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?

....or... from the other direction
$|mx- n| \le 5$
$-5 < mx - n < 5$
Stretch those so the end points are only one apart
$\frac {-5}{10} < \frac {mx-n}{10} < \frac {5}{10}$.
Then shift so that the endpoints line up where we want.
$\frac {-5}{10} + 2\frac 12 < \frac {mx-n}{10}+2\frac 12 < \frac 5{10} +2\frac 12$.
We need $2<\frac m{10}x + (\frac 52 -\frac n{10}) < 3 \iff 2< \frac{6-4x}5 < 3$
.... I suppose we can give a handwavy argument for why that would require $\frac m{10} = -\frac 45$ and for $\frac 52-\frac n{10} = \frac 65$
that is to say why $a < kx + j < b \iff a< wx + v < b$ must imply $k=w$ and (if $k\ne 0$) $j=v$.  I think your method of solving directly does that.  I could argue that if $j\ne v$ or $k\ne w$ there will always be wiggle room to slide an $x$ so that one inequality is true and the other not.  Handwavy.... but true.
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An astute observer might notice that once I chose the assymetrce shift of adding $2\frac 12$ to every term I committed $m$ to equaling $-8$ and $n$ to equalling $13$ (the exact opposite of your values).
That is because as long as we have $-M < something < M$ we could just as well have $-M < -something < M$.  But once we commit to one and we "shift the origin" ... well, the die is cast.
