Решите систему уравнений: x^2-3x-2y=4; x^2+x-3y=18.

Ответ:

\[2x^{2} + 2x - 3x^{2} + 9x + 12 - 36 =\]

\[= 0\]

\[- x^{2} + 11x - 24 = 0\]

\[x^{2} - 11x + 24 = 0\]

\[D = 121 - 96 = 25\]

\[x_{1} = \frac{11 + 5}{2} = 8;\ \ \ \]

\[x_{2} = \frac{11 - 5}{2} = 3.\]

\[y_{1} = \frac{64 - 24 - 4}{2} = \frac{36}{2} = 18;\]

\[y_{2} = \frac{9 - 9 - 4}{2} = - 2.\]

\[Ответ:(8;18);\ \ (3; - 2).\]