Решите систему уравнений: 4x-y-24=2(5x-2y); 3y-2=4-(x-y).

\[\left\{ \begin{matrix} 4x - y - 24 = 2 \cdot (5x - 2y) \\ 3y - 2 = 4 - (x - y)\text{\ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 4x - y - 24 = 10x - 4y \\ 3y - 2 = 4 - x + y\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 4x - 10x - y + 4y = 24 \\ x + 3y - y = 4 + 2\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} - 6x + 3y = 24\ \ |\ :3 \\ x + 2y = 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\left\{ \begin{matrix} - 2x + y = 8\ \ \ \ \ \ \\ x + 2y = 6\ \ | \cdot 2 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} - 2x + y = 8\ \\ 2x + 4y = 12 \\ \end{matrix} \right.\ \Longrightarrow (1) + (2) \Longrightarrow 5y = 20,\ \ \]

\[y = 4\]

\[\left\{ \begin{matrix} y = 4\ \ \ \ \ \ \ \ \ \\ x = 6 - 2y \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} y = 4\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x = 6 - 2 \cdot 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} y = 4\ \ \ \\ x = - 2 \\ \end{matrix} \right.\ \]

\[Ответ:( - 2;4).\]