Решите систему уравнений: 3(2a-5)+4(7-3b)=7, 2(4+b)-7(1+8a)=-53.

\[\left\{ \begin{matrix} 3 \cdot (2a - 5) + 4 \cdot (7 - 3b) = 7\ \ \ \\ 2 \cdot (4 + b) - 7 \cdot (1 + 8a) = - 53 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 6a - 15 + 28 - 12b = 7 \\ 8 + 2b - 7 - 56a = - 53 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 6a - 12b = - 6\ \ \ \ \ \ \ |\ :6 \\ - 56a + 2b = - 54\ |\ :2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} a - 2b = - 1\ \ \ | \cdot 28 \\ - 28a + b = - 27\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 28a - 56b = - 28 \\ - 28a + b = - 27\ \ \\ \end{matrix} + \right.\ \]

\[\left\{ \begin{matrix} - 55b = - 55 \\ a = \frac{- 1 + 2b}{1} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} b = 1 \\ a = 1 \\ \end{matrix} \right.\ \]

\[Ответ:(1;1).\]