## Precalculus (6th Edition) Blitzer

The solution is $\left\{ \left( -1,-\frac{1}{3} \right),\left( -1,\frac{1}{3} \right),\left( 1,\frac{-1}{3} \right),\left( 1,\frac{1}{3} \right) \right\}$
The provided equations are, \begin{align} & \frac{2}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}=11\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\ & \frac{4}{{{x}^{2}}}-\frac{2}{{{y}^{2}}}=-14\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \end{align} Multiply the equation (I) by (II) and add the equation (I) and equation (II): \begin{align} & \frac{4}{{{x}^{2}}}+\frac{2}{{{y}^{2}}}=22 \\ & \frac{4}{{{x}^{2}}}-\frac{2}{{{y}^{2}}}=-14 \\ & \overline{\begin{align} & \,\,\,\,\,\,\frac{8}{{{x}^{2}}}=8\,\,\,\,\,\,\,\,\,\,\,\,\, \\ & x=\pm 1 \end{align}} \\ \end{align} Substitute the value of $x=1,x=-1$ in equation (I) For \begin{align} & x=-1; \\ & \frac{2}{{{\left( -1 \right)}^{2}}}+\frac{1}{{{y}^{2}}}=11 \\ & 2+\frac{1}{{{y}^{2}}}=11 \\ & {{y}^{2}}=\frac{1}{9} \end{align} Then, $y=\pm \frac{1}{3}$ For \begin{align} & x=1; \\ & \frac{2}{{{\left( 1 \right)}^{2}}}+\frac{1}{{{y}^{2}}}=11 \\ & 2+\frac{1}{{{y}^{2}}}=11 \\ & {{y}^{2}}=\frac{1}{9} \end{align} Then, $y=\pm \frac{1}{3}$ Thus, the solution set of system is $\left\{ \left( -1,-\frac{1}{3} \right),\left( -1,\frac{1}{3} \right),\left( 1,\frac{-1}{3} \right),\left( 1,\frac{1}{3} \right) \right\}$.