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Algebra Difficulty 4.3 AIME Prove it Philippines

Problem:
If x+y+xy=1x + y + x y = 1, where x,yx, y are nonzero real numbers, find the value of
xy+1xy−yx−xy x y + \frac{1}{x y} - \frac{y}{x} - \frac{x}{y}

Solution

Solution:
Observe that
xy+1xy−yx−xy=(xy)2+1−x2−y2xy=(x2−1)(y2−1)xy=(x+1)(y+1)(x−1)(y−1)xy=(xy+x+y+1)(xy−x−y+1)xy \begin{aligned} x y + \frac{1}{x y} - \frac{y}{x} - \frac{x}{y} &= \frac{(x y)^2 + 1 - x^2 - y^2}{x y} \\ &= \frac{\left(x^2 - 1\right)\left(y^2 - 1\right)}{x y} \\ &= \frac{(x + 1)(y + 1)(x - 1)(y - 1)}{x y} \\ &= (x y + x + y + 1) \frac{(x y - x - y + 1)}{x y} \end{aligned}
Since xy+x+y=1x y + x + y = 1, the first term will equal 22. Moreover, dividing both sides of the equation xy+x+y=1x y + x + y = 1 by xyx y, we obtain
1+1y+1x=1xy 1 + \frac{1}{y} + \frac{1}{x} = \frac{1}{x y}
which is equivalent to
1=1xy−1y−1x 1 = \frac{1}{x y} - \frac{1}{y} - \frac{1}{x}
Hence, (xy+x+y+1)(xy−x−y+1)xy=2⋅2=4(x y + x + y + 1) \frac{(x y - x - y + 1)}{x y} = 2 \cdot 2 = 4.

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