3. Solve the system of equations $\left\{\begin{array}{l}x^{2} y-x y^{2}-5 x+5 y+3=0, \\ x^{3} y-x y^{3}-5 x^{2}+5 y^{2}+15=0 .\end{array}\right.$
Answer: $(4 ; 1)$.

Solution. The given system is equivalent to the following:

$$
\left\{\begin{array} { l } 
{ x y ( x - y ) - 5 ( x - y ) + 3 = 0 } \\
{ x y ( x ^ { 2 } - y ^ { 2 } ) - 5 ( x ^ { 2 } - y ^ { 2 } ) + 1 5 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
((x y-5)(x-y)=-3 \\
(x y-5)(x-y)(x+y)=-15
\end{array}\right.\right.
$$

Dividing the second equation of the last system term by term by the first, we get $x+y=5$, from which $y=-x+5$. Substituting this into the first equation: $\left(-x^{2}+5 x-5\right)(2 x-5)=-3,2 x^{3}-15 x^{2}+35 x-28=0$. By trial, we find the integer root of this equation - $x=4$. Factoring out $(x-4)$, we get $(x-4)\left(2 x^{2}-7 x+7\right)=0$. Therefore, the equation has only one root $x=4$. Then $y=1$, and the pair of numbers $(4 ; 1)$ is the only solution to the system.