Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1421

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\[1)\ \left\{ \begin{matrix} 5x - 7y = 3\ \ \\ 6x + 5y = 17 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} 5x = 3 + 7y\ \ \\ 6x = 17 - 5y \\ \end{matrix} \right.\ \text{\ \ }\]

\[\left\{ \begin{matrix} x = \frac{3}{5} + \frac{7}{5}\text{y\ \ } \\ x = \frac{17}{6} - \frac{5}{6}y \\ \end{matrix} \right.\ \]

\[\frac{3}{5} + \frac{7}{5}y = \frac{17}{6} - \frac{5}{6}y\ \ \ \ \ | \bullet 30\]

\[18 + 42y = 85 - 25y\]

\[67y = 67\]

\[y = 1;\]

\[x = \frac{3}{5} + \frac{7}{5} = \frac{10}{5} = 2.\]

\[Ответ:\ \ (2;\ 1).\]

\[2)\ \left\{ \begin{matrix} 2x - y - 13 = 0 \\ x + 2y + 1 = 0\ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} 2x = 13 + y\ \\ x = - 1 - 2y \\ \end{matrix} \right.\ \text{\ \ \ }\]

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

\[6,5 + 0,5y = - 1 - 2y\]

\[2,5y = - 7,5\]

\[y = - 3;\]

\[x = - 1 - 2 \bullet ( - 3) = - 1 + 6 = 5.\]

\[Ответ:\ \ (5;\ - 3).\]