10) \( \begin{array}{l}3 x^{2}+4 x+15 y+15=0 \\ x-3 y-3=0\end{array} \)

To solve the given system of equations, we can use substitution or elimination. Here’s a fun twist! First, let’s solve the second equation for \(x\): From \(x - 3y - 3 = 0\), we get: \(x = 3y + 3\). Now, we can substitute this expression for \(x\) into the first equation. Plugging \(x\) into \(3x^2 + 4x + 15y + 15 = 0\) will yield a new equation purely in terms of \(y\). After substitution: \[ 3(3y + 3)^2 + 4(3y + 3) + 15y + 15 = 0, \] which simplifies further to: \[ 27y^2 + 54y + 27 + 12y + 12 + 15y + 15 = 0. \] Combine like terms to obtain a quadratic equation in \(y\). Once you’ve done that, you can apply the quadratic formula to find the values of \(y\), then substitute those values back into \(x = 3y + 3\) to get the corresponding \(x\) values. This is problem-solving at its finest!