Given 3x – 1 ≥ 5 and x + 2 > –1

Let us consider the first inequality.

3x – 1 ≥ 5

⇒ 3x – 1 + 1 ≥ 5 + 1

⇒ 3x ≥ 6

⇒ x ≥ 2

∴ x ∈ (2, ∞) (1)

Now, let us consider the second inequality.

x + 2 > –1

⇒ x + 2 – 2 > –1 – 2

⇒ x > –3

∴ x ∈ (–3, ∞) (2)

From (1) and (2), we get

x ∈ (2, ∞) ∩ (–3, ∞)

∴ x ∈ (2, ∞)

Thus, the solution of the given system of inequations is (2, ∞).