Posted by ALEX on Tuesday, April 2, 2013 at 11:28am.
as with all fractions , cross-multiply to get rid of them:
(x+5)(x) = (2x+7)(x-3) assuming x≠0,3
x^2+5x = 2x^2+x-21
x^2-4x-21 = 0
(x-7)(x+3) = 0
x = -3,7
check:
(-3+5)/(-3-3) = -1/3 = (-6+7)/(-3)
(7+5)/(7-3) = 3 = (14+7)/7