Find the range of values of x, which satisfy 7 ≤ –4x + 2 < 12, x ∈ R. Graph these values of x on the real number line.

Given,
7 ≤ –4x + 2 < 12

$\Rightarrow 7 \le - 4x + 2 \text{ and } - 4x + 2 \lt 12 \\[0.5em]$

Solving left side,

$7 \le - 4x + 2 \\[0.5em] \Rightarrow 4x \le 2-7 \\[0.5em] \Rightarrow 4x \le -5 \\[0.5em] \Rightarrow x \le \dfrac{-5}{4}$

Solving right side,

$-4x + 2 \lt 12 \\[0.5em] \Rightarrow -4x \lt 12 - 2 \\[0.5em] \Rightarrow -4x \lt 10 \\[0.5em] \Rightarrow 4x \gt -10 \\[0.5em] \Rightarrow x \gt \dfrac{-5}{2}$

∴ Solution set = {x : x ∈ R, $-\dfrac{5}{2}$ < x ≤ $-\dfrac{5}{4}$}.

The graph of the inequation is represented by thick black line starting from $-\dfrac{5}{2}$ (excluding $-\dfrac{5}{2}$) till $-\dfrac{5}{4}$ (including $-\dfrac{5}{4}$).