2. The system of equations
$$
\left\{\begin{array}{l}
x^{2}-y^{2}+x-2 y-6=0, \\
2 x y+2 x+y-4=0
\end{array}\right.
$$

has the real solutions $(x, y)$ as $\qquad$ .
2. $(2,0),(-3,-2)$.
(2) $\times \mathrm{i}+$ (1) gives
$$
\begin{array}{l}
(x+y \mathrm{i})^{2}+(1+2 \mathrm{i})(x+y \mathrm{i})-(6+4 \mathrm{i})=0 \\
\Rightarrow((x+y \mathrm{i})-2)((x+y \mathrm{i})+3+2 \mathrm{i})=0 \\
\Rightarrow x+y \mathrm{i}=2 \text { or }-3-2 \mathrm{i} .
\end{array}
$$

Since $x, y$ are real numbers, we have
$$
(x, y)=(2,0),(-3,-2) .
$$