Solve the system. Use any method you wish. \[ \begin{array}{l}y=x+9 \\ 14 x^{2}+y^{2}=81\end{array} \] List all the solutions. Select the correct choice below and fill in any answer boxes in your choice. A. The solution(s) is/are (Type an ordered pair. Use a comma to separate answers as needed.) B. There is no solution.

Solve the system of equations \( y=x+9;14x^2+y^2=81 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=x+9\\14x^{2}+y^{2}=81\end{array}\right.\) - step1: Substitute the value of \(y:\) \(14x^{2}+\left(x+9\right)^{2}=81\) - step2: Simplify: \(15x^{2}+18x+81=81\) - step3: Cancel equal terms: \(15x^{2}+18x=0\) - step4: Factor the expression: \(3x\left(5x+6\right)=0\) - step5: Separate into possible cases: \(\begin{align}&3x=0\\&5x+6=0\end{align}\) - step6: Solve the equation: \(\begin{align}&x=0\\&x=-\frac{6}{5}\end{align}\) - step7: Calculate: \(x=0\cup x=-\frac{6}{5}\) - step8: Rearrange the terms: \(\left\{ \begin{array}{l}x=0\\y=x+9\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{6}{5}\\y=x+9\end{array}\right.\) - step9: Calculate: \(\left\{ \begin{array}{l}x=0\\y=9\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{6}{5}\\y=\frac{39}{5}\end{array}\right.\) - step10: Calculate: \(\left\{ \begin{array}{l}x=-\frac{6}{5}\\y=\frac{39}{5}\end{array}\right.\cup \left\{ \begin{array}{l}x=0\\y=9\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=-\frac{6}{5}\\y=\frac{39}{5}\end{array}\right.\cup \left\{ \begin{array}{l}x=0\\y=9\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(-\frac{6}{5},\frac{39}{5}\right)\cup \left(x,y\right) = \left(0,9\right)\) The system of equations has two solutions: 1. \( (x,y) = \left(-\frac{6}{5}, \frac{39}{5}\right) \) or \( (x,y) = (-1.2, 7.8) \) 2. \( (x,y) = (0,9) \) Therefore, the correct choice is: A. The solution(s) is/are \( (-\frac{6}{5}, \frac{39}{5}) \) or \( (-1.2, 7.8) \) and \( (0,9) \)