4B. Solve, in the set of real numbers, the system of equations

$$
\left\{\begin{array}{l}
x+y+x y=11 \\
x^{2}+x y+y^{2}=19
\end{array}\right.
$$
Solution. If we add the two equations, we get

$$
x^{2}+2 x y+y^{2}+x+y=30
$$

or equivalently

$$
(x+y)^{2}+(x+y)=30
$$

By introducing the substitution $t=x+y$, we obtain the quadratic equation

$$
t^{2}+t-30=0
$$

whose solutions are $t_{1}=5$ and $t_{2}=-6$. From $x+y=5$, we get $y=5-x$, and substituting into one of the given equations, we get $x_{1}=3, x_{2}=2$ and $y_{1}=2, y_{2}=3$. From $x+y=-6$ we get $y=-6-x$, and substituting into one of the equations, we get complex solutions. Therefore, the given system has two pairs of real solutions $(3,2)$ and $(2,3)$.

## III year