Based on the first inequality, we know that $-9\textless3 x+12\textless9$ because $3 x+2$ must be less than 9 units away from zero. We can subtract 12 from all three parts of the inequality to arrive at $-21\textless3 x\textless-3 \rightarrow-7\textless x\textless-1$. From the second inequality we can rewrite $\textbar-3 x-6\textbar$ as $\textbar3 x+6\textbar=3\textbar x+2\textbar$ because they must be equal. The second inequality must be true for all numbers except for when $\textbar x+2\textbar=0$, or when $x=-2$. Thus the answer includes all numbers from $-7$ to $-1$ with the exception of $-2$. The answer is $\mathbf{D}$. 
Find all values of $x$ such that $\textbar3 x+12\textbar\textless9$ and $\textbar x+2\textbar\textless\textbar-3 x-6\textbar$.~\uline{~~~~~~~~~~}~ 
Options:
- $x\textless-2$ 
- $-7\textless x\textless-1$ 
- $-7\textless x\textless-2$ 
- $-7\textless x\textless-1 ; x 
eq-2$ 