Solve each of the following system of inequations in R x + 5 > 2(x +

1.	Solve each of the following system of inequations in R x + 5 > 2(x + 1), 2 – x < 3(x + 2)
Answer» 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 ⇒ \(\frac{4x}{4}\) > \(\frac{-4}{4}\) ⇒ 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).

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

Answer»

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

⇒ \(\frac{4x}{4}\) > \(\frac{-4}{4}\)

⇒ 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).