ii) \( a+b=3 \) and \( a b=20 \).

Given: \( a+b=3 \) and \( a b=20 \). We are asked to find the values of \( a \) and \( b \) that satisfy these equations. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}a+b=3\\ab=20\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}a=3-b\\ab=20\end{array}\right.\) - step2: Substitute the value of \(a:\) \(\left(3-b\right)b=20\) - step3: Simplify: \(b\left(3-b\right)=20\) - step4: Expand the expression: \(3b-b^{2}=20\) - step5: Move the expression to the left side: \(3b-b^{2}-20=0\) - step6: Rewrite in standard form: \(-b^{2}+3b-20=0\) - step7: Multiply both sides: \(b^{2}-3b+20=0\) - step8: Solve using the quadratic formula: \(b=\frac{3\pm \sqrt{\left(-3\right)^{2}-4\times 20}}{2}\) - step9: Simplify the expression: \(b=\frac{3\pm \sqrt{-71}}{2}\) - step10: The expression is undefined: \(b \notin \mathbb{R}\) - step11: The system of equations has no solution in the set of real numbers: \(\left(a,b\right) \notin \mathbb{R}^{2}\) - step12: Alternative Form: \(\textrm{No real solution}\) The system of equations \( a+b=3 \) and \( a b=20 \) has no real solution. This means that there are no real values of \( a \) and \( b \) that satisfy both equations simultaneously.