Solve the following equation \(\frac{2}{2r - 1}\) - \(\frac{5}{3}\) = \(\frac{1}{r + 2}\)
 

  • A (\(\frac{5}{2}\), 1)
  • B (5, -4)
  • C (2, 1)
  • D (1, \(\frac{-5}{2}\))
  • E (1,-2)
jamb 1985mathematics

The correct answer is D. (1, \(\frac{-5}{2}\))

\(\frac{2}{2r - 1}\) - \(\frac{5}{3}\) = \(\frac{1}{r + 2}\)
\(\frac{2}{2r - 1}\) - \(\frac{1}{r + 2}\) = \(\frac{5}{3}\)
\(\frac{2r + 4 - 2r + 1}{2r - 1 (r + 2)}\) = \(\frac{5}{3}\)
\(\frac{5}{(2r + 1)(r + 2)}\) = \(\frac{5}{3}\)
5(2r - 1)(r + 2) = 15
(10r - 5)(r + 2) = 15
10r2 + 20r - 5r - 10 = 15
10r2 + 15r = 25
10r2 + 15r - 25 = 0
2r2 + 3r - 5 = 0
(2r2 + 5r)(2r + 5) = r(2r + 5) - 1(2r + 5)
(r - 1)(2r + 5) = 0
r = 1 or \(\frac{-5}{2}\)

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