Solve the following system of equations:

$$
\begin{aligned}
x^{2}+y^{2}+x+y & =50 \ldots \\
x y & =20 \ldots
\end{aligned}
$$
Given that the product $xy$ is known, our goal is to calculate $x+y=z$. From (1):

$$
(x+y)^{2}-2 x y+(x+y)=50, \quad \text { therefore } \quad z^{2}+z-90=0 \ldots
$$

The roots of equation (3) are: $z_{1}=\frac{-1+\sqrt{361}}{2}=9$ and $z_{2}=\frac{-1-\sqrt{361}}{2}=-10$.

Accordingly, $x$ and $y$ are the roots of two quadratic equations.

I. If $x+y=z_{1}=9$ and $x y=20$, then $x$ and $y$ satisfy the following equation:

$$
u^{2}-9 u+20=0 \quad \text { and from this } \quad u_{1}=5, \quad u_{2}=4
$$

Thus

$$
x=5 \quad \text { and } \quad y=4 \quad \text { or } \quad x=4 \text { and } y=5 \text {. }
$$

II. If $x+y=-10, \quad x y=20$, then

$$
v^{2}+10 v+20=0 \quad \text { and from this } \quad v_{1}=-5+\sqrt{5}, \quad v_{2}=-5-\sqrt{5}
$$

Accordingly

$$
x=v_{1}, \quad \text { and } \quad y=v_{2} \quad \text { or } \quad x=v_{2} \quad \text { and } \quad y=v_{1} \text {. }
$$

Our system of equations is satisfied by 4 value pairs:

| $x$ | 5 | 4 | $-5+\sqrt{5}$ | $-5-\sqrt{5}$ |
| :--- | :--- | :--- | :--- | :--- |
| $y$ | 4 | 5 | $-5-\sqrt{5}$ | $-5+\sqrt{5}$. |

(The corresponding $(x, y)$ value pairs are listed one below the other.)