6.075. $\left\{\begin{array}{l}x^{2} y+x y^{2}=6, \\ x y+x+y=5 .\end{array}\right.$
## Solution.

From the condition we have $\left\{\begin{array}{l}x y(x+y)=6, \\ x y+(x+y)=5 .\end{array}\right.$

Let $\left\{\begin{array}{l}x y=u, \\ x+y=v .\end{array}\right.$

For $u$ and $v$, we get the system $\left\{\begin{array}{l}u v=6, \\ u+v=5\end{array}\right.$, from which, by Vieta's theorem, we find $u_{1}=2, u_{2}=3, v_{1}=3, v_{2}=2$.

Then $\left\{\begin{array}{l}x y=2, \\ x+y=3\end{array}\right.$ or $\left\{\begin{array}{l}x y=3, \\ x+y=2\end{array}\right.$, from which $\left\{\begin{array}{l}x_{1}=1, \\ y_{1}=2\end{array}\right.$ and $\left\{\begin{array}{l}x_{2}=2, \\ y_{2}=1 .\end{array}\right.$

Answer: $(1 ; 2),(2 ; 1)$