Решить систему: x+2y^2=8; x-y=7.

\[\left\{ \begin{matrix}
x + 2y^{2} = 8 \\
x - y = 7\ \ \ \ \ \\
\end{matrix} \right.\ \]

\[\left\{ \begin{matrix}
x = 7 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \\
7 + y + 2y^{2} = 8 \\
\end{matrix} \right.\ \]

\[2y^{2} + y - 1 = 0\]

\[D = 1 + 8 = 9\]

\[y_{1} = \frac{- 1 + 3}{4} = \frac{2}{4} = 0,5;\]

\[y_{2} = \frac{- 1 - 3}{4} = - 1;\]

\[x_{1} = 7 + 0,5 = 7,5;\]

\[x_{2} = 7 - 1 = 6.\]

\[Ответ:(7,5;0,5);\ \ (6; - 1).\]