Without using a calculator, solve

1. Rearrange the inequality to form a single rational expression compared to zero.

$1 + \frac{6}{x} - x \ge 0$
$\frac{x + 6 - x^2}{x} \ge 0$
$\frac{x^2 - x - 6}{x} \le 0$

2. Factorise the numerator to identify all critical values.

$\frac{(x - 3)(x + 2)}{x} \le 0$

The critical values are $x = -2$, $x = 0$, and $x = 3$.

3. Use a sign test to determine the intervals that satisfy the inequality, ensuring the denominator is not zero.

Testing the intervals separated by the critical values $-2$, $0$, and $3$ for the sign of $\frac{(x - 3)(x + 2)}{x}$:

• For $x < -2$: $\frac{(-)(-)}{(-)} < 0$
• For $-2 < x < 0$: $\frac{(-)(+)}{(-)} > 0$
• For $0 < x < 3$: $\frac{(-)(+)}{(+)} < 0$
• For $x > 3$: $\frac{(+)(+)}{(+)} > 0$

Including the boundaries where the expression equals zero (but excluding $x=0$ where it is undefined), we get:
$x \le -2 \text{ or } 0 < x \le 3$

1. Solve the rational inequality from part (i) to establish the base solution.

To solve part (ii) using the intended sequence, we first explicitly state the solution to part (i):
$1 + \frac{6}{x} \ge x$
$\frac{x + 6 - x^2}{x} \ge 0$
$\frac{(x-3)(x+2)}{x} \le 0$
Using a sign diagram for the critical values $-2$, $0$, and $3$, the solution to (i) is:
$x \le -2 \quad \text{or} \quad 0 < x \le 3$

2. Substitute $x$ with $|x|$ using the relationship between the two inequalities and reject the invalid case.

The inequality in (ii) is identical to (i) but with $x$ replaced by $|x|$. We substitute $|x|$ into the solution from (i):
$|x| \le -2 \quad \text{or} \quad 0 < |x| \le 3$
Since the absolute value of a real number is always non-negative ($|x| \ge 0$), the first inequality $|x| \le -2$ yields no real solutions.

3. Solve the valid modulus inequality to find the final range of values for $x$.

We solve the remaining inequality $0 < |x| \le 3$:
The condition $|x| > 0$ means $x \neq 0$.
The condition $|x| \le 3$ gives $-3 \le x \le 3$.
Combining these results, the final solution is:
$-3 \le x < 0 \quad \text{or} \quad 0 < x \le 3$