WAEC 2008 · Paper 2 · Q2

(a)
Simplify $\dfrac{x^2 - 8x + 16}{x^2 - 7x + 12}$ .
(b)
If $\frac12$ , $\frac1x$ , $\frac13$ are successive terms of an arithmetic progression (A.P.), show that $\dfrac{2 - x}{x - 3} = \dfrac23$ .
Model answer
Equal common differences: $\frac1x - \frac12 = \frac13 - \frac1x$ , so $\frac{2 - x}{2x} = \frac{x - 3}{3x}$ . Multiply both sides by $6x$ : $3(2 - x) = 2(x - 3)$ . Divide both sides by $3(x - 3)$ : $\frac{2 - x}{x - 3} = \frac23$ , as required.

Worked solution (try it first)

Factorise the top: $x^2 - 8x + 16 = (x - 4)^2$ .
Factorise the bottom: $x^2 - 7x + 12 = (x - 3)(x - 4)$ .
Cancel the common factor $(x - 4)$ : the fraction simplifies to $\frac{x - 4}{x - 3}$ .

In an A.P. the common difference is the same between neighbouring terms, so $\frac1x - \frac12 = \frac13 - \frac1x$ .
Write each side as one fraction: $\frac{2 - x}{2x} = \frac{x - 3}{3x}$ .
Multiply both sides by $6x$ : $3(2 - x) = 2(x - 3)$ .
Divide both sides by $3(x - 3)$ : $\frac{2 - x}{x - 3} = \frac23$ , which is what we had to show.