7} over {"10"} } - { {3} over {"10"} } + { {"11"} over {"10"} } } {}

16 5 1 5 2 5 size 12{ { {"16"} over {5} } - { {1} over {5} } - { {2} over {5} } } {}

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

21 35 17 35 + 31 35 size 12{ { {"21"} over {"35"} } - { {"17"} over {"35"} } + { {"31"} over {"35"} } } {}

5 2 + 16 2 1 2 size 12{ { {5} over {2} } + { {"16"} over {2} } - { {1} over {2} } } {}

10

1 18 + 3 18 + 1 18 + 4 18 5 18 size 12{ { {1} over {"18"} } + { {3} over {"18"} } + { {1} over {"18"} } + { {4} over {"18"} } - { {5} over {"18"} } } {}

6 22 2 22 + 4 22 1 22 + 11 22 size 12{ { {6} over {"22"} } - { {2} over {"22"} } + { {4} over {"22"} } - { {1} over {"22"} } + { {"11"} over {"22"} } } {}

9 11 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.

3 10 3 10 size 12{ { {3} over {"10"} } - { {3} over {"10"} } } {}

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

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

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.

Two months ago, a woman paid off 3 24 size 12{ { {3} over {"24"} } } {} of a loan. One month ago, she paid off 5 24 size 12{ { {5} over {"24"} } } {} of the total loan. This month she will again pay off 5 24 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?

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

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

( [link] ) Round 2,650 to the nearest hundred.

2700

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

( [link] ) Find the prime factors of 495.

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

( [link] ) Find the value of 3 4 16 25 5 9 size 12{ { {3} over {4} } cdot { {"16"} over {"25"} } cdot { {5} over {9} } } {} .

( [link] ) 8 3 size 12{ { {8} over {3} } } {} of what number is 1 7 9 size 12{1 { {7} over {9} } } {} ?

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

Questions & Answers

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Source:  OpenStax, Contemporary math applications. OpenStax CNX. Dec 15, 2014 Download for free at http://legacy.cnx.org/content/col11559/1.6
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