Answer
The solution is $\left\{ \left( 2,2 \right),\left( -2,-2 \right),\left( 4,1 \right),\left( -4,-1 \right) \right\}$
Work Step by Step
The given equations are,
${{x}^{2}}+4{{y}^{2}}=20$ (I)
And
$ xy=4$ (II)
From equation (II):
$ y=\frac{4}{x}$
Substitute $ y=\frac{4}{x}$ in equation (I):
$\begin{align}
& {{x}^{2}}-4{{\left( \frac{4}{x} \right)}^{2}}=20 \\
& {{x}^{4}}-20{{x}^{2}}+64=0 \\
& \left( {{x}^{2}}-4 \right)\left( {{x}^{2}}-16 \right)=0 \\
& x=\pm 2,\pm 4\,
\end{align}$
Substitute $ x=2$ in equation (II):
$\begin{align}
& 2y=4 \\
& y=2
\end{align}$
Substitute $ x=-2$ in equation (II):
$\begin{align}
& -2y=4 \\
& y=-2
\end{align}$
Substitute $ x=4$ in equation (II):
$ y=1$
Substitute $ x=-4$ in equation (II):
$ y=-1$
Thus, the solution is $\left\{ \left( 2,2 \right),\left( -2,-2 \right),\left( 4,1 \right),\left( -4,-1 \right) \right\}$.
