Solve the inequality $\frac{x-1}{x-2} \le \frac{3}{2}$.

1. Problem Description

Solve the inequality $\frac{x-1}{x-2} \le \frac{3}{2}$.

2. Solution Steps

First, subtract $\frac{3}{2}$ from both sides of the inequality:

$$\frac{x-1}{x-2} - \frac{3}{2} \le 0$$

Find a common denominator and combine the fractions:

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

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

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

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

$$\frac{x-4}{2(x-2)} \ge 0$$

$$\frac{x-4}{x-2} \ge 0$$

Now, we analyze the sign of the expression. The critical points are $x=2$ and $x=4$.

We have three intervals to consider: $(-\infty, 2)$, $(2, 4]$, and $[4, \infty)$.

Case 1: $x < 2$. Then $x-4 < 0$ and $x-2 < 0$. Thus, $\frac{x-4}{x-2} > 0$. So, $x \in (-\infty, 2)$.

Case 2: $2 < x \le 4$. Then $x-4 \le 0$ and $x-2 > 0$. Thus, $\frac{x-4}{x-2} \le 0$. So, $x \in (2, 4]$.

Case 3: $x > 4$. Then $x-4 > 0$ and $x-2 > 0$. Thus, $\frac{x-4}{x-2} > 0$. So, $x \in (4, \infty)$.

Also, we need to check the endpoints.

If $x = 4$, $\frac{4-4}{4-2} = \frac{0}{2} = 0 \ge 0$, which is true.

If $x = 2$, the expression is undefined.

Therefore, the solution is $x \in (-\infty, 2) \cup [4, \infty)$.

3. Final Answer

$x \in (-\infty, 2) \cup [4, \infty)$