For the second one....just multiply everything through by x (x + 1).....this leaves
2(x + 1) + 3x(x) = 2x + 5 simplify
2x + 2 + 3x^2 = 2x + 5 subtract 2x and 5 from both sides
3x^2 - 3 = 0 factor
3(x^2 -1) = 0 divide both sides by 3
x^2 - 1 = 0 factor again
(x + 1)(x - 1) = 0 set both factors to 0 and x = -1 and x = 1
Reject x = -1 (it makes two denominators in the original problem = 0)
So....x = 1 is the solution
Be sure and check this in the original problem........
