Starting with: 15/x - 15/(x-2) = -2 You must begin by stipulating that x must not equal 0 and x must not equal 2. If x = 0 or x = 2, then this will cause a division by 0 in one of the terms of the original equation. It does not matter what you do to the equations to solve them these requirements must be retained throughout the duration of the problem as follows: 15/x -15/(x-2)=-2; x >< 0, x >< 2 (Note: I am using >< to mean NOT equal to) By inspection one can see that the common denominator is x(x-2). To make the first term, (15/x), have a denominator of {x(x-2)}, you multiply it by 1 in the form of (x-2)/(x-2) as follows: {(15/x)}{(x-2)/(x-2)} -15/(x-2)=-2; x >< 0, x >< 2 To make the second term, {-15/(x-2)}, have a denominator of {x(x-2)} you multiply it by 1 in the form of (x/x) as follows: {(15/x)}{(x-2)/(x-2)} - {15/(x-2)}{(x/x)} = -2; x >< 0, x >< 2 Now simplify the numerators: {15(x-2)}/{x(x-2)} - (15x)/{x(x-2)} = -2; x >< 0, x >< 2 (15x-30)/{x(x-2)} - (15x)/{x(x-2)} = -2; x >< 0, x >< 2 Write the numberators over the common denominator: {15x - 30 - 15x}/{x(x-2)} = -2; x >< 0, x >< 2 -30/{x(x-2)} = -2; x >< 0, x >< 2 Divide both sides of the equation by -2: 15/{x(x-2)} = 1; x >< 0, x >< 2 Multiply both sides by x(x-2): 15 = {x(x-2)}; x >< 0, x >< 2 Multiply x(x-2): 15 = x² - 2x; x >< 0, x >< 2 Subtract 15 from both sides and flip the equation: x² - 2x - 15 = 0; x >< 0, x >< 2 Factor the quadratic: (x - 5)(x + 3) = 0; x >< 0, x >< 2 At this point you can drop the x >< 0, x >< 2 requirements, because it is obvious that solution set does not contain 0 or 2. x = 5 or x = -3