Assume that a pair $(x, y)$ is a solution to (1) and (2). Then it follows from (1)

$$
(x+1)(y+1)=0
$$

which means at least one of the numbers $x$ or $y$ must be equal to -1.

For $x=-1$, equation (2) yields $y= \pm 2$, and for $y=-1$, equation (2) gives $x= \pm 2$. Therefore, the only possible pairs that can be solutions to the system (1), (2) are $(-1,-2),(-1,2),(-2,-1),(2,-1)$. Indeed, these four pairs satisfy equations (1), (2), as a check shows.

## Adopted from $[5]$
Task 1 - 271211

Determine all pairs $(x, y)$ of real numbers $x, y$ that satisfy the following system of equations (1), (2):

$$
\begin{aligned}
x + xy + y & = -1 \\
x^2 + y^2 & = 5!
\end{aligned}
$$