$1^{0}$. Solve the following system of equations:

$$
\begin{gathered}
\frac{x}{2}-\frac{2 x-3 y}{5}=a-1 \ldots \\
x+3=\frac{y}{3} \ldots
\end{gathered}
$$

$2^{0}$. For which values of $a$ will $x$ be simultaneously greater than $0$?
$1^{0}$. By eliminating the fractions, we get:

$$
\begin{gathered}
x+6 y=10(a-1) \ldots \\
y=3 x+9 \ldots
\end{gathered}
$$

Substituting the expression for $y$ from $2 a)$ into $1 a)$:

$$
19 x=10 a-64, \text { hence } x=\frac{10 a-64}{19}
$$

and

$$
y=3 \frac{10 a-64}{19}+9, \text { that is } y=\frac{30 a-21}{19}
$$

$2^{0} . x0 \quad, \quad a>\frac{7}{10}$.

These two conditions can be satisfied simultaneously if

$$
0.7<a<6.4
$$

Hugo Martonfalvay (Rákóczi Ferenc r. int. 5th grade, Sopron)

Note. There is no restriction on the number $a$ - according to the problem - that it must be an integer!