Решите систему уравнений: x^2-3xy+y^2=-1; 8y^2-3xy=2.

\[\left\{ \begin{matrix} (2x - 5y)(x - 2y) = 0 \\ 8y^{2} - 3xy = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix}\text{\ \ \ \ } \right.\ \]

\[8y^{2} - 3 \cdot 2,5y \cdot y = 2\]

\[8y^{2} - 7,5y^{2} = 2\]

\[0,5y^{2} = 2\]

\[y^{2} = 4\]

\[y_{1} = 2 \Longrightarrow \ \ \ \ x_{1} = 2,5 \cdot 2 = 5.\]

\[y_{2} = - 2 \Longrightarrow \ \ \ x_{2} = 2,5 \cdot ( - 2) =\]

\[= - 5.\]

\[8y^{2} - 3 \cdot 2y \cdot y = 2\]

\[8y^{2} - 6y^{2} = 2\]

\[2y^{2} = 2\]

\[y^{2} = 1\]

\[y_{3} = 1 \Longrightarrow x_{3} = 2 \cdot 1 = 2.\]

\[y_{4} = - 1 \Longrightarrow x_{4} = 2 \cdot ( - 1) = - 2.\]

\[Ответ:(5;2),\ \ \ ( - 5;\ - 2),\ \ \ (2;\ 1),\ \ \]

\[\ ( - 2;\ - 1).\]