Решите систему уравнений: x²-4y=8; y²+8x=-28.

\[\left\{ \begin{matrix} x^{2} - 4y = 8\ \ \ \ \ \ \ (1) \\ y^{2} + 8x = - 28\ \ (2) \\ \end{matrix} \right.\ \]

\[(1) + (2):\ \ x^{2} + 8x + y^{2} - 4y =\]

\[= - 20\]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }(x + 4)^{2} + (y - 2)^{2} = 0\]

\[\left\{ \begin{matrix} (x + 4)^{2} + (y - 2)^{2} = 0 \\ x^{2} - 4y = 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x = - 4;\ \ \ y = 2:\]

\[( - 4)^{2} - 4 \cdot 2 = 8\]

\[16 - 8 = 8\]

\[8 = 8 \Longrightarrow ( - 4;2) \Longrightarrow решение.\]

\[Ответ:( - 4;2).\]