Решите систему уравнений y^2+2x=2; x+y=1.

\[\left\{ \begin{matrix} y^{2} + 2x = 2 \\ x + y = 1\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 1 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + 2 \cdot (1 - y) = 2 \\ \end{matrix} \right.\ \]

\[y^{2} + 2 - 2y - 2 = 0\]

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

\[y(y - 2) = 0\]

\[y = 0;\ \ y = 2.\]

\[\left\{ \begin{matrix} y = 0 \\ x = 1 \\ \end{matrix} \right.\ \ \ \ \ \ или\ \ \ \left\{ \begin{matrix} y = 2\ \ \ \\ x = - 1 \\ \end{matrix} \right.\ \]

\[Ответ:(1;0);( - 1;2).\]