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The problem provided asks you to solve the inequality for \( x \), where \( x \) is in the set of real numbers excluding 1:

\[
\frac{3x + 1}{x - 1} \leq 6
\]

### Step-by-Step Solution:

1. **Start by isolating the fraction:**

   \[
   \frac{3x + 1}{x - 1} \leq 6
   \]

   Subtract 6 from both sides to bring all terms to one side:

   \[
   \frac{3x + 1}{x - 1} - 6 \leq 0
   \]

2. **Combine the terms on the left side:**

   Rewrite 6 as \( \frac{6(x - 1)}{x - 1} \) to combine the fractions:

   \[
   \frac{3x + 1 - 6(x - 1)}{x - 1} \leq 0
   \]

   Simplify the numerator:

   \[
   3x + 1 - 6x + 6 = -3x + 7
   \]

   So the inequality becomes:

   \[
   \frac{-3x + 7}{x - 1} \leq 0
   \]

3. **Determine the critical points:**

   The inequality can change sign at the points where the numerator or the denominator is zero.

   - The numerator \(-3x + 7 = 0\) gives \(x = \frac{7}{3}\).
   - The denominator \(x - 1 = 0\) gives \(x = 1\).

4. **Analyze the sign of the expression:**

   We analyze the expression \( \frac{-3x + 7}{x - 1} \) in the intervals determined by the critical points \(x = 1\) and \(x = \frac{7}{3}\).

   - For \( x < 1 \): Both numerator and denominator are positive, so the expression is negative.
   - For \( 1 < x < \frac{7}{3} \): Numerator is positive, denominator is positive, so the expression is positive.
   - For \( x > \frac{7}{3} \): Numerator is negative, denominator is positive, so the expression is negative.

5. **Determine where the expression is less than or equal to zero:**

   The expression \( \frac{-3x + 7}{x - 1} \leq 0 \) holds in the intervals where the expression is negative or zero. The intervals where this happens are:

   \[
   (-\infty, 1) \cup \left[ \frac{7}{3}, \infty \right)
   \]

   However, \( x = 1 \) is excluded since the denominator becomes zero, causing the expression to be undefined at that point.

   Therefore, the solution set is:

   \[
   (-\infty, 1) \cup \left[ \frac{7}{3}, \infty \right)
   \]

### Final Answer:

The solution set for the inequality is:

\[
x \in (-\infty, 1) \cup \left[ \frac{7}{3}, \infty \right)
\]

Would you like more details, or do you have any questions?

Here are five related questions:

1. How would the solution change if the inequality was \( \frac{3x + 1}{x - 1} > 6 \)?
2. What would happen if \( x = 1 \) were allowed in the domain?
3. How do you solve the inequality \( \frac{ax + b}{cx + d} \leq k \) in general?
4. Can this inequality be solved graphically, and if so, how?
5. What are the implications if the inequality sign changes from \( \leq \) to \( \geq \)?

**Tip:** Always check the critical points where the numerator or denominator equals zero, as these can change the sign of the inequality.

Solve the following inequality for x ∈ ℝ, x ≠ 1: (3x + 1)/(x - 1) ≤ 6.

Explore a detailed step-by-step solution for solving the inequality (3x + 1)/(x - 1) ≤ 6, including identifying critical points and analyzing the sign of the rational function. This algebraic problem is suitable for students in Grades 10-12.

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