# Subtracting algebraic fraction

In this chapter we will learn to subtract algebraic fractions with solved examples.

To understand the chapter, you should have basic understanding of the concept of LCM and algebraic fractions.

## How to subtract fractions with variables ?

In order to subtract the algebraic fractions, you should first try to convert the fractions into common denominator.

Given below are the steps for subtraction;

(a) Factorize the denominator of each fraction into smaller components.

(b) Find LCM of denominator

(c) Multiply each fraction to make denominator equal to LCM.

(d) Now we have fraction with same denominator. Simply subtract the numerator and retain the denominator.

I hope you understood the above process. Let us see some examples for further understanding.

## Subtracting algebraic fractions – Solved examples

Example 01
Subtract the below fractions.

\mathtt{\Longrightarrow \frac{5}{7x^{2}} -\ \frac{3}{x}}

Solution
Note that both the algebraic fractions have different denominators.

First we have to make the denominator same and then do the subtraction.

(a) Find LCM of denominator.

\mathtt{LCM\ \left( 7x^{2} ,x\right) =7x^{2}}

(b) Multiply each fraction to make denominator equals \mathtt{7x^{2}}

Fraction \mathtt{\frac{5}{7x^{2}}}

The denominator is already equals to LCM. Hence, don’t need to do anything here.

Fraction \mathtt{\frac{3}{x}}

Multiply numerator and denominator by 7x.

\mathtt{\Longrightarrow \frac{3\times 7x}{x\times 7x}}\\\ \\ \mathtt{\Longrightarrow \ \frac{21x}{7x^{2}}}

(c) Now we have fractions with same denominator, simply subtract the numerator and retain the denominator.

\mathtt{\Longrightarrow \ \frac{5}{7x^{2}} -\frac{21x}{7x^{2}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{5-21x}{7x^{2}}}

Hence, \mathtt{\frac{5-21x}{7x^{2}}} is the solution of given subtraction.

Example 02
Subtract the below fractions.

\mathtt{\Longrightarrow \frac{10}{x-1} -\ \frac{11}{\left( x^{2} -2x+1\right)}}

Solution

(a) Factorize each of the fraction into smaller components.

(i) Fraction \mathtt{\frac{10}{x-1}}

It is already factorized. Hence, don’t need to do anything.

(ii) Fraction \mathtt{\frac{11}{\left( x^{2} -2x+1\right)}}

Referring the formula;
\mathtt{( a-b)^{2} =a^{2} -2ab+b^{2}}

Using the formula, we get;

\mathtt{\Longrightarrow \ \frac{11}{\left( x^{2} -2x+1\right)}}\\\ \\ \mathtt{\Longrightarrow \ \frac{11}{( x-1)^{2}}}

Now the subtraction can be expressed as;

\mathtt{\Longrightarrow \frac{10}{x-1} -\ \frac{11}{( x-1)^{2}}}

Note that both fractions have different denominator. Now we will try to make denominator same.

(b) Find LCM of denominators.

\mathtt{\Longrightarrow \ LCM\ \left(( x-1) ,( x-1)^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ ( x-1)^{2}}

(c) Multiply each fraction to make denominator equal to LCM.

Fraction \mathtt{\frac{10}{x-1}}

Multiply numerator and denominator by (x – 1)

\mathtt{\Longrightarrow \ \frac{10\ \times ( x-1)}{( x-1) \times ( x-1)}}\\\ \\ \mathtt{\Longrightarrow \frac{10x-10}{( x-1)^{2}}}

Fraction \mathtt{\frac{11}{( x-1)^{2}}}

The denominator is already equals to LCM. So don’t need to do anything.

(d) Now we have fractions with same denominator. Simply subtract the numerator and retain the denominator.

\mathtt{\Longrightarrow \frac{10x-10}{( x-1)^{2}} -\frac{11}{( x-1)^{2}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{10x-10-11}{( x-1)^{2}}}\\\ \\ \mathtt{\Longrightarrow \frac{10x-21}{( x-1)^{2}} \ }

Hence, \mathtt{\frac{10x-21}{( x-1)^{2}} \ } is the solution of given subtraction.

Example 03
Subtract the below fractions.

\mathtt{\Longrightarrow \frac{5x}{x^{2} -4x} -\ \frac{3( x-1)}{( x-4)}}

Solution
(a) First factorize each algebraic fraction into simple terms.

Simplifying Fraction \mathtt{\frac{5x}{x^{2} -4x}}\\\ \\

\mathtt{\Longrightarrow \ \frac{5x}{x( x-4)}}

Fraction \mathtt{\frac{3( x-1)}{( x-4)}}

The fraction is already factorized into simpler components. Hence, don’t need to do anything.

Now the subtraction can be expressed as;

\mathtt{\Longrightarrow \frac{5x}{x( x-4)} -\ \frac{3( x-1)}{( x-4)}}

In the below steps we will try to make both fractions with same denominators.

(b) Find LCM of denominators.

LCM ( x (x – 4), (x – 4) ) = x (x – 4)

(c) Multiply each fraction to make denominator equals to LCM.

Fraction \mathtt{\frac{5x}{x( x-4)}}

Denominator is already equals to LCM. So don’t need to do anything.

Fraction \mathtt{\frac{3( x-1)}{( x-4)}}

Multiply numerator and denominator by x.

\mathtt{\Longrightarrow \ \frac{3( x-1) \times x}{( x-4) \times x}}\\\ \\ \mathtt{\Longrightarrow \ \frac{3x( x-1)}{x( x-4)}}

(d) Now we have got fractions with same denominator. Simply subtract the numerator and retain the denominator.

\mathtt{\Longrightarrow \ \frac{5x}{x( x-4)} -\frac{3x( x-1)}{x( x-4)}}\\\ \\ \mathtt{\Longrightarrow \ \frac{5x-3x^{2} -3x}{x( x-4)}}\\\ \\ \mathtt{\Longrightarrow \ \frac{2x-3x^{2}}{x( x-4)}}

Hence, the above expression is solution of given subtraction.

Example 04
Subtract the below algebraic fraction.

\mathtt{\Longrightarrow \frac{5}{3xy} -\ \frac{2}{5y}}

Solution
Note that both the algebraic fraction are present in simplified form. Hence, further factorization is not necessary.

(a) Find LCM of denominators.

LCM (3xy, 5y) = 15xy

(b) Multiply each fraction to make denominator 15xy.

Fraction \mathtt{\frac{5}{3xy}}

Multiply numerator and denominator by 5.

\mathtt{\Longrightarrow \ \frac{5\times 5}{3xy\times 5}}\\\ \\ \mathtt{\Longrightarrow \ \frac{25}{15xy}}

Fraction \mathtt{\frac{2}{5y}}

Multiply numerator and denominator by 3x.

\mathtt{\Longrightarrow \ \frac{2\times 3x}{5y\times 3x}}\\\ \\ \mathtt{\Longrightarrow \ \frac{6x}{15xy}}

Now the subtraction can be expressed as;

\mathtt{\Longrightarrow \frac{25}{15xy} -\ \frac{6x}{15xy}}

Since we have fractions with same denominator, you can simply subtract the numerator while retaining the same denominator.

\mathtt{\Longrightarrow \frac{25}{15xy} -\ \frac{6x}{15xy}}\\\ \\ \mathtt{\Longrightarrow \ \frac{25-6x}{15xy}}

Hence, the above expression is the solution.

Example 05
Solve the below expression.

\mathtt{\Longrightarrow \frac{2}{\left( x^{2} -81\right)} -\ \frac{4}{\left( x^{2} +18x+81\right)}}

Solution
(a) Factorize each fraction into simplified form.

Fraction \mathtt{\frac{2}{\left( x^{2} -81\right)}} \\\ \\ \mathtt{\Longrightarrow \frac{2}{\left( x^{2} -81\right)}}\\\ \\ \mathtt{\Longrightarrow \ \frac{2}{( x-9)( x+9)}}

Fraction \mathtt{\frac{4}{\left( x^{2} +18x+81\right)}}

Referring the formula;

\mathtt{( a+b)^{2} =\ a^{2} +b^{2} +2ab}

Factorizing the algebraic fraction;

\mathtt{\Longrightarrow \ \frac{4}{\left( x^{2} +2.x.9+9^{2}\right)}}\\\ \\ \mathtt{\Longrightarrow \ \frac{4}{( x+9)^{2}}}

Now the subtraction can be expressed as;

\mathtt{\Longrightarrow \frac{2}{( x-9)( x+9)} -\ \frac{4}{( x+9)^{2}}}

(b) Find the LCM of denominator.

\mathtt{\Longrightarrow \ LCM\ \left(( x-9)( x+9) ,( x+9)^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ ( x-9)( x+9)^{2}}

(c) Multiply each fraction to make denominator equal to LCM.

Fraction \mathtt{\frac{2}{( x-9)( x+9)}}

Multiply numerator and denominator by (x + 9)

\mathtt{\Longrightarrow \frac{2\times ( x+9)}{( x-9)( x+9) \times ( x+9)}}\\\ \\ \mathtt{\Longrightarrow \ \frac{2x+18}{( x-9)( x+9)^{2}}}

Fraction \mathtt{\frac{4}{( x+9)^{2}}}

Multiply numerator and denominator by (x – 9)

\mathtt{\Longrightarrow \frac{4\times ( x-9)}{( x+9)^{2}( x-9)}}\\\ \\ \mathtt{\Longrightarrow \ \frac{4x-36}{( x-9)( x+9)^{2}}}

(d) Now we have fractions with same denominator. Simply subtract the numerator and retain the denominator.

\mathtt{\Longrightarrow \frac{2x+18}{( x+9)^{2}( x-9)} -\frac{4x-36}{( x-9)( x+9)^{2}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{2x+18-4x+36}{( x-9)( x+9)^{2}}}\\\ \\ \mathtt{\Longrightarrow \ \frac{-2x+54}{( x-9)( x+9)^{2}}}

Hence, the above expression is the solution.