Answer
The solution is $\left\{ \left( 2,2 \right),\left( 4,1 \right) \right\}$
Work Step by Step
The given equations are,
$\begin{align}
& ~{{x}^{2}}+4{{y}^{2}}=20~~~~~~~~~~~~~~~~~~\left( \text{I} \right) \\
& x+2y=6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right)
\end{align}$
From equation (II):
$ x=6-2y $
Substitute $ x=6-2y $ in equation (I):
$\begin{align}
& {{\left( 6-2y \right)}^{2}}+4{{y}^{2}}=20 \\
& 8{{y}^{2}}-24{{y}^{{}}}+16=0 \\
& \left( y-2 \right)\left( y-1 \right)=0 \\
& y=1,2\,
\end{align}$
Substitute $ y=2$ in equation (II):
$\begin{align}
& x+2\left( 2 \right)=6 \\
& x=2
\end{align}$
Substitute $ x=1$ in equation (II):
$\begin{align}
& x+2\left( 1 \right)=6 \\
& x=4
\end{align}$
Thus, the solution is $\left\{ \left( 2,2 \right),\left( 4,1 \right) \right\}$.
