Let's work through the rational inequalities from the image one by one.

Problem 1: $\frac{x - 7}{x - 1} < 0$

1. Find the critical points: These occur when the numerator and denominator are zero.

   • $x - 7 = 0 \Rightarrow x = 7$ (numerator)
   • $x - 1 = 0 \Rightarrow x = 1$ (denominator)

2. Sign analysis: Break the real number line into intervals using the critical points: $(-\infty, 1), (1, 7), (7, \infty)$.

   • Test each interval to determine the sign of $\frac{x - 7}{x - 1}$:
     • For $x = 0$ (in $(- \infty, 1)$): $\frac{0 - 7}{0 - 1} = \frac{-7}{-1} = 7$, which is positive.
     • For $x = 2$ (in $(1, 7)$): $\frac{2 - 7}{2 - 1} = \frac{-5}{1} = -5$, which is negative.
     • For $x = 8$ (in $(7, \infty)$): $\frac{8 - 7}{8 - 1} = \frac{1}{7}$, which is positive.

3. Solution: We want the expression to be negative, so the solution is $1 < x < 7$.

Problem 2: $\frac{x + 5}{x - 4} \leq 0$

1. Find the critical points:

   • $x + 5 = 0 \Rightarrow x = -5$
   • $x - 4 = 0 \Rightarrow x = 4$

2. Sign analysis: Use intervals $(- \infty, -5), (-5, 4), (4, \infty)$.

   • For $x = -6$ (in $(- \infty, -5)$): $\frac{-6 + 5}{-6 - 4} = \frac{-1}{-10} = 0.1$, positive.
   • For $x = 0$ (in $(-5, 4)$): $\frac{0 + 5}{0 - 4} = \frac{5}{-4} = -1.25$, negative.
   • For $x = 5$ (in $(4, \infty)$): $\frac{5 + 5}{5 - 4} = \frac{10}{1} = 10$, positive.

3. Solution: Since the inequality is $\leq 0$, we include the points where it equals zero, giving the solution $[-5, 4)$.

Problem 3: $\frac{x + 32}{x + 6} \leq 3$

1. Rewrite inequality: Move all terms to one side: $\frac{x + 32}{x + 6} - 3 \leq 0$ $\frac{x + 32 - 3(x + 6)}{x + 6} \leq 0$ $\frac{x + 32 - 3x - 18}{x + 6} \leq 0$ $\frac{-2x + 14}{x + 6} \leq 0$

2. Find the critical points:

   • $-2x + 14 = 0 \Rightarrow x = 7$
   • $x + 6 = 0 \Rightarrow x = -6$

3. Sign analysis: Use intervals $(- \infty, -6), (-6, 7), (7, \infty)$.

   • For $x = -7$: $\frac{-2(-7) + 14}{-7 + 6} = \frac{14 + 14}{-1} = -28$, negative.
   • For $x = 0$: $\frac{-2(0) + 14}{0 + 6} = \frac{14}{6} \approx 2.33$, positive.
   • For $x = 8$: $\frac{-2(8) + 14}{8 + 6} = \frac{-16 + 14}{14} = \frac{-2}{14} \approx -0.14$, negative.

4. Solution: Since the inequality is $\leq 0$, the solution is $(- \infty, -6) \cup [7, \infty)$.

Would you like me to continue solving the rest, or would you like a detailed explanation for any of these steps?

Here are 5 related questions to consider:

1. How can you determine the intervals for testing the sign in rational inequalities?
2. What is the significance of the critical points in solving rational inequalities?
3. How do you solve inequalities involving complex rational expressions?
4. How would the solutions change if we had a strict inequality instead of a non-strict one?
5. Can inequalities be solved by graphing? If so, how would that approach differ?

Tip: Always watch for points where the denominator equals zero; these are excluded from the solution since division by zero is undefined.