https://file.aiquickdraw.com/mathbot/file/17254791418339ej7cdyu.jpg

The inequality in the image is:

\[
\frac{1}{x} > 2
\]

To solve this inequality for \( x \), we proceed as follows:

### Step 1: Isolate \( x \)
Multiply both sides of the inequality by \( x \) to eliminate the fraction. However, since \( x \) could be positive or negative, we must consider two cases.

**Case 1: \( x > 0 \)**
- If \( x > 0 \), then multiplying by \( x \) doesn't change the inequality sign:
  \[
  1 > 2x \implies x < \frac{1}{2}
  \]

**Case 2: \( x < 0 \)**
- If \( x < 0 \), then multiplying by \( x \) will reverse the inequality sign:
  \[
  1 < 2x \implies x > \frac{1}{2}
  \]
  However, this would contradict our initial assumption that \( x < 0 \). Thus, there is no valid solution in this case.

### Step 2: Combine the results
From Case 1, the solution is \( x < \frac{1}{2} \), provided \( x > 0 \).

### Final Solution
The solution to the inequality is:
\[
0 < x < \frac{1}{2}
\]

This means \( x \) must be positive and less than \( \frac{1}{2} \).

Would you like any additional details or have any further questions?

### Follow-up Questions:
1. What is the solution to \(\frac{1}{x} < -2\)?
2. How does the solution change if the inequality is \(\frac{1}{x} \geq 2\)?
3. What happens if \( \frac{1}{x} > -2 \)?
4. Can you solve \(\frac{1}{x} > \frac{1}{y}\) where \(x\) and \(y\) are both positive?
5. How do you solve a quadratic inequality such as \(x^2 - 4x + 3 < 0\)?

### Tip:
When solving inequalities involving fractions, always consider the sign of the variable to correctly determine the direction of the inequality.

Learn how to solve the inequality 1/x > 2 with a detailed step-by-step solution. This problem involves key algebraic concepts, such as inequalities and the multiplication property of inequalities. Suitable for students in Grades 9-11.

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