Solve the following system of equations.
\[
\left\{\begin{array}{l}
x^{2}+y^{2}=8 \\
y+3 x=-4
\end{array}\right.
\]

If there is more than one solution, enter additional solutions with the "or" button. If there is no real solution, use the "No solution" button.
\[
(x, y)=\left(\prod, \square\right)
\]

Step 1 ：First, let's express y from the second equation: \(y = -3x - 4\).

Step 2 ：Then we substitute this into the first equation: \(x^2 + (-3x - 4)^2 = 8\). This is a quadratic equation, which we can solve for x.

Step 3 ：After finding the values of x, we substitute them back into the second equation to find the corresponding values of y.

Step 4 ：The solutions to the quadratic equation are \(x = -2\) and \(x = -\frac{2}{5}\).

Step 5 ：Substituting these values into the second equation, we find the corresponding y values: \(y = 2\) when \(x = -2\), and \(y = -\frac{14}{5}\) when \(x = -\frac{2}{5}\).

Step 6 ：\(\boxed{\text{Final Answer: The solutions to the system of equations are } (x, y) = \left(-2, 2\right) \text{ or } (x, y) = \left(-\frac{2}{5}, -\frac{14}{5}\right)}\)