Solve the following system of equations:

$$
\left.\begin{array}{r}
\sqrt{7 x+y}+\sqrt{x+y}=6, \\
\sqrt{x+y}-y+x=2 .
\end{array}\right\}
$$
Solution. Let's rearrange equation (1) and then square it:

$$
\begin{aligned}
& \sqrt{7 x+y}=6-\sqrt{x+y} \\
& 7 x+y=36-12 \sqrt{x+y}+x+y
\end{aligned}
$$

From here, $\sqrt{x+y}=\frac{6-x}{2}$. By the definition of the square root, $\frac{6-x}{2} \geq 0$, which means $x \leq 6$. Substitute the obtained value of $\sqrt{x+y}$ into equation (2):

$$
\frac{6-x}{2}-y+x=2
$$

From this, $x=2 y-2$. Substitute back into (2):

$$
\sqrt{2 y-2+y}-y+2 y-2=2
$$

After rearranging and squaring, we get the following quadratic equation:

$$
y^{2}-11 y+18=0
$$

From this, $y_{1}=9$ and $x_{1}=16$, or $y_{2}=2$ and $x_{2}=2$. According to the condition, $x \leq 6$, so only $x_{2}=2, y_{2}=2$ can be a solution to the equation. We can verify by substitution that this is indeed a solution.