1. In the field of real numbers, solve the system of equations

$$
\begin{aligned}
a x+y & =2 \\
x-y & =2 a \\
x+y & =1
\end{aligned}
$$

for the unknowns $x$ and $y$ and the real parameter $a$.
1. By adding the second and third equations, we get $2 x=2 a+1$, and by subtracting the second equation from the third, we get $2 y=-2 a+1$. From this, we can express

$$
x=a+\frac{1}{2}, \quad y=-a+\frac{1}{2}
$$

and substitute these into the first equation of the original system. After simplification, we obtain the quadratic equation

$$
a^{2}-\frac{1}{2} a-\frac{3}{2}=0,
$$

which has roots $a_{1}=-1$ and $a_{2}=\frac{3}{2}$. For each of these two (and only possible) values of the parameter $a$, we can easily determine the unknowns $x$ and $y$ by substituting into formulas (1).

The given system of equations has solutions only for two values of the parameter $a$, namely for $a=-1$, when its only solution is $(x, y)=\left(-\frac{1}{2}, \frac{3}{2}\right)$, and for $a=\frac{3}{2}$, when $(x, y)=$ $=(2,-1)$.

Checking by substitution is straightforward and can be omitted with the following justification: The system of two equations that we obtained (and solved) by adding and subtracting the second and third equations is equivalent to this pair of original equations. The remaining (first) equation of the system is then equivalent to the quadratic equation (2), whose solution we used to find the possible values of the parameter $a$.

For a complete solution, award 6 points, of which 2 points for correctly expressing $x$ and $y$ from the second and third equations, 2 points for solving the quadratic equation that arises from substituting these values into the first equation, and 1 point each for the correct answer and verification. Deduct no more than 1 point for numerical errors in the calculation.