How To Simplify Algebraic Expressions Involving Complex Fractions?

By Robert O

How do we define complex fractions?

If a fraction has a fraction or fractions either on the numerator, denominator, or both, then it becomes a complex fraction (aka compound fraction). Simplifying compound fractions is no different from solving other fractions. Such fractions may seem complex, but following the necessary steps makes everything easier.

What are methods for simplifying compound fractions?

We use two methods in simplifying any complex fraction. These are the division and least common multiple methods.

How to solve complex fractions using the division method?

We follow steps to solve complex fractions using this method. Use these steps:

  • Re-write the fraction by changing the multiplication with division.
  • Find the reciprocal of the fraction on the write and change it to a product problem.
  • Factor numerator and the denominator if applicable
  • Simplify the resulting expression if applicable.

We will use two examples to illustrate how to simplify complex fractions using these steps.

Example 1:




Using the first step, we rewrite the expression as a division :

    \[\frac{4}{x-3} \div \frac{8}{x^{2}-9}\]

Finding the reciprocal and changing to a multiplication problem:

    \[\frac{4}{x-3} \times \frac{x^{2}-9}{8}\]

Getting the product and factorizing:


Simplifying the resulting expression:


Example 2:

Simplify :



We use the division method to solve this problem too, but it is not all that direct as we would expect. First, we perform the addition operation on the numerator and subtraction operation on the denominator. We can then proceed with the rest of the steps.


Rewriting as a division:

    \[\frac{\frac{y+1}{y}}{\frac{y^{2}-1}{y^{2}}}=\frac{1+y}{y} \div \frac{y^{2}-1}{y^{2}}\]

    \[\frac{1+y}{y} \div \frac{y^{2}-1}{y^{2}}=\frac{1+y}{y} \times \frac{y^{2}}{y^{2}-1}=\frac{y^{2}(1+y)}{y\left(y^{2}-1\right)}\]

We factorize the denominator and then simplify the resulting expression:


How do I simplify complex fractions using the LCM method?


  • Determine the LCM of denominators of fraction at the top and fraction at the bottom.
  • Multiply both fractions by this LCM
  • Simplify the fraction.

Example 3:


    \[\frac{\frac{5 x-10}{5}}{\frac{x-2}{x}}\]


The fraction at the top is

    \[\frac{5 x-10}{5}\]

and the denominator fraction is


Our two denominators are x and 5. So, the LCM of 5 and x is 5x. To simplify, multiply both fractions by 5x.

    \[\frac{\frac{5 x-10}{5}}{\frac{x-2}{x}}=\frac{\left(\frac{5 x-10}{5}\right) 5 x}{\left(\frac{x-2}{x}\right) 5 x}=\frac{5 x^{2}-10 x}{5 x-10}\]


    \[\frac{5 x^{2}-10 x}{5 x-10}=\frac{5 x(x-2)}{5(x-2)}=x\]

Example 4:




Let us work with the numerator and denominator separately and reduce them to single fractions:

    \[1-\frac{7}{x+1}=\frac{(x+1)-7}{x+1}=\frac{x-6}{x+1}, \text { denominator is } x+1\]

    \[\frac{4}{x+1}+1=\frac{4+x+1}{x+1}=\frac{5+x}{x+1}, \text { denominato is } x+1\]

We can readily see that the LCM of the two denominators is x+1. So, we multiply both fractions by x+1.



The following identities can help you to factorize algebraic expressions:

    \[(a+b)(a+b)=a^{2}+a b+b a+b^{2}=a^{2}+2 a b+b^{2} \ldots \ldots \ldots . i\]

    \[(a-b)(a-b)=a^{2}-a b-b a+b^{2}=a^{2}-2 a b+b^{2} \ldots \ldots \ldots i i\]

    \[(a-b)(a+b)=a^{2}+a b-b a-b^{2}=a^{2}-b^{2} \ldots \ldots \ldots \ldots . . i i i\]


By using either the division or LCM method, you can simplify any complex algebraic fractions. All you need to do is to follow the steps without skipping any. If the denominator or numerator, or both involve other operations, you must carry out those operations first.

About the Author

This lesson was prepared by Robert O. He holds a Bachelor of Engineering (B.Eng.) degree in Electrical and electronics engineering. He is a career teacher and headed the department of languages and assumed various leadership roles. He writes for Full Potential Learning Academy.