Two real numbers $x$ and $y$ satisfy $x-y=4$ and $x^3-y^3=28$.  Compute $xy$.
I notice that the second equation is a difference of cubes, which can be factored as $(x-y)(x^2+xy+y^2)$. I also notice that I already know the value of $x-y$, which is 4. So I can substitute that into the second equation and get $4(x^2+xy+y^2)=28$. Dividing both sides by 4, I get $x^2+xy+y^2=7$. Now I want to find a way to use the first equation, $x-y=4$, to simplify this. One idea is to square both sides of the first equation and see what happens. If I do that, I get $x^2-2xy+y^2=16$. I notice that this has some terms in common with the other equation, so I can subtract them and eliminate the squared terms. If I subtract $x^2-2xy+y^2=16$ from $x^2+xy+y^2=7$, I get $3xy=-9$. Solving for $xy$, I get $xy=-3$. So that's my answer. 