\( x^{2}-5 x y+y^{2}=7 \) and \( x-4 y=5 \)

Solve the system of equations \( x^{2}-5x*y+y^{2}=7;x-4y=5 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x^{2}-5xy+y^{2}=7\\x-4y=5\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x^{2}-5xy+y^{2}=7\\x=5+4y\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left(5+4y\right)^{2}-5\left(5+4y\right)y+y^{2}=7\) - step3: Simplify: \(25+15y-3y^{2}=7\) - step4: Move the expression to the left side: \(25+15y-3y^{2}-7=0\) - step5: Subtract the numbers: \(18+15y-3y^{2}=0\) - step6: Factor the expression: \(3\left(6-y\right)\left(1+y\right)=0\) - step7: Divide the terms: \(\left(6-y\right)\left(1+y\right)=0\) - step8: Separate into possible cases: \(\begin{align}&6-y=0\\&1+y=0\end{align}\) - step9: Solve the equation: \(\begin{align}&y=6\\&y=-1\end{align}\) - step10: Calculate: \(y=6\cup y=-1\) - step11: Rearrange the terms: \(\left\{ \begin{array}{l}x=5+4y\\y=6\end{array}\right.\cup \left\{ \begin{array}{l}x=5+4y\\y=-1\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=29\\y=6\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=-1\end{array}\right.\) - step13: Calculate: \(\left\{ \begin{array}{l}x=1\\y=-1\end{array}\right.\cup \left\{ \begin{array}{l}x=29\\y=6\end{array}\right.\) - step14: Check the solution: \(\left\{ \begin{array}{l}x=1\\y=-1\end{array}\right.\cup \left\{ \begin{array}{l}x=29\\y=6\end{array}\right.\) - step15: Rewrite: \(\left(x,y\right) = \left(1,-1\right)\cup \left(x,y\right) = \left(29,6\right)\) The solutions to the system of equations \(x^{2}-5xy+y^{2}=7\) and \(x-4y=5\) are \((x,y) = (1,-1)\) and \((x,y) = (29,6)\).