Given

The above inequality can be split into two inequalities.

and

Let us consider the first inequality.

As x > 0, we have x + 1 > 0.

⇒ 4 ≤ 3(x + 1)

⇒ 4 ≤ 3x + 3

⇒ 3x + 3 ≥ 4

⇒ 3x + 3 – 3 ≥ 4 – 3

⇒ 3x ≥ 1

(1)

Now, let us consider the second inequality.

As x > 0, we have x + 1 > 0.

⇒ 3(x + 1) ≤ 6

⇒ 3x + 3 ≤ 6

⇒ 3x + 3 – 3 ≤ 6 – 3

⇒ 3x ≤ 3

⇒ x ≤ 1

∴ x ∈ (–∞, 1] (2)

From (1) and (2), we get

Thus, the solution of the given system of inequations is