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.