Решите систему уравнений x+y=2; 2x^2+xy+y^2=16.

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

\[\left\{ \begin{matrix} y = 2 - x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x² + x(2 - x) + (2 - x)^{2} = 16 \\ \end{matrix} \right.\ \]

\[2x² + 2x - x^{2} + 4 - 4x + x^{2} - 16 = 0\ \ \ |\ :2\]

\[x^{2} - x - 6 = 0\]

\[x_{1} + x_{2} = 1;\ \ x_{1} \cdot x_{2} = - 6\]

\[x_{1} = 3;\ \ x_{2} = - 2\]

\[\left\{ \begin{matrix} x = 3\ \ \ \ \\ y = - 1 \\ \end{matrix} \right.\ или\ \left\{ \begin{matrix} x = - 2 \\ y = 4\ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:(3;\ - 1);\ \ ( - 2;4).\]