Without using a calculator, solve the inequality $\dfrac{4x}{x-3} \ge 1. Hence find the range of values of $x$ that satisfies $\dfrac{4|x|}{|x|-3} \ge 1$.

1. Rearrange the inequality to compare with zero and simplify into a single fraction.

$\frac{4x}{x-3} - 1 \ge 0$\n$\frac{4x - (x - 3)}{x-3} \ge 0$\n$\frac{3x + 3}{x - 3} \ge 0$

2. Identify critical values and determine the solution range for $x$.

The critical values are found by equating the numerator and denominator to zero:
$3x + 3 = 0 \implies x = -1$
$x - 3 = 0 \implies x = 3$
Since the inequality is $\ge 0$ and $x=3$ makes the fraction undefined, we test the intervals:

• For $x < -1$, $\frac{3x+3}{x-3} > 0$
• For $-1 < x < 3$, $\frac{3x+3}{x-3} < 0$
• For $x > 3$, $\frac{3x+3}{x-3} > 0$

Thus, the solution is $x \le -1$ or $x > 3$.

3. Use the result to solve the 'hence' part by substituting $x$ with $|x|$.

The inequality $\frac{4|x|}{|x|-3} \ge 1$ is identical to the original inequality, with $x$ replaced by $|x|$.
Substituting $|x|$ into our earlier solution gives:
$|x| \le -1 \quad \text{or} \quad |x| > 3$

4. Evaluate the modulus inequalities to find the final range of values.

The inequality $|x| \le -1$ has no real solutions because the absolute value of a real number is always non-negative ($|x| \ge 0$).

For the second inequality:
$|x| > 3 \implies x < -3 \quad \text{or} \quad x > 3$