subtraction of fractions with solved examples

By | November 11, 2021

Subtraction of fractions simply means the removal of one fraction from the other.

At times solving subtraction fraction problems in mathematics is disturbing for basic school children. Some of them will begin to solve subtraction maths problems with different denominators, by subtracting the top numbers and down numbers together as shown below. [latex]\frac{6}{4} – \frac{5}{3} = \frac{1}{1}[/latex] what is wrong with this?

This triggered me to compile this note to help all basic school children with this problem.

Subtraction of fractions

In subtracting fractions with different denominators, it is just the same procedure as solving the addition problem of fractions. Click here to see how the addition of fractions is done. There are two ways you can solve subtraction problems of fractions. Let’s tackle it one after another by demonstrating with examples.

Example1. subtract [latex]2\frac{1}{5}[/latex] from [latex]3\frac{2}{3}[/latex]


solving [latex]3\frac{2}{3} – 2\frac{1}{5}[/latex]

first change the mixed numbers to improper fraction

[latex]\frac{3\times3 + 2}{3} – \frac{2\times5 + 1}{5}[/latex]

=[latex]\frac{9+2}{3} – \frac{10 + 1}{5}[/latex]

=[latex]\frac{11}{3} – \frac{11}{5}[/latex]

the lcm of the denominators 3 and 5 is 15

we find the equivalent fractions of the two given fractions

=[latex]\frac{11}{3} = \frac{11\times5}{3\times5} = \frac{55}{15}[/latex]

=[latex]\frac{11}{5} = \frac{11\times3}{5\times3} = \frac{33}{15}[/latex]

since the two fraction now have a common denominator, subtract their numerators and pick the common denominator

=[latex]\frac{55 – 33}{15}[/latex]



alternatively, we could have solved it by subtracting the whole number part and the fraction part together as shown below

=[latex]3\frac{2}{3} – 2\frac{1}{5}[/latex]

=[latex](3 – 2)(\frac{2}{3} – \frac{1}{5})[/latex]

=[latex](1)\frac{(2\times5) – (1\times3)}{15)}[/latex]

since the the lcm of 3 and 5 is 15

=[latex](1)\frac{10 – 3}{15}[/latex]

=[latex](1)\frac{7}{15}[/latex] as required

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Example2 simplify [latex]3\frac{2}{5} – 1\frac{1}{3} – 2\frac{1}{5}[/latex]


first change the mixed numbers to improper fraction

=[latex]\frac{(3\times5)+2}{5} – \frac{(+1\times3)+1}{3} – \frac{(2\times5)+1}{5}[/latex]

=[latex]\frac{17}{5} – \frac{4}{3} – \frac{11}{5}[/latex]

the lcm of 3, 5,and 5 is 15

=[latex]\frac{(17\times3) – (4\times5) – (11\times3)}{15}[/latex]

=[latex]\frac{51 – 20 – 33}{15}[/latex]

=[latex]\frac{31 – 33}{15}[/latex]


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