11.2. Solve the system of equations

$$
\left\{\begin{array}{l}
x y+x=15 \\
x^{2}+y^{2}+x y+y=41
\end{array}\right.
$$
Answer: (3;4), (5;2). Solution. By adding the equations, we get $(x+y)^{2}+(x+y)=56$. Let $x+y=t$, then $t^{2}+t-56=0 \Leftrightarrow t_{1}=-8, t_{2}=7$. Thus, we have a combination of two systems equivalent to the original system: 1) $\left\{\begin{array}{l}x+y=-8 \\ x y+x=15\end{array}\right.$ and 2) $\left\{\begin{array}{l}x+y=7 \\ x y+x=15\end{array}\right.$. Expressing $y$ from the first equation and substituting it into the second, we get a quadratic equation. In case 1) $x^{2}+7 x+15=0$; in case 2) $x^{2}-8 x+15=0$. In the first case $D<0$, and there is no solution; in the second case $x_{1}=3, x_{2}=5$. The corresponding values are $y_{1}=7-x_{1}=4, y_{2}=7-x_{2}=2$.