Let $x,y,z$ be integers such that

$xy+x+y=20\\ zy+z+y=6\\ xz+x+z=2$

Given that $x,y,z\neq 0$ , find $z-x-y$

The answer is 10.

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We get $(x+1)(y+1) = 21$ , $(y+1)(z+1) = 7$ , $(x+1)(z+1) = 3$ . Multiplying the last two and comparing with the first equation shows that $(z+1)^2 = 1$ . Since $z+1 \ne 1$ , we get $z+1 = -1$ and hence $x+1 = -3$ and $y+1 = -7$ . So $z - x - y = -2 - (-4) - (-8) = \fbox{10}$ .