Play around with it. See if some other arrangement, equality, squaring, etc, leads somewhere potentially useful. Try stuff!
For instance, if you factor the difference of squares for the third "half", can you substitute from either of the other two "halves"? Can you create a new equation that can factor helpfully? And so forth.
x y = (x + y) (x - y) and x y = x + y (given) Subst. this in the first eqn. gives: x y = x y (x - y) so x - y = 1 and x = y + 1 (same as : x - y - 1 = 0, see earlier by Martingale). Subst. this in: x y = x + y (with x = y + 1) gives the quadratic eqn.: y² - y - 1 = 0 with solutions y = -0.618 or y = 1.618 Knowing that x = y + 1 gives: x = .382 and y = - 0.618 x = 2.618 and y = 1.618 Seems that both results are solutions of the problem when worked out. So thanks a lot for helping me out! Great stuff!!! Luke.