Suppose that $f(x)$ is a function such that
\[f(xy) + x = xf(y) + f(x)\]for all real numbers $x$ and $y.$  If $f(-1) = 5$ then compute $f(-1001).$
For example, what if we plug in $y = 1?$ Then the equation becomes $f(x) + x = xf(1) + f(x),$ which simplifies to $x = xf(1).$ This means that either $x = 0$ or $f(1) = 1.$ But we can't have $x = 0$ for all $x,$ so we must have $f(1) = 1.$ What if we plug in $y = -1?$ Then the equation becomes $f(-x) + x = xf(-1) + f(x),$ which simplifies to $f(-x) - f(x) = x(f(-1) - 1).$ We already know that $f(-1) = 5,$ so we can substitute that and get $f(-x) - f(x) = 4x.$ What if we plug in $x = -1?$ Then the equation becomes $f(-y) - 1 = -f(y) + 5,$ which simplifies to $f(-y) + f(y) = 6.$ We can use the relation $f(-x) - f(x) = 4x$ with $x = 1001$ and get $f(-1001) - f(1001) = 4004.$ We can also use the relation $f(-y) + f(y) = 6$ with $y = 1001$ and get $f(-1001) + f(1001) = 6.$ Adding these two equations, we get $2f(-1001) = 4010,$ so $f(-1001) = 2005.$ 