Answer
The values are $ x=\ 5\ \text{m and }y=\text{ 2}\ \text{m}$.
Work Step by Step
We start with:
$\begin{align}
& {{x}^{2}}-{{y}^{2}}=21\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\
& 4x+2y=24\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right)
\end{align}$
The substitution method is used to solve the system. Divide each term in equation (II) by 2 and then the value of $ y $ is,
$\begin{align}
& 2x+y=12 \\
& y=12-2x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{III} \right) \\
\end{align}$
Substitute the value of $ y=12-2x $ into equation (I). Then,
$\begin{align}
& {{x}^{2}}-{{\left( 12-2x \right)}^{2}}=21 \\
& {{x}^{2}}-\left( 144-48x+4{{x}^{2}} \right)=21 \\
& {{x}^{2}}-144+48x-4{{x}^{2}}-21=0 \\
& 3{{x}^{2}}+48x-165=0
\end{align}$
Now, further solve for $ x $.
Then,
$\begin{align}
& {{x}^{2}}-16x+55=0 \\
& \left( x-5 \right)\left( x-11 \right)=0 \\
& x=5,11
\end{align}$
If $ x=5$, then,
$\begin{align}
& y=12-2(5) \\
& =12-10 \\
& =2
\end{align}$
Therefore, the solution set is $\left\{ \left( 5,5 \right),\left( 2,2 \right) \right\}$
Hence, the solution set of the system is $\left\{ \left( 5,5 \right),\left( 2,2 \right) \right\}$. Hence, the value of $ x=\ 5\ \text{m and }y=\text{ 2}\ \text{m}$.
