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?