Решите систему уравнений 3x^2-2y=14; 2x+y=3.

Ответ:

\[\left\{ \begin{matrix} 3x^{2} - 2y = 14 \\ 2x + y = 3\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \ \]

\[\left\{ \begin{matrix} y = 3 - 2x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3x^{2} - 2(3 - 2x) = 14 \\ \end{matrix} \right.\ \]

\[3x^{2} - 6 + 4x - 14 = 0\]

\[3x^{2} + 4x - 20 = 0\]

\[D = 4 + 60 = 64\]

\[x_{1} = \frac{- 2 + 8}{3} = 2;\]

\[x_{2} = \frac{- 2 - 8}{3} = - \frac{10}{3} = - 3\frac{1}{3}\]

\[\left\{ \begin{matrix} x = 2\ \ \ \\ y = - 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = - 3\frac{1}{3} \\ y = 9\frac{2}{3}\text{\ \ \ } \\ \end{matrix} \right.\ \]

\[Ответ:(2;\ - 1)\ и\ \left( - 3\frac{1}{3};9\frac{2}{3} \right).\]