2. Solve the system of equations $\left\{\begin{array}{l}2 x+y+2 x y=11, \\ 2 x^{2} y+x y^{2}=15 .\end{array}\right.$ Answer: $\left(\frac{1}{2} ; 5\right),(1 ; 3),\left(\frac{3}{2} ; 2\right),\left(\frac{5}{2} ; 1\right)$.
Solution. Let's make the substitution $2 x+y=u, x y=w$. Then the system takes the form

$$
\left\{\begin{array} { l } 
{ u + 2 w = 1 1 , } \\
{ u w = 1 5 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ u = 1 1 - 2 w , } \\
{ w ( 1 1 - 2 w ) = 1 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
u=11-2 w \\
2 w^{2}-11 w+15=0
\end{array}\right.\right.\right.
$$

From the second equation of the last system, we find that $w=3$ (and then $u=5$) or $w=\frac{5}{2}$ (and then $u=6$). We return to the original variables.

If $u=6, w=\frac{5}{2}$, then $\left\{\begin{array}{l}2 x+y=6, \\ x y=\frac{5}{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}y=6-2 x, \\ x^{2}-3 x+\frac{5}{4}=0 .\end{array} \quad\right.\right.$ This means $x=\frac{5}{2}$ (in which case $y=1$) or $x=\frac{1}{2}$ (in which case $y=5$).

If $u=5, w=3$, then $\left\{\begin{array}{l}2 x+y=5, \\ x y=3\end{array} \Leftrightarrow\left\{\begin{array}{l}y=5-2 x, \\ 2 x^{2}-5 x+3=0 .\end{array} \quad\right.\right.$ This means $x=\frac{3}{2}$ (in which case $y=2$) or $x=1$ (in which case $y=3$).