doesn't look easy at all. Absolute value algebra is a nightmare!

Lets see, if x < 1 then the expression |x -1| will be - (x-1). If x > 4, then |8 - x| will be -(8 - 2x). Fortunatly, x can never be both less then 1 and greater then 4 so its either one or the other. Not both. Of course, if x is neither less then 1, nor greater then 4, then both the abolute value symbols can be eliminated. So lets check all three situations.

if x < 1

-(x - 1) - ( 8 - 2x) = 3 solved x = 10. 10 is not less then 1 so we discard this solution.

if (x > 4)

(x - 1) + (8 - 2x) = 3 solved x = 4. 4 is not less then 4 so we discard this solution.

if (1 <= x <=4)

(x - 1) - (8 - 2x) = 3. solved x = 4. 4 is greater then 1 and less then 4 so this is an acceptable solution.

So if my thinking is correct, x = 4 should be the solution.