Let's solve the rational inequality:

$\frac{x + 12}{x + 2} \leq 2$

Step 1: Rewrite the inequality

We want to express this inequality in a standard form. Subtract 2 from both sides:

$\frac{x + 12}{x + 2} - 2 \leq 0$

Now, get a common denominator on the left-hand side:

$\frac{x + 12}{x + 2} - \frac{2(x + 2)}{x + 2} \leq 0$

This simplifies to:

$\frac{x + 12 - 2(x + 2)}{x + 2} \leq 0$

Distribute and combine like terms in the numerator:

$\frac{x + 12 - 2x - 4}{x + 2} \leq 0$

$\frac{-x + 8}{x + 2} \leq 0$

Step 2: Solve the inequality

We now solve the inequality:

$\frac{-x + 8}{x + 2} \leq 0$

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

1. Numerator: $-x + 8 = 0 \Rightarrow x = 8$
2. Denominator: $x + 2 = 0 \Rightarrow x = -2$ (Note: $x = -2$ is a vertical asymptote, so it cannot be included in the solution).

Step 3: Determine the sign changes

We need to test the sign of the expression $\frac{-x + 8}{x + 2}$ in the intervals determined by $x = -2$ and $x = 8$. These intervals are:

• $(- \infty, -2)$
• $(-2, 8)$
• $(8, \infty)$

Test a point in each interval:

• In $(- \infty, -2)$, pick $x = -3$: $\frac{-(-3) + 8}{-3 + 2} = \frac{3 + 8}{-1} = \frac{11}{-1} = -11 \quad (\text{negative})$

• In $(-2, 8)$, pick $x = 0$: $\frac{-(0) + 8}{0 + 2} = \frac{8}{2} = 4 \quad (\text{positive})$

• In $(8, \infty)$, pick $x = 9$: $\frac{-(9) + 8}{9 + 2} = \frac{-1}{11} = -\frac{1}{11} \quad (\text{negative})$

Step 4: Write the solution

The inequality $\frac{-x + 8}{x + 2} \leq 0$ holds where the expression is negative or zero. This happens in the intervals $(- \infty, -2)$ and $(8, \infty)$, as well as at $x = 8$ where the expression is zero.

Thus, the solution is:

$(-\infty, -2) \cup [8, \infty)$

However, $x = -2$ is excluded because it makes the denominator zero, so it is not included in the solution.

Would you like more details on any of these steps?

Here are five related questions:

1. How do you solve rational inequalities involving strict inequalities?
2. What is the difference between solving rational and polynomial inequalities?
3. How can vertical asymptotes affect the solution of rational inequalities?
4. How do you find the critical points for rational expressions?
5. Can you explain why we test intervals around critical points in inequalities?

Tip: Always exclude values that make the denominator zero when solving rational inequalities.