Solve the following system of equations:

$$
\left\{\begin{array}{l}
\frac{7}{2 x-3}-\frac{2}{10 z-3 y}+\frac{3}{3 y-8 z}=8 \\
\frac{2}{2 x-3 y}-\frac{3}{10 z-3 y}+\frac{1}{3 y-8 z}=0 \\
\frac{5}{2 x-3 y}-\frac{4}{10 z-3 y}+\frac{7}{3 y-8 z}=8
\end{array}\right.
$$
I. Solution: We must exclude the values of the denominators being 0, so $2 x \neq 3 y, 10 z \neq 3 y, 8 z \neq 3 y$. By adding the second and third equations,

$$
\frac{7}{2 x-3 y}-\frac{7}{10 z-3 y}+\frac{8}{3 y-8 z}=8
$$

comparing this with the first equation:

$$
\frac{5}{10 z-3 y}=\frac{5}{3 y-8 z}
$$

This can only hold if the denominators are equal, from which

$$
y=3 z
$$

Substituting this into the second equation:

$$
\frac{2}{2 x-9 z}-\frac{3}{z}+\frac{1}{z}=0
$$

from which, after rearrangement, we get

$$
x=5 z .
$$

Substituting the values of $y$ and $x$ in terms of $z$ into the first equation

$$
\frac{7}{z}-\frac{2}{z}+\frac{3}{z}=8
$$

from which

$$
z=1
$$

Thus, $x=5$ and $y=3$.

By substitution, we can verify that the above values indeed satisfy our system of equations.

Gergely Csaba (Bp. I., Toldy g. II. o. t.)

II. Solution: Let

$$
\frac{1}{2 x-3 y}=a, \quad \frac{1}{10 z-3 y}=b \quad \text { and } \quad \frac{1}{3 y-8 z}=c
$$

With these notations, our system of equations becomes:

$$
\begin{aligned}
& 7 a-2 b+3 c=8 \\
& 2 a-3 b+c=0 \\
& 5 a-4 b+7 c=8
\end{aligned}
$$

We eliminate $c$ from these equations by subtracting 3 times (2) from (1), and 7 times (2) from (3):

$$
\begin{aligned}
a+7 b & =8 \\
-9 a+17 b & =8
\end{aligned}
$$

Adding 9 times (4) to (5):

$$
80 b=80, \quad b=1
$$

Using this in (4), and then in (2), we get

$$
a=1, \quad c=1
$$

Using the meaning of $a, b, c$ and immediately switching to their reciprocals, we get

$$
\begin{aligned}
2 x-3 y & =1 \\
10 z-3 y & =1 \\
3 y-8 z & =1
\end{aligned}
$$

If we solve for $z$ and $y$ from the last two equations, and then for $x$ from the first, we obtain the values found in the previous solution:

$$
x=5, \quad y=3 \quad \text { and } \quad z=1
$$

Szatmári Attila (Bp. V., Eötvös g. I. o. t.)