Adding and subtracting with reversed factors

Denominators such as \(x-3\) and \(3-x\) are reversed: \(3-x=-(x-3)\). Rewrite the reversed one so both denominators match, and move the minus sign to the front of the fraction.

Why: with numbers, \(3-5=-2=-(5-3)\). So \(\dfrac{a}{3-x}=\dfrac{a}{-(x-3)}=-\dfrac{a}{x-3}\): a plus sign in front becomes minus, and a minus becomes plus.

Method

1. Spot a reversed bracket, including one inside a factorised quadratic such as \(16-x^2=(4-x)(4+x)\).
2. Rewrite it with \(b-a=-(a-b)\) and change the sign in front of that fraction.
3. Use the LCD and combine as before.
4. Simplify, keeping any minus sign, and state the restrictions.

Example

Express \(\dfrac{2x}{x-5}+\dfrac{10}{5-x}\) as a single fraction.

\(\begin{aligned}\dfrac{2x}{x-5}+\dfrac{10}{5-x}&=\dfrac{2x}{x-5}-\dfrac{10}{x-5} \\ &=\dfrac{2x-10}{x-5} \\ &=\dfrac{2(x-5)}{x-5} \\ &=2,\qquad x\ne5\end{aligned}\)

Express \(\dfrac{x}{x^2-7x+10}+\dfrac{2}{5-x}\) as a single fraction.

\(\begin{aligned}\dfrac{x}{x^2-7x+10}+\dfrac{2}{5-x}&=\dfrac{x}{(x-2)(x-5)}-\dfrac{2}{x-5} \\ &=\dfrac{x-2(x-2)}{(x-2)(x-5)} \\ &=\dfrac{x-2x+4}{(x-2)(x-5)} \\ &=-\dfrac{x-4}{(x-2)(x-5)},\qquad x\ne2,\ x\ne5\end{aligned}\)

Check: changing the bracket without changing the sign in front changes the value. Test with \(x=6\) in the first example: \(\dfrac{12}{1}+\dfrac{10}{-1}=2\).