\( \therefore x=-4 \) Solve \( =0 \) or \( x \) and \( y \) simultanesus \( 2 x+y=5 \) \( x y=6 \)

Solve the system of equations \( 2x+y=5;x*y=6 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2x+y=5\\xy=6\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}y=5-2x\\xy=6\end{array}\right.\) - step2: Substitute the value of \(y:\) \(x\left(5-2x\right)=6\) - step3: Expand the expression: \(5x-2x^{2}=6\) - step4: Move the expression to the left side: \(5x-2x^{2}-6=0\) - step5: Rewrite in standard form: \(-2x^{2}+5x-6=0\) - step6: Multiply both sides: \(2x^{2}-5x+6=0\) - step7: Solve using the quadratic formula: \(x=\frac{5\pm \sqrt{\left(-5\right)^{2}-4\times 2\times 6}}{2\times 2}\) - step8: Simplify the expression: \(x=\frac{5\pm \sqrt{\left(-5\right)^{2}-4\times 2\times 6}}{4}\) - step9: Simplify the expression: \(x=\frac{5\pm \sqrt{-23}}{4}\) - step10: The expression is undefined: \(x \notin \mathbb{R}\) - step11: The system of equations has no solution in the set of real numbers: \(\left(x,y\right) \notin \mathbb{R}^{2}\) - step12: Alternative Form: \(\textrm{No real solution}\) The system of equations \(2x+y=5\) and \(xy=6\) does not have any real solutions.