Clearing denominators

Consider the equation

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

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

${\displaystyle 5xz+2y=30z.\,}$

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.

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

${\displaystyle \sum _{i=1}^{n}{\frac {P_{i}}{Q_{i}}}=0.}$

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 Q_i.

This means that each Q_i is a factor of D, so D = R_iQ_i for some expression R_i that is not a fraction. Then

${\displaystyle {\frac {P_{i}}{Q_{i}}}={\frac {R_{i}P_{i}}{R_{i}Q_{i}}}={\frac {R_{i}P_{i}}{D}}\,,}$

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

So we have now

${\displaystyle \sum _{i=1}^{n}{\frac {P_{i}}{Q_{i}}}=\sum _{i=1}^{n}{\frac {R_{i}P_{i}}{D}}={\frac {1}{D}}\sum _{i=1}^{n}R_{i}P_{i}=0.}$

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

${\displaystyle \sum _{i=1}^{n}R_{i}P_{i}=0\,,}$

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.

Consider the equation

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

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

Following the method as described above results in

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

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.