Clearing denominators

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.

Example
Consider the equation


 * $$ \frac x 6 + \frac y {15z} = 1.$$

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


 * $$ 5xz + 2y = 30z. \, $$

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 $E$1 = $E$2 may equivalently be rewritten in the form $E$1 − $E$2 = 0.

So let the equation have the form
 * $$\sum_{i=1}^n \frac{P_i}{Q_i} = 0.$$

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 $Q_{i}$.

This means that each $Q_{i}$ is a factor of $D$, so $D$ = $R_{i}Q_{i}$ for some expression $R_{i}$ that is not a fraction. Then


 * $$\frac{P_i}{Q_i} = \frac{R_i P_i}{R_i Q_i} = \frac{R_i P_i} D \,,$$

provided that $R_{i}Q_{i}$ does not assume the value 0 – in which case also $D$ equals 0.

So we have now


 * $$\sum_{i=1}^n \frac{P_i}{Q_i} = \sum_{i=1}^n \frac{R_i P_i} D = \frac 1 D \sum_{i=1}^n R_i P_i = 0.$$

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


 * $$\sum_{i=1}^n R_i P_i = 0\,,$$

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


 * $$\frac{1}{x(x+1)}+\frac{1}{x(x+2)}-\frac{1}{(x+1)(x+2)} = 0.$$

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

Following the method as described above results in


 * $$(x+2)+(x+1)-x = 0.$$

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.