( -1,\(\frac{5}{2}\) )
( 1, - \(\frac{5}{2}\) )
( \(\frac{5}{2}\), 1 )
(2,1)
Correct answer is B
\(\frac{2}{(2r - 1)}\) - \(\frac{5}{3}\) = \(\frac{1}{(r + 2)}\)
\(\frac{2}{(2r - 1)}\) - \(\frac{1}{(r + 2)}\) = \(\frac{5}{3}\)
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 - \(\frac{5}{2}\)