Before multiplying away a denominator containing the unknown, exclude the values that make it zero. Then multiply every term on both sides by a common denominator. Check the answer in the original equation.
Example
Solve \(\dfrac3{x-2}=6\), where \(x\ne2\).
\(\begin{aligned}3&=6(x-2)\\3&=6x-12\\x&=\dfrac52.\end{aligned}\)
Check: \(\dfrac3{\dfrac52-2}=6\). Clearing the denominator is valid because the excluded value \(2\) is not used.
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