Solve the following equation: \(\frac{2}{(2r - 1)}\) - \(\frac{5}{3}\) = \(\frac{1}{(r + 2)}\)
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Correct Answer: Option B
Explanation:
\(\frac{2}{(2r - 1)}\) - \(\frac{5}{3}\) = \(\frac{1}{(r + 2)}\)
\(\frac{2}{(2r - 1)}\) - \(\frac{1}{(r + 2)}\) = \(\frac{5}{3}\)
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 = (2r - 1) (r + 2) or 2r\(^2\) + 3r - 2 (when expanded)
collect like terms
2r\(^2\) + 3r - 2 - 3 = 0
2r\(^2\) + 3r - 5 = 0
Factorize to get x = 1 or - \(\frac{5}{2}\)
\(\frac{2}{(2r - 1)}\) - \(\frac{5}{3}\) = \(\frac{1}{(r + 2)}\)
\(\frac{2}{(2r - 1)}\) - \(\frac{1}{(r + 2)}\) = \(\frac{5}{3}\)
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 = (2r - 1) (r + 2) or 2r\(^2\) + 3r - 2 (when expanded)
collect like terms
2r\(^2\) + 3r - 2 - 3 = 0
2r\(^2\) + 3r - 5 = 0
Factorize to get x = 1 or - \(\frac{5}{2}\)