Решите систему уравнений: x^2+y^2-2xy=36; x+y=-4.

\[\left\{ \begin{matrix} x^{2} + y^{2} - 2xy = 36 \\ x + y = - 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} (x - y)^{2} = 36 \\ x + y = - 4\ \ \ \\ \end{matrix}\text{\ \ \ \ \ } \right.\ \]

\[1)\ \left\{ \begin{matrix} x - y = - 6 \\ x + y = - 4 \\ \end{matrix} \right.\ \ ( + )\text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} 2x = 2\ \ \ \ \ \ \ \ \\ y = - 4 - x \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 1\ \ \ \ \\ y = - 5 \\ \end{matrix} \right.\ \right.\ \]

\[2)\ \left\{ \begin{matrix} x - y = - 6 \\ x + y = - 4 \\ \end{matrix}\text{\ \ }( + ) \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} 2x = - 10\ \ \ \\ y = - 4 - x \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ } \right.\ \left\{ \begin{matrix} x = - 5 \\ y = 1\ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:\ \ (1;\ - 5),\ ( - 5;1).\]