JAMB Mathematics 2021 Paper

$\frac{2}{2r-1}$ - $\frac{5}{3}$ = $\frac{1}{r+2}$

$\frac{2}{2r-1}$ - $\frac{1}{r+2}$ = $\frac{5}{3}$

The L.C.M.: (2r - 1) (r + 2)

$\frac{2(r+2)-1(2r-1)}{(2r-1)(r+2)}$ = $\frac{5}{3}$

$\frac{2r+4-2r+1}{(2r-1)(r+2)}$ = $\frac{5}{3}$

cross multiply the solution

3 * 5 = (2r - 1) (r + 2) * 5

divide both sides 5

3 = 2r$^2$ + 3r - 2 (when expanded)

collect like terms

2r$^2$ + 3r - 2 - 3 = 0

2r$^2$ + 3r - 5 = 0

Factors are -2r and +5r

2r$^2$ -2r + 5r - 5 = 0

[2r$^2$ -2r] [+ 5r - 5] = 0

2r(r-1) + 5(r-1) = 0

(2r+5) (r-1) = 0

r = 1 or - $\frac{5}{2}$