\(\frac{2}{2r - 1}\) - \(\frac{5}{3}\) = \(\frac{1}{r + 2}\) \(\frac{2}{2r - 1}\) - \(\frac{1}{r + 2}\) = \(\frac{5}{3}\) \(\frac{2r + 4 - 2r + 1}{2r - 1 (r + 2)}\) = \(\frac{5}{3}\) \(\frac{5}{(2r + 1)(r + 2)}\) = \(\frac{5}{3}\) 5(2r - 1)(r + 2) = 15 (10r - 5)(r + 2) = 15 \(10r^2\) + 20r - 5r - 10 = 15 \(10r^2\) + 15r = 25 \(10r^2\) + 15r - 25 = 0 \(2r^2\) + 3r - 5 = 0 (\(2r^2 + 5r)\)(2r + 5) = r(2r + 5) - 1(2r + 5) (r - 1)(2r + 5) = 0 r = 1 or \(\frac{-5}{2}\)