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

Ответ:

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

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

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

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

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

\[x_{1} = 0;\ \ \ x_{2} = 2;\]

\[y_{1} = - 1 - 0 = - 1;\]

\[y_{2} = - 1 - 2 = - 3.\]

\[Ответ:(0; - 1);(2; - 3).\]