\( x^{2}-3 x y+2 y^{2}=4 \) and \( 3 x-2 y=0 \)

Solve the system of equations \( x^{2}-3x*y+2*y^{2}=4;3x-2y=0 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x^{2}-3xy+2y^{2}=4\\3x-2y=0\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x^{2}-3xy+2y^{2}=4\\x=\frac{2y}{3}\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left(\frac{2y}{3}\right)^{2}-3\times \frac{2y}{3}y+2y^{2}=4\) - step3: Simplify: \(\frac{4}{9}y^{2}=4\) - step4: Multiply by the reciprocal: \(\frac{4}{9}y^{2}\times \frac{9}{4}=4\times \frac{9}{4}\) - step5: Multiply: \(y^{2}=9\) - step6: Simplify the expression: \(y=\pm \sqrt{9}\) - step7: Simplify: \(y=\pm 3\) - step8: Separate into possible cases: \(y=3\cup y=-3\) - step9: Rearrange the terms: \(\left\{ \begin{array}{l}x=\frac{2y}{3}\\y=3\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{2y}{3}\\y=-3\end{array}\right.\) - step10: Calculate: \(\left\{ \begin{array}{l}x=2\\y=3\end{array}\right.\cup \left\{ \begin{array}{l}x=-2\\y=-3\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}x=-2\\y=-3\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=3\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=-2\\y=-3\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=3\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(-2,-3\right)\cup \left(x,y\right) = \left(2,3\right)\) The solutions to the system of equations \(x^{2}-3xy+2y^{2}=4\) and \(3x-2y=0\) are \((-2,-3)\) and \((2,3)\).