\( 2 x-y=7 \) and \( x^{2}+x y=21-y^{2} \)

Solve the system of equations \( 2x-y=7;x^{2}+xy=21-y^{2} \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2x-y=7\\x^{2}+xy=21-y^{2}\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}y=-7+2x\\x^{2}+xy=21-y^{2}\end{array}\right.\) - step2: Substitute the value of \(y:\) \(x^{2}+x\left(-7+2x\right)=21-\left(-7+2x\right)^{2}\) - step3: Simplify: \(3x^{2}-7x=-28+28x-4x^{2}\) - step4: Move the expression to the left side: \(3x^{2}-7x-\left(-28+28x-4x^{2}\right)=0\) - step5: Subtract the terms: \(7x^{2}-35x+28=0\) - step6: Factor the expression: \(7\left(x-4\right)\left(x-1\right)=0\) - step7: Divide the terms: \(\left(x-4\right)\left(x-1\right)=0\) - step8: Separate into possible cases: \(\begin{align}&x-4=0\\&x-1=0\end{align}\) - step9: Solve the equation: \(\begin{align}&x=4\\&x=1\end{align}\) - step10: Calculate: \(x=4\cup x=1\) - step11: Rearrange the terms: \(\left\{ \begin{array}{l}x=4\\y=-7+2x\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=-7+2x\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=4\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=1\\y=-5\end{array}\right.\) - step13: Calculate: \(\left\{ \begin{array}{l}x=1\\y=-5\end{array}\right.\cup \left\{ \begin{array}{l}x=4\\y=1\end{array}\right.\) - step14: Check the solution: \(\left\{ \begin{array}{l}x=1\\y=-5\end{array}\right.\cup \left\{ \begin{array}{l}x=4\\y=1\end{array}\right.\) - step15: Rewrite: \(\left(x,y\right) = \left(1,-5\right)\cup \left(x,y\right) = \left(4,1\right)\) The solutions to the system of equations \(2x-y=7\) and \(x^{2}+xy=21-y^{2}\) are \((x,y) = (1,-5)\) and \((x,y) = (4,1)\).