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

Ответ:

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

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

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

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

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

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

\[\left\{ \begin{matrix} x = 0\ \ \ \\ y = - 1 \\ \end{matrix} \right.\ \ \ \ \ \ или\ \ \ \ \left\{ \begin{matrix} x = 2\ \ \ \ \\ y = - 3 \\ \end{matrix} \right.\ \]

\[Ответ:(0; - 1);(2; - 3).\]