Answer
The solution of the system is $\underline{\left\{ \left( 3,2 \right),\left( -3,2 \right),\left( 3,-2 \right),\left( -3,-2 \right) \right\}}$
Work Step by Step
Let us consider the system of the given equations:
$2{{x}^{2}}-5{{y}^{2}}=-2$ (I)
$3{{x}^{2}}+2{{y}^{2}}=35$ (II)
Rewrite equation (I) as given below:
$\begin{align}
& 2{{x}^{2}}-5{{y}^{2}}=-2 \\
& 2{{x}^{2}}=5{{y}^{2}}-2 \\
& {{x}^{2}}=\frac{5{{y}^{2}}-2}{2}
\end{align}$
Put the value of x in equation (II) to obtain the value of y as follows:
$\begin{align}
& 3\left( \frac{5{{y}^{2}}-2}{2} \right)+2{{y}^{2}}=35 \\
& 15{{y}^{2}}-6+4{{y}^{2}}=70 \\
& 19{{y}^{2}}=76 \\
& {{y}^{2}}=4
\end{align}$
$ y=\pm 2$
Put the value of ${{y}^{2}}$ in equation (II) to obtain the value of x as follows:
$\begin{align}
& 3{{x}^{2}}+2\left( 4 \right)=35 \\
& 3{{x}^{2}}+8=35 \\
& 3{{x}^{2}}=27 \\
& {{x}^{2}}=9
\end{align}$
$ x=\pm 3$
Thus, the solutions of the equations are $\left( x,y \right)=\left\{ \left( 3,2 \right),\left( -3,2 \right),\left( 3,-2 \right),\left( -3,-2 \right) \right\}$.
