JAMB 1994 · UME · Q15

Solve for $r$ in the equation $\dfrac{1}{r - 1} + \dfrac{2}{r + 1} = \dfrac3r$ .

Worked solution (try it first)

Add the left side over the common denominator $(r - 1)(r + 1) = r^2 - 1$ : the top is $(r + 1) + 2(r - 1) = 3r - 1$ .
So $\dfrac{3r - 1}{r^2 - 1} = \dfrac3r$ .
Cross-multiply: $r(3r - 1) = 3(r^2 - 1)$ .
Expand: $3r^2 - r = 3r^2 - 3$ .
The $3r^2$ terms cancel, leaving $-r = -3$ .
So $r = 3$ , option A.