To solve the rational inequality:

$\frac{x+2}{x-3} \geq 0$

Steps:

1. Find critical points: These are values of $x$ that make the numerator or denominator equal to zero.

   • Set $x+2 = 0 \Rightarrow x = -2$ (numerator).
   • Set $x-3 = 0 \Rightarrow x = 3$ (denominator).

   So, the critical points are $x = -2$ and $x = 3$. Notice that $x = 3$ will make the denominator zero, which means the function is undefined at this point.

2. Create test intervals: Divide the real line based on the critical points:

   • Interval 1: $(-\infty, -2)$
   • Interval 2: $(-2, 3)$
   • Interval 3: $(3, \infty)$

3. Test the sign in each interval: Pick a test point from each interval and determine whether the expression is positive or negative.

   • For $x = -3$ (from interval $(-∞, -2)$): $\frac{-3+2}{-3-3} = \frac{-1}{-6} = \frac{1}{6} > 0$ The expression is positive in this interval.

   • For $x = 0$ (from interval $(-2, 3)$): $\frac{0+2}{0-3} = \frac{2}{-3} = -\frac{2}{3} < 0$ The expression is negative in this interval.

   • For $x = 4$ (from interval $(3, ∞)$): $\frac{4+2}{4-3} = \frac{6}{1} = 6 > 0$ The expression is positive in this interval.

4. Analyze the critical points:

   • At $x = -2$, the numerator is zero, so $\frac{x+2}{x-3} = 0$. This satisfies the inequality because we are looking for $\geq 0$.
   • At $x = 3$, the denominator is zero, so the expression is undefined. Hence, $x = 3$ is excluded from the solution.

5. Conclusion: The solution includes the intervals where the expression is positive or zero: $(-\infty, -2] \cup (3, \infty)$

Final answer:

$x \in (-\infty, -2] \cup (3, \infty)$

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Would you like more details or have any questions?

Here are 5 related questions:

1. How do we solve rational inequalities with quadratic expressions?
2. What happens if both numerator and denominator contain factors?
3. Can we solve rational inequalities using a sign chart method?
4. How do we deal with strict inequalities (e.g., $> 0$) in rational functions?
5. How do we interpret the graph of a rational function with inequality constraints?

Tip: Always check critical points where the expression is undefined by setting the denominator to zero!