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1.7 Fractions: addition

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This module discusses how to add and subtract fractions with like denominators and how to find the least common denominator to allow addition and subtraction of fractions with unlike denominators.

Adding fractions with 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.

Find the following sums.

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

The denominators are the same.

Add the numerators and place the sum over the common denominator, 7.

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

When necessary, reduce the result.

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

We do not add denominators.

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

1 2 size 12{ { {1} over {2} } } {} and 1 2 size 12{ { {1} over {2} } } {} .

Adding the numerators and mistakenly adding the denominators produces:

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

This means that 1 2 size 12{ { {1} over {2} } } {} + 1 2 size 12{ { {1} over {2} } } {} is the same as 1 2 size 12{ { {1} over {2} } } {} , which is preposterous! We do not add denominators .

Adding fractions with like denominators - exercises

Find the following sums.

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

6 8 size 12{ { {6} over {8} } } {} = 3 4 size 12{ { {3} over {4} } } {}

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

11 11 size 12{ { {"11"} over {"11"} } } {} = 1

15 20 size 12{ { {"15"} over {"20"} } } {} + 1 20 size 12{ { {1} over {"20"} } } {} + 2 20 size 12{ { {2} over {"20"} } } {}

18 20 size 12{ { {"18"} over {"20"} } } {} = 9 10 size 12{ { {9} over {"10"} } } {}

Subtracting fractions with like denominators

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

Find the following differences.

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

The denominators are the same.

Subtract the numerators and place the difference over the common denominator, 5.

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

When necessary, reduce the result.

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

We do not subtract denominators.

To see what happens if we mistakenly subtract the denominators as well as the numerators, let’s subtract

7 15 size 12{ { {7} over {"15"} } } {} - 4 15 size 12{ { {4} over {"15"} } } {} .

Subtracting the numerators and mistakenly subtracting the denominators produces:

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

We end up dividing by zero, which is undefined. We do not subtract denominators.

Subtracting fractions with like denominators - exercises

Find the following differences.

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

4 12 size 12{ { {4} over {"12"} } } {} = 1 3 size 12{ { {1} over {3} } } {}

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

Result is 0

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

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

Adding and subtracting fractions with unlike denominators

Basic Rule: Fractions can only be added or subtracted conveniently if they have like denominators.

To see why this rule makes sense, let’s consider the problem of adding a quarter and a dime.

A quarter is 1 4 size 12{ { {1} over {4} } } {} of a dollar.

A dime is 1 10 size 12{ { {1} over {"10"} } } {} of a dollar.

We know that 1 quarter + 1 dime = 35 cents. How do we get to this answer by adding 1 4 size 12{ { {1} over {4} } } {} and 1 10 size 12{ { {1} over {"10"} } } {} ?

We convert them to quantities of the same denomination.

A quarter is equivalent to 25 cents, or 25 100 size 12{ { {"25"} over {"100"} } } {} .

A dime is equivalent to 10 cents, or 10 100 size 12{ { {"10"} over {"100"} } } {} .

By converting them to quantities of the same denomination, we can add them easily:

25 100 size 12{ { {"25"} over {"100"} } } {} + 10 100 size 12{ { {"10"} over {"100"} } } {} = 35 100 size 12{ { {"35"} over {"100"} } } {} .

Same denomination → size 12{ rightarrow } {} same denominator

If the denominators are not the same, make them the same by building up the fractions so that they both have a common denominator. A common denominator can always be found by multiplying all the denominators, but it is not necessarily the Least Common Denominator.

Least common denominator (lcd)

The LCD is the smallest number that is evenly divisible by all the denominators.

It is the least common multiple of the denominators.

The LCD is the product of all the prime factors of all the denominators, each factor taken the greatest number of times that it appears in any single denominator.

Finding the lcd

Find the sum of these unlike fractions.

1 12 size 12{ { {1} over {"12"} } } {} + 4 15 size 12{ { {4} over {"15"} } } {}

Factor the denominators:

12 = 2 × 2 × 3

15 = 3 × 5

What is the greatest number of times the prime factor 2 appear in any single denominator? Answer: 2 times. That is the number of times the prime factor 2 will appear as a factor in the LCD.

What is the greatest number of times the prime factor 3 appear in any single denominator? Answer: 1 time. That is the number of times the prime factor 3 will appear as a factor in the LCD.

What is the greatest number of times the prime factor 5 appear in any single denominator? Answer: 1 time. That is the number of times the prime factor 5 will appear as a factor in the LCD.

So we assemble the LCD by multiplying each prime factor by the number of times it appears in a single denominator, or:

2 × 2 × 3 × 5 = 60

60 is the Least Common Denominator (the Least Common Multiple of 12 and 15) .

Building up the fractions

To create fractions with like denominators, we now multiply the numerators by whatever factors are missing when we compare the original denominator to the new LCD.

In the fraction 1 12 size 12{ { {1} over {"12"} } } {} , we multiply the denominator 12 by 5 to get the LCD of 60. Therefore we multiply the numerator 1 by the same factor (5).

1 12 size 12{ { {1} over {"12"} } } {} × 5 5 size 12{ { {5} over {5} } } {} = 5 60 size 12{ { {5} over {"60"} } } {}

Similarly,

4 15 size 12{ { {4} over {"15"} } } {} × 4 4 size 12{ { {4} over {4} } } {} = 16 60 size 12{ { {"16"} over {"60"} } } {}

Adding the built up fractions

We can now add the two fractions because they have like denominators:

5 60 size 12{ { {5} over {"60"} } } {} + 16 60 size 12{ { {"16"} over {"60"} } } {} = 21 60 size 12{ { {"21"} over {"60"} } } {}

Reduce the result: 21 60 size 12{ { {"21"} over {"60"} } } {} = 7 20 size 12{ { {7} over {"20"} } } {}

Adding and subtracting fractions with unlike denominators - exercises

Find the following sums and differences.

1 6 size 12{ { {1} over {6} } } {} + 3 4 size 12{ { {3} over {4} } } {}

Result is 11 12 size 12{ { {"11"} over {"12"} } } {}

5 9 size 12{ { {5} over {9} } } {} - 5 12 size 12{ { {5} over {"12"} } } {}

Result is 5 36 size 12{ { {5} over {"36"} } } {}

15 16 size 12{ { {"15"} over {"16"} } } {} + 1 2 size 12{ { {1} over {2} } } {} - 3 4 size 12{ { {3} over {4} } } {}

Result is 35 16 size 12{ { {"35"} over {"16"} } } {}

Module review exercises

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

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

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

Result is 15 10 size 12{ { {"15"} over {"10"} } } {} (reduce to 1 1 2 size 12{ { {1} over {2} } } {} )

Find the total length of the screw in this diagram:

Total length is 19 32 size 12{ { {"19"} over {"32"} } } {} in.

5 2 size 12{ { {5} over {2} } } {} + 16 2 size 12{ { {"16"} over {2} } } {} - 3 2 size 12{ { {3} over {2} } } {}

Result is 18 2 size 12{ { {"18"} over {2} } } {} (reduce to 9)

3 4 size 12{ { {3} over {4} } } {} + 1 3 size 12{ { {1} over {3} } } {}

Result is 13 12 size 12{ { {"13"} over {"12"} } } {}

Two months ago, a woman paid off 3 24 size 12{ { {3} over {"24"} } } {} of a loan. One month ago, she paid off 4 24 size 12{ { {4} over {"24"} } } {} of the loan. This month she will pay off 5 24 size 12{ { {5} over {"24"} } } {} of the total loan. At the end of this month, how much of her total loan will she have paid off?

She will have paid off 12 24 size 12{ { {"12"} over {"24"} } } {} , or 1 2 size 12{ { {1} over {2} } } {} of the total loan.

8 3 size 12{ { {8} over {3} } } {} - 1 4 size 12{ { {1} over {4} } } {} + 7 36 size 12{ { {7} over {"36"} } } {}

Result is 94 36 size 12{ { {"94"} over {"36"} } } {} (reduce to 47 18 size 12{ { {"47"} over {"18"} } } {} )
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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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