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The inequality you provided is:

\[
3 \geq \frac{a}{33} + 2
\]

To solve this inequality, follow these steps:

1. **Isolate** the variable term \(\frac{a}{33}\):
   \[
   3 - 2 \geq \frac{a}{33}
   \]
   \[
   1 \geq \frac{a}{33}
   \]

2. **Multiply** both sides by 33 to eliminate the denominator:
   \[
   33 \times 1 \geq a
   \]
   \[
   33 \geq a
   \]

   This simplifies to:
   \[
   a \leq 33
   \]

The solution to the inequality is \(a \leq 33\). Now, let's check which values from the options provided are solutions:

- \(a = 132\): This does **not** satisfy \(a \leq 33\), so it's not a solution.
- \(a = 66\): This does **not** satisfy \(a \leq 33\), so it's not a solution.
- \(a = 99\): This does **not** satisfy \(a \leq 33\), so it's not a solution.
- \(a = 33\): This **does** satisfy \(a \leq 33\), so it's a solution.

Thus, **the correct solution** is \(a = 33\).

---

Do you want more details or have any further questions about this problem? Here are 5 related questions to deepen your understanding:

1. What is the process for solving inequalities with fractions?
2. How do you handle inequalities when multiplying or dividing by negative numbers?
3. Can you describe what happens when both sides of an inequality contain the variable?
4. How do you graph the solution of an inequality on a number line?
5. What are the differences between solving equalities and inequalities?

**Tip:** Always reverse the inequality symbol when multiplying or dividing both sides of an inequality by a negative number.

Which of the following are solutions to the inequality 3 ≥ a/33 + 2?

Learn how to solve the inequality 3 ≥ a/33 + 2 by isolating the variable and determining valid solutions. This detailed solution covers algebraic concepts such as inequality manipulation and multiplication rules. Suitable for students in Grades 6-8.

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