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.

By arrange the terms, you can note that 
$$\begin{cases}
5a^{2}-4ab-b^2+9=0\\
-21a^{2}-10ab+40a-b^{2}+8b-12=0 
\end{cases}
\Leftrightarrow
\begin{cases}
9(a^{2}+1)=(2a+b)^{2}\\
4(a^{2}+1)=(5a+b-4)^{2} 
\end{cases}$$
So we get 
$$\begin{cases}
9(a^{2}+1)=(2a+b)^{2}\\
4(2a+b)^{2}=9(5a+b-4)^{2} 
\end{cases}$$
According to the second equation, we can get two cases of first relation between $a$ and $b$, then substitute them into the first equation respectively, we can get all the cases of values of pairs $(a,b)$.
