Решите систему уравнений: xy=-10; x^2+y^2=29.

\[(1) + (2):\ \ \ \ {\text{\ \ }(x + y)}^{2} = 9\]

\[3y - y^{2} + 10 = 0\]

\[y^{2} - 3y - 10 = 0\]

\[D = ( - 3)^{2} - 4 \cdot 1 \cdot ( - 10) =\]

\[= 9 + 40 = 49;\ \ \ \sqrt{D} = 7.\]

\[y_{1} = \frac{3 + 7}{2} = \frac{10}{2} = 5;\ \ \ \]

\[\text{\ \ }y_{2} = \frac{3 - 7}{2} = \frac{- 4}{2} = - 2\]

\[x_{1} = 3 - 5 = - 2;\ \ \ \ \ \ \ \]

\[\text{\ \ \ \ \ }x_{2} = 3 - ( - 2) = 3 + 2 = 5.\]

\[- y^{2} - 3y + 10 = 0\]

\[y^{2} + 3y - 10 = 0\]

\[D = 3^{2} - 4 \cdot 1 \cdot ( - 10) =\]

\[= 9 + 40 = 49;\ \ \ \ \ \sqrt{D} = 7.\]

\[y_{1} = \frac{- 3 + 7}{2} = \frac{4}{2} = 2;\ \ \ \ \ \]

\[\text{\ \ \ \ }y_{2} = \frac{- 3 - 7}{2} = \frac{- 10}{2} = - 5\]

\[x_{1} = - 2 - 3 = - 5;\ \ \ \ \ \ \ \ \]

\[\text{\ \ \ \ }x_{2} = - ( - 5) - 3 = 5 - 3 = 2\]

\[Ответ:( - 2;5);\ \ (5;2);\ \ ( - 5;2);\ \ \]

\[(2;\ - 5).\]