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

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

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

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

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

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

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

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

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

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