Solve $5a^2 - 4ab - b^2 + 9 = 0$, $ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0$ 
Solve $\left\{\begin{matrix} 5a^2 - 4ab - b^2 + 9 = 0\\ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0. \end{matrix}\right.$

I know that we can use quadratic equation twice, but then we'll get some very complicated steps. Are there any elegant way to solve this? Thank you.

You can notice that many terms of 
$$(b+5a-4)^2=b^2+10ab-8b-40a+25a^2+16$$
appear in the first equation. Similarly, in the first one, you can notice $(b+2a)^2$.
By algebraic manipulation you get that the original equations are equivalent to
$$
\begin{align}
(b+5a-4)^2&=4(a^2+1)\\
(b+2a)^2&=9(a^2+1)
\end{align}
$$
which implies $4(b+2a)^2=9(b+5a-4)^2$ and $2(b+2a)=\pm 3(b+5a-4)$. This should simplify things a little.
(In each of the two possibilities you can express $b$ using $a$ as a linear expression.
Then you will get a quadratic equation in $a$. Or you can start by eliminating $a$.)
