You are dealing with an inequality of the form:

\[
\frac{x - 9}{2(x - 6)} \geq 0
\]

To solve this inequality, let's follow these steps:

### 1. **Find the critical points:**
   - The critical points occur where the numerator or the denominator is zero.
   - **Numerator:** \(x - 9 = 0 \Rightarrow x = 9\)
   - **Denominator:** \(2(x - 6) = 0 \Rightarrow x = 6\)

### 2. **Determine the sign of the expression:**
   We need to check the sign of the expression \(\frac{x - 9}{2(x - 6)}\) in the intervals defined by the critical points. The intervals are:
   - \(x < 6\)
   - \(6 < x < 9\)
   - \(x > 9\)

### 3. **Test the sign in each interval:**
   - **Interval 1:** \(x < 6\) (e.g., \(x = 0\))  
     \(\frac{0 - 9}{2(0 - 6)} = \frac{-9}{-12} > 0\)  
     The expression is positive.
   
   - **Interval 2:** \(6 < x < 9\) (e.g., \(x = 7\))  
     \(\frac{7 - 9}{2(7 - 6)} = \frac{-2}{2} < 0\)  
     The expression is negative.

   - **Interval 3:** \(x > 9\) (e.g., \(x = 10\))  
     \(\frac{10 - 9}{2(10 - 6)} = \frac{1}{8} > 0\)  
     The expression is positive.

### 4. **Include critical points:**
   - At \(x = 9\), the expression equals 0 (which satisfies the inequality).
   - At \(x = 6\), the denominator is zero, so the expression is undefined (exclude this value).

### 5. **Conclusion:**
   The solution includes the intervals where the expression is positive or zero:

\[
x \in (-\infty, 6) \cup [9, \infty)
\]

Would you like more details or have any questions?

Here are some related questions to consider:

1. How does the inequality change if it becomes strict, i.e., \( \frac{x - 9}{2(x - 6)} > 0 \)?
2. What if the denominator was not multiplied by 2?
3. How would the intervals change if the inequality was \( \leq 0 \)?
4. Can you graph the solution on a number line?
5. What if the expression had a quadratic numerator?

**Tip:** Always remember to check where the expression is undefined, as these points may create intervals in the solution.

Learn how to solve rational inequalities of the form (x - 9) / (2(x - 6)) >= 0 with detailed steps. This problem involves identifying critical points, analyzing intervals, and determining when the expression is positive or zero. Suitable for students in Grades 10-12.

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