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