Solve the following system of equations:

$$
\frac{x+y+1}{x+y-1}=p, \quad \frac{x-y+1}{x-y-1}=q
$$
We can immediately state at the beginning that neither $p$ nor $q$ can be 1, because in the fractions on the left side, the numerator is greater than the denominator. - From the two equations, the first step is to express $x+y$ and $x-y$:

\[
\begin{aligned}
& x+y=\frac{p+1}{p-1} \\
& x-y=\frac{q+1}{q-1}
\end{aligned}
\]

From here, by adding and subtracting, we get

\[
x=\frac{p q-1}{(p-1)(q-1)} \quad y=\frac{q-p}{(p-1)(q-1)}
\]

No further constraints on the parameters are necessary, because the right-hand sides of (1) and (2) make sense, and neither of their values is 1, so none of the denominators in the original equations is 0.

Bellay Ágnes (Budapest, Fazekas M. Gyak. G. II. o. t.)

Note. Recognizing that in the given equations, the difference between the numerator and the denominator is 2, we can also arrive at the following steps for (1):

\[
1+\frac{2}{x+y-1}=p, \quad \frac{2}{x+y-1}=p-1, \quad x+y-1=\frac{2}{p-1}
\]

Benczúr András (Budapest, Fazekas M. Gyak. G. II. o. t.)