Find the set of values of x, satisfying :

7x + 3 ≥ 3x - 5 and $\dfrac{x}{4} - 5 \le \dfrac{5}{4} - x$, where x ∈ N.

Solving,

⇒ 7x + 3 ≥ 3x - 5

⇒ 7x - 3x ≥ -5 - 3

⇒ 4x ≥ -8

⇒ x ≥ -2 …….(i)

Solving,

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

From (i) and (ii) we get,

-2 ≤ x ≤ 5

Since, x ∈ N

∴ Solution set = {1, 2, 3, 4, 5}.