Worksheet for practicing graphing compound inequalities with number lines.

I'll solve this worksheet on graphing compound inequalities. Let me go through each problem step by step.

First, I need to understand what compound inequalities are and how to graph them. Compound inequalities involve two inequality statements joined by "and" or "or". When graphing, we represent the solution set on a number line.

Let me start with problem 21: x + 2 ≤ 6 or x/3 ≥ 0

For x + 2 ≤ 6:
- Subtract 2 from both sides: x ≤ 4

For x/3 ≥ 0:
- Multiply both sides by 3: x ≥ 0

Since this is an "or" statement, the solution is all x values that satisfy either condition: x ≤ 4 OR x ≥ 0.
This actually covers all real numbers since every number is either ≤ 4 or ≥ 0 (in fact, many numbers satisfy both).

Let me move to problem 22: -6 < x + 7 or x/2 > 0

For -6 < x + 7:
- Subtract 7 from both sides: -13 < x, or x > -13

For x/2 > 0:
- Multiply both sides by 2: x > 0

Since this is an "or" statement, the solution is x > -13 OR x > 0.
The combined solution is simply x > -13 (since if x > 0, it's automatically > -13).

Let me continue with problem 23: -2 < 3x ≤ 0

This is a compound inequality with "and" implied between the two parts.
- For -2 < 3x: divide by 3: -2/3 < x
- For 3x ≤ 0: divide by 3: x ≤ 0

So the solution is -2/3 < x ≤ 0.

Problem 24: 6x < -18 or x + 3 ≥ 2

For 6x < -18:
- Divide by 6: x < -3

For x + 3 ≥ 2:
- Subtract 3: x ≥ -1

Since this is "or", the solution is x < -3 OR x ≥ -1.

Problem 25: 1 ≤ x + 3 < 6

This is another compound inequality with "and" implied.
- For 1 ≤ x + 3: subtract 3: -2 ≤ x
- For x + 3 < 6: subtract 3: x < 3

So the solution is -2 ≤ x < 3.

Problem 26: -1 ≤ x + 1 ≤ 5

This is also a compound inequality with "and" implied.
- For -1 ≤ x + 1: subtract 1: -2 ≤ x
- For x + 1 ≤ 5: subtract 1: x ≤ 4

So the solution is -2 ≤ x ≤ 4.

Problem 27: x + 3 > 0 and x - 5 > -11

For x + 3 > 0: subtract 3: x > -3
For x - 5 > -11: add 5: x > -6

Since this is "and", we need both conditions to be true: x > -3 AND x > -6.
The more restrictive condition is x > -3, so the solution is x > -3.

Problem 28: -11 ≤ 2x < -9

This is a compound inequality with "and" implied.
- For -11 ≤ 2x: divide by 2: -11/2 ≤ x, or -5.5 ≤ x
- For 2x < -9: divide by 2: x < -4.5

So the solution is -5.5 ≤ x < -4.5.

Problem 29: x/2 < -1 or x + 4 ≥ 1

For x/2 < -1: multiply by 2: x < -2
For x + 4 ≥ 1: subtract 4: x ≥ -3

Since this is "or", the solution is x < -2 OR x ≥ -3.
This actually covers all real numbers since every number is either < -2 or ≥ -3 (in fact, many numbers satisfy both).

Problem 30: x + 5 > 8 and x/2 ≤ 2

For x + 5 > 8: subtract 5: x > 3
For x/2 ≤ 2: multiply by 2: x ≤ 4

Since this is "and", we need both conditions: x > 3 AND x ≤ 4.
So the solution is 3 < x ≤ 4.

Problem 31: x + 2 > 3 and x - 6 ≤ 2

For x + 2 > 3: subtract 2: x > 1
For x - 6 ≤ 2: add 6: x ≤ 8

Since this is "and", we need both conditions: x > 1 AND x ≤ 8.
So the solution is 1 < x ≤ 8.

Problem 32: 2/3x - 6 < 5

This appears to be a single inequality, not a compound one.
- Add 6 to both sides: 2/3x < 11
- Multiply both sides by 3/2: x < 33/2, or x < 16.5

To summarize the solutions for graphing:

21. All real numbers (x ≤ 4 OR x ≥ 0)
22. x > -13
23. -2/3 < x ≤ 0
24. x < -3 OR x ≥ -1
25. -2 ≤ x < 3
26. -2 ≤ x ≤ 4
27. x > -3
28. -5.5 ≤ x < -4.5
29. All real numbers (x < -2 OR x ≥ -3)
30. 3 < x ≤ 4
31. 1 < x ≤ 8
32. x < 16.5

For graphing these on a number line:
- Use open circles for strict inequalities (<, >)
- Use closed circles for inclusive inequalities (≤, ≥)
- Shade the appropriate regions based on the solution sets above.