1. Note that when $a=0$, the system is undefined. Let the solution of the system be of the form

$\left(x_{0} ; x_{0}\right)$. Then we get $\left\{\begin{array}{l}3 x_{0}+2 x_{0}=15 a \\ \frac{1}{a} x_{0}+x_{0}=9\end{array}\right.$, which is $\left\{\begin{array}{l}5 x_{0}=15 a \\ \frac{1+a}{a} \cdot x_{0}=9\end{array}\right.$, from which $x_{0}=3 a$.

Transform the second equation of the system by substitution. We get $\frac{1+a}{a} \cdot 3a=9$. Solving the equation with the restriction $a$ not equal to 0, we get that $a=2$. In this case, the solution of the system is the point $(6 ; 6)$
Answer: when $a=2$ the solution to the system is $(6 ; 6)$.

| Points | Conditions for Awarding |
| :--- | :--- |
| 15 points | The problem is completely correctly solved. |
| 10 points | The value of the parameter is correctly found, but an error is made in calculating the coordinates of the point - the solution to the system. |
| 5 points | The domain of the parameter is correctly found. |
| 0 points | In all other cases. |