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).

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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!