5. Solve the system of equations $\left\{\begin{array}{l}x^{2} y+x y^{2}+3 x+3 y+24=0, \\ x^{3} y-x y^{3}+3 x^{2}-3 y^{2}-48=0 .\end{array}\right.$
Answer: $(-3, -1)$.

Solution. The given system is equivalent to the following:

$$
\left\{\begin{array} { l } 
{ x y ( x + y ) + 3 ( x + y ) + 24 = 0 , } \\
{ x y ( x ^ { 2 } - y ^ { 2 } ) + 3 ( x ^ { 2 } - y ^ { 2 } ) - 48 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x y+3)(x+y)=-24 \\
(x y+3)(x-y)(x+y)=48
\end{array}\right.\right.
$$

"Phystech-2016", mathematics, solution to ticket 27

Dividing the second equation of the last system term by term by the first, we get $x-y=-2$, from which $y=x+2$. Substituting this into the first equation: $\left(x^{2}+2 x+3\right)(2 x+2)=-24, x^{3}+3 x^{2}+5 x+15=0$. By trial, we find the integer root of this equation $-x=-3$. Factoring out $(x+3)$, we get $(x+3)\left(x^{2}+5\right)=0$. Therefore, the equation has a unique root $x=-3$. Then $y=-1$, and the pair of numbers $(-3, -1)$ is the only solution to the system.