Given x + 5 > 2(x + 1) and 2 – x < 3(x + 2)

Let us consider the first inequality.

x + 5 > 2(x + 1)

⇒ x + 5 > 2x + 2

⇒ x + 5 – 5 > 2x + 2 – 5

⇒ x > 2x – 3

⇒ 2x – 3 < x

⇒ 2x – 3 + 3 < x + 3

⇒ 2x < x + 3

⇒ 2x – x < x + 3 – x

⇒ x < 3

∴ x ∈ (–∞, 3) (1)

Now, let us consider the second inequality.

2 – x < 3(x + 2)

⇒ 2 – x < 3x + 6

⇒ 2 – x – 2 < 3x + 6 – 2

⇒ –x < 3x + 4

⇒ 3x + 4 > –x

⇒ 3x + 4 – 4 > –x – 4

⇒ 3x > –x – 4

⇒ 3x + x > –x + x – 4

⇒ 4x > –4

⇒ x > –1

∴ x ∈ (–1, ∞) (2)

From (1) and (2), we get

x ∈ (–∞, 3) ∩ (–1, ∞)

∴ x ∈ (–1, 3)

Thus, the solution of the given system of inequations is (–1, 3).