What do you mean by open modulus?
The way I would solve this problem is by evaluating how I would calculate the absolute values. I would note down the points at which the "opening of the absolute functions" changes.
|x+3| : changes when x < -3;
|4-x| : changes when x > 4;
|8+x| : changes when x < -8
So you have to evaluate the equality in the cases below:
Case1: if x < -8;
|x+3| would "open" as -(x+3); say x =-9, x+3 = -9+3 = -6 Since outcome is negative, it opens with negative multiplier
|4-x| would "open" as (4-x); say x = -9, 4-x = 4 - (-9) = 13 Since outcome is positive, it opens with positive multiplier
|8+x| would "open" as -(8+x); you can use the above logic for this as well.
Putting all together:
-(x+3) - (4-x) = -(8+x)
-x-3-4+x=-8-x
x=-1; this is not a valid answer since we started with the assumption that x<-8.
Case2: if -8<=x<-3
|x+3| would "open" as -(x+3)
|4-x| would "open" as (4-x)
|8+x| would "open" as (8+x)
Putting all together:
-(x+3) - (4-x) = (8+x)
-x-3-4+x=8+x
-7=8+x
x=-15; this is not a valid answer since we started with the assumption that -8<=x<-3
Case3: if -3<=x<4;
|x+3| would "open" as (x+3)
|4-x| would "open" as (4-x)
|8+x| would "open" as (8+x)
Putting all together:
(x+3)-(4-x) = (8+x)
x+3-4+x=8+x
2x-1=8+x
x=9; this is not a valid answer since we started with the assumption that -3<=x<=4
Case4: if 4<=x
|x+3| would "open" as (x+3)
|4-x| would "open" as -(4-x)
|8+x| would "open" as (8+x)
Putting it all together:
(x+3)- (-(4-x))=(8+x)
x+3+4-x=8+x
x=-1; this is not a valid answer since we started with the assumption that 4<=x
Result:
The answer is 0 since we could not find any solutions.
By the way, in answer to your first question about picking roots and plugging them in: No, you cannot. This is not an inequality problem. You have to solve the x.
The answer of your second question is in the Case1.
Hope this helps
