\( x+y=3 \) and \( x^{2}+y^{2}=0 \)

Solve the system of equations \( x+y=3;x^{2}+y^{2}=0 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+y=3\\x^{2}+y^{2}=0\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=3-y\\x^{2}+y^{2}=0\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left(3-y\right)^{2}+y^{2}=0\) - step3: Simplify: \(9-6y+2y^{2}=0\) - step4: Rewrite in standard form: \(2y^{2}-6y+9=0\) - step5: Solve using the quadratic formula: \(y=\frac{6\pm \sqrt{\left(-6\right)^{2}-4\times 2\times 9}}{2\times 2}\) - step6: Simplify the expression: \(y=\frac{6\pm \sqrt{\left(-6\right)^{2}-4\times 2\times 9}}{4}\) - step7: Simplify the expression: \(y=\frac{6\pm \sqrt{-36}}{4}\) - step8: The expression is undefined: \(y \notin \mathbb{R}\) - step9: The system of equations has no solution in the set of real numbers: \(\left(x,y\right) \notin \mathbb{R}^{2}\) - step10: Alternative Form: \(\textrm{No real solution}\) The system of equations \(x+y=3\) and \(x^{2}+y^{2}=0\) does not have any real solutions.