Решите систему уравнений: 3x-y=-10; x^2-4xy-y^2=-20.

\[- 20x^{2} - 100x - 80 = 0\ \ \ \ \ |\ :( - 20)\]

\[x^{2} + 5x + 4 = 0\]

\[D = 5^{2} - 4 \cdot 1 \cdot 4 = 25 - 16 =\]

\[= 9;\ \ \ \ \sqrt{D} = 3.\]

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

\[\text{\ \ \ \ }x_{2} = \frac{- 5 - 3}{2} = \frac{- 8}{2} = - 4\]

\[y_{1} = 3 \cdot ( - 1) + 10 = - 3 + 10 =\]

\[= 7;\]

\[y_{2} = 3 \cdot ( - 4) + 10 =\]

\[= - 12 + 10 = 2.\]

\[Ответ:( - 1;7);\ \ \ ( - 4;\ - 2).\]