Решите систему способом подстановки: 3(x+y)+1=x+4y; 7-2(x-y)=x-8y.

\[\left\{ \begin{matrix} 3(x + y) + 1 = x + 4y \\ 7 - 2(x - y) = x - 8y \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 3x + 3y + 1 = x + 4y \\ 7 - 2x + 2y = x - 8y \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 2x - y = - 1\text{\ \ \ } \\ 3x - 10y = 7\ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = 2x + 1\ \text{\ \ }\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 3x - 10(2x + 1) = 7 \\ \end{matrix} \right.\ \]

\[3x - 20x - 10 = 7\]

\[- 17x = 17\]

\[x = - 1.\]

\[y = 2x + 1 = - 2 + 1 = - 1.\]

\[Ответ:( - 1;\ - 1).\]