Name: Zach Who is asking: Student Level of the question: Secondary Question: I can solve easy problems such as (x/2)+3=2+(3x/4). That is easy because the Lowest Common Denominator is 4. But what really gets me stuck is a problem like this one. (6/x-2) = ( 21/(x-2)(x+2) )+ 1. Please help me understand this!!!!! Hi Zach, I want to change the problem to  6/(x-2)= 21/(x-2)(x+2) + 1 The denominators here are (x - 2) and (x - 2)(x + 2) so the least common denominator is (x - 2)(x + 2). Thus the problem becomes  6(x+2)/(x-2)(x+2)= 21/(x-2)(x+2) + (x-2)(x+2)/(x-2)(x+2) Thus, as long as (x - 2)(x + 2) is not zero, 6(x + 2) = 21 + (x - 2)(x + 2) Expanding and simplifying I got x2 - 6x + 5 = 0 This factors as (x - 1)(x - 5) = 0 so x = 1 or x = 5. If the problem is as you sent, 6/x - 2 = 21/(x-2)(x+2) + 1 Then the least common denominator is x(x - 2)(x + 2) and the equation becomes 6(x-2)(x+2)/x(x-2)(x+2) - 2x(x-2)(x+2)/x(x-2)(x+2) = 21x/x(x-2)(x+2) + x(x-2)(x+2)/x(x-2)(x+2) If x(x - 2)(x + 2) is not zero then 6(x - 2)(x + 2) - 2x(x - 2)(x + 2) = 21 x + x(x - 2)(x + 2) Simplifying this expression gives a cubic that doesn't seem to factor so I think that the intended problem was  6/(x-2)= 21/(x-2)(x+2) + 1 Penny