<s>[INST] Simultaneous solution(s) to $a^2+4b^2+4ab=0$ and $a^2+4b^2+32+16a-8b=0$? Could you tell me just how should I solve this system:
$$
a^2+4b^2+4ab=0\\
a^2+4b^2+32+16a-8b=0
$$
I can't remember the proceeding and it's driving me crazy.
Thanks a lot
 [/INST] We have: 
$$
a^2+4b^2+4ab=0 \tag{1}$$
$$
a^2+4b^2+32+16a-8b=0\tag{2}
$$
$\bf (I)$ Subtract equation $(2)$ from $(1)$:
$$\begin{align}
& a^2+4b^2+4ab & =0\\
- & a^2+4b^2+32+16a-8b &=0 \\
& \hline \\
= & 4ab -16a + 8b - 32 & = 0 \\
4 & (ab - 4a + 2b - 8) & = 0 \\ \\
= & ab - 4a + 2b -8 = 0 \tag{3}\\
\end{align}
$$
$\bf (II)$ Factor equation $(1)$ $$
a^2+4b^2+4ab=0 \iff (a + 2b)^2 = 0 \iff a = -2b \tag{4}
$$
$\bf (III)$
Substitute $a = -2b$ into equation $(3)$. Then solve for $a$ the solution for $b$ to obtain $a = -2b$.
</s>