Simplifying by common factors and grouping

Only a factor of the whole numerator and of the whole denominator can be cancelled. A term joined to the rest by \(+\) or \(-\) is not a factor, so it can never be cancelled on its own. To find common factors in a fraction with sums, factorise the numerator and the denominator first.

Why terms cannot be cancelled: test with numbers

The middle columns disagree, so cancelling the \(x\) terms in \(\dfrac{x+6}{x+2}\) is wrong. In \(\dfrac{3x+6}{x+2}\) the bracket \(x+2\) is a factor of both parts, so it cancels and the value is \(3\) for every allowed \(x\).

Method

1. Take out the common factor of each numerator and denominator, for example \(3ab+6a=3a(b+2)\).
2. With four terms, group them in pairs so that the same bracket appears twice, then take the bracket out.
3. Cancel any bracket or letter that is a factor of both the whole numerator and the whole denominator.
4. State the restrictions from the original denominator.

Example

Simplify \(\dfrac{3ab+6a}{9a^2}\).

\(\begin{aligned}\dfrac{3ab+6a}{9a^2}&=\dfrac{3a(b+2)}{3a\times3a} \\ &=\dfrac{b+2}{3a},\qquad a\ne0\end{aligned}\)

The \(b\) in \(b+2\) cannot cancel with anything: \(b+2\) is a sum, and the denominator has no factor \(b\).

Simplify \(\dfrac{ab+2a+3b+6}{b^2+2b}\).

\(\begin{aligned}\dfrac{ab+2a+3b+6}{b^2+2b}&=\dfrac{a(b+2)+3(b+2)}{b(b+2)} \\ &=\dfrac{(a+3)(b+2)}{b(b+2)} \\ &=\dfrac{a+3}{b},\qquad b\ne0,\ b\ne-2\end{aligned}\)

Check: in \(\dfrac{a+3}{b}\) nothing more cancels, because \(a+3\) is a sum. Cancelling \(2\) from \(\dfrac{2x+5}{2}\) to get \(x+5\) is the same error: the \(2\) divides only the first term.