Clearing denominators

Short description: Method for simplifying equations

In mathematics, the method of clearing denominators, also called clearing fractions, is a technique for simplifying an equation equating two expressions that each are a sum of rational expressions – which includes simple fractions.

## Contents

* 1 Example
* 2 Description
* 3 Example 2
* 4 References

## Example

Consider the equation

[math]\displaystyle{ \frac x 6 + \frac y {15z} = 1. }[/math]

The smallest common multiple of the two denominators 6 and 15z is 30z, so one multiplies both sides by 30z:

[math]\displaystyle{ 5xz + 2y = 30z. \, }[/math]

The result is an equation with no fractions.

The simplified equation is not entirely equivalent to the original. For when we substitute y = 0 and z = 0 in the last equation, both sides simplify to 0, so we get 0 = 0, a mathematical truth. But the same substitution applied to the original equation results in x/6 + 0/0 = 1, which is mathematically meaningless.

## Description

Without loss of generality, we may assume that the right-hand side of the equation is 0, since an equation E1 = E2 may equivalently be rewritten in the form E1 − E2 = 0.

So let the equation have the form

[math]\displaystyle{ \sum_{i=1}^n \frac{P_i}{Q_i} = 0. }[/math]

The first step is to determine a common denominator D of these fractions – preferably the least common denominator, which is the least common multiple of the Qi.

This means that each Qi is a factor of D, so D = RiQi for some expression Ri that is not a fraction. Then

[math]\displaystyle{ \frac{P_i}{Q_i} = \frac{R_i P_i}{R_i Q_i} = \frac{R_i P_i} D \,, }[/math]

provided that RiQi does not assume the value 0 – in which case also D equals 0.

So we have now

[math]\displaystyle{ \sum_{i=1}^n \frac{P_i}{Q_i} = \sum_{i=1}^n \frac{R_i P_i} D = \frac 1 D \sum_{i=1}^n R_i P_i = 0. }[/math]

Provided that D does not assume the value 0, the latter equation is equivalent with

[math]\displaystyle{ \sum_{i=1}^n R_i P_i = 0\,, }[/math]

in which the denominators have vanished.

As shown by the provisos, care has to be taken not to introduce zeros of D – viewed as a function of the unknowns of the equation – as spurious solutions.

## Example 2

Consider the equation

[math]\displaystyle{ \frac{1}{x(x+1)}+\frac{1}{x(x+2)}-\frac{1}{(x+1)(x+2)} = 0. }[/math]

The least common denominator is x(x \+ 1)(x \+ 2).

Following the method as described above results in

[math]\displaystyle{ (x+2)+(x+1)-x = 0. }[/math]

Simplifying this further gives us the solution x = −3.

It is easily checked that none of the zeros of x(x \+ 1)(x \+ 2) – namely x = 0, x = −1, and x = −2 – is a solution of the final equation, so no spurious solutions were introduced.

## References

* Richard N. Aufmann; Joanne Lockwood (2012). Algebra: Beginning and Intermediate (3 ed.). Cengage Learning. p. 88. ISBN 978-1-133-70939-8.

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