Precalculus (6th Edition) Blitzer

The values are $x=\ 5\ \text{m and }y=\text{ 2}\ \text{m}$.
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 $x=\ 5\ \text{m and }y=\text{ 2}\ \text{m}$.