Answer: 0 ≤ x ≤ 2 or 8 ≤ x ≤ 10
=========================================
Work Shown
The rule I'll use is |x| < k becomes -k < x < k when k is positive
||x-5| - 4| ≤ 1
-1 ≤ |x-5| - 4 ≤ 1 ........ use the rule mentioned above
-1+4 ≤ |x-5| - 4+4 ≤ 1+4
3 ≤ |x-5| ≤ 5
3 ≤ |x-5| and |x-5| ≤ 5
We'll solve each piece separately in the next two sections.
------------
3 ≤ |x-5|
|x-5| ≥ 3
Use the rule that |x| ≥ k becomes x ≥ k or x ≤ -k when k is positive
So |x-5| ≥ 3 becomes x-5 ≥ 3 or x-5 ≤ -3
x-5 ≥ 3 solves to x ≥ 8
x-5 ≤ -3 solves to x ≤ 2
------------
use the rule that |x| ≤ k leads to -k ≤ x ≤ k when k is positive.
|x-5| ≤ 5
-5 ≤ x-5 ≤ 5
-5+5 ≤ x-5+5 ≤ 5+5
0 ≤ x ≤ 10
------------
Solving 3 ≤ |x-5| led to x ≤ 2 or x ≥ 8
Solving |x-5| ≤ 5 led to 0 ≤ x ≤ 10
Overlap the intervals to see that we get the disjoint separate intervals of 0 ≤ x ≤ 2 or 8 ≤ x ≤ 10