An inequality compares two sides with <, >, ≤ or ≥ instead of =. Solving one works almost exactly like solving an equation, with one extra rule. The answer is usually a whole range of numbers rather than a single value, so you also need to know how to graph it and write it in interval notation.
The symbols
| Symbol | Meaning | On a number line |
|---|---|---|
| x < 3 | x is less than 3 | open circle at 3, shade left |
| x > 3 | x is greater than 3 | open circle at 3, shade right |
| x ≤ 3 | x is less than or equal to 3 | closed circle at 3, shade left |
| x ≥ 3 | x is greater than or equal to 3 | closed circle at 3, shade right |
An open circle means the endpoint is not included. A closed (filled) circle means it is.
Solving a linear inequality
Use the same steps as for linear equations: clear parentheses, collect the variable on one side, then divide. You can add or subtract anything from both sides, and you can multiply or divide by a positive number, without changing the inequality.
Example Solve 4x − 5 < 11
- Add 5 to both sides: 4x < 16.
- Divide by 4: x < 4.
- Every number less than 4 works. Check one: x = 0 gives −5 < 11. ✓
The one extra rule: flip when you multiply or divide by a negative
Multiplying or dividing both sides by a negative number reverses the direction of the inequality.
You can see why with plain numbers. 2 < 5 is true. Multiply both sides by −1 and you get −2 and −5, and on the number line −2 is to the right of −5. So −2 > −5. The order swapped.
Example Solve 7 − 3x ≥ 19
- Subtract 7 from both sides: −3x ≥ 12.
- Divide by −3 and flip the sign: x ≤ −4.
- Check x = −5: 7 − 3(−5) = 22, and 22 ≥ 19. ✓ Check x = 0: 7 ≥ 19 is false, so 0 is correctly excluded.
Only flip for multiplying or dividing by a negative
Subtracting a number, or having a negative answer, does not flip anything. In x − 6 > −10, add 6 to get x > −4. No flip.
If you would rather never flip, move the variable to the side where its coefficient is positive. For 7 − 3x ≥ 19, add 3x to both sides and subtract 19: −12 ≥ 3x, so −4 ≥ x. That is the same answer as x ≤ −4, read from the other side.
Compound inequalities: “and”
An “and” inequality needs both conditions at once. It is often written as one chain, like −2 ≤ x < 3, which you read as “x is between −2 and 3, including −2”.
To solve a chain, do the same thing to all three parts.
Example Solve −5 < 2x + 3 ≤ 9
- Subtract 3 from all three parts: −8 < 2x ≤ 6.
- Divide all three parts by 2: −4 < x ≤ 3.
Example Solve 1 ≤ 5 − 2x < 11
- Subtract 5 from all parts: −4 ≤ −2x < 6.
- Divide all parts by −2 and flip both signs: 2 ≥ x > −3.
- Rewrite from smallest to largest: −3 < x ≤ 2.
Compound inequalities: “or”
An “or” inequality is true when at least one part is true. The graph is usually two separate rays pointing away from each other. Solve each part on its own.
Example Solve 3x + 1 < −8 or 2x − 4 ≥ 6
- First part: 3x < −9, so x < −3.
- Second part: 2x ≥ 10, so x ≥ 5.
- The solution is x < −3 or x ≥ 5. Any number between −3 and 5 (including −3) fails both parts.
Absolute value inequalities
∣x∣ is the distance from x to 0. That gives a quick way to read these:
- |x| < a means x is within a units of 0: −a < x < a. This becomes an “and” chain.
- |x| > a means x is more than a units away from 0: x < −a or x > a. This becomes an “or”.
A memory aid: “less than” is “less thAND”, “greater than” is “greatOR”.
Example Solve |2x − 1| ≤ 7
- Less than, so write a chain: −7 ≤ 2x − 1 ≤ 7.
- Add 1: −6 ≤ 2x ≤ 8.
- Divide by 2: −3 ≤ x ≤ 4.
Example Solve |x + 4| > 2
- Greater than, so split into “or”: x + 4 < −2 or x + 4 > 2.
- Solve each: x < −6 or x > −2.
Special cases
Get the absolute value alone first. Then, if the right side is negative: |x − 3| < −1 has no solution, because a distance is never negative. |x − 3| > −1 is true for every x.
Interval notation
Interval notation writes a range with its endpoints. Use a parenthesis ( ) when the endpoint is excluded and a bracket [ ] when it is included. Infinity always gets a parenthesis.
| Inequality | Interval notation |
|---|---|
| x < 4 | (−∞, 4) |
| x ≥ −2 | [−2, ∞) |
| −4 < x ≤ 3 | (−4, 3] |
| x < −3 or x ≥ 5 | (−∞, −3) ∪ [5, ∞) |
The ∪ symbol means “union”, which is how “or” is written in interval notation.
Word problems
Phrases translate directly: “at least” is ≥, “at most” and “no more than” are ≤, “more than” is >, “fewer than” is <.
Example A school dance costs $300 to run, and tickets are $8 each. How many tickets must be sold to make at least $500 profit?
- Let t be the number of tickets. Profit is 8t − 300.
- “At least $500”: 8t − 300 ≥ 500.
- Add 300: 8t ≥ 800, so t ≥ 100.
- At least 100 tickets must be sold.
When the answer counts whole things, round in the direction that still satisfies the inequality. If you got t ≥ 99.4, you would need 100 tickets, not 99.
Practice problems
Solve 5x + 2 > 3x − 8.
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x > −5. Subtract 3x: 2x + 2 > −8. Subtract 2: 2x > −10. Divide by 2.
Solve −4(x − 1) ≤ 20.
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x ≥ −4. Distribute: −4x + 4 ≤ 20, so −4x ≤ 16. Dividing by −4 flips the sign.
Solve −1 < (x + 5)/2 < 4 and write the answer in interval notation.
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(−7, 3). Multiply all parts by 2: −2 < x + 5 < 8, then subtract 5: −7 < x < 3.
Solve |3x + 2| < 10.
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−4 < x < 8/3. Write −10 < 3x + 2 < 10, subtract 2 to get −12 < 3x < 8, then divide by 3.
Solve |x − 1| ≥ 5.
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x ≤ −4 or x ≥ 6. Split into x − 1 ≤ −5 or x − 1 ≥ 5.
A phone has 64 GB of storage and 22 GB is used. Each video takes 1.5 GB. How many videos can be added?
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28 videos. 22 + 1.5v ≤ 64 gives v ≤ 28.