می‌دانیم $\left| x \right| = \left\{ \begin{gathered}
  x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \geqslant 0 \hfill \\
   - x\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \lt 0 \hfill \\ 
\end{gathered}  \right.$

برای این منظور قدرمطلق‌ها را تعیین علامت می‌کنیم:

$\left| {a - b - 2} \right| = $ و $a \lt b \Rightarrow a - b \lt 0 \Rightarrow a - b - 2 \Rightarrow $ کل عدد $ \lt 0 \Rightarrow \left| {a - b - 2} \right| =  - (a - b - 2)$

$\begin{gathered}
  \left| { - a - b} \right| = \left| { - (a + b)} \right| \Rightarrow a + b > 0 \Rightarrow  - (a + b) < 0 \Rightarrow \left| { - (a + b)} \right| =  - \left( { - (a + b)} \right) = a + b \hfill \\
  \left| { - 4a} \right| = a > 0 \Rightarrow  - 4a < 0 \Rightarrow \left| { - 4a} \right| =  - ( - 4a) = 4a \hfill \\ 
\end{gathered} $

$\begin{gathered}
   \Rightarrow \left| {a - b - 2} \right| + \left| { - a - b} \right| - 3\left| { - 4a} \right| =  \hfill \\
  \, - (a - b - 2) + (a + b) - 3(4a) = \cancel{{ - a}} + b + 2\cancel{{ + a}} + b - 12a = 2b - 12a + 2 = 2(b - 6a + 1) \hfill \\ 
\end{gathered} $