Найдите решение системы уравнений x+y+xy=7; x^2+y^2=10.

\[\left\{ \begin{matrix}
x + y + xy = 7 \\
x^{2} + y^{2} = 10\ \ \ \ \\
\end{matrix} \right.\ \]

\[x^{2} + 2xy + y^{2} - 2xy = 10\]

\[(x + y)^{2} - 2xy = 10\]

\[\left\{ \begin{matrix}
(x + y)^{2} - 2xy = 10 \\
x + y = 7 - xy\ \ \ \ \ \ \ \ \ \ \ \\
\end{matrix} \right.\ \]

\[(7 - xy)^{2} - 2xy = 10\]

\[49 - 14xy + \left( \text{xy} \right)^{2} = 10\]

\[\left( \text{xy} \right)^{2} - 16xy + 39 = 0\]

\[xy = t:\]

\[t^{2} - 16t + 39 = 0\]

\[D = 64 - 39 = 25\]

\[t_{1} = 8 + 5 = 13;\]

\[t_{2} = 8 - 5 = 3\]

\[1)\ xy = 9:\]

\[y = \frac{13}{x}\]

\[x^{\backslash x} + \frac{13}{x} + 13^{\backslash x} = 7^{\backslash x}\]

\[x^{2} + 6x + 13 = 0\]

\[D = 9 - 13 < 0\]

\[нет\ корней.\]

\[2)\ xy = 3:\]

\[y = \frac{3}{x}\]

\[x + \frac{3}{x} + 3 = 7\]

\[x^{\backslash x} + \frac{3}{x} - 4^{\backslash x} = 0\ \ \ \ \]

\[x^{2} - 4x + 3 = 0\]

\[D = 4 - 3 = 1\]

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

\[x_{2} = 2 + 1 = 3\]

\[\left\{ \begin{matrix}
x = 1 \\
y = 3 \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix}
x = 3 \\
y = 1 \\
\end{matrix} \right.\ \]

\[Ответ:(1;3);\ \ (3;1).\]