Clearing denominators

Template:Short description 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 Template:Nowrap and Template:Nowrap in the last equation, both sides simplify to 0, so we get Template:Nowrap, a mathematical truth. But the same substitution applied to the original equation results in Template:Nowrap, 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 Template:Nowrap may equivalently be rewritten in the form Template:Nowrap.

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 Template:Mvar of these fractions – preferably the least common denominator, which is the least common multiple of the Template:Mvar.

This means that each Template:Mvar is a factor of Template:Mvar, so Template:Nowrap for some expression Template:Mvar 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 Template:Mvar does not assume the value 0 – in which case also Template:Mvar 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 Template:Mvar 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 Template:Mvar – 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 Template:Nowrap.

Following the method as described above results in

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

Simplifying this further gives us the solution Template:Nowrap.

It is easily checked that none of the zeros of Template:Nowrap – namely Template:Nowrap, Template:Nowrap, and Template:Nowrap – is a solution of the final equation, so no spurious solutions were introduced.

References

• Template:Cite book