Procedure

Clear the denominators of an equation

Algebra • Equations

Statement

Multiplying both sides of an equation by a common multiple of all its denominators leaves an equivalent equation free of fractions, provided the multiplier is not zero; when it contains the unknown, the values that make it zero are excluded from the start and any solution equal to one of them is rejected.

How to use it

1Multiply EVERY term on both sides, including the terms that were not fractions, by the common denominator.
2Before multiplying out, group the terms of highest degree on one side: in a literal equation they often cancel and leave something linear.
3Write down the excluded values (those making a denominator zero) before starting, and check the answer against them at the end.

Worked example

Solve

\frac{x}{2} + \frac{x - 1}{3} = \frac{5}{6}

Step-by-step

1
$6 \times \frac{x}{2} + 6 \times \frac{x - 1}{3} = 6 \times \frac{5}{6}$
Why: The denominators are $2$ , $3$ and $6$ , whose LCM is $6$ .
2
$3 x + 2 (x - 1) = 5$
3
$5 x - 2 = 5 \Rightarrow x = \frac{7}{5}$

Answer

x = \frac{7}{5}

Watch out

Multiplying by an expression that contains the unknown can create a solution the original equation never had:

\frac{1}{x - 2} = \frac{1}{x - 2}

becomes

1 = 1

once cleared, which hides that

x = 2

was never allowed.

Builds on

Comparing two fractions

Also known as

clear fractions, multiply through by the LCM, cross-multiply, equation with algebraic fractions, common denominator in an equation

Appears in past papers

Statement

Worked example

Appears in past papers

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