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

Ответ:

\[\left\{ \begin{matrix} 2x + 3y = 3\ |x2 \\ 3y^{2} - 4x = 18 \\ \end{matrix}\text{\ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} 4x + 6y = 6 \\ - 4x + 3y^{2} = 18 \\ \end{matrix}\text{\ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} 3y62 + 6y - 24 = 0\ |\ :3 \\ x = \frac{3 - 3y}{2} \\ \end{matrix} \right.\ \]

\[y^{2} + 2y - 8 = 0\]

\[y_{1} + y_{2} - 2,\ \ \ \ \ \ \ y_{1} = - 4\]

\[y_{1}y_{2} = - 8,\ \ \ \ \ y_{2} = 2\]

\[\left\{ \begin{matrix} x = 7,5 \\ y = - 4 \\ \end{matrix}\ \ \ \ \ или\ \ \right.\ \left\{ \begin{matrix} x = - 1,5 \\ y = 2 \\ \end{matrix} \right.\ \]

\[Ответ:(7,5;\ - 4),\ \ ( - 1,5;2).\]