Solve each of the following system of inequations in R
x + 5 > 2(x + 1), 2 – x < 3(x + 2)
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).