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

Solution. The given system is equivalent to the following:

$$
\left\{\begin{array} { l } 
{ x y ( x - y ) - 3 ( x - y ) + 1 = 0 , } \\
{ x y ( x ^ { 2 } - y ^ { 2 } ) - 3 ( x ^ { 2 } - y ^ { 2 } ) + 3 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x y-3)(x-y)=-1, \\
(x y-3)(x-y)(x+y)=-3 .
\end{array}\right.\right.
$$

Dividing the second equation of the last system term by term by the first, we get $x+y=3$, from which $y=-x+3$. Substituting this into the first equation: $\left(-x^{2}+3 x-3\right)(2 x-3)=-1,2 x^{3}-9 x^{2}+15 x-10=0$. By trial, we find the integer root of this equation - $x=2$. Factoring out $(x-2)$, we get $(x-2)\left(2 x^{2}-5 x+5\right)=0$. Therefore, the equation has only one root $x=2$. Then $y=1$, and the pair of numbers $(2 ; 1)$ is the only solution to the system.