<s>[INST] Denoting $2\le\frac{6-4x}5\le3$ by $|mx-n|\le5$. What is the value of $|n-m|$? 
If we denote solution set of the inequality $2\le\frac{6-4x}5\le3$ by
$|mx-n|\le5$, what is the value of $|n-m|$?
$1)7\qquad\qquad2)5\qquad\qquad3)21\qquad\qquad4)23$

I solved this problem with following approach:
$$2\le\frac{6-4x}5\le3\quad\Rightarrow-\frac94\le x\le-1$$
And $|mx-n|\le5$ is equivalent to $\frac{-5+n}m\le x\le\frac{5+n}{m}$. Hence we have $\frac{-5+n}m=-\frac94$ and $\frac{5+n}m=-1$ and by solving system of equations I got  $m=8$ and $n=-13$. So the final answer is $21$.
I wonder, can we solve this problem with other approahces?
 [/INST] $$2 \le \frac{6-4x}{5} \le 3$$
I first centralize it,
$$2-\frac52 \le  \frac{6-4x}{5}-\frac52\le 3 - \frac52$$
$$-\frac12 \le \frac{12-8x-25}{10}\le \frac12$$
$$-1 \le \frac{-8x -13}{5}\le 1$$
$$|-8x-(13)|\le 5$$
Hence $|n-m|=|13-(-8)|=21.$
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