Answer
When solving ${{x}^{2}}+4{{y}^{2}}=20\text{ and }xy=4$ by the substitution method, one can eliminate $y$ by solving the second equation for $y$ to obtain $y=\frac{4}{x}$; then we substitute $\frac{4}{x}$ for $y$ in the first equation.
Work Step by Step
Let us consider the provided system:
$\begin{align}
& {{x}^{2}}+4{{y}^{2}}=20 \\
& xy=4
\end{align}$
So, from the second equation:
$y=\frac{4}{x}$
Substitute $y=\frac{4}{x}$ in the first equation to obtain,
$\begin{align}
& {{x}^{2}}+4{{\left( \frac{4}{x} \right)}^{2}}=20 \\
& {{\left( {{x}^{2}} \right)}^{2}}+64=2{{x}^{2}} \\
& {{\left( {{x}^{2}} \right)}^{2}}-2{{x}^{2}}+64=0
\end{align}$
Thus, $y=\frac{4}{x}$ was obtained from the second equation and the quadratic expression was obtained by substituting $\frac{4}{x}$ for $y$ in the first equation.
