Решите систему уравнений:3x-2(3y+1)=-2; 2(x+1)-1=3y-1.

\[\left\{ \begin{matrix}
3x - 2 \cdot (3y + 1) = - 2\ \ \\
2 \cdot (x + 1) - 1 = 3y - 1 \\
\end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix}
3x - 6y - 2 = - 2\ \ \ \ \\
2x + 2 - 1 = 3y - 1 \\
\end{matrix} \right.\ \]

\[\left\{ \begin{matrix}
3x - 6y = 0\ \ \ |\ :3 \\
2x - 3y = - 1 - 1 \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix}
x - 2y = 0\ \ \ | \cdot ( - 2) \\
2x - 3y = - 2\ \ \ \ \ \ \ \ \ \ \ \ \\
\end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix}
- 2x + 4y = 0\ \ (1) \\
2x - 3y = - 2\ \ \ (2) \\
\end{matrix} \right.\ \]

\[y = - 2\]

\[\left\{ \begin{matrix}
y = - 2\ \ \ \ \ \ \ \\
x - 2y = 0 \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix}
y = - 2 \\
x = 2y \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix}
y = - 2\ \ \ \ \ \ \ \ \ \ \\
x = 2 \cdot ( - 2) \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix}
y = - 2 \\
x = - 4 \\
\end{matrix} \right.\ \]

\[Ответ:( - 4; - 2).\]