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The inequality shown is:

\[
\frac{3x + 1}{x - 2} \geq 5
\]

Let's solve this inequality step by step.

### Step 1: Eliminate the denominator
Multiply both sides by \( x - 2 \), but note that \( x - 2 \) could be positive or negative. We need to handle both cases separately:

1. When \( x - 2 > 0 \) (i.e., \( x > 2 \)), the inequality remains the same.
2. When \( x - 2 < 0 \) (i.e., \( x < 2 \)), we reverse the inequality sign.

#### Case 1: \( x > 2 \)
\[
\frac{3x + 1}{x - 2} \geq 5 \implies 3x + 1 \geq 5(x - 2)
\]

Now, expand and simplify:
\[
3x + 1 \geq 5x - 10
\]
\[
3x - 5x \geq -10 - 1
\]
\[
-2x \geq -11
\]
Divide both sides by \(-2\), which reverses the inequality:
\[
x \leq \frac{11}{2}
\]

So for \( x > 2 \), we have \( 2 < x \leq \frac{11}{2} \).

#### Case 2: \( x < 2 \)
\[
\frac{3x + 1}{x - 2} \geq 5 \implies 3x + 1 \leq 5(x - 2)
\]

Now, expand and simplify:
\[
3x + 1 \leq 5x - 10
\]
\[
3x - 5x \leq -10 - 1
\]
\[
-2x \leq -11
\]
Divide both sides by \(-2\) (again reversing the inequality):
\[
x \geq \frac{11}{2}
\]

Since this contradicts \( x < 2 \), there is **no solution** in this case.

### Final Solution:
Thus, the solution is \( 2 < x \leq \frac{11}{2} \).

Would you like further details or have any questions?

Here are some related questions:
1. How do you solve rational inequalities in general?
2. What happens when you multiply or divide by negative numbers in inequalities?
3. How can we graph the solution of a rational inequality?
4. What if the inequality was \( \frac{3x + 1}{x - 2} < 5 \)? How would the solution change?
5. What are other methods to solve rational inequalities, such as using a sign chart?

**Tip:** Always check the domain of rational expressions. In this case, \( x \neq 2 \), since the denominator would be zero!

Solve the inequality \( \frac{3x + 1}{x - 2} \geq 5 \)

Learn how to solve the rational inequality \( \frac{3x + 1}{x - 2} \geq 5 \) step-by-step. This solution involves key algebraic concepts, including simplifying rational expressions, handling inequalities, and correctly applying the sign rule when multiplying or dividing by negative numbers. Suitable for students in Grades 9-11.

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