JAMB 1983 · UME · Q15

Simplify $\dfrac{x - 7}{x^2 - 9} \times \dfrac{x^2 - 3x}{x^2 - 49}$ .

Worked solution (try it first)

Factorise each part: $x^2 - 9 = (x - 3)(x + 3)$ , $x^2 - 3x = x(x - 3)$ and $x^2 - 49 = (x - 7)(x + 7)$ .
So the product is $\dfrac{x - 7}{(x - 3)(x + 3)} \times \dfrac{x(x - 3)}{(x - 7)(x + 7)}$ .
Cancel the common factors $x - 7$ and $x - 3$ , top and bottom.
This leaves $\dfrac{x}{(x + 3)(x + 7)}$ , option D.