Solve the following system of equations:

$$
\begin{gathered}
\log _{x}(x+y)+\log _{y}(x+y)=4 \\
(x-1)(y-1)=1 .
\end{gathered}
$$
$$
\begin{gathered}
\log _{x}(x+y)+\log _{y}(x+y)=4 \\
(x-1)(y-1)=1
\end{gathered}
$$

To ensure that the logarithms in (1) are defined, we must exclude the cases where

$$
x>0, \quad y>0, \quad x \neq 1, \quad y \neq 1
$$

which means that only pairs $(x, y)$ that satisfy (3) can be solutions to the system of equations. Simplifying the operations in (2);

$$
x+y=x y
$$

substituting this into (1) gives

$$
\begin{gathered}
\log _{x}(x y)+\log _{y}(x y)=2+\log _{x} y+\log _{y} x=4 \\
\log _{x} y+\frac{1}{\log _{x} y}=2
\end{gathered}
$$

From this, $\left(\log _{x} y-1\right)^{2}=0$, which means $\log _{x} y=1$, and thus $x=y$. Substituting $x=y$ into (2) gives $(x-1)^{2}=1$, from which $x=0$ or $x=2$. Since $x=0$ is not a valid solution due to (3), the only possible solution to the system of equations is $x=y=2$. That this is indeed a solution can be verified by substitution.

Juhász Ágnes (Tata, Eötvös J. Gymn., I. o. t.)