Adding and subtracting fractions with like and unlike denominators

Find the following sums.

$\frac{3}{7}$ + $\frac{2}{7}$

The denominators are the same.

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

$\frac{3}{7}$ + $\frac{2}{7}$ = $\frac{3+2}{7}$ = $\frac{5}{7}$

$\frac{1}{8}$ + $\frac{3}{8}$ = $\frac{1+3}{8}$ = $\frac{4}{8}$ = $\frac{1}{2}$

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

$\frac{1}{2}$ and $\frac{1}{2}$ .

Adding the numerators and mistakenly adding the denominators produces:

$\frac{1}{2}$ + $\frac{1}{2}$ = $\frac{1+1}{2+2}$ = $\frac{2}{4}$ = $\frac{1}{2}$

This means that $\frac{1}{2}$ + $\frac{1}{2}$ is the same as $\frac{1}{2}$ , which is preposterous! We do not add denominators .

Find the following differences.

$\frac{3}{5}$ - $\frac{1}{5}$

The denominators are the same.

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

$\frac{3}{5}$ - $\frac{1}{5}$ = $\frac{3-1}{5}$ = $\frac{2}{5}$

$\frac{8}{6}$ - $\frac{2}{6}$ = $\frac{6}{6}$ = 1

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

$\frac{7}{\text{15}}$ - $\frac{4}{\text{15}}$ .

Subtracting the numerators and mistakenly subtracting the denominators produces:

$\frac{7}{\text{15}}$ - $\frac{4}{\text{15}}$ = $\frac{7-4}{\text{15}-\text{15}}$ = $\frac{3}{0}$

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

Find the sum of these unlike fractions.

$\frac{1}{\text{12}}$ + $\frac{4}{\text{15}}$

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) .

In the fraction $\frac{1}{\text{12}}$ , we multiply the denominator 12 by 5 to get the LCD of 60. Therefore we multiply the numerator 1 by the same factor (5).

$\frac{1}{\text{12}}$ × $\frac{5}{5}$ = $\frac{5}{\text{60}}$

Similarly,

$\frac{4}{\text{15}}$ × $\frac{4}{4}$ = $\frac{\text{16}}{\text{60}}$

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

$\frac{5}{\text{60}}$ + $\frac{\text{16}}{\text{60}}$ = $\frac{\text{21}}{\text{60}}$

Reduce the result: $\frac{\text{21}}{\text{60}}$ = $\frac{7}{\text{20}}$