Absolute Value is confusing.

HERE IS A VERY LONG AND DETAILED SOLUTION. GO THROUGH IT LINE BY LINE SLOWLY AND CAREFULLY. GOOD LUCK:

Solve for x over the real numbers:
abs(x-9) = abs(x-5)+abs(x-1)

Split the equation into two possible cases:
x-9 = abs(x-5)+abs(x-1) or x-9 = -abs(x-5)-abs(x-1)

Subtract abs(x-5)+abs(x-1) from both sides:
-9+x-abs(x-5)-abs(x-1) = 0 or x-9 = -abs(x-5)-abs(x-1)

Subtract -9+x-abs(x-1) from both sides:
-abs(x-5) = 9-x+abs(x-1) or x-9 = -abs(x-5)-abs(x-1)

Multiply both sides by -1:
abs(x-5) = -9+x-abs(x-1) or x-9 = -abs(x-5)-abs(x-1)

Split the equation into two possible cases:
x-5 = -9+x-abs(x-1) or x-5 = 9-x+abs(x-1) or x-9 = -abs(x-5)-abs(x-1)

Subtract -9+x-abs(x-1) from both sides:
4+abs(x-1) = 0 or x-5 = 9-x+abs(x-1) or x-9 = -abs(x-5)-abs(x-1)

Subtract 4 from both sides:
abs(x-1) = -4 or x-5 = 9-x+abs(x-1) or x-9 = -abs(x-5)-abs(x-1)

By definition, absolute value is positive. Therefore abs(x-1) = -4 has no solution:
x-5 = 9-x+abs(x-1) or x-9 = -abs(x-5)-abs(x-1)

Subtract 9-x+abs(x-1) from both sides:
-14+2 x-abs(x-1) = 0 or x-9 = -abs(x-5)-abs(x-1)

Subtract 2 x-14 from both sides:
-abs(x-1) = 14-2 x or x-9 = -abs(x-5)-abs(x-1)

Multiply both sides by -1:
abs(x-1) = 2 x-14 or x-9 = -abs(x-5)-abs(x-1)

Split the equation into two possible cases:
x-1 = 2 x-14 or x-1 = 14-2 x or x-9 = -abs(x-5)-abs(x-1)

Subtract 2 x-1 from both sides:
-x = -13 or x-1 = 14-2 x or x-9 = -abs(x-5)-abs(x-1)

Multiply both sides by -1:
x = 13 or x-1 = 14-2 x or x-9 = -abs(x-5)-abs(x-1)

Add 2 x+1 to both sides:
x = 13 or 3 x = 15 or x-9 = -abs(x-5)-abs(x-1)

Divide both sides by 3:
x = 13 or x = 5 or x-9 = -abs(x-5)-abs(x-1)

Add abs(x-5)+abs(x-1) to both sides:
x = 13 or x = 5 or -9+x+abs(x-5)+abs(x-1) = 0

Subtract -9+x+abs(x-1) from both sides:
x = 13 or x = 5 or abs(x-5) = 9-x-abs(x-1)

Split the equation into two possible cases:
x = 13 or x = 5 or x-5 = 9-x-abs(x-1) or x-5 = -9+x+abs(x-1)

Subtract 9-x-abs(x-1) from both sides:
x = 13 or x = 5 or -14+2 x+abs(x-1) = 0 or x-5 = -9+x+abs(x-1)

Subtract 2 x-14 from both sides:
x = 13 or x = 5 or abs(x-1) = 14-2 x or x-5 = -9+x+abs(x-1)

Split the equation into two possible cases:
x = 13 or x = 5 or x-1 = 14-2 x or x-1 = 2 x-14 or x-5 = -9+x+abs(x-1)

Add 2 x+1 to both sides:
x = 13 or x = 5 or 3 x = 15 or x-1 = 2 x-14 or x-5 = -9+x+abs(x-1)

Divide both sides by 3:
x = 13 or x = 5 or x = 5 or x-1 = 2 x-14 or x-5 = -9+x+abs(x-1)

Subtract 2 x-1 from both sides:
x = 13 or x = 5 or x = 5 or -x = -13 or x-5 = -9+x+abs(x-1)

Multiply both sides by -1:
x = 13 or x = 5 or x = 5 or x = 13 or x-5 = -9+x+abs(x-1)

Subtract -9+x+abs(x-1) from both sides:
x = 13 or x = 5 or x = 5 or x = 13 or 4-abs(x-1) = 0

Subtract 4 from both sides:
x = 13 or x = 5 or x = 5 or x = 13 or -abs(x-1) = -4

Multiply both sides by -1:
x = 13 or x = 5 or x = 5 or x = 13 or abs(x-1) = 4

Split the equation into two possible cases:
x = 13 or x = 5 or x = 5 or x = 13 or x-1 = 4 or x-1 = -4

Add 1 to both sides:
x = 13 or x = 5 or x = 5 or x = 13 or x = 5 or x-1 = -4

Add 1 to both sides:
x = 13 or x = 5 or x = 5 or x = 13 or x = 5 or x = -3

abs(x-9) ⇒ abs(-9-3) = 12
abs(x-5)+abs(x-1) ⇒ abs(-5-3)+abs(-1-3) = 12:
So this solution is correct

abs(x-9) ⇒ abs(5-9) = 4
abs(x-5)+abs(x-1) ⇒ abs(5-5)+abs(5-1) = 4:
So this solution is correct

abs(x-9) ⇒ abs(13-9) = 4
abs(x-5)+abs(x-1) ⇒ abs(13-5)+abs(13-1) = 20:
So this solution is incorrect

The solutions are:
Answer: | x = 5  or   x = -3

 |x-1| + |x-5| = |x-9|         we can write this as

√[(x -1)^2] + √ [(x - 5)^2]   = √ [(x - 9)^2 ]         square both sides

(x - 1)^2 + 2√[ (x -5)^2 * (x -1)^2] + (x -5)^2   =  (x - 9)^2           simplify

x^2 - 2x + 1 + 2√[ (x -5)^2 * (x -1)^2]  + x^2 - 10x + 25   =  x^2 - 18x + 81

 2√[ (x -5)^2 * (x -1)^2]  =   -x^2 - 6x + 55

 2√[ (x -5)^2 * (x -1)^2]  =   - [x^2 + 6x - 55]   factor the right side

 2√[ (x -5)^2 * (x -1)^2] =  - (x + 11)(x - 5)        square both sides

4 (x - 5)^2 * (x - 1)^2   = (x + 11)^2 (x - 5)^2

4(x - 5)^2 * (x - 1)^2  - (x - 5)^2 (x + 11)^2   = 0         factor out (x - 5)^2

(x - 5)^2   [ 4(x - 1)^2 - (x + 11)^2 ]   =  0        simplify

(x - 5)^2 [ 4x^2 - 8x + 4  - x^2 - 22x - 121]   = 0

(x - 5)^2 [ 3x^2 - 30x - 117]  = 0

(x - 5)^2 [ x^2 - 10x - 39] * 3  = 0        divide through by 3   and factor

( x - 5)^2 [ (x - 13) ( x + 3) ]  =  0

So.........setting each factor to 0 we have that    x = 5  or x = 13   or x = -3

x = 13   is an extraneous solution produced by squaring both sides........reject it

The other two solutions x = 5   or x = -3      are good

I think it almost always pays to plot a graph to get a handle on these sorts of questions.

Here's a quicker analytic method.

As a lead in, consider the first term, I x - 1 I.

If x is greater than 1, it would be calculated as  x - 1,

while if x is less than 1 it would calculated as 1 - x.

Same sort of thing for the others at x = 5 and x = 9.

(i) If x is less than 1, the equation becomes (1 - x) + (5 - x) = (9 - x), solution x = -3.

(ii) If x is less than 5 but greater than 1,  (x - 1) + (5 - x) = (9 - x), solution x = 5.

(iii) If x is greater than 9, (x - 1) + (x - 5) = (x - 9), solution x = -3.