So let's say I want to solve $$|x-a|\pm|x-b|=c$$ Using the classic multiple cases approach, one can show that the solutions are given by $$x=\frac{a+b\pm c}2 $$ But how can one make sense of this geometrically?
(I think it has to do with the same thing as the solution to $|x-a|=c$, however here we add another distance term and so we take half of that quantity...)
First, let's look at $|x-a|+|x-b|=c$.
On the number line, $|x-a|$ is the distance between points $x$ and $a$. So $|x-a|+|x-b|$ is the sum of the distances from $x$ to $a$ and from $x$ to $b$. If $x$ is to the right of $a$ and $b$, we get this diagram (where the thick blue segment is $|x-a|$ and the thinner red segment is $|x-b|$):
We can shorten the segment between $a$ and $x$ and lengthen the segment between $b$ and $x$ by half the distance between $a$ and $b$, and we get this (where we now consider the green segment twice):
If we let $c=|x-a|+|x-b|$ from the first diagram, the second diagram shows that also $$c=2\left(x-\frac{a+b}2\right)$$ which leads us directly to $$x=\frac{a+b-c}2$$ If we do the diagrams with $x$ to the left of $a$ and $b$, we get the sign of $c$ reversed and get instead $$x=\frac{a+b+c}2$$ Thus, $$x=\frac{a+b \pm c}2$$
Your other equation $|x-a|-|x-b|=c$ would be handled with $x$ between $a$ and $b$. The diagrams are similar, but I get the solution $$x=\frac{a+b+c}2$$ without the minus before $c$. So if I am right, your offered formula is not quite right.