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

Solve the equation by following steps: - step0: Solve for \(y\): \(y^{2}=16-x^{2}\) - step1: Simplify the expression: \(y=\pm \sqrt{16-x^{2}}\) - step2: Separate into possible cases: \(\begin{align}&y=\sqrt{16-x^{2}}\\&y=-\sqrt{16-x^{2}}\end{align}\) Solve the system of equations \( x^{2}+y^{2}=16; \frac{y^{2}}{4}-\frac{x^{2}}{3}=1 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x^{2}+y^{2}=16\\\frac{y^{2}}{4}-\frac{x^{2}}{3}=1\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\sqrt{16-y^{2}}\cup x=-\sqrt{16-y^{2}}\\\frac{y^{2}}{4}-\frac{x^{2}}{3}=1\end{array}\right.\) - step2: Evaluate: \(\left\{ \begin{array}{l}x=\sqrt{16-y^{2}}\\\frac{y^{2}}{4}-\frac{x^{2}}{3}=1\end{array}\right.\cup \left\{ \begin{array}{l}x=-\sqrt{16-y^{2}}\\\frac{y^{2}}{4}-\frac{x^{2}}{3}=1\end{array}\right.\) - step3: Calculate: \(\left\{ \begin{array}{l}x=\frac{2\sqrt{63}}{7}\\y=-\frac{2\sqrt{133}}{7}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{2\sqrt{63}}{7}\\y=\frac{2\sqrt{133}}{7}\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{2\sqrt{63}}{7}\\y=-\frac{2\sqrt{133}}{7}\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{2\sqrt{63}}{7}\\y=\frac{2\sqrt{133}}{7}\end{array}\right.\) - step4: Rearrange the terms: \(\left\{ \begin{array}{l}x=-\frac{2\sqrt{63}}{7}\\y=-\frac{2\sqrt{133}}{7}\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{2\sqrt{63}}{7}\\y=\frac{2\sqrt{133}}{7}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{2\sqrt{63}}{7}\\y=-\frac{2\sqrt{133}}{7}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{2\sqrt{63}}{7}\\y=\frac{2\sqrt{133}}{7}\end{array}\right.\) - step5: Check the solution: \(\left\{ \begin{array}{l}x=-\frac{2\sqrt{63}}{7}\\y=-\frac{2\sqrt{133}}{7}\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{2\sqrt{63}}{7}\\y=\frac{2\sqrt{133}}{7}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{2\sqrt{63}}{7}\\y=-\frac{2\sqrt{133}}{7}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{2\sqrt{63}}{7}\\y=\frac{2\sqrt{133}}{7}\end{array}\right.\) - step6: Rewrite: \(\left(x,y\right) = \left(-\frac{2\sqrt{63}}{7},-\frac{2\sqrt{133}}{7}\right)\cup \left(x,y\right) = \left(-\frac{2\sqrt{63}}{7},\frac{2\sqrt{133}}{7}\right)\cup \left(x,y\right) = \left(\frac{2\sqrt{63}}{7},-\frac{2\sqrt{133}}{7}\right)\cup \left(x,y\right) = \left(\frac{2\sqrt{63}}{7},\frac{2\sqrt{133}}{7}\right)\) To solve the given system of equations: \[ \left\{ \begin{array}{l} x^{2}+y^{2}=4^{2} \\ \frac{y^{2}}{4}-\frac{x^{2}}{3}=1 \end{array} \right. \] we will follow the hint provided and solve for \( y \) first. ### Step 1: Solve for \( y \) From the first equation, we can express \( y \) in terms of \( x \): \[ y^{2} = 16 - x^{2} \] Thus, we have two expressions for \( y \): \[ y = \sqrt{16 - x^{2}} \quad \text{and} \quad y = -\sqrt{16 - x^{2}} \] ### Step 2: Substitute \( y \) into the second equation Now, we substitute \( y^{2} \) into the second equation: \[ \frac{16 - x^{2}}{4} - \frac{x^{2}}{3} = 1 \] ### Step 3: Clear the fractions To eliminate the fractions, we can multiply through by 12 (the least common multiple of 4 and 3): \[ 12 \left( \frac{16 - x^{2}}{4} \right) - 12 \left( \frac{x^{2}}{3} \right) = 12 \] This simplifies to: \[ 3(16 - x^{2}) - 4x^{2} = 12 \] ### Step 4: Expand and simplify Expanding gives: \[ 48 - 3x^{2} - 4x^{2} = 12 \] Combining like terms results in: \[ 48 - 7x^{2} = 12 \] ### Step 5: Solve for \( x^{2} \) Rearranging gives: \[ -7x^{2} = 12 - 48 \] \[ -7x^{2} = -36 \] \[ x^{2} = \frac{36}{7} \] ### Step 6: Solve for \( x \) Taking the square root gives: \[ x = \pm \sqrt{\frac{36}{7}} = \pm \frac{6}{\sqrt{7}} = \pm \frac{6\sqrt{7}}{7} \] ### Step 7: Substitute back to find \( y \) Now we substitute \( x \) back into the expression for \( y \): \[ y = \sqrt{16 - x^{2}} = \sqrt{16 - \frac{36}{7}} = \sqrt{\frac{112}{7} - \frac{36}{7}} = \sqrt{\frac{76}{7}} = \frac{\sqrt{76}}{\sqrt{7}} = \frac{2\sqrt{19}}{\sqrt{7}} \] Thus, we have two values for \( y \): \[ y = \frac{2\sqrt{19}}{\sqrt{7}} \quad \text{and} \quad y = -\frac{2\sqrt{19}}{\sqrt{7}} \] ### Final Solutions The complete set of solutions for the system is: \[ (x,y) = \left(-\frac{6\sqrt{7}}{7}, -\frac{2\sqrt{19}}{\sqrt{7}}\right) \cup \left(-\frac{6\sqrt{7}}{7}, \frac{2\sqrt{19}}{\sqrt{7}}\right) \cup \left(\frac{6\sqrt{7}}{7}, -\frac{2\sqrt{19}}{\sqrt{7}}\right) \cup \left(\frac{6\sqrt{7}}{7}, \frac{2\sqrt{19}}{\sqrt{7}}\right) \]