4)) Solve the system of equations. \[ y=x^{2}-6 x+19 \] \( y=-3 x+47 \) 4)) Write the coordinates in exact form. Simplify all fractions and radicals. (

Solve the system of equations \( y=x^{2}-6x+19;y=-3x+47 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=x^{2}-6x+19\\y=-3x+47\end{array}\right.\) - step1: Substitute the value of \(y:\) \(x^{2}-6x+19=-3x+47\) - step2: Move the expression to the left side: \(x^{2}-6x+19-\left(-3x+47\right)=0\) - step3: Calculate: \(x^{2}-3x-28=0\) - step4: Factor the expression: \(\left(x-7\right)\left(x+4\right)=0\) - step5: Separate into possible cases: \(\begin{align}&x-7=0\\&x+4=0\end{align}\) - step6: Solve the equation: \(\begin{align}&x=7\\&x=-4\end{align}\) - step7: Calculate: \(x=7\cup x=-4\) - step8: Rearrange the terms: \(\left\{ \begin{array}{l}x=7\\y=x^{2}-6x+19\end{array}\right.\cup \left\{ \begin{array}{l}x=-4\\y=x^{2}-6x+19\end{array}\right.\) - step9: Calculate: \(\left\{ \begin{array}{l}x=7\\y=26\end{array}\right.\cup \left\{ \begin{array}{l}x=-4\\y=59\end{array}\right.\) - step10: Calculate: \(\left\{ \begin{array}{l}x=-4\\y=59\end{array}\right.\cup \left\{ \begin{array}{l}x=7\\y=26\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=-4\\y=59\end{array}\right.\cup \left\{ \begin{array}{l}x=7\\y=26\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(-4,59\right)\cup \left(x,y\right) = \left(7,26\right)\) The solution to the system of equations is \( (x,y) = (-4,59) \) or \( (x,y) = (7,26) \). Therefore, the coordinates in exact form are \( (-4,59) \) or \( (7,26) \).