solve for \( x \& y \) \( x-2 y-3=0 \& x^{2}-3 x y+y^{2}=11 \)

Solve the system of equations \( x-2y-3=0;x^{2}-3xy+y^{2}=11 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x-2y-3=0\\x^{2}-3xy+y^{2}=11\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=2y+3\\x^{2}-3xy+y^{2}=11\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left(2y+3\right)^{2}-3\left(2y+3\right)y+y^{2}=11\) - step3: Simplify: \(-y^{2}+3y+9=11\) - step4: Move the expression to the left side: \(-y^{2}+3y+9-11=0\) - step5: Subtract the numbers: \(-y^{2}+3y-2=0\) - step6: Factor the expression: \(\left(-y+2\right)\left(y-1\right)=0\) - step7: Separate into possible cases: \(\begin{align}&-y+2=0\\&y-1=0\end{align}\) - step8: Solve the equation: \(\begin{align}&y=2\\&y=1\end{align}\) - step9: Calculate: \(y=2\cup y=1\) - step10: Rearrange the terms: \(\left\{ \begin{array}{l}x=2y+3\\y=2\end{array}\right.\cup \left\{ \begin{array}{l}x=2y+3\\y=1\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}x=7\\y=2\end{array}\right.\cup \left\{ \begin{array}{l}x=5\\y=1\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=5\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=7\\y=2\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=5\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=7\\y=2\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(5,1\right)\cup \left(x,y\right) = \left(7,2\right)\) The solutions to the system of equations \(x-2y-3=0\) and \(x^{2}-3xy+y^{2}=11\) are: 1. \(x = 5\) and \(y = 1\) 2. \(x = 7\) and \(y = 2\)