Решите систему уравнений: x-5y=3; x^2-2xy-y^2=-1.

Ответ:

\[\left\{ \begin{matrix} x - 5y = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 2xy - y^{2} = - 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[14y² + 24y + 10 = 0\ \ \ |\ :2\]

\[7y^{2} + 12y + 5 = 0\]

\[D = 144 - 140 = 4\]

\[y_{1} = \frac{- 12 + 2}{14} = - \frac{5}{7},\ \ \]

\[y_{2} = \frac{- 12 - 2}{14} = - 1\]

\[\left\{ \begin{matrix} x = - 2 \\ y = - 1 \\ \end{matrix} \right.\ \ \ \ \ \ или\ \ \ \ \left\{ \begin{matrix} x = - \frac{4}{7} \\ y = - \frac{5}{7} \\ \end{matrix} \right.\ \]

\[Ответ:( - 2;\ - 1);\ \ \left( - \frac{4}{7};\ - \frac{5}{7} \right).\]