Addition & Subtraction of Fractions

Fractions with the Same Denominators:

Worked Example 8.8.1:

Solve the following:

i. \(\frac{3}{7} \: + \: \frac{5}{7}\)

ii. \(\frac{4}{9}\: – \: \frac{2}{9}\)

iii. \(\scriptsize 1 \: + \: \normalsize \frac{3}{4} \)

Solution:

i. \(\frac{3}{7} \: + \: \frac{5}{7}\\ = \frac{3\:+\:5}{7}\\ = \frac{8}{7} \\ =\scriptsize 1 \normalsize \frac{1}{7}\)

ii. \(\frac{4}{9}\: – \: \frac{2}{9} \\ = \frac{4 \: – \: 2}{9} \\ = \frac{2}{9}\)

iii. \(\scriptsize 1 \: + \: \normalsize \frac{3}{4} \\ = \frac{4}{4} \: + \: \frac{3}{4}\\ = \frac{4 \: + \: 3}{4} \\ = \frac{7}{4} \\ \scriptsize = 1 \normalsize \frac{3}{4}\)

(Note:   ⁴/₄   =   4   ÷  4   =  1)

iv. \(\scriptsize 1 \: – \: \normalsize \frac{3}{7} \\= \frac{7}{7} \: – \: \frac{3}{7} \\ = \frac{7 \: – \: 3}{7} \\ = \frac{4}{7}\)

Fractions with Different Denominators:

To add or subtract fractions with different denominators, first, change them into equivalent fractions.

Worked Example 8.8.2:

Solve the following:

i. \(\frac{3}{5} \: + \: \frac{1}{2} \)

ii. \(\frac{1}{4} + \frac{2}{3} + \frac{4}{6} \)

iii. \(\frac{2}{3} \: – \: \frac{5}{9} \)

iv. \(\frac{8}{9} \: – \: \frac{4}{5} \)

i. \(\frac{3}{5} \: + \: \frac{1}{2} \)

Solution

The denominators are 5 and 2.

The LCM of 5 and 2  =  10.

Change \(\frac{3}{5} \scriptsize \: and \: \normalsize \frac{1}{2} \) to equivalent fraction using the 

LCM 10 as new denominator.

\(\frac{3}{5}\\ = \frac{3\: \times \: 2 }{5 \: \times \: 2} \\= \frac{6}{10}\)

also,

\(\frac{1}{2} \\ = \frac{1\: \times \: 5 }{2 \: \times \: 5} \\ = \frac{5}{10}\)

So, \(\frac{3}{5} \: +\: \frac{1}{2} \\ = \frac{6}{10} + \frac{5}{10} \\ = \frac{11}{10} \\ \scriptsize = 1 \normalsize \frac{1}{10}\)

ii. \(\frac{1}{4} + \frac{2}{3} + \frac{4}{6} \)

Solution

The LCM of 4, 3 and 6  =  12

So, \(\frac{1}{4} \\= \frac{1 \: \times \: 3 }{4 \: \times \: 3} \\= \frac{3}{12}\)

\(\frac{2}{3} \\ = \frac{2 \: \times \: 4 }{3 \: \times \: 4} \\= \frac{8}{12}\)

\(\frac{4}{6} \\ = \frac{4 \: \times\: 2 }{6 \: \times \: 2} = \frac{8}{12}\)

Therefore, \(\frac{1}{4} \: + \: \frac{2}{3} \: + \: \frac{4}{6} \)

= \(\frac{3}{12} \: + \: \frac{8}{12} \: + \: \frac{8}{12}\\ = \frac{19}{12} \\ = \scriptsize 1 \normalsize \frac{7}{12} \)

iii. \(\frac{2}{3} \: – \: \frac{5}{9} \)

Solution

The LCM of 3 and 9  =   9

Because 3 goes in 9 and 9 goes in 9.

So, \(\frac{2}{3} = \frac{2 \: \times \: 3 }{3 \: \times \: 3} = \frac{6}{9}\)

⇒ \(\frac{5}{9} = \frac{5 \: \times \: 1 }{9 \: \times \: 1} = \frac{5}{9}\)

So, \(\frac{2}{3} \: – \: \frac{5}{9} \\ = \frac{6}{9}\: – \: \frac{5}{9} \\ = \frac{1}{9} \)

iv. \(\frac{8}{9} \: – \: \frac{4}{5} \)

Solution

The LCM of 9 and 5   =   45

So, \(\frac{8}{9} = \frac{8 \: \times \: 5 }{9 \: \times \: 5} = \frac{40}{45}\)

⇒ \(\frac{4}{5} = \frac{4 \: \times \: 9 }{5 \: \times \: 9} = \frac{36}{45}\)

So, \(\frac{8}{9} \: – \: \frac{4}{5} = \frac{40}{45}\: – \: \frac{36}{45} \)

= \(\frac{4}{45}\)