# Solving Algebraic Fractions

An algebraic fraction is a type of fraction where the numerator, the denominator, or both contain at least one variable.

${\dfrac{x}{5}}$, ${\dfrac{3}{x}}$, ${\dfrac{x-7}{5}}$, and ${\dfrac{2}{x-11}}$ are a few examples of algebraic fractions.

Now, we will learn how to solve algebraic fractions with various operations here.

## Reducing

Like reducing numeric fractions, to reduce algebraic fractions to their lowest term, we will eliminate the common factor of the numerator and denominator.

Let us reduce ${\dfrac{4x^{8}}{20x}}$ to its simplest form

${\dfrac{4x^{8}}{20x}}$

= ${\dfrac{4}{20}\times \dfrac{x^{8}}{x}}$

= ${\dfrac{1}{5}x^{7}}$ or ${\dfrac{x^{7}}{5}}$

We can also simplify complex algebraic fractions through the basic mathematical operations: addition, subtraction, multiplication, and division.

To add algebraic fractions, we need to have a common denominator. Once we have the same denominator, we can add the numerators.

Let us add the algebraic fractions ${\dfrac{3}{6x+18}}$ and ${\dfrac{10}{x+3}}$

${\dfrac{3}{6x+18}+\dfrac{10}{x+3}}$

Finding the Common Denominator

First, we will find the common denominator for (6x + 18) and (x + 3)

On multiplying each fraction by the denominator of the other fraction,

${\dfrac{3}{6x+18}+\dfrac{10}{x+3}}$

= ${\dfrac{3\left( x+3\right) }{6x+18}+\dfrac{10\left( 6x+18\right) }{x+3}}$

= ${\dfrac{3\left( x+3\right) +10\left( 6x+18\right) }{\left( 6x+18\right) \left( x+3\right) }}$

Taking Out the Common Factors (If Needed)

Now, we will take out the common factors from the numerator and denominator

= ${\dfrac{3\left( x+3\right) +10\times 6\left( x+3\right) }{6\left( x+3\right) \left( x+3\right) }}$

= ${\dfrac{\left( x+3\right) \left( 3+60\right) }{6\left( x+3\right) \left( x+3\right) }}$

= ${\dfrac{63}{6\left( x+3\right) }}$

= ${\dfrac{21}{2\left( x+3\right) }}$

Thus, the sum is ${\dfrac{21}{2\left( x+3\right) }}$

## Subtracting

${\dfrac{3}{6x+18}-\dfrac{10}{x+3}}$

Finding the Common Denominator

Multiplying each fraction by the denominator of the other fraction,

= ${\dfrac{3\left( x+3\right) }{6x+18}-\dfrac{10\left( 6x+18\right) }{x+3}}$

Subtracting the Numerators

= ${\dfrac{3\left( x+3\right) -10\left( 6x+18\right) }{\left( 6x+18\right) \left( x+3\right) }}$

Taking Out the Common Factors (If Needed)

= ${\dfrac{3\left( x+3\right) -10\times 6\left( x+3\right) }{6\left( x+3\right) \left( x+3\right) }}$

= ${\dfrac{\left( x+3\right) \left( 3-60\right) }{6\left( x+3\right) \left( x+3\right) }}$

= ${\dfrac{-57}{6\left( x+3\right) }}$

= ${\dfrac{-19}{2\left( x+3\right) }}$

Thus, the difference is ${\dfrac{-19}{2\left( x+3\right) }}$

## Multiplying

Multiplying algebraic fractions follows the same rules as multiplying numerical fractions.

Let us multiply ${\dfrac{5x+20}{4}}$ and ${\dfrac{16}{\left( x+4\right) \left( x+2\right) }}$

${\dfrac{5x+20}{4}\times \dfrac{16}{\left( x+4\right) \left( x+2\right) }}$

Factoring the Numerator and Denominator

= ${\dfrac{5\left( x+4\right) }{4}\times \dfrac{16}{\left( x+4\right) \left( x+2\right) }}$

Taking Out the Common Factors

= ${5\times \dfrac{4}{\left( x+2\right) }}$

Multiplying the Numerators and Denominators

= ${\dfrac{20}{\left( x+2\right) }}$

Thus, the product is ${\dfrac{20}{\left( x+2\right) }}$

## Dividing

It is similar to dividing numerical fractions.

Dividing ${\dfrac{8x^{12}}{3}}$ by ${\dfrac{2}{7}x^{8}}$

${\dfrac{8x^{12}}{3}\div \dfrac{2}{7}x^{8}}$

Multiplying the First Fraction by the Reciprocal of the Second Fraction

= ${\dfrac{8x^{12}}{3}\times \dfrac{7}{2x^{8}}}$

Factoring the Numerator and Denominator

= ${\dfrac{4\times 2\times x^{12}}{3}\times \dfrac{7}{2\times x^{8}}}$

Taking Out the Common Factors

= ${\dfrac{\left( 4\times 7\right) x^{\left( 12-8\right) }}{3}}$

Multiplying the Numerators and Denominators

= ${\dfrac{28x^{4}}{3}}$

Thus, the quotient is ${\dfrac{28x^{4}}{3}}$

## Solved Examples

Simplify: ${\dfrac{x^{2}+7x+12}{2\left( x^{2}+3x\right) }}$

Solution:

Thus, the algebraic fraction ${\dfrac{x^{2}+7x+12}{2\left( x^{2}+3x\right) }}$ is simplified to ${\dfrac{x+4}{2x}}$

Multiply:
${\dfrac{7u^{3}v^{3}}{9xy^{2}}\times \dfrac{18x^{3}y^{2}}{15u^{2}v^{3}}}$

Solution:

Thus, the product is ${\dfrac{14}{15}x^{2}u}$