Solve the compound inequality.

3 – 2x ≥ 5 or 3(x – 2) + 1 > 7

Let's solve the compound inequality step by step.

### Inequality 1: \( 3 - 2x \geq 5 \)
1. Subtract 3 from both sides:
   \[
   -2x \geq 2
   \]
2. Divide both sides by -2, and remember to reverse the inequality sign when dividing by a negative:
   \[
   x \leq -1
   \]

So, from the first inequality, \( x \leq -1 \).

### Inequality 2: \( 3(x - 2) + 1 > 7 \)
1. Distribute the 3 on the left side:
   \[
   3x - 6 + 1 > 7
   \]
2. Simplify the left-hand side:
   \[
   3x - 5 > 7
   \]
3. Add 5 to both sides:
   \[
   3x > 12
   \]
4. Divide both sides by 3:
   \[
   x > 4
   \]

So, from the second inequality, \( x > 4 \).

### Conclusion
The solution to the compound inequality \( 3 - 2x \geq 5 \) or \( 3(x - 2) + 1 > 7 \) is:
\[
x \leq -1 \quad \text{or} \quad x > 4
\]

Would you like any further details or have any questions?

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Here are 5 related questions to explore:
1. How do we handle inequalities involving fractions?
2. What happens if there is an "and" instead of "or" in a compound inequality?
3. Can you solve a compound inequality where both parts include absolute values?
4. How do compound inequalities differ from single inequalities?
5. What is the graphical representation of compound inequalities?

**Tip**: Always reverse the inequality when multiplying or dividing both sides by a negative number!

Learn how to solve the compound inequality 3 – 2x ≥ 5 or 3(x – 2) + 1 > 7 with step-by-step instructions. This problem involves key algebraic concepts such as solving inequalities, distributing terms, and reversing inequality signs when dealing with negative coefficients. Suitable for students in Grades 9-10.

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