Solve the following system of equations:

$$
\begin{aligned}
& x^{2}+x y+y=1 \\
& y^{2}+x y+x=5
\end{aligned}
$$
Solve the following system of equations:

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

Solution I. Factor out the common factors of the first two terms in both equations, and replace $x+y$ with $S$:

$$
\begin{aligned}
& x S+y=1 \\
& y S+x=5
\end{aligned}
$$

By adding the two equations and again replacing $x+y$ with $S$, we obtain a quadratic equation in $S$:

$$
S^{2}+S=6
$$

From this, the value of $S$ can be found by completing the square:

$$
(S+0.5)^{2}=6.25
$$

Thus, $S=-0.5 \pm 2.5$. Knowing the value of $S$, we can use the relationship

$$
x+y=S
$$

Subtracting (1a) from (2a), we get:

$$
x=\frac{1-S}{S-1}=-1, \quad y=\frac{5-S}{S-1}
$$

Finally, based on the latter, $y_{1}=\frac{5-2}{2-1}=3, y_{2}=\frac{5+3}{-3-1}=-2$. By substitution, we can verify that both pairs of roots satisfy the original system of equations.

Solution II. The difference between the two sides of equation (1) can be factored as:

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

Thus, either $x=-1$ or $x+y=1$. If $x=-1$, then according to equation (2), $y^{2}-y=6$, which can also be solved by completing the square:

$$
(y-0.5)^{2}=6.25
$$

Thus, $y=0.5 \pm 2.5$. If $x+y=1$, then according to the form (2a) of equation (2), $x+y=5$, which is an obvious contradiction. Therefore, the only possible roots of the system (1)-(2) are $x=-1, y=0.5 \pm 2.5$. In this case, there is no need to verify the obtained roots, since, as we have seen, when $x=-1$, equation (1) holds regardless of the value of $y$, and the value of $y$ is precisely determined based on equation (2).