2(2r−1) - 53 = 1(r+2)
2(2r−1) - 1(r+2) = 53
The L.C.M.: (2r - 1) (r + 2)
2(r+2)−1(2r−1)(2r−1)(r+2) = 53
2r+4−2r+1(2r−1)(r+2) = 53
cross multiply the solution
3 * 5 = (2r - 1) (r + 2) * 5
divide both sides 5
3 = 2r2 + 3r - 2 (when expanded)
collect like terms
2r2 + 3r - 2 - 3 = 0
2r2 + 3r - 5 = 0
Factors are -2r and +5r
2r2 -2r + 5r - 5 = 0
[2r2 -2r] [+ 5r - 5] = 0
2r(r-1) + 5(r-1) = 0
(2r+5) (r-1) = 0
r = 1 or - 52