how to find $\frac{a}{b}$ given that $2a^2 + 2007a + 3 = 0$ and $3b^2 + 2007b + 2 = 0$ Given that
$$\begin{cases}
2a^2 + 2007a + 3 = 0 \\ 
3b^2 + 2007b + 2 = 0
\end{cases}
$$
and $ab \ne 1$, how to solve for $\frac{a}{b}$?
My try:
\begin{align}
2007a = -3 - 2a^2 \\
2007b = -2 - 3b^2 \\
\frac{a}{b} = \frac{-3 - 2a^2}{-2 - 3b^2}
\end{align}
Also,
\begin{align}
2a^2 = -2007a - 3 \\
3b^2 = -2007b - 2 \\
\frac{a^2}{b^2} = \frac{-2007a - 3}{-2007b - 2}\frac{3}{2} \\
\frac{a}{b} = \pm\sqrt{\frac{-6021a - 9}{-4014b - 4}}
\end{align}
Well, it doesn't look very promising. So I want to seek help here. How to find $\frac{a}{b}$ without actually solving for the roots? Hints are welcomed.

$\mathcal{Hint}$
Consider $2x^2+2007x+3=0$
Note that $a$ is one the roots.
Now, replacing $x$ by $\frac{1}{x}$
How does that affect the equation and its roots?
