Solve the system. Use any method you wish. \[ \begin{array}{l}y=x+3 \\ 3 x^{2}+y^{2}=9\end{array} \] 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.

To solve the given system of equations, we start with substituting the first equation into the second: 1. First, we have \( y = x + 3 \). 2. Now substitute \( y \) into the second equation: \[ 3x^2 + (x + 3)^2 = 9 \] 3. Expand \( (x + 3)^2 \): \[ (x + 3)^2 = x^2 + 6x + 9 \] 4. Substitute this back into the equation: \[ 3x^2 + x^2 + 6x + 9 = 9 \] 5. Simplifying gives: \[ 4x^2 + 6x + 9 - 9 = 0 \Rightarrow 4x^2 + 6x = 0 \] 6. Factor the equation: \[ 2x(2x + 3) = 0 \] 7. This gives us the solutions: \[ x = 0 \quad \text{or} \quad 2x + 3 = 0 \Rightarrow x = -\frac{3}{2} \] 8. Now substitute back to find \( y \): - For \( x = 0 \), \( y = 0 + 3 = 3 \) → solution (0, 3) - For \( x = -\frac{3}{2} \), \( y = -\frac{3}{2} + 3 = \frac{3}{2} \) → solution \((- \frac{3}{2}, \frac{3}{2})\) Thus, the solutions are \((0, 3)\) and \((- \frac{3}{2}, \frac{3}{2})\). A. The solution(s) is/are (0, 3), (-3/2, 3/2)