To solve the inequality:

\[
\frac{2x}{x - 1} \leq 0
\]

We need to follow these steps:

### Step 1: Analyze the function
The inequality is of the form \(\frac{A}{B} \leq 0\), where \(A = 2x\) and \(B = x - 1\). The fraction \(\frac{A}{B}\) will be negative or zero when:

- \(A\) and \(B\) have opposite signs, or
- The fraction is equal to zero when the numerator is zero.

### Step 2: Find when the fraction is equal to zero
The fraction \(\frac{2x}{x - 1} = 0\) when the numerator \(2x = 0\). Solving for \(x\):
\[
2x = 0 \implies x = 0
\]

Thus, \(x = 0\) is a solution to the equality part of the inequality.

### Step 3: Determine when the fraction is negative
We need to figure out when \(\frac{2x}{x - 1}\) is negative. This depends on the signs of \(2x\) and \(x - 1\):

- \(2x > 0\) when \(x > 0\)
- \(2x < 0\) when \(x < 0\)

- \(x - 1 > 0\) when \(x > 1\)
- \(x - 1 < 0\) when \(x < 1\)

Now, divide the real number line into intervals based on the critical points \(x = 0\) and \(x = 1\).

#### Interval 1: \(x < 0\)
In this interval:
- \(2x < 0\)
- \(x - 1 < 0\)

Since both the numerator and denominator are negative, the fraction \(\frac{2x}{x - 1}\) is **positive** in this interval.

#### Interval 2: \(0 < x < 1\)
In this interval:
- \(2x > 0\)
- \(x - 1 < 0\)

Since the numerator is positive and the denominator is negative, the fraction \(\frac{2x}{x - 1}\) is **negative** in this interval.

#### Interval 3: \(x > 1\)
In this interval:
- \(2x > 0\)
- \(x - 1 > 0\)

Since both the numerator and denominator are positive, the fraction \(\frac{2x}{x - 1}\) is **positive** in this interval.

### Step 4: Combine the results
From the analysis:
- The fraction is negative when \(0 < x < 1\).
- The fraction equals zero when \(x = 0\).

Thus, the solution to the inequality \(\frac{2x}{x - 1} \leq 0\) is:

\[
x \in [0, 1)
\]

### Final Answer:
\[
x \in [0, 1)
\]

This means that \(x\) can be any value from 0 (inclusive) to 1 (exclusive).

---

Do you want more details or have any questions?

Here are 5 related questions to practice:
1. Solve \(\frac{3x}{x + 2} \leq 0\).
2. Solve \(\frac{4x}{x - 3} > 0\).
3. Solve \(\frac{5x}{2x + 1} < 0\).
4. Solve \(\frac{x^2 - 4}{x - 2} \geq 0\).
5. Solve \(\frac{x - 3}{x^2 - 4x} \leq 0\).

**Tip:** Always check for points where the denominator equals zero; these are critical points for solving rational inequalities!

Learn how to solve the rational inequality 2x / (x - 1) ≤ 0 with detailed steps. This solution includes an analysis of critical points and sign intervals. Suitable for students in Grades 10-12 studying inequalities and rational expressions.

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