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

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

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

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

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

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

\[y = - 2\]

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

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

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

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

\[Ответ:( - 4; - 2).\]