Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 25

\[\boxed{\mathbf{25}.}\]

\[1)\ \left\{ \begin{matrix} 2x + 5y = 28 \\ 5x + y = 1\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} y = 1 - 5x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x + 5 \cdot (1 - 5x) = 28 \\ \end{matrix} \right.\ \]

\[2x + 5 - 25x = 28\]

\[- 23x = 23\]

\[x = - 1.\]

\[\left\{ \begin{matrix} x = - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y = 1 - 5 \cdot ( - 1) \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = - 1 \\ y = 6\ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:( - 1;6).\]

\[2)\ \left\{ \begin{matrix} 3x - y = - 1\ \\ 2x - 3y = 11 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} y = 3x + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x - 3 \cdot (3x + 1) = 11 \\ \end{matrix} \right.\ \]

\[2x - 9x - 3 = 11\]

\[- 7x = 14\]

\[x = - 2.\]

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

\[Ответ:( - 2;\ - 5).\]