Answer :

Subtracting 3 from both sides, we get–


|x+1|+|x|–3>0


For this, we have 3 cases,


Case 1:–∞ <x<–1


For this, |x+1|=–(x+1) and |x|=–x


–(x+1)–x–3>0


–2x–1–3>0


2x+4<0


x<–2


x ϵ (–∞ , –2) …(1)


Case 2: –1<x<0


For this, |x+1|=x+1 and |x|=–x


x+1–x–3>0


–2>0


Which is absurd, thus it leads to no solution.


Case 3: 0<x<∞


For this, |x+1|=x+1 and |x|=x


x+1+x–3>0


2x–2>0


x>1


x ϵ (1 , ∞ ) …(2)


xϵ (–∞ , –2) (1 , ∞ ) (from 1 and 2)


We can verify the answers using graph as well.



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