We need to solve the inequality $\frac{x-1}{x-2} \le \frac{3}{2}$.

1. Problem Description

We need to solve the inequality $\frac{x-1}{x-2} \le \frac{3}{2}$.

2. Solution Steps

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

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

Next, we find a common denominator and combine the fractions:

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

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

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

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

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

Now we analyze the sign of the expression $\frac{x-4}{x-2}$. We have two critical points, $x = 2$ and $x = 4$.

We consider three intervals:

1. $x < 2$: In this interval, $x-4 < 0$ and $x-2 < 0$, so $\frac{x-4}{x-2} > 0$.

2. $2 < x < 4$: In this interval, $x-4 < 0$ and $x-2 > 0$, so $\frac{x-4}{x-2} < 0$.

3. $x > 4$: In this interval, $x-4 > 0$ and $x-2 > 0$, so $\frac{x-4}{x-2} > 0$.

Since we want $\frac{x-4}{x-2} \ge 0$, we need to consider the intervals where the expression is positive or zero.

$x < 2$ gives us $\frac{x-4}{x-2} > 0$.

$x > 4$ gives us $\frac{x-4}{x-2} > 0$.

When $x = 4$, $\frac{x-4}{x-2} = \frac{4-4}{4-2} = \frac{0}{2} = 0$, so $x=4$ is a solution.

When $x=2$, the expression $\frac{x-4}{x-2}$ is undefined, so $x=2$ is not a solution.

Thus, the solution is $x < 2$ or $x \ge 4$.

3. Final Answer

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