\section*{Problem 4 - 141244}

Determine all pairs \((x, y)\) of real numbers \(x\) and \(y\) for which the equations hold:

\[
\begin{array}{r}
24 x^{2}-25 x y-73 x+25 y-35=0 \\
x^{2}-y^{2}-2 x-2 y-7=0
\end{array}
\]
}
Assuming for a pair \((x ; y)\) conditions (1) and (2) hold, we set \(u=x-1\) and \(v=y+1\), so that (1) is equivalent to

\[
\begin{aligned}
24(u+1)^{2}-25(u+1)(v-1)-73(u+1)+25(v+1)-35 & =0 \\
24 u^{2}-84-25 u v & =0
\end{aligned}
\]

while (2) is equivalent to

\[
u^{2}-v^{2}=7
\]

Now we set \(a=u+v, b=u-v\) and thus \(u=\frac{a+b}{2}, v=\frac{a-b}{2}\). Then (4) transforms into the equivalent equation \(a \cdot b=7\) (5), while from (83) the equation

\[
6 a^{2}+12 a b+6 b^{2}-84-\frac{25}{4} a^{2}+\frac{25}{4} b^{2}=0
\]

follows. From this, in conjunction with (5),

\[
-\frac{1}{4} a^{2}+84+\frac{49}{4} b^{2}-84=0 \quad ; \quad a^{2}=49 b^{2}
\]

is obtained. According to (5), \(a\) and \(b\) have the same sign, and thus from (6) \(a=7 b\) and from this \(a^{2}=7 a b=49\) (7). Therefore, at most \((a ; b)=(7 ; 1)\) and \((a ; b)=(-7 ;-1)\) can be solutions of (7) and (5).

Since \(x=u+1=\frac{a+b}{2}+1, y=v-1=\frac{a-b}{2}-1\), at most \((x ; y)=(5 ; 2)\) and \((x ; y)=(-3 ;-4)\) can satisfy equations (1) and (2). The verification confirms the found solutions.

Adapted from \([5]\)