clearing denominators

Consider the equation

: $\backslash frac\; x\; 6\; +\; \backslash frac\; y\; \{15z\}\; =\; 1.$

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

: $5xz\; +\; 2y\; =\; 30z.\; \backslash ,$

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.

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

:$\backslash sum\_\{i=1\}^n\; \backslash frac\{P\_i\}\{Q\_i\}\; =\; 0.$

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

: $\backslash frac\{P\_i\}\{Q\_i\}\; =\; \backslash frac\{R\_i\; P\_i\}\{R\_i\; Q\_i\}\; =\; \backslash frac\{R\_i\; P\_i\}\; D\; \backslash ,,$

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

So we have now

: $\backslash sum\_\{i=1\}^n\; \backslash frac\{P\_i\}\{Q\_i\}\; =\; \backslash sum\_\{i=1\}^n\; \backslash frac\{R\_i\; P\_i\}\; D\; =\; \backslash frac\; 1\; D\; \backslash sum\_\{i=1\}^n\; R\_i\; P\_i\; =\; 0.$

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

: $\backslash sum\_\{i=1\}^n\; R\_i\; P\_i\; =\; 0\backslash ,,$

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.

Consider the equation

:$\backslash frac\{1\}\{x(x+1)\}+\backslash frac\{1\}\{x(x+2)\}-\backslash frac\{1\}\{(x+1)(x+2)\}\; =\; 0.$

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

Following the method as described above results in

:$(x+2)+(x+1)-x\; =\; 0.$

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.

• {{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