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We are given the inequality:

\[
\frac{-5x - 6}{x - 8} \leq -6
\]

### Step-by-Step Solution:

1. **Multiply both sides by \((x - 8)\)**:
   To eliminate the denominator, we multiply both sides by \(x - 8\), but we need to consider two cases: whether \(x - 8 > 0\) or \(x - 8 < 0\), because multiplying by a negative number flips the inequality sign.

   \[
   \frac{-5x - 6}{x - 8} \cdot (x - 8) \leq -6 \cdot (x - 8)
   \]
   This simplifies to:
   \[
   -5x - 6 \leq -6(x - 8)
   \]
   Now, distribute the \(-6\) on the right-hand side:
   \[
   -5x - 6 \leq -6x + 48
   \]

2. **Combine like terms**:
   Add \(6x\) to both sides to bring the terms involving \(x\) together:
   \[
   x - 6 \leq 48
   \]
   Now, add \(6\) to both sides:
   \[
   x \leq 54
   \]

3. **Consider restrictions on \(x\)**:
   The denominator \(x - 8\) cannot be zero, so \(x \neq 8\).

4. **Solution intervals**:
   Since \(x - 8 > 0\) when \(x > 8\), and \(x - 8 < 0\) when \(x < 8\), the inequality is valid for the solution \(x \leq 54\) with the restriction that \(x \neq 8\).

   Therefore, the solution is:

   \[
   x \in (-\infty, 8) \cup (8, 54]
   \]

Would you like any further details or explanations on these steps? 

Here are five related questions for further exploration:
1. What happens to the inequality if we change the right-hand side to a different number?
2. How do we handle inequalities involving absolute values?
3. Can inequalities like this be solved graphically? How would that work?
4. What if the numerator or denominator was quadratic? How would that affect the solution process?
5. How can we check the solution by plugging in values for \(x\)?

**Tip**: Always consider the domain of the inequality, especially when there's a variable in the denominator, as it can lead to restrictions on possible solutions.

Learn how to solve the rational inequality -5x - 6 / (x - 8) ≤ -6 with a step-by-step solution. This problem covers key algebraic concepts such as inequalities, rational expressions, and domain restrictions. Suitable for students in Grades 9-12.

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