Given,
|x+2|/x < 2
The equation can be re–written as
\(\frac{|x+2|}{x}\) - 1 < 2
Adding 1 both the sides, we get,
\(\frac{|x+2|}{x}\) < 3
Subtracting 3 both the sides,
⇒ \(\frac{|x+2|}{x}\) - 3 < 0
Clearly,
x≠0,
As it will lead equation unmeaningful.
Now,
Two case arise :
Case1: x+2>0
⇒ x>–2
In this case,
|x+2| = x+2
Considering Numerator,
2x–2>0
⇒ x>1
⇒ x ∈ (1, ∞) ….(1)
Case 2 : x+2<0
⇒ x<–2
In this case,
|x+2| = –(x+2)
Considering Numerator,
4x+2>0
⇒ x > \(-\frac{1}{2}\)
But x<–2
Now,
From Denominator, we have–
⇒ x ∈ (–∞ , 0) …(2)
⇒ x ∈ (–∞ , 0) ∪ (1, ∞) (from 1 and 2)
We can verify the answers using graph as well.
