Решите систему уравнений 16x^2+8xy+y^2=36; 3x-y=8.

Ответ:

\[\left\{ \begin{matrix} 16x^{2} + 8xy + y^{2} = 36 \\ 3x - y = 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = 3x - 8\ \ \ \ \ \ \ \ \\ (4x + y)^{2} = 36 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = 3x - 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (4x + 3x - 8)^{2} = 36 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = 3x - 8\ \ \ \ \ \ \ \\ (7x - 8)^{2} = 36 \\ \end{matrix} \right.\ \]

\[7x - 8 = 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ 7x - 8 = - 6\]

\[7x = 14\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 7x = 2\]

\[x = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = \frac{2}{7}\]

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

\[Ответ:(2;\ - 2);\left( \frac{2}{7};\ - 7\frac{1}{7} \right).\]