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

\[(1) + (2):\ \ \ \ {\ (x + y)}^{2} = 16\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x + y = 4\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x + y = - 4\]

\[1)\ \left\{ \begin{matrix} x + y = 4 \\ 2xy = 6\ \ \ \\ \end{matrix}\text{\ \ \ \ \ } \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} x = 4 - y \\ 2xy = 6\ \ \\ \end{matrix} \right.\ \]

\[2 \bullet (4 - y)y = 6\]

\[8y - 2y^{2} = 6\]

\[2y^{2} - 8y + 6 = 0\ \ \ \ \ \ \ |\ :2\]

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

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

\[= 16 - 12 = 4\]

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

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

\[y_{1} = 3 \Longrightarrow x_{1} = 4 - 3 = 1.\]

\[y_{2} = 1 \Longrightarrow x_{2} = 4 - 1 = 3.\]

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

\[2 \bullet ( - 4 - y) \bullet y = 6\]

\[- 8y - 2y^{2} = 6\]

\[2y^{2} + 8y + 6 = 0\ \ \ \ \ \ \ \ \ \ |\ :2\]

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

\[D = 4^{2} - 4 \cdot 1 \cdot 3 = 16 - 12 = 4\]

\[y_{1} = \frac{- 4 + \sqrt{4}}{2} = \frac{- 4 + 2}{2} =\]

\[= \frac{- 2}{2} = - 1\]

\[y_{2} = \frac{- 4 - \sqrt{4}}{2} = \frac{- 4 - 2}{2} = \frac{- 6}{2} =\]

\[= - 3\]

\[y_{1} = - 1 \Longrightarrow \ \ \]

\[\Longrightarrow x_{1} = - 4 - ( - 1) = - 4 + 1 =\]

\[= - 3.\]

\[y_{2} = - 3 \Longrightarrow \ \ \ \ \ \ \]

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

\[= - 1.\]

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

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