Task B-2.6. Solve the following system of equations in the set $\mathbb{C}$

$$
\begin{aligned}
& (x+2 y)^{2}-x-2 y-12=0 \\
& x \cdot y=2
\end{aligned}
$$
Solution. Let $x+2 y=t$.

Then $t^{2}-t-12=0, t_{1}=-3, t_{2}=4$.
a) $x+2 y=-3, x \cdot y=2$.

We reduce the system to the quadratic equation $2 y^{2}+3 y+2=0$,

(1 point)

whose solutions are $y_{1,2}=\frac{-3 \pm \sqrt{7} i}{4}$.

(1 point)

Then the solutions of the system are the ordered pairs $(x, y)$:

$$
\left(\frac{-3+i \sqrt{7}}{2}, \frac{-3-i \sqrt{7}}{4}\right),\left(\frac{-3-i \sqrt{7}}{2}, \frac{-3+i \sqrt{7}}{4}\right)
$$

b) $x+2 y=4, x \cdot y=2$.

We reduce the system to the quadratic equation $y^{2}-2 y+1=0$,

which has the only solution $y=1$.

The solution of this system is the ordered pair $(x, y)=(2,1)$.