Solve the following system of equations:

$$
x y+x+y=34, \quad x^{2}+y^{2}-x-y=42 .
$$
Add twice (2) to (1), thus we get a quadratic equation for the unknown sum $x+y=z$:

$$
\left(x^{2}+2 x y+y^{2}\right)+x+y=110, \quad \text { or } \quad z^{2}+z-110=0
$$

From this, $z_{1}=10, z_{2}=-11$.

Using these and (I) again, we get two systems of equations for $x$ and $y$:

(I.)

$$
\begin{array}{r}
x+y=10 \\
x y=24
\end{array}
$$

(II.)

$$
\begin{gathered}
x+y=-11 \\
x y=45
\end{gathered}
$$

According to these, $x$ and $y$ are the roots of the following quadratic equations:

$$
u^{2}-10 u+24=0, \quad v^{2}+11 v+45=0
$$

From the first, $u_{1}=4, u_{2}=6$, and the second has no real roots. Thus, our system of equations has two real solutions:

$$
\left\{\begin{array} { l } 
{ x _ { 1 } = u _ { 1 } = 4 , } \\
{ y _ { 1 } = u _ { 2 } = 6 , }
\end{array} \quad \text { and } \quad \left\{\begin{array}{l}
x_{2}=u_{2}=6 \\
y_{2}=u_{1}=4
\end{array}\right.\right.
$$

Katalin Karsai (Makó, József A. Gymnasium, 2nd grade)