Решите систему уравнений: 2y^2-3x^2=1; 3x^2+2y^2=19.

\[\left\{ \begin{matrix} 2y^{2} - 3x^{2} = 1\ \ \ \\ 3x^{2} + 2y^{2} = 19 \\ \end{matrix}\ \right.\ ( + )\text{\ \ \ }\]

\[\left\{ \begin{matrix} 4y^{2} = 20\ \ \ \ \ \ \ \ \ \ \ \\ 2y^{2} - 3x^{2} = 1 \\ \end{matrix}\text{\ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} y^{2} = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2y^{2} - 3x^{2} = 1 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = \pm \sqrt{5}\text{\ \ \ \ \ \ \ \ \ \ } \\ x^{2} = \frac{2y^{2} - 1}{3}\text{\ \ } \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} y = \pm \sqrt{5} \\ x^{2} = 3\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} y = \pm \sqrt{5} \\ x = \pm \sqrt{3} \\ \end{matrix} \right.\ \]

\[Ответ:\left( \sqrt{3};\sqrt{5} \right);\ \left( \sqrt{3};\ - \sqrt{5} \right);\ \]

\[\left( - \sqrt{3}\ ;\sqrt{5} \right);\ \ \left( - \sqrt{3};\ - \sqrt{5} \right).\]