2.4Compound Inequalities
Two inequalities joined by and or or — one describes a bounded range, the other an unbounded union. Learn to solve, graph, and write both types.
Why This Matters
Compound inequalities describe ranges with two boundaries — like acceptable blood pressure levels or safe operating temperatures in engineering. They're also the foundation for absolute value inequalities in Algebra 2.
Workbook
Lesson, vocabulary, worked examples, and practice problems.
Essential Question
How does joining two inequalities with AND versus OR change the solution set, and how do you represent each type on a number line?
Lesson Overview
A compound inequality combines two inequalities using the word AND or the word OR. The connecting word determines the shape of the solution set entirely. An AND compound inequality (also called a conjunction) requires both inequalities to be true at the same time — the solution is the intersection of the two solution sets, which forms a bounded segment on the number line. It is often written as a three-part (double) inequality such as −2 < x ≤ 5. An OR compound inequality (also called a disjunction) requires at least one inequality to be true — the solution is the union of the two solution sets, which forms two rays pointing in opposite directions. To solve either type, isolate the variable in each inequality separately, then combine the results according to the connecting word.
AND vs. OR — At a Glance
AND — Intersection
- Both inequalities must be true simultaneously.
- Solution is the overlap of both sets.
- Graph: a segment between two endpoints.
- Interval notation: (a, b) or [a, b] etc.
- If the two sets don't overlap → no solution.
Graph: 1 < x ≤ 5
Open at 1, closed at 5 → (1, 5]
OR — Union
- At least one inequality must be true.
- Solution is the combination of both sets.
- Graph: two rays pointing away from each other.
- Interval notation: (−∞, a) ∪ (b, ∞)
- If the two sets cover everything → all real numbers.
Graph: x < −2 OR x ≥ 4
Open at −2, closed at 4 → (−∞, −2) ∪ [4, ∞)
Shortcut — Solving a Double Inequality
When an AND inequality is already written in three-part form (e.g., −1 ≤ 2x + 3 < 9), apply the same operation to all three sections at once.
−1 ≤ 2x + 3 < 9
subtract 3 from all three parts:
−4 ≤ 2x < 6
divide all three parts by 2 (positive → no flip):
−2 ≤ x < 3 → [−2, 3)
If you divide all three parts by a negative, flip BOTH inequality symbols.
Worked Examples
Solve, graph, and write in interval notation: x > −1 AND x ≤ 4
Both inequalities are already solved — no algebra needed.
AND → find the intersection: values that satisfy BOTH x > −1 and x ≤ 4.
Intersection: −1 < x ≤ 4
Graph: open circle at −1, closed circle at 4, shade the segment between them.
Interval notation: (−1, 4]
Check: Test x = 2: 2 > −1 ✓ and 2 ≤ 4 ✓ — in solution.
Check: Test x = 5: 5 > −1 ✓ but 5 ≤ 4 ✗ — not in solution (correct).
Graph: −1 < x ≤ 4
Open at −1, closed at 4 · segment → (−1, 4]
Solve, graph, and write in interval notation: x < −3 OR x ≥ 2
Both inequalities are already solved.
OR → find the union: values that satisfy EITHER x < −3 or x ≥ 2.
Graph: open circle at −3, shade left; closed circle at 2, shade right.
These two rays do not overlap — the solution is two separate pieces.
Interval notation: (−∞, −3) ∪ [2, ∞)
Check: Test x = −5: −5 < −3 ✓ — in solution.
Check: Test x = 0: 0 < −3 ✗ and 0 ≥ 2 ✗ — not in solution (correct).
Graph: x < −3 OR x ≥ 2
Open at −3, closed at 2 · two rays → (−∞, −3) ∪ [2, ∞)
Solve and graph: 2x − 1 > 3 AND 2x − 1 ≤ 9
Solve each inequality separately.
Left: 2x − 1 > 3 → 2x > 4 → x > 2
Right: 2x − 1 ≤ 9 → 2x ≤ 10 → x ≤ 5
AND → intersection: x > 2 AND x ≤ 5 → 2 < x ≤ 5
Graph: open circle at 2, closed circle at 5, shade between.
Interval notation: (2, 5]
Check: Test x = 4: 2(4) − 1 = 7 > 3 ✓ and 7 ≤ 9 ✓.
Graph: 2 < x ≤ 5
Open at 2, closed at 5 · segment → (2, 5]
Solve and graph: 3x + 2 < −4 OR 3x + 2 ≥ 11
Solve each inequality separately.
Left: 3x + 2 < −4 → 3x < −6 → x < −2
Right: 3x + 2 ≥ 11 → 3x ≥ 9 → x ≥ 3
OR → union: x < −2 OR x ≥ 3
Graph: open circle at −2, shade left; closed circle at 3, shade right.
Interval notation: (−∞, −2) ∪ [3, ∞)
Check: Test x = −5: 3(−5) + 2 = −13 < −4 ✓.
Check: Test x = 0: 2 < −4? No. 2 ≥ 11? No. Not in solution ✓.
Graph: x < −2 OR x ≥ 3
Open at −2, closed at 3 · two rays → (−∞, −2) ∪ [3, ∞)
Solve the double inequality: −5 < −2x + 1 ≤ 9
Subtract 1 from all three parts:
−5 − 1 < −2x + 1 − 1 ≤ 9 − 1
−6 < −2x ≤ 8
Divide all three parts by −2 (NEGATIVE → flip BOTH symbols):
−6 ÷ (−2) > −2x ÷ (−2) ≥ 8 ÷ (−2)
3 > x ≥ −4
Rewrite with smaller value on the left: −4 ≤ x < 3
Graph: closed circle at −4, open circle at 3, shade between.
Interval notation: [−4, 3)
Check: Test x = 0: −5 < −2(0) + 1 = 1 ≤ 9 ✓.
Graph: −4 ≤ x < 3
Closed at −4, open at 3 · segment → [−4, 3)
Guided Practice
Answers are in the Answer Key section.
Guided Practice Video: Compound Inequalities
Watch the guided practice walkthrough for compound inequalities, then complete the problems below.
Video by Sang Real Math
Watch on YouTube ↗Solve and graph: x ≥ −2 AND x < 5
Hint: AND → intersection. Both must be true. Overlap: −2 ≤ x < 5. Closed circle at −2, open circle at 5, shade between. Interval: [−2, 5).
Graph your answer (x ≥ −2 AND x < 5):
Solve and graph: x ≤ 1 OR x > 6
Hint: OR → union. Combine both sets. Graph: closed circle at 1 shade left; open circle at 6 shade right. Interval: (−∞, 1] ∪ (6, ∞).
Graph your answer (x ≤ 1 OR x > 6):
Solve and graph: 2x − 3 ≥ 1 AND 2x − 3 < 9
Hint: Solve each: 2x ≥ 4 → x ≥ 2; 2x < 12 → x < 6. AND → intersection: 2 ≤ x < 6. Interval: [2, 6).
Graph your answer (2x − 3 ≥ 1 AND 2x − 3 < 9):
Solve and graph: x + 4 < 2 OR x + 4 ≥ 8
Hint: Solve each: x < −2; x ≥ 4. OR → union. Interval: (−∞, −2) ∪ [4, ∞).
Graph your answer (x + 4 < 2 OR x + 4 ≥ 8):
Solve the double inequality: 1 ≤ 3x − 2 < 10
Hint: Add 2 to all parts: 3 ≤ 3x < 12. Divide by 3 (positive, no flip): 1 ≤ x < 4. Interval: [1, 4).
Graph your answer (1 ≤ 3x − 2 < 10):
Key Vocabulary
Compound Inequality
Two inequalities joined by AND or OR.
Example: x > 1 AND x ≤ 5 | x < −2 OR x ≥ 4
AND (Conjunction)
Both must be true. Solution is the intersection — a segment on the number line.
x > 1 AND x ≤ 5 → (1, 5]
OR (Disjunction)
At least one must be true. Solution is the union — two outward rays.
x < −2 OR x ≥ 4 → (−∞, −2) ∪ [4, ∞)
Double Inequality
A compact AND inequality written as a three-part expression: a < x ≤ b.
Example: −2 < x ≤ 5 means x > −2 AND x ≤ 5
Intersection (∩)
The overlap of two sets — values that appear in both solution sets.
Example: x > 1 and x ≤ 5 intersect at (1, 5]
Union (∪)
The combination of two sets — all values that appear in either solution set.
Example: x < −2 or x ≥ 4 → (−∞, −2) ∪ [4, ∞)
Interval Notation — Visual Reference
Interactive Practice — 5 Questions
Which word produces a solution set that is the intersection of two sets?
The solution to x < −2 OR x > 5 in interval notation is:
What is the solution of x > 7 AND x < 3?
Solve the double inequality: −3 ≤ x + 2 < 7
When solving −6 ≤ −2x + 2 < 4, what happens to both inequality symbols when you divide by −2?
Independent Practice
Answers are in the Answer Key section.
Independent Practice
Solve and graph: x > 0 AND x ≤ 7
Solve and graph: x < 2 OR x ≥ 5
Solve and graph: x + 3 > 1 AND x + 3 ≤ 8
Solve and graph: 3x + 1 < −5 OR 3x + 1 ≥ 10
Solve and graph: −2 ≤ 2x + 4 < 10
Problem 1 graph:
Problem 2 graph:
Problem 3 graph:
Problem 4 graph:
Problem 5 graph:
Common Mistakes
Confusing 'AND' and 'OR' — writing a union when the problem means an intersection.
'AND' means both conditions must be true at the same time (intersection, ∩). 'OR' means at least one is true (union, ∪).
Graphing an 'OR' inequality as if it were 'AND' — shading only the middle section instead of both outer regions.
For 'OR', shade both solution regions. The graph has two separate arrows pointing outward.
Forgetting to flip BOTH symbols when dividing a double inequality by a negative.
When dividing all three parts by a negative, flip BOTH inequality symbols simultaneously.
Writing an impossible 'AND' solution (like x > 5 AND x < 2) as if it has solutions.
If no value satisfies both conditions simultaneously, the solution is the empty set — no solution (∅).
Visual: AND vs. OR — Same Numbers, Different Graphs
AND — segment (intersection)
x > −2 AND x ≤ 3
Only the overlap is shaded.
OR — two rays (union)
x < −2 OR x > 3
Both outer regions are shaded.
Visual: No Solution — AND with No Overlap
x > 5 AND x < 2 — No Solution
x > 5
x < 2
No number is both greater than 5 AND less than 2. Solution: ∅ (empty set).
Math Tips
AND = overlap, OR = everything: Think of AND as a Venn diagram center; OR as the full Venn diagram.
Double inequality direction: in a < x < b, both symbols must point the same way. If they don't, there's no solution.
No solution vs. all reals: AND with no overlap → no solution (∅). OR where the sets cover the whole line → all real numbers (−∞, ∞).
SAT/ACT tip: interval notation for OR always uses ∪ between the two intervals — never write them as one piece.