## Precalculus (6th Edition) Blitzer

The solution set is $\left\{ \left( 2,\sqrt{3} \right),\left( 2,-\sqrt{3} \right),\left( -2,\sqrt{3} \right),\left( -2.-\sqrt{3} \right) \right\}$.
Let us consider the provided system: \begin{align} & {{x}^{2}}+4{{y}^{2}}=16 \\ & {{x}^{2}}-{{y}^{2}}=1 \end{align} And subtract the second equation form the first to obtain, \begin{align} & {{x}^{2}}+4{{y}^{2}}=16 \\ & {{x}^{2}}-{{y}^{2}}=1 \\ & \underline{-+-} \\ \end{align} \begin{align} & 5{{y}^{2}}=16 \\ & {{y}^{2}}=3 \\ & y=\pm \sqrt{3} \end{align} Substitute $y=\sqrt{3}$ in the second equation and solve for x: \begin{align} & {{x}^{2}}-{{\left( \sqrt{3} \right)}^{2}}=1 \\ & {{x}^{2}}=4 \\ & x=\pm 2 \end{align} Substitute $y=-\sqrt{3}$ in the second equation and solve for x: \begin{align} & {{x}^{2}}+{{\left( -\sqrt{3} \right)}^{2}}=1 \\ & {{x}^{2}}=4 \\ & x=\pm 2 \end{align} Thus, the solution set is $\left\{ \left( 2,\sqrt{3} \right),\left( 2,-\sqrt{3} \right),\left( -2,\sqrt{3} \right),\left( -2.-\sqrt{3} \right) \right\}$.