Question

Solve for x, |x + 1| + |x| >3 

Answer 1

Two terms, each having modulus, are present on the LHS of the stated inequality. We obtain x = -1,0 as the crucial points by equating the equation contained within the modulus to zero. These pivotal spots split the actual line into three segments as follows:
(−∞, −1),[−1,0),[0. ∞).
Case - I : 
When −∞ < x < −1 
|x + 1| + |x| > 3  
⇒ −x− 1 − x > 3  
⇒ x < −2. 
Case - II  
When −1 ≤ x < 0 
x + 1| + |x| > 3  
⇒ x + 1 − x > 3
⇒ 1 > 3 
Case - III  
When 0 ≤ < ∞, 
x + 1| + |x| > 3 
⇒ x + 1 + x > 3 
⇒ x > 1. 
Combining the results of cases (I) (II) and (III), we get,
x ∈ (−∞, −2),∪ [1, ∞)