## Precalculus (6th Edition) Blitzer

The solution is $\left\{ \left( -1,-\frac{1}{2} \right),\left( -1,\frac{1}{2} \right),\left( 1,\frac{-1}{2} \right),\left( 1,\frac{1}{2} \right) \right\}$
We have the equations, \begin{align} & \frac{3}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}=7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\ & \frac{5}{{{x}^{2}}}-\frac{2}{{{y}^{2}}}=-3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \end{align} And multiply the equation (I) by 2 and then add the equation (I) and equation (II): \begin{align} & \frac{6}{{{x}^{2}}}+\frac{2}{{{y}^{2}}}=14 \\ & \frac{5}{{{x}^{2}}}-\frac{2}{{{y}^{2}}}=-3 \\ & \overline{\begin{align} & \,\,\,\,\,\,\frac{11}{{{x}^{2}}}=11\,\,\,\,\,\,\,\,\,\,\,\,\, \\ & {{x}^{2}}=1 \\ \end{align}} \\ \end{align} Finally, $x=\pm 1$ Substitute the value of x as $x=1,x=-1$ into equation (I): For \begin{align} & x=-1; \\ & \frac{3}{{{\left( -1 \right)}^{2}}}+\frac{1}{{{y}^{2}}}=7 \\ & 3+\frac{1}{{{y}^{2}}}=7 \\ & {{y}^{2}}=\frac{1}{4} \end{align} Then, $y=\pm \frac{1}{2}$ For \begin{align} & x=1; \\ & \frac{3}{{{\left( 1 \right)}^{2}}}+\frac{1}{{{y}^{2}}}=7 \\ & 3+\frac{1}{{{y}^{2}}}=7 \\ & {{y}^{2}}=\frac{1}{4} \end{align} Then, $y=\pm \frac{1}{2}$ Therefore, the solution set of system is $\left\{ \left( -1,-\frac{1}{2} \right),\left( -1,\frac{1}{2} \right),\left( 1,\frac{-1}{2} \right),\left( 1,\frac{1}{2} \right) \right\}$.