Решите систему уравнений: 2-3x=2(1-y); 4(x+y)=x-1,5.

\[\left\{ \begin{matrix}
2 - 3x = 2 \cdot (1 - y) \\
4 \cdot (x + y) = x - 1,5 \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix}
2 - 3x = 2 - 2y\ \ \ \\
4x + 4y = x - 1,5 \\
\end{matrix} \right.\ \]

\[\left\{ \begin{matrix}
3x - 2y = 0\ \ \ \ \ \ \ \ \ \ \ (1) \\
3x + 4y = - 1,5\ \ \ \ (2) \\
\end{matrix} \right.\ \]

\[(1) + (2) \Longrightarrow 6y = - 1,5\]

\[\left\{ \begin{matrix}
6y = - 1,5\ \ \ \\
3x - 2y = 0 \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix}
y = - 0,25 \\
3x = 2y\ \ \ \ \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix}
y = - 0,25\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
x = \frac{2}{3} \cdot ( - 0,25) = \frac{2}{3} \cdot \left( - \frac{1}{4} \right) \\
\end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix}
y = - 0,25 \\
x = - \frac{1}{6}\text{\ \ \ \ } \\
\end{matrix} \right.\ \]

\[Ответ:\left( - \frac{1}{6}; - 0,25 \right).\]