Algebra
How to Solve Compound Inequalities: AND/OR Examples
Solve "and" and "or" inequalities step by step, handle negative coefficients, and write every answer in interval notation
By AI Picture Answer Education Team · · Updated September 25, 2026
What is a compound inequality?
A compound inequality is two inequalities joined by the word and or the word or. The connecting word decides everything about the final answer: "and" keeps only the numbers that satisfy both inequalities, while "or" keeps the numbers that satisfy at least one of them.
You can check any answer in this guide with the compound inequality calculator, which solves both pieces and shows the combined interval.
Quick definition:
AND = intersection (the overlap of both solutions). OR = union (everything in either solution).
AND vs. OR inequalities
"And" inequalities (intersection)
An "and" inequality such as x > −1 and x ≤ 6 is often written as a single double inequality: −1 < x ≤ 6. The solution is the part of the number line where the two shaded regions overlap. On a graph, that usually looks like one segment between two endpoints.
"Or" inequalities (union)
An "or" inequality such as x < −3 or x ≥ 2 keeps both regions. On a graph it typically looks like two rays pointing away from each other, and in interval notation you join them with ∪: (−∞, −3) ∪ [2, ∞).
How to solve a compound inequality
- Identify the connector. Is it "and" (or a double inequality like a < … < b), or is it "or"?
- Isolate the variable. For a double inequality, do the same operation to all three parts. For two separate inequalities, solve each one on its own.
- Flip the sign when needed. Whenever you multiply or divide by a negative number, reverse every inequality symbol involved.
- Combine the solutions. Take the overlap for "and," or the union for "or."
- Write the answer as an inequality, a graph, or interval notation, and test a value to confirm.
Worked examples
Example 1: A double ("And") inequality
Solve −4 < 2x + 2 ≤ 10.
Step 1: Subtract 2 from all three parts: −4 − 2 < 2x + 2 − 2 ≤ 10 − 2, which gives −6 < 2x ≤ 8.
Step 2: Divide all three parts by 2 (positive, so no flip): −3 < x ≤ 4.
Answer: −3 < x ≤ 4, or in interval notation (−3, 4].
Check: x = 0 gives 2(0) + 2 = 2, and −4 < 2 ≤ 10 is true. x = 4 gives 10, and 10 ≤ 10 is true, so 4 is included. x = −3 gives −4, and −4 < −4 is false, so −3 is excluded.
Example 2: An "Or" inequality
Solve 3x − 1 < −7 or 2x + 5 ≥ 11.
Step 1: Solve the first inequality. Add 1: 3x < −6. Divide by 3: x < −2.
Step 2: Solve the second inequality. Subtract 5: 2x ≥ 6. Divide by 2: x ≥ 3.
Step 3: "Or" means keep both regions: x < −2 or x ≥ 3.
Answer: (−∞, −2) ∪ [3, ∞)
Check: x = −5 gives 3(−5) − 1 = −16 < −7 (true). x = 3 gives 2(3) + 5 = 11 ≥ 11 (true). x = 0 gives −1 < −7 (false) and 5 ≥ 11 (false), so 0 is correctly left out.
Example 3: A negative coefficient (flip both signs)
Solve 1 ≤ 5 − 2x < 9.
Step 1: Subtract 5 from all three parts: −4 ≤ −2x < 4.
Step 2: Divide all three parts by −2 and reverse both symbols: 2 ≥ x > −2.
Step 3: Rewrite with the smaller number on the left: −2 < x ≤ 2.
Answer: (−2, 2]
Check: x = 2 gives 5 − 4 = 1, and 1 ≤ 1 < 9 is true. x = −2 gives 5 + 4 = 9, and 9 < 9 is false, so −2 is excluded. x = 0 gives 5, which lies between 1 and 9.
Example 4: When "And" has no solution
Solve x + 3 < 1 and x − 2 > 4.
Step 1: First inequality: subtract 3 to get x < −2.
Step 2: Second inequality: add 2 to get x > 6.
Step 3: "And" needs a number that is both less than −2 and greater than 6. No such number exists; the regions do not overlap.
Answer: No solution (∅)
Example 5: When "Or" covers every number
Solve x > 1 or x < 5.
Step 1: The first region is everything to the right of 1. The second is everything to the left of 5.
Step 2: Together they cover the whole number line: any number is either greater than 1 or less than 5 (numbers between 1 and 5 are both).
Answer: All real numbers, (−∞, ∞)
Connection to absolute value inequalities
Absolute value inequalities turn into compound inequalities. For a positive number a:
- |x| < a becomes the "and" inequality −a < x < a.
- |x| > a becomes the "or" inequality x < −a or x > a.
For example, |x − 1| ≤ 3 becomes −3 ≤ x − 1 ≤ 3. Adding 1 to every part gives −2 ≤ x ≤ 4, or [−2, 4]. Our guide on how to solve absolute value equations covers the equation version of this idea.
Common mistakes to avoid
Watch out for these:
- Only operating on two parts of a double inequality. Whatever you do to the middle, you must do to the left and right sides too.
- Flipping only one symbol. Dividing a double inequality by a negative number reverses both symbols.
- Mixing up intersection and union. "And" means overlap; "or" means combine. Using the wrong one changes the whole answer.
- Writing an impossible double inequality. Something like 5 < x < 2 has no solution; it usually signals a sign error earlier in the work.
- Squeezing an "or" answer into one interval. x < −2 or x ≥ 3 cannot be written as (−2, 3]. It needs two intervals joined by ∪.
Practice problems
Try these on your own:
- Solve 2 ≤ x − 3 < 7.
- Solve −3x > 12 or x − 1 ≥ 2.
- Solve −1 < (x + 1)/2 ≤ 3.
Click to see answers
- Add 3 to every part: 5 ≤ x < 10, so [5, 10)
- x < −4 or x ≥ 3, so (−∞, −4) ∪ [3, ∞)
- Multiply by 2: −2 < x + 1 ≤ 6. Subtract 1: −3 < x ≤ 5, so (−3, 5]
Frequently asked questions
What is the difference between "and" and "or" compound inequalities?
An "and" inequality keeps only the values that satisfy both parts (the overlap). An "or" inequality keeps values that satisfy either part (the union).
When do I flip the inequality sign?
Flip every inequality symbol whenever you multiply or divide by a negative number. Adding or subtracting never flips the sign.
Can a compound inequality have no solution?
Yes. An "and" inequality has no solution when its two regions do not overlap, such as x < −2 and x > 6.
Need help with compound inequalities?
Writing the final answer correctly is half the battle, so review how to write interval notation if brackets and unions still feel shaky. To verify your work on homework sets, try the compound inequality calculator or the interval notation calculator.
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