Решите систему уравнений: 2x-3(2y+1)=15; 3(x+1)+3y=2y-2.

\[\left\{ \begin{matrix}
2x - 3 \cdot (2y + 1) = 15\ \ \ \ \ \\
3 \cdot (x + 1) + 3y = 2y - 2 \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix}
2x - 6y - 3 = 15\ \ \ \ \ \ \ \ \\
3x + 3 + 3y = 2y - 2 \\
\end{matrix} \right.\ \]

\[\left\{ \begin{matrix}
2x - 6y = 18\ \ \ \ |\ :2 \\
3x + y = - 5\ \ \ \ \ \ \ \ \ \ \ \ \ \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix}
x - 3y = 9\ \ \ \ | \cdot ( - 3) \\
3x + y = - 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
\end{matrix} \right.\ \]

\[\left\{ \begin{matrix}
- 3x + 9y = - 27\ \ \ (1) \\
3x + y = - 5\ \ \ \ \ \ \ \ \ \ \ (2) \\
\end{matrix} \right.\ \]

\[(1) + (2) \Longrightarrow 10y = - 32\]

\[\left\{ \begin{matrix}
10y = - 32 \\
x - 3y = 9 \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix}
y = - 3,2\ \ \ \ \\
x = 9 + 3y \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix}
y = - 3,2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
x = 9 + 3 \cdot ( - 3,2) \\
\end{matrix}\ \right.\ \text{\ \ }\]

\[\left\{ \begin{matrix}
y = - 3,2 \\
x = - 0,6 \\
\end{matrix} \right.\ \]

\[Ответ:( - 0,6;\ - 3,2).\]