Solve exactly the inequality $\frac{3x-2}{x-1} \le \frac{4}{x+2}.$ Hence solve $\frac{3|x|-2}{|x|-1} \le \frac{4}{|x|+2}.$

1. Rearrange the inequality to have zero on one side and combine the terms into a single algebraic fraction.

$\frac{3x-2}{x-1} - \frac{4}{x+2} \le 0$
$\frac{(3x-2)(x+2) - 4(x-1)}{(x-1)(x+2)} \le 0$

2. Expand and simplify the numerator.

$\frac{(3x^2 + 6x - 2x - 4) - 4x + 4}{(x-1)(x+2)} \le 0$
$\frac{3x^2 + 4x - 4 - 4x + 4}{(x-1)(x+2)} \le 0$
$\frac{3x^2}{(x-1)(x+2)} \le 0$

3. Determine the critical values and solve the inequality using a sign test.

The critical values are $x = -2$, $x = 0$, and $x = 1$.
Since $x-1$ and $x+2$ are in the denominator, the expression is undefined at $x = 1$ and $x = -2$.
Testing the sign of $\frac{3x^2}{(x-1)(x+2)}$ in the intervals between critical values:

• For $x < -2$: positive ($+$)
• For $-2 < x < 0$: negative ($-$)
• At $x = 0$: zero ($0$)
• For $0 < x < 1$: negative ($-$)
• For $x > 1$: positive ($+$)

Thus, the inequality $\le 0$ is satisfied for $-2 < x < 1$.

1. Substitute $|x|$ for $x$ using the solution from part (a).

To use the 'Hence' command, we first recall the solution to part (a).
Solving $\frac{3x-2}{x-1} \le \frac{4}{x+2}$ gives:
$-2 < x < 1$
Substituting $|x|$ for $x$ in the inequality, we get:
$-2 < |x| < 1$

2. Solve the resulting modulus inequality for $x$.

Since the absolute value $|x|$ is always non-negative ($|x| \ge 0$), the left part of the inequality ($-2 < |x|$) is always satisfied for all real $x$.
Therefore, we only need to consider the right part:
$|x| < 1$
This gives the solution:
$-1 < x < 1$