Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 26

\[\boxed{\mathbf{26}.}\]

\[1)\ \left\{ \begin{matrix} 3x - 8y = - 9\ \ \ \ \ \ \ \ \ \ \ \ \\ - 5x + 2y = 19\ \ \ | \cdot 4 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} 3x - 8y = - 9\ \ \ \ \\ - 20x + 8y = 76 \\ \end{matrix} \right.\ ( + )\]

\[- 17x = 67\]

\[x = - \frac{67}{17}.\]

\[- 5x + 2y = 19\]

\[2y = 19 + 5x\]

\[2y = 19 + 5 \cdot \left( - \frac{67}{17} \right)\text{\ \ }\]

\[2y = 19 - \frac{335}{17}\]

\[2y = 19 - 19\frac{12}{17}\]

\[2y = - \frac{12}{17}\]

\[y = - \frac{6}{17}.\]

\[\left\{ \begin{matrix} x = - \frac{67}{17} \\ y = - \frac{6}{17} \\ \end{matrix} \right.\ \]

\[Ответ:\left( - \frac{67}{17};\ - \frac{6}{17} \right).\]

\[2)\ \left\{ \begin{matrix} - 4x + 6y = 1\ \ | \cdot 3 \\ 3x - 8y = - 6\ \ | \cdot 4 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} - 12x + 18y = 3\ \ \\ 12x - 32y = - 24 \\ \end{matrix} \right.\ ( + )\]

\[- 14y = - 21\]

\[y = \frac{21}{14} = \frac{3}{2} = 1,5.\]

\[3x - 8y = - 6\]

\[3x = 8y - 6\]

\[3x = 8 \cdot \frac{3}{2} - 6\]

\[3x = 12 - 6\]

\[3x = 6\]

\[x = 2.\]

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

\[Ответ:(2;1,5).\]