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

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 * 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 - \(\frac{5}{2}\)

There is an explanation video available .