The equation can be re–written as

Adding 1 both the sides, we get,

Subtracting 3 both the sides

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 ϵ (1, ∞) ….(1)

Case 2: x+2<0

⇒ x<–2

In this case, |x+2|=–(x+2)

Considering Numerator,

4x+2>0

ut x<–2

Now, from Denominator, we have–

⇒ x ϵ (–∞ , 0) …(2)

(1, ∞) (from 1 and 2)

We can verify the answers using graph as well.