Answer:
Step-by-step explanation:
||x-3|-2|≤ 1
-1≤|x-3|-2≤1
-1+2≤|x-3|≤1+2
1≤|x-3|≤2
we break in two parts.
|x-3|≥1
so x-3≤-1
x≤3-1
x≤2 ...(1)
or x-3≥1
x≥4 ...(2)
|x-3|≤2
-2≤x-3≤2
-2+3≤x≤2+3
1≤x≤5 ...(3)
combining all three
1≤x≤2 ∪ 4≤x≤5
