Solve: `(|x+3|+x)/(x+2)>1`

Explanation

When `x + 3 ≥ 0, i.e., x ≥ – 3`.
Then
⇒ `{(x + 1) > 0 and x + 2 > 0} or {x + 1 < 0 and x + 2 < 0}`

⇒ `{x > –1 and x > –2} or {x < – 1 and x < – 2}`

⇒ `x > –1 or x < – 2 ⇒ x ∈ (–1, ∞) or x ∈ (– ∞, – 2) ⇒ x ∈ (–3, –2) ∪ ( – 1, ∞)`

[Since x ≥ – 3] ... (1)
Case II

When `x + 3 < 0, i.e., x < – 3`

`⇒ (x + 5 < 0 and x + 2 > 0) or (x + 5 > 0 and x + 2 < 0)`
`⇒ (x < – 5 and x > – 2) or (x > – 5 and x < –2)` it is not possible.

`⇒ x ∈ (– 5 , – 2) ... (2)`
Combining (I) and (II), the required solution is x ∈ (– 5 , – 2) ∪ (– 1, ∞)