To solve the inequality $\frac{|2x - 1|}{|x + 5|} \leq 0$, let's analyze it step by step.

Step 1: Consider the nature of absolute values

• The absolute value $|2x - 1|$ is always non-negative, i.e., $|2x - 1| \geq 0$.
• The absolute value $|x + 5|$ is also always non-negative, i.e., $|x + 5| \geq 0$.

Step 2: Analyze the inequality $\frac{|2x - 1|}{|x + 5|} \leq 0$

• A fraction is less than or equal to zero only when the numerator is zero and the denominator is non-zero.

Step 3: Set the numerator to zero

$|2x - 1| = 0$ $2x - 1 = 0 \implies x = \frac{1}{2}$

Step 4: Ensure the denominator is non-zero at $x = \frac{1}{2}$

$|x + 5| \neq 0 \implies x + 5 \neq 0 \implies x \neq -5$

Step 5: Combine the results

• The inequality $\frac{|2x - 1|}{|x + 5|} \leq 0$ holds only when $x = \frac{1}{2}$, since this makes the numerator zero and the entire expression zero.

Final Answer:

$x = \frac{1}{2}$

This is the only solution.

Would you like more details or have any questions? Here are five related questions for further exploration:

1. What happens if the inequality is strictly less than zero, $\frac{|2x - 1|}{|x + 5|} < 0$?
2. How would the solution change if the absolute values were not present?
3. What if we consider a similar inequality but with a different function, like $\frac{|x^2 - 4|}{|x + 5|} \leq 0$?
4. Can you solve the inequality $\frac{|2x - 1|}{x + 5} \leq 0$ without absolute value in the denominator?
5. How does the solution process differ if the inequality is non-strict, such as $\frac{|2x - 1|}{|x + 5|} \geq 0$?

Tip: When working with absolute values in inequalities, carefully consider the cases where the expression inside the absolute value is zero, as this often defines critical points for the solution.