Compare the process of solving |x – 1| + 1 < 15 to that of solving |x – 1| + 1 > 15.

For |x - 1 | + 1 < 15 , we should first break the modulus function .

two cases are possible for this inequality ,
case 1 : - when x ≥ 1
x - 1 + 1 < 15 ⇒ x < 15
putting the number line both x ≥ 1 and x≤ 15
Then, x∈ [1, 15)

case 2:- when x < 1
-x + 1 + 1 < 15
⇒ -x + 2 < 15
⇒ -x < 13 ⇒ x > -13
putting the number line both x > -13 and x < 1
then, x∈ (-13, 1)

now, answer is x∈ (-13,1) ∪ [1, 15 ] or, x∈(-13, 15]

Again, for |x - 1| + 1 > 15
Similarly z there are two cases possible for this
case 1 :- when x ≥ 1
x - 1 + 1 > 15 ⇒ x > 15
Putting the number line both x ≥ 1 and x > 15
then, x ∈ [1, ∞)

case 2 :- when x < 1
-x + 1 + 1 > 15 ⇒ x < -13
putting the number line both x < 1 and x < -13
Then, x∈ (-∞, 1)
hence, final answer is x∈ (-∞ , 1) ∪ [1, ∞) or, x∈ (-∞, ∞) or x∈ R

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