Find possible distances between C and E 
A, B, C, D and E lies on a line, not necessarily in that order.
  Distance between A and B is $2$, distance between B and C is $3$, distance
  between C and D is $5$ and distance between D and E is $4$.  Find  the possible 
  distances between C and E?

I find this type of problems difficult to solve. I don't understand clearly about the orders. In this particular case, I think if A, B, C, and D are placed in order and E is placed in both sides of D, then two possible distances are possible between C and E, which is $9$ and $1$. But what if possible distances between A and E are to be found? Can anyone clearly elaborate the process and concepts behind these line problems?
 A: Let us denote by lower case letters $a,b,c,d,e$ the abscissas of points $A,B,C,D,E$ resp.
Two basic observations about this issue:


*

*it is "up to a translation" which allows to take $E$ as the origin, i.e., $e=0$.

*it is "up to a symmetry", which allows $D$ to be set at abscissa $d \geq 0$. 
Consider the two last constraints, the other constraints being useless.
$$\begin{cases}|c-d|&=&5\\d&=&4\end{cases} \ \ \Rightarrow \ \ |c-4|=5.$$
Thus 2 cases:


*

*$c=-1$; thus $CE=1$.

*$c=9$; thus $CE=9$.
Edit: Here is a tree explaining how the other possible positions of points $A,B,C$ can be found (truly speaking,  it is not exactly a tree because two branches meet in a "leaf"...). Taking a branch in the upwards direction like $(A_3,B_1,C_1,D)$ provides a solution ($E$ being fixed).

A: given
equation1 $b-a = 2$
equation2 $c-b = 3$
equation3 $d-c = 5$
equation4 $e-d = 4$
need to find $e - c$ and $e - a$
solution:
add equation3 and equation4 and you get $e-c = 9$
add all of them and you get $e - a = 14$
