Answer
The solution is $\left\{ \left( 2,1 \right),\left( 2,-1 \right)\left( -2,1 \right)\text{ and }\left( -2,-1 \right) \right\}$.
Work Step by Step
Let us consider the given equation:
$3{{x}^{2}}+4{{y}^{2}}=16$ …… (I)
$2{{x}^{2}}-3{{y}^{2}}=5$ …… (II)
By multiplying equation (I) by $3$, equation (II) by $4$ and add equation (I) and equation (II): we get,
$\begin{align}
& 9{{x}^{2}}+12{{y}^{2}}=48 \\
& \underline{4{{x}^{2}}-12{{y}^{2}}=20} \\
& 17{{x}^{2}}=64 \\
& {{x}^{2}}=4 \\
& x=\pm 2 \\
\end{align}$
Substitute $ x=2$ in equation (I):
$\begin{align}
& 3{{\left( 2 \right)}^{2}}+4{{y}^{2}}=16 \\
& {{y}^{2}}=2 \\
& y=\pm 1
\end{align}$
Substitute $ x=-2$ in equation (II):
$\begin{align}
& 3{{\left( -2 \right)}^{2}}+4{{y}^{2}}=16 \\
& {{y}^{2}}=2 \\
& y=\pm 1
\end{align}$
Thus, the solution is $\left\{ \left( 2,1 \right),\left( 2,-1 \right)\left( -2,1 \right)\text{ and }\left( -2,-1 \right) \right\}$.
