PDF Ex. 13.8, Video Ex following system. Leave answers in radical form. \( \left\{\begin{array}{l}x^{2}+y^{2}=4^{2} \\ \text { Hint: Solve for } y \text { first, and leave it in unsimplified, radical } \\ \text { form until after you solve for } x \text {. } \\ \text { Optional: use a GC to check. }\end{array} \frac{y^{2}}{4}-\frac{x^{2}}{3}=1\right. \)

To solve the system of equations, we start by isolating \(y\) in the first equation, \(x^{2} + y^{2} = 4^{2}\) or \(y^{2} = 16 - x^{2}\). Next, we can substitute this expression into the second equation to solve for \(x\): \[ \frac{16 - x^{2}}{4} - \frac{x^{2}}{3} = 1 \] Multiplying through by 12 to eliminate the denominators, we get: \[ 3(16 - x^{2}) - 4x^{2} = 12 \] Distributing gives: \[ 48 - 3x^{2} - 4x^{2} = 12 \] Combining like terms results in: \[ 48 - 7x^{2} = 12 \] Rearranging yields: \[ 7x^{2} = 36 \] Thus, we find: \[ x^{2} = \frac{36}{7} \implies x = \pm \sqrt{\frac{36}{7}} = \pm \frac{6}{\sqrt{7}} = \pm \frac{6\sqrt{7}}{7} \] Now we substitute these \(x\) values back into \(y^{2} = 16 - x^{2}\) to find \(y\): For \(x = \frac{6\sqrt{7}}{7}\): \[ y^{2} = 16 - \frac{36}{7} = \frac{112 - 36}{7} = \frac{76}{7} \implies y = \pm\sqrt{\frac{76}{7}} = \pm\frac{2\sqrt{19}}{\sqrt{7}} = \pm\frac{2\sqrt{133}}{7} \] So the solutions to the system are \( \left( \frac{6\sqrt{7}}{7}, \frac{2\sqrt{19}}{\sqrt{7}} \right), \left( \frac{6\sqrt{7}}{7}, -\frac{2\sqrt{19}}{\sqrt{7}} \right), \left( -\frac{6\sqrt{7}}{7}, \frac{2\sqrt{19}}{\sqrt{7}} \right), \text{ and } \left( -\frac{6\sqrt{7}}{7}, -\frac{2\sqrt{19}}{\sqrt{7}} \right).\)