\( x+y=2 \) and \( x^{2}+y^{2}=10 \)

Solve the system of equations \( x+y=2;x^{2}+y^{2}=10 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+y=2\\x^{2}+y^{2}=10\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=2-y\\x^{2}+y^{2}=10\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left(2-y\right)^{2}+y^{2}=10\) - step3: Simplify: \(4-4y+2y^{2}=10\) - step4: Move the expression to the left side: \(4-4y+2y^{2}-10=0\) - step5: Subtract the numbers: \(-6-4y+2y^{2}=0\) - step6: Factor the expression: \(2\left(-3+y\right)\left(1+y\right)=0\) - step7: Divide the terms: \(\left(-3+y\right)\left(1+y\right)=0\) - step8: Separate into possible cases: \(\begin{align}&-3+y=0\\&1+y=0\end{align}\) - step9: Solve the equation: \(\begin{align}&y=3\\&y=-1\end{align}\) - step10: Calculate: \(y=3\cup y=-1\) - step11: Rearrange the terms: \(\left\{ \begin{array}{l}x=2-y\\y=3\end{array}\right.\cup \left\{ \begin{array}{l}x=2-y\\y=-1\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=-1\\y=3\end{array}\right.\cup \left\{ \begin{array}{l}x=3\\y=-1\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=-1\\y=3\end{array}\right.\cup \left\{ \begin{array}{l}x=3\\y=-1\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(-1,3\right)\cup \left(x,y\right) = \left(3,-1\right)\) The solutions to the system of equations \(x+y=2\) and \(x^{2}+y^{2}=10\) are \((x,y) = (-1,3)\) and \((x,y) = (3,-1)\).