Решите систему уравнений: x^2+y^2=10; xy=-3.

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

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x + y = 2\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x + y = - 2\]

\[1)\ \left\{ \begin{matrix} x + y = 2 \\ xy = - 3\ \\ \end{matrix}\text{\ \ \ \ \ } \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} x = 2 - y \\ xy = - 3\ \ \\ \end{matrix} \right.\ \]

\[(2 - y)y = - 3\]

\[2y - y^{2} = - 3\]

\[y^{2} - 2y - 3 = 0\]

\[D = ( - 2)^{2} - 4 \cdot 1 \cdot ( - 3) =\]

\[= 4 + 12 = 16\]

\[y_{1} = \frac{2 + \sqrt{16}}{2} = \frac{2 + 4}{2} = \frac{6}{2} = 3\]

\[y_{2} = \frac{2 - \sqrt{16}}{2} = \frac{2 - 4}{2} = \frac{- 2}{2} =\]

\[= - 1\]

\[y_{1} = 3 \Longrightarrow \text{\ \ \ \ \ }x_{1} = 2 - 3 = - 1.\]

\[y_{2} = - 1 \Longrightarrow \text{\ \ }x_{2} = 2 - ( - 1) =\]

\[= 2 + 1 = 3.\]

\[( - 2 - y)y = - 3\]

\[- 2y - y^{2} = - 3\]

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

\[D = 2^{2} - 4 \cdot 1 \cdot ( - 3) =\]

\[= 4 + 12 = 16\]

\[y_{3} = \frac{- 2 + \sqrt{16}}{2} = \frac{- 2 + 4}{2} = \frac{2}{2} =\]

\[= 1\]

\[y_{4} = \frac{- 2 - \sqrt{16}}{2} = \frac{- 2 - 4}{2} =\]

\[= \frac{- 6}{2} = - 3\]

\[y_{3} = 1 \Longrightarrow \text{\ \ \ \ \ }x_{3} = - 2 - 1 = - 3.\]

\[y_{4} = - 3 \Longrightarrow \text{\ \ \ \ }x_{4} = - 2 - ( - 3) =\]

\[= - 2 + 3 = 1.\]

\[Ответ:( - 1;3),\ (3;\ - 1),\ ( - 3;1),\ \]

\[(1;\ - 3).\]