<s>[INST] 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.
 [/INST] 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 $(a,b)$.
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