## Precalculus (6th Edition) Blitzer

The solution is $\left\{ \left( 1,4 \right),\left( -1,-4 \right),\left( 2\sqrt{2},\sqrt{2} \right),\left( -2\sqrt{2},-\sqrt{2} \right) \right\}$
The given equations are, $2{{x}^{2}}+{{y}^{2}}=18$ (I) And $xy=4$ (II) From equation (II): $y=\frac{4}{x}$ Substitute $y=\frac{4}{x}$ in equation (I): \begin{align} & 2{{x}^{2}}-{{\left( \frac{4}{x} \right)}^{2}}=18 \\ & {{x}^{4}}-9{{x}^{2}}+8=0 \\ & \left( {{x}^{2}}-8 \right)\left( {{x}^{2}}-1 \right)=0 \\ & x=\pm 2\sqrt{2},\pm 1\, \end{align} Substitute $x=2\sqrt{2}$ in equation (II): \begin{align} & 2\sqrt{2}y=4 \\ & y=\sqrt{2} \end{align} Substitute $x=-2\sqrt{2}$ in equation (II): \begin{align} & -2\sqrt{2}y=4 \\ & y=-\sqrt{2} \end{align} Substitute $x=1$ in equation (II): $y=4$ Substitute $x=-1$ in equation (II): $y=-4$ Thus, the solution is, $\left\{ \left( 1,4 \right),\left( -1,-4 \right),\left( 2\sqrt{2},\sqrt{2} \right),\left( -2\sqrt{2},-\sqrt{2} \right) \right\}$.