Worksheet for solving compound inequalities with problems and number line graphs.

Let's solve each compound inequality step by step and then graph the solution on the number line. I'll go through each one carefully.

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1) $ m - 2 < 8 $ or $ \frac{m}{8} > 1 $

Solve each part:

- $ m - 2 < 8 $
Add 2 to both sides:
$ m < 10 $

- $ \frac{m}{8} > 1 $
Multiply both sides by 8:
$ m > 8 $

Now combine with "or":
We want values where $ m < 10 $ or $ m > 8 $.
This is true for all real numbers, because:

- If $ m > 8 $, it satisfies the second inequality.
- If $ m < 10 $, it satisfies the first.
- The only value not covered would be between 8 and 10? Wait — let’s think carefully.

Actually, let’s find the union of:
- $ m < 10 $
- $ m > 8 $

So:
- $ m > 8 $ covers (8, ∞)
- $ m < 10 $ covers (-∞, 10)

Union: $ (-\infty, 10) \cup (8, \infty) = (-\infty, \infty) $? No!

Wait:
- $ (-\infty, 10) \cup (8, \infty) $ → This covers everything except possibly between 8 and 10?

No — actually:
- From $ m < 10 $: all numbers less than 10
- From $ m > 8 $: all numbers greater than 8

So the union is all real numbers because:
- For $ m \leq 8 $: covered by $ m < 10 $
- For $ m \geq 10 $: covered by $ m > 8 $
- And in between: $ 8 < m < 10 $: covered by both

So union is all real numbers: $ (-\infty, \infty) $

But wait — what about $ m = 8 $?
- $ m = 8 $:
- $ m - 2 = 6 < 8 $ → true
- $ \frac{8}{8} = 1 > 1 $? No → false
But since it's OR, as long as one is true → yes, so included.

$ m = 10 $:
- $ m - 2 = 8 < 8 $? No
- $ \frac{10}{8} = 1.25 > 1 $? Yes → true → included

So even though $ m = 10 $ doesn't satisfy first, it does second → included

So overall: All real numbers are solutions.

✔ Solution: $ (-\infty, \infty) $

Graph: A number line shaded entirely from left to right.

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2) $ -1 < 9 + n < 17 $

This is a compound inequality with "and" implied.

Solve by subtracting 9 from all parts:

$ -1 - 9 < n < 17 - 9 $
$ -10 < n < 8 $

✔ Solution: $ (-10, 8) $

Graph: Open circle at -10, open circle at 8, shade in between.

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3) $ 2x < 10 $ or $ \frac{x}{2} \geq 3 $

Solve each:

- $ 2x < 10 $ → divide by 2 → $ x < 5 $
- $ \frac{x}{2} \geq 3 $ → multiply by 2 → $ x \geq 6 $

Now, OR: $ x < 5 $ or $ x \geq 6 $

So the solution is all numbers less than 5 or greater than or equal to 6.

That means: $ (-\infty, 5) \cup [6, \infty) $

Graph: Shade left of 5 (open circle), and from 6 onwards (closed circle at 6).

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4) $ x + 8 \geq 9 $ and $ \frac{x}{7} \leq 1 $

Solve each:

- $ x + 8 \geq 9 $ → $ x \geq 1 $
- $ \frac{x}{7} \leq 1 $ → $ x \leq 7 $

Now, AND: both must be true → $ x \geq 1 $ and $ x \leq 7 $

So: $ 1 \leq x \leq 7 $

✔ Solution: $ [1, 7] $

Graph: Closed circles at 1 and 7, shade between.

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5) $ -3 \leq \frac{p}{2} < 0 $

Multiply all parts by 2:

$ -6 \leq p < 0 $

✔ Solution: $ [-6, 0) $

Graph: Closed circle at -6, open circle at 0, shade in between.

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6) $ r + 5 \geq 12 $ or $ \frac{r}{9} \leq 0 $

Solve each:

- $ r + 5 \geq 12 $ → $ r \geq 7 $
- $ \frac{r}{9} \leq 0 $ → $ r \leq 0 $

OR: $ r \geq 7 $ or $ r \leq 0 $

So: $ (-\infty, 0] \cup [7, \infty) $

Graph: Shade left up to 0 (closed), and from 7 onward (closed at 7). Gap between 0 and 7.

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7) $ 7v - 5 \geq 65 $ or $ -3v - 2 \geq -2 $

Solve each:

- $ 7v - 5 \geq 65 $ → $ 7v \geq 70 $ → $ v \geq 10 $
- $ -3v - 2 \geq -2 $ → add 2: $ -3v \geq 0 $ → divide by -3 (flip inequality): $ v \leq 0 $

So: $ v \geq 10 $ or $ v \leq 0 $

Solution: $ (-\infty, 0] \cup [10, \infty) $

Graph: Shade left to 0 (closed), and from 10 onward (closed at 10). Gap between 0 and 10.

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8) $ -10b + 3 \leq -37 $ or $ 3b - 10 \leq -25 $

Solve each:

- $ -10b + 3 \leq -37 $ → subtract 3: $ -10b \leq -40 $ → divide by -10 (flip): $ b \geq 4 $
- $ 3b - 10 \leq -25 $ → add 10: $ 3b \leq -15 $ → divide by 3: $ b \leq -5 $

So: $ b \geq 4 $ or $ b \leq -5 $

Solution: $ (-\infty, -5] \cup [4, \infty) $

Graph: Shade left to -5 (closed), and from 4 onward (closed at 4). Gap between -5 and 4.

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✔ Final Answers Summary:

| Problem | Solution |
|--------|---------|
| 1 | All real numbers: $ (-\infty, \infty) $ |
| 2 | $ (-10, 8) $ |
| 3 | $ (-\infty, 5) \cup [6, \infty) $ |
| 4 | $ [1, 7] $ |
| 5 | $ [-6, 0) $ |
| 6 | $ (-\infty, 0] \cup [7, \infty) $ |
| 7 | $ (-\infty, 0] \cup [10, \infty) $ |
| 8 | $ (-\infty, -5] \cup [4, \infty) $ |

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📊 Graphing Instructions (for each):

- Use open circle for strict inequalities ($<$ or $>$)
- Use closed circle for inclusive inequalities ($\leq$ or $\geq$)
- Shade the region that satisfies the inequality
- For OR, combine both regions
- For AND, find intersection (overlap)

If you'd like, I can generate text-based graphs for each (using ASCII art), but since this is a printed worksheet, you can sketch them based on the intervals above.

Let me know if you'd like the graphs drawn out!