Solve the following system of equations:

$$
\begin{gathered}
\frac{10}{2 x+3 y-29}+\frac{9}{7 x-8 y+24}=8 \\
\frac{2 x+3 y-29}{2}=\frac{7 x-8 y}{3}+8
\end{gathered}
$$
Let

$$
2 x+3 y-29=u \text { and } 7 x-8 y+24=v
$$

then

$$
\frac{10}{u}+\frac{9}{v}=8 \text { and } \frac{u}{2}=\frac{v}{3}
$$

from the second equation, $u=\frac{2 v}{3}$, substituting this into the first equation, we get:

$$
\frac{15}{v}+\frac{9}{v}=8
$$

from which

$$
v=3 \text{ and thus } u=2
$$

Therefore,

$$
2 x+3 y-29=2 \text { and } 7 x-8 y+24=3
$$

from this system of equations:

$$
x=5, \quad y=7
$$

(Tóth Balázs, Eger.)

The problem was also solved by: Ehrenstein P., Felhössy J., Freund E., Füstös P., Harsányi Z., Heimlich P., Jánosy Gy., Kiss J., Makó E., Paunz A., Petrik S., Schuster Gy., Schwarz Gy., Schlesinger Gy., Steiger J., Szabó J., Vámossy L.