346. Solve the system of equations:

$$
\left\{\begin{array}{l}
(x+y)^{2}-8(x+y)=33 \\
(x-y)^{2}+2(x-y)=80
\end{array}\right.
$$
$\triangle$ Let's introduce the substitutions:

$$
x+y=z, \quad x-y=t .
$$

We have the system of equations:

$$
z^{2}-8 z-33=0, \quad t^{2}+2 t-80=0
$$

Let's solve each of the quadratic equations in this system:

$$
z_{1}=11, \quad z_{2}=-3 ; \quad t_{1}=8, \quad t_{2}=-10
$$

Now, to find $x$ and $y$, we need to combine each value of $z$ with each value of $t$. This results in four systems of linear equations. Complete the solution on your own.

Answer: ( $11 / 2 ; 3 / 2$), ( $21 / 2 ; 1 / 2$), ( $5 / 2 ;-11 / 2$), ( $-13 / 2 ; 7 / 2$).