Answer
The values of x, y are $\left( 3,2 \right),\left( -3,2 \right),\left( 3,-2 \right),\left( -3,-2 \right)$
Work Step by Step
$4{{x}^{2}}+3{{y}^{2}}=48$ (I)
$3{{x}^{2}}+2{{y}^{2}}=35$ (II)
And subtract $3$ times (I) from $4$ times (II), to get:
$\begin{align}
& 12{{x}^{2}}+8{{y}^{2}}-12{{x}^{2}}-9{{y}^{2}}=140-144 \\
& -{{y}^{2}}=-4 \\
& y=\sqrt{4} \\
& y=\pm 2
\end{align}$
To get ${{x}^{2}}$ substitute the value of ${{y}^{2}}=4$ in (I):
$\begin{align}
& 4{{x}^{2}}+3\left( 4 \right)=48 \\
& 4{{x}^{2}}=48-12 \\
& {{x}^{2}}=\frac{36}{4} \\
& {{x}^{2}}=9
\end{align}$
Therefore,
$ x=\pm 3$
So, the pairs of $\left( x,y \right)$ are as follows:
$\left( 3,2 \right),\left( -3,2 \right),\left( 3,-2 \right),\left( -3,-2 \right)$
Thus, the values of x, y are $\left( 3,2 \right),\left( -3,2 \right),\left( 3,-2 \right),\left( -3,-2 \right)$.
