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

Ответ:

\[- x^{2} + x = - 6\]

\[- x^{2} + x + 6 = 0\]

\[x^{2} - x - 6 = 0\]

\[x_{1} + x_{2} = 1;\ \ \ x_{1} \cdot x_{2} = - 6\]

\[x_{1} = 3:\]

\[y_{1} = \frac{4 - x^{2} - x}{2} = \frac{4 - 9 - 3}{2} =\]

\[= - 4;\]

\[x_{2} = - 2:\]

\[y_{2} = \frac{4 - x^{2} - x}{2} = \frac{4 - 4 + 2}{2} = 1.\]

\[Ответ:(3;\ - 4);( - 2;1).\]