JAMB Mathematics 2021 Paper

Concept:Rearrange the rational equation, use the lowest common multiple, then solve the resulting quadratic by factorisation.Explanation:Start with $\frac{2}{2r-1} - \frac{5}{3} = \frac{1}{r+2}$.Rearrange so the terms containing $r$ are on one side:$\frac{2}{2r-1} - \frac{1}{r+2} = \frac{5}{3}$.The L.C.M. of $(2r-1)$ and $(r+2)$ is $(2r-1)(r+2)$.Combine the left-hand side:$\frac{2(r+2) - 1(2r-1)}{(2r-1)(r+2)} = \frac{5}{3}$.Simplify the numerator: $2r+4-2r+1 = 5$.So $\frac{5}{(2r-1)(r+2)} = \frac{5}{3}$.Cancel the common factor 5:$(2r-1)(r+2) = 3$.Expand the product:$2r^2 + 3r - 2 = 3$.Bring all terms to one side:$2r^2 + 3r - 5 = 0$.Factorise the quadratic:$(2r+5)(r-1) = 0$.Therefore $r = 1$ or $r = -\frac{5}{2}$.Answer:$r = 1$ or $r = -\frac{5}{2}$.The correct option is B.