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

\[(1) + (2):\ \ \ 2x^{2} - 9xy + 9y^{2} = 0\]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }(2x - 3y)(x - 3y) = 0\]

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

\[y^{2} - 1,5y \cdot y = - 2\]

\[y^{2} - 1,5y^{2} = - 2\]

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

\[y^{2} = 4\]

\[y_{1} = 2;\ \ \ \ y_{2} = - 2.\]

\[y_{1} = 2 \Longrightarrow \ \ \ \ \ \ \ \ \ x_{1} = 1,5 \cdot 2 = 3.\]

\[y_{2} = - 2 \Longrightarrow \ \]

\[\Longrightarrow \ x_{2} = 1,5 \cdot ( - 2) = - 3.\]

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

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

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

\[y^{2} = 1\]

\[y_{3} = 1;\ \ \ y_{4} = - 1.\]

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

\[y_{4} = - 1 \Longrightarrow \ \ \ \ \ \ \]

\[\Longrightarrow x_{4} = 3 \bullet ( - 1) = - 3.\]

\[Ответ:\ \ (3;2),\ \ \ ( - 3;\ - 2),\ \ \ (3;1),\ \]

\[\text{\ \ }( - 3;\ - 1).\]