Solve for \( x \) and \( y \) : \[ \begin{array}{l} x=-3 y \\ \text { and } \\ x^{2}+2 x y=48\end{array} \]

Solve the system of equations \( x=-3y; x^{2}+2xy=48 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x=-3y\\x^{2}+2xy=48\end{array}\right.\) - step1: Substitute the value of \(x:\) \(\left(-3y\right)^{2}+2\left(-3y\right)y=48\) - step2: Simplify: \(3y^{2}=48\) - step3: Divide both sides: \(\frac{3y^{2}}{3}=\frac{48}{3}\) - step4: Divide the numbers: \(y^{2}=16\) - step5: Simplify the expression: \(y=\pm \sqrt{16}\) - step6: Simplify: \(y=\pm 4\) - step7: Separate into possible cases: \(y=4\cup y=-4\) - step8: Rearrange the terms: \(\left\{ \begin{array}{l}x=-3y\\y=4\end{array}\right.\cup \left\{ \begin{array}{l}x=-3y\\y=-4\end{array}\right.\) - step9: Calculate: \(\left\{ \begin{array}{l}x=-12\\y=4\end{array}\right.\cup \left\{ \begin{array}{l}x=12\\y=-4\end{array}\right.\) - step10: Check the solution: \(\left\{ \begin{array}{l}x=-12\\y=4\end{array}\right.\cup \left\{ \begin{array}{l}x=12\\y=-4\end{array}\right.\) - step11: Rewrite: \(\left(x,y\right) = \left(-12,4\right)\cup \left(x,y\right) = \left(12,-4\right)\) The solutions to the system of equations are \( (x,y) = (-12,4) \) and \( (x,y) = (12,-4) \).