Sorry guys but second line is false its only true when x=1 or 2 x = y. Then x2 = xy is wrong if x=3 , xy=9 x2=6 6 !=9 you can't divide this way if its proper algebra left side is x2 - y2   2(x-y) ------- = ------ = 2 <- left side of the equation (x-y)        (x-y) and obviously you cant separate (x-y) from the right side (xy - y2)/(x-y) != y sorry guys example is flawed and 1 != 0