Find the set of values of x satisfying 7x + 3 ≥ 3x - 5 and $\dfrac{x}{4} - 5 \le \dfrac{5}{4} - x, x$ ∈ N.

Given,

1st inequation : 7x + 3 ≥ 3x - 5

⇒ 7x + 3 ≥ 3x - 5

⇒ 7x - 3x ≥ -5 - 3

⇒ 4x ≥ -8

⇒ x ≥ $-\dfrac{8}{4}$

⇒ x ≥ -2 …………..(1)

2nd inequation : $\dfrac{x}{4} - 5 \le \dfrac{5}{4} - x$

$\Rightarrow \dfrac{x}{4} - 5 \le \dfrac{5}{4} - x \\[1em] \Rightarrow \dfrac{x}{4} + x \le \dfrac{5}{4} + 5 \\[1em] \Rightarrow \dfrac{x + 4x}{4} \le \dfrac{5 + 20}{4} \\[1em] \Rightarrow \dfrac{5x}{4} \le \dfrac{25}{4} \\[1em] \Rightarrow x \le \dfrac{25}{4} \times \dfrac{4}{5} \\[1em] \Rightarrow x \le 5 \text{ …………..(2)}$

From (1) and (2),

-2 ≤ x ≤ 5

Since, x ∈ N

Hence, solution set = {1, 2, 3, 4, 5}.