## Precalculus (6th Edition) Blitzer

The solution is $\left\{ \left( 2,2 \right),\left( -2,-2 \right),\left( 4,1 \right),\left( -4,-1 \right) \right\}$
The given equations are, ${{x}^{2}}+4{{y}^{2}}=20$ (I) And $xy=4$ (II) From equation (II): $y=\frac{4}{x}$ Substitute $y=\frac{4}{x}$ in equation (I): \begin{align} & {{x}^{2}}-4{{\left( \frac{4}{x} \right)}^{2}}=20 \\ & {{x}^{4}}-20{{x}^{2}}+64=0 \\ & \left( {{x}^{2}}-4 \right)\left( {{x}^{2}}-16 \right)=0 \\ & x=\pm 2,\pm 4\, \end{align} Substitute $x=2$ in equation (II): \begin{align} & 2y=4 \\ & y=2 \end{align} Substitute $x=-2$ in equation (II): \begin{align} & -2y=4 \\ & y=-2 \end{align} Substitute $x=4$ in equation (II): $y=1$ Substitute $x=-4$ in equation (II): $y=-1$ Thus, the solution is $\left\{ \left( 2,2 \right),\left( -2,-2 \right),\left( 4,1 \right),\left( -4,-1 \right) \right\}$.