3. Solve the system of equations $\left\{\begin{array}{l}x^{2} y+x y^{2}-2 x-2 y+10=0, \\ x^{3} y-x y^{3}-2 x^{2}+2 y^{2}-30=0 .\end{array}\right.$
Answer: $(-4, -1)$.

Solution. The given system is equivalent to the following:

$$
\left\{\begin{array} { l } 
{ x y ( x + y ) - 2 ( x + y ) + 10 = 0 , } \\
{ x y ( x ^ { 2 } - y ^ { 2 } ) - 2 ( x ^ { 2 } - y ^ { 2 } ) - 30 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x y-2)(x+y)=-10 \\
(x y-2)(x-y)(x+y)=30
\end{array}\right.\right.
$$

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: $\left(x^{2}+3 x-2\right)(2 x+3)=-10, 2 x^{3}+9 x^{2}+5 x+4=0$. By trial, we find the integer root of this equation $-x=-4$. Factoring out the factor $(x+4)$, we get $(x+4)\left(2 x^{2}+x+1\right)=0$. Therefore, the equation has a unique root $x=-4$. Then $y=-1$, and the pair of numbers $(-4, -1)$ is the only solution to the system.