Question

`a<b<c<d`. How do you find the solution(s) of |x-a|+|x-c|=|x-b|+|x-d|? It is verifiable that, for `(a, b, c, d)=(1, 2, 3, 5), x=7/2` is a solution.

Answer 1

You need case analysis.

Explanation:

There are several cases, in the point view of the value of `x`.

[Case1]
If `x< a`, the equation will be

`-(x-a)-(x-c)=-(x-b)-(x-d)`
`-2x+a+c=-2x+b+d`
`a+c=b+d`

This result is inconsistent with the fact `a< b< c< d`. So there is no solutions in this case.

[Case2]
If `a<= x< b`,

`(x-a)-(x-c)=-(x-b)-(x-d)`
`-a+c=-2x+b+d`
`2x=a+b-c+d`
`x=(a+b-c+d)/2`

For example, if `(a,b,c,d)=(1,2,3,5)`, `x=(1+2-3+5)/2=5/2`.
But this is an inappropriate solution as the result is inconsistent with `1<= x <2`.

Then, proceed to the following cases.
[Case3] `b<= x< c`

`(x-a)-(x-c)=(x-b)-(x-d)`
`-a+c=-b+d`
`a+d=b+c`

If `a+d` equals to `b+c`, every `x` that satisfy `b<= x< c` is the solution. Otherwise, no `x` in this domain will be the answer.

[Case4] `c<= x< d`

`(x-a)+(x-c)=(x-b)-(x-d)`
`2x-a-c=-b+d`
`2x=a-b+c+d`
`x=(a-b+c+d)/2`

When `(a,b,c,d)=(1,2,3,5)`, the solution is
`x=(1-2+3+5)/2=7/2` and this is the appropriate solution.(`3<= 7/2 <5`

[Case5] `d<= x`
In this case, there is no solution. The reason is same as [Case1].

Answer 2

Here is the alternative way(drawing the graph)

Explanation:

Let `f(x)=abs(x-a)+abs(x-c)`.

This can be written as a piecewise function:
`f(x)=-2x+a+c` `(x< a)`
`f(x)=-a+c` `(a <= x < c)`
`f(x)=2x-a-c` `(c<= x)`

You can write `g(x)=abs(x-b)+abs(x-d)` as a piecewise function in the same way.

Then, draw the two graphs: `y=f(x)` and `y=g(x)`.
The graphs below is for `(a,b,c,d)=(1,2,3,5)`. You can see that the two graphs crosses at `(7/2,3)`.
graph{(abs(x-1)+abs(x-3)-y)(abs(x-2)+abs(x-5)-y)=0 [-6.05, 13.95, -1.64, 8.36]}