Clearing denominators explained

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 .

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

5 xz + 2 y = 30 z.

The result is an equation with no fractions.

The simplified equation is not entirely equivalent to the original. For when we substitute and in the last equation, both sides simplify to 0, so we get, a mathematical truth. But the same substitution applied to the original equation results in, 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 may equivalently be rewritten in the form .

So let the equation have the form

\sum ^{n} _{i =1} (P _{i})/(Q _{i}) = 0 .

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

This means that each is a factor of, so for some expression that is not a fraction. Then

(P _{i})/(Q _{i}) = (R _{i} P _{i})/(R _{i} Q _{i}) = (R _{i} P _{i})/(D) ,

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

So we have now

\sum ^{n} _{i =1} (P _{i})/(Q _{i}) = \sum ^{n} _{i =1} (R _{i} P _{i})/(D) = (1)/(D) \sum ^{n} _{i =1} R _{i} P _{i} = 0 .

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

\sum ^{n} _{i =1} 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 – 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 .

The least common denominator is .

Following the method as described above results in

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

Simplifying this further gives us the solution .

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

References

• Book: Algebra: Beginning and Intermediate . 3 . Richard N. Aufmann . Joanne Lockwood . 88 . Cengage Learning . 2012 . 978-1-133-70939-8.