Clearing denominators

{{Short description|Method for simplifying equations}} {{Multiple issues| {{No footnotes|date=December 2025}} {{One source|date=December 2025}} }} 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

: <math> \frac x 6 + \frac y {15z} = 1.</math>

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

: <math> 5xz + 2y = 30z. \, </math>

The result is an equation with no fractions.

The simplified equation is not entirely equivalent to the original. For when we substitute {{nowrap|1=''y'' = 0}} and {{nowrap|1=''z'' = 0}} in the last equation, both sides simplify to 0, so we get {{nowrap|1=0 = 0}}, a mathematical truth. But the same substitution applied to the original equation results in {{nowrap|1=''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 {{nowrap|1={{mvar|E}}₁ = {{mvar|E}}₂}} may equivalently be rewritten in the form {{nowrap|1={{mvar|E}}₁ − {{mvar|E}}₂ = 0}}.

So let the equation have the form :<math>\sum_{i=1}^n \frac{P_i}{Q_i} = 0.</math>

The first step is to determine a common denominator {{mvar|D}} of these fractions – preferably the least common denominator, which is the least common multiple of the {{mvar|Q_i}}.

This means that each {{mvar|Q_i}} is a factor of {{mvar|D}}, so {{nowrap|1={{mvar|D}} = {{mvar|R_iQ_i}}}} for some expression {{mvar|R_i}} that is not a fraction. Then

: <math>\frac{P_i}{Q_i} = \frac{R_i P_i}{R_i Q_i} = \frac{R_i P_i} D \,,</math>

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

So we have now

: <math>\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.</math>

Provided that {{mvar|D}} does not assume the value 0, the latter equation is equivalent with

: <math>\sum_{i=1}^n R_i P_i = 0\,,</math>

in which the denominators have vanished.

As shown by the provisos, care has to be taken not to introduce zeros of {{mvar|D}} – viewed as a function of the unknowns of the equation – as spurious solutions.

==Example 2==

Consider the equation

:<math>\frac{1}{x(x+1)}+\frac{1}{x(x+2)}-\frac{1}{(x+1)(x+2)} = 0.</math>

The least common denominator is {{nowrap|{{mvar|x}}({{mvar|x}} + 1)({{mvar|x}} + 2)}}.

Following the method as described above results in

:<math>(x+2)+(x+1)-x = 0.</math>

Simplifying this further gives us the solution {{nowrap|1={{mvar|x}} = −3}}.

It is easily checked that none of the zeros of {{nowrap|{{mvar|x}}({{mvar|x}} + 1)({{mvar|x}} + 2)}} – namely {{nowrap|1= {{mvar|x}} = 0}}, {{nowrap|1= {{mvar|x}} = −1}}, and {{nowrap|1= {{mvar|x}} = −2}} – is a solution of the final equation, so no spurious solutions were introduced.

==References== * {{cite book |title=Algebra: Beginning and Intermediate |edition=3 |author=Richard N. Aufmann |author2=Joanne Lockwood |page=88 |publisher=Cengage Learning |year=2012 |isbn=978-1-133-70939-8}}

Category:Elementary algebra Category:Equations