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

x 6 + y 15 z = 1. {\displaystyle {\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:

5 x z + 2 y = 30 z . {\displaystyle 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 E1 = E2 may equivalently be rewritten in the form E1 − E2 = 0.

So let the equation have the form

∑ i = 1 n P i Q i = 0. {\displaystyle \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 Qi.

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

P i Q i = R i P i R i Q i = R i P i D , {\displaystyle {\frac {P_{i}}{Q_{i}}}={\frac {R_{i}P_{i}}{R_{i}Q_{i}}}={\frac {R_{i}P_{i}}{D}}\,,}

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

So we have now

∑ i = 1 n P i Q i = ∑ i = 1 n R i P i D = 1 D ∑ i = 1 n R i P i = 0. {\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.}

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

∑ i = 1 n R i P i = 0 , {\displaystyle \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

1 x ( x + 1 ) + 1 x ( x + 2 ) − 1 ( x + 1 ) ( x + 2 ) = 0. {\displaystyle {\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. {\displaystyle (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.

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