You can put this solution on YOUR website! Assume the problem is + 1 =
Factor the 2nd denominator, this will be the common denominator + 1 =
multiply by the common denominator and you have
2(x+1) +(x+1)(x-1) = 4
2x + 2 + x^2 - 1 = 4
combine on the left to form a quadratic equation
x^2 + 2x + 2 - 1 - 4 = 0
x^2 + 2x - 3 = 0
factors to
(x+3)(x-1) = 0
Two solutions
x = 1
x = -3
:
You can check both of these in the original problem, you should note that
x = 1 is not a valid solution, it results in a denominator of 0