Adding and subtracting with linear denominators

When two denominators are different linear expressions with no common factor, such as \(x+1\) and \(x-4\), the LCD is their product \((x+1)(x-4)\).

Why: \(x+1\) and \(x-4\) share no factor, just as \(3\) and \(5\) share none, and \(\dfrac13+\dfrac15=\dfrac{5+3}{15}\) uses the product \(15\). Each fraction is multiplied top and bottom by the other denominator.

Method

1. Write the LCD as the product of the two brackets. Leave it in factorised form.
2. Multiply each numerator by the bracket its denominator is missing.
3. Write one fraction with each numerator in brackets.
4. Expand the numerator, collect like terms, and check whether the result factorises and cancels.
5. State the values that make an original denominator zero.

Example

Express \(\dfrac{x}{x+4}+\dfrac{2}{3x+2}\) as a single fraction in its simplest form.

\(\begin{aligned}\dfrac{x}{x+4}+\dfrac{2}{3x+2}&=\dfrac{x(3x+2)}{(x+4)(3x+2)}+\dfrac{2(x+4)}{(x+4)(3x+2)} \\ &=\dfrac{x(3x+2)+2(x+4)}{(x+4)(3x+2)} \\ &=\dfrac{3x^2+2x+2x+8}{(x+4)(3x+2)} \\ &=\dfrac{3x^2+4x+8}{(x+4)(3x+2)},\qquad x\ne-4,\ x\ne-\dfrac23\end{aligned}\)

The numerator \(3x^2+4x+8\) has no factor \(x+4\) or \(3x+2\), so nothing cancels.

Check: never add the denominators; \(\dfrac{3}{x+1}+\dfrac{2}{x-4}\) is not \(\dfrac{5}{2x-3}\) (try \(x=0\): \(3-\dfrac12\ne-\dfrac53\)). Expanding the denominator is unnecessary and hides factors that might cancel.