Absolute value problem $|mx+\alpha|+|mx+\beta|=c$ $x=2,4$ are the solutions of the equation 
$|mx+\alpha|+|mx+\beta|=c$
where $m>0$ and $\alpha,\beta,c$ are non-zero constants, then find the value of $$\left|\frac{\alpha+\beta}{m}\right|$$
My Attempt:
I am trying to use triangle inequality but not able to get the answer which is $6$.
 A: We have
$$|2m+\alpha|+|2m+\beta|=|4m+\alpha|+|4m+\beta|\tag1$$
We may suppose that $\alpha\le\beta$.


*

*Supposing that $\alpha\le\beta\le -4m$ gives $$(1)\iff -(2m+\alpha)-(2m+\beta)=-(4m+\alpha)-(4m+\beta)\iff m=0$$ which is impossible.

*Supposing that $\alpha\le -4m\le\beta\le -2m$ gives $$(1)\iff -(2m+\alpha)-(2m+\beta)=-(4m+\alpha)+(4m+\beta)\iff \beta=-2m$$Then, $|mx+\alpha|+|mx+\beta|=c$ has a solution $x=3$ since
$$|3m+\alpha|+|3m-2m|=-(3m+\alpha)+m=-(2m+\alpha)=|2m+\alpha|=c$$which contradicts that $x=2,4$ are the solutions.

*Supposing that $\alpha\le -4m\le -2m\le\beta$ gives that $|mx+\alpha|+|mx+\beta|=c$ has a solution $x=3$ since$$|3m+\alpha|+|3m+\beta|=-(3m+\alpha)+(3m+\beta)=\beta-\alpha=|2m+\alpha|+|2m+\beta|=c$$which contradicts that $x=2,4$ are the solutions.

*Supposing that $-4m\le \alpha\le -2m\le \beta$ gives $$(1)\iff -(2m+\alpha)+(2m+\beta)=(4m+\alpha)+(4m+\beta)\iff \alpha=-4m$$Then, $|mx+\alpha|+|mx+\beta|=c$ has a solution $x=3$ since$$|3m-4m|+|3m+\beta|=m+(3m+\beta)=4m+\beta=|2m-4m|+|2m+\beta|=c$$which contradicts that $x=2,4$ are the solutions.

*Supposing that $-2m\le\alpha\le\beta$ gives $$(1)\iff (2m+\alpha)+(2m+\beta)=(4m+\alpha)+(4m+\beta)\iff m=0$$ which is impossible.
Therefore, we have to have
$$-4m\lt \alpha\le\beta\lt -2m$$
for which we have $$(1)\iff -(2m+\alpha)-(2m+\beta)=(4m+\alpha)+(4m+\beta)\iff \alpha+\beta=-6m$$
It follows from this that$$\left|\frac{\alpha+\beta}{m}\right|=6$$
