b) \( \left\{\begin{array}{l}x^{2}+y^{2}=13 \\ y=x+1\end{array}\right. \)

Solve the system of equations \( x^{2}+y^{2}=13;y=x+1 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x^{2}+y^{2}=13\\y=x+1\end{array}\right.\) - step1: Substitute the value of \(y:\) \(x^{2}+\left(x+1\right)^{2}=13\) - step2: Simplify: \(2x^{2}+2x+1=13\) - step3: Move the expression to the left side: \(2x^{2}+2x+1-13=0\) - step4: Subtract the numbers: \(2x^{2}+2x-12=0\) - step5: Factor the expression: \(2\left(x-2\right)\left(x+3\right)=0\) - step6: Divide the terms: \(\left(x-2\right)\left(x+3\right)=0\) - step7: Separate into possible cases: \(\begin{align}&x-2=0\\&x+3=0\end{align}\) - step8: Solve the equation: \(\begin{align}&x=2\\&x=-3\end{align}\) - step9: Calculate: \(x=2\cup x=-3\) - step10: Rearrange the terms: \(\left\{ \begin{array}{l}x=2\\y=x+1\end{array}\right.\cup \left\{ \begin{array}{l}x=-3\\y=x+1\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}x=2\\y=3\end{array}\right.\cup \left\{ \begin{array}{l}x=-3\\y=-2\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=-3\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=3\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=-3\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=2\\y=3\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(-3,-2\right)\cup \left(x,y\right) = \left(2,3\right)\) The solution to the system of equations is \( (x,y) = (-3,-2) \) or \( (x,y) = (2,3) \).