If the replacement set is the set of whole numbers, solve :

(i) x + 7 ≤ 11

(ii) 3x - 1 > 8

(iii) 8 - x > 5

(iv) 7 - 3x ≥ $-\dfrac{1}{2}$

(v) $x - \dfrac{3}{2} \lt \dfrac{3}{2} - x$

(vi) $18 \le 3x - 2$

(i) x + 7 ≤ 11

⇒ x ≤ 11 - 7

⇒ x ≤ 4

Since, x ∈ W

∴ Solution set = {0, 1, 2, 3, 4}.

(ii) 3x - 1 > 8

⇒ 3x > 8 + 1

⇒ 3x > 9

Dividing both sides by 3 we get,

⇒ x > 3

Since, x ∈ W

∴ Solution set = {4, 5, 6, ……..}.

(iii) 8 - x > 5

⇒ -x > 5 - 8

⇒ -x > -3

Multiplying both sides by (-1) we get,

⇒ x < 3 (As on multiplying by negative no. the sign reverses.)

Since, x ∈ W

∴ Solution set = {0, 1, 2}.

(iv) 7 - 3x ≥ $-\dfrac{1}{2}$

⇒ -3x ≥ $-\dfrac{1}{2} - 7$

⇒ -3x ≥ $-\dfrac{15}{2}$

Dividing both sides by (-3) we get,

⇒ x ≤ $\dfrac{5}{2}$ (As on dividing by negative no. the sign reverses.)

Since, x ∈ W

∴ Solution set = {0, 1, 2}.

(v) Given,

$\Rightarrow x - \dfrac{3}{2} \lt \dfrac{3}{2} - x \\[1em] \Rightarrow x + x \lt \dfrac{3}{2} + \dfrac{3}{2} \\[1em] \Rightarrow 2x \lt \dfrac{6}{2} \\[1em] \Rightarrow 2x \lt 3 \\[1em] \Rightarrow x \lt \dfrac{3}{2} \\[1em] \Rightarrow x \lt 1.5$

Since, x ∈ W

∴ Solution set = {0, 1}.

(vi) 18 ≤ 3x - 2

⇒ 3x ≥ 18 + 2

⇒ 3x ≥ 20

⇒ x ≥ $\dfrac{20}{3}$

⇒ x ≥ $6\dfrac{2}{3}$

Since, x ∈ W

∴ Solution set = {7, 8, 9, …..}.