Determine all pairs $(x, y)$ of real numbers such that: $|x+y|=3$ and $x y=-10$.
The hypothesis $|x+y|=3$ means that $x+y$ is either 3 or -3. In both cases, $(x+y)^{2}=9$. Now, $(x+y)^{2}=x^{2}+y^{2}+2xy$ and $(x-y)^{2}=x^{2}+y^{2}-2xy=(x+y)^{2}-4xy$. Given the hypothesis $xy=-10$, we have $(x-y)^{2}=9-4(-10)=49$, which means that $x-y$ is either 7 or -7. Therefore, if $x+y=3, 2x=(x+y)+(x-y)$ can be 10 (in which case $x=5$ and $y=-2$) or -4 (in which case $x=-2$ and $y=5$, which is the same because $x$ and $y$ play symmetric roles). Similarly, if $x+y=-3, x$ can be 2 (in which case $y=-5$) or -5 (in which case $y=2$). The four pairs of solutions are therefore: $(5,-2),(-2,5),(2,-5)$ and $(-5,2)$.