What is an inequality?
An equation says two sides are equal. An inequality says they are not equal, and tells which side is bigger.
- a < b: a is less than b.
- a > b: a is greater than b.
- a ≤ b: a is less than or equal to b.
- a ≥ b: a is greater than or equal to b.
On a number line, numbers grow from left to right. So the number on the left is always the smaller one. −7 < −2, because −7 is further left.
Every two real numbers can be compared: either a < b, a = b or a > b. Exactly one is true.
The rules (properties) of inequalities
These rules are like the rules for equations, with one big warning.
- Add or subtract the same number on both sides: the sign stays. If a < b then a + c < b + c.
- Multiply or divide by a positive number: the sign stays. If a < b and c > 0 then ac < bc.
- Multiply or divide by a negative number: the sign flips. If a < b and c < 0 then ac > bc.
- Chain rule (transitive): if a < b and b < c, then a < c.
- Adding two inequalities that point the same way is allowed: a < b and c < d give a + c < b + d. Subtracting them is NOT safe.
- Taking reciprocals of two positive numbers flips the sign: 2 < 5 but 1/2 > 1/5.
Why does a negative flip the sign? Multiplying by −1 mirrors every point across 0. The one on the left jumps to the right. So the order swaps.
Solving linear inequalities and interval notation
Solve just like an equation, using the rules above. The answer is a set of numbers.
Example: 5 − 2x > 11 → −2x > 6 → divide by −2 and flip → x < −3.
Showing the answer
- On a number line: a filled dot if the end number is included (≤, ≥); a hollow dot if it is not (<, >). Shade the part that works.
- Interval notation: square bracket [ ] = included, round bracket ( ) = not included. ∞ always gets a round bracket.
| Inequality | Interval |
|---|---|
| x ≤ 4 | (−∞, 4] |
| x > −1 | (−1, ∞) |
| −2 < x ≤ 5 | (−2, 5] |
| all real numbers | (−∞, ∞) = ℝ |
Double inequalities like −1 ≤ 2x + 3 < 9: do the same thing to all three parts. −4 ≤ 2x < 6, so −2 ≤ x < 3.
Systems: solve each inequality, then keep only the numbers that satisfy both (the overlap, or intersection).
Quadratic, rational and modulus inequalities
Quadratic inequality (ax² + bx + c > 0 or < 0):
- Move everything to one side so the other side is 0.
- Find the roots (factorise or use the formula).
- Think of the graph. If a > 0 the parabola opens upward: it is below zero between the roots and above zero outside them.
x² − x − 6 < 0 → (x − 3)(x + 2) < 0 → −2 < x < 3.
x² − x − 6 ≥ 0 → x ≤ −2 or x ≥ 3.
Sign chart method: mark the roots on a line; test one number in each part; keep the parts with the right sign. This also works for rational inequalities like (x − 1)/(x + 4) > 0 (never divide by zero, so x ≠ −4 always gets a hollow dot). Do not multiply both sides by (x + 4), because you do not know its sign.
Modulus (absolute value): |x| is the distance of x from 0.
- |x| < a (a > 0) means −a < x < a (close to 0).
- |x| > a means x < −a or x > a (far from 0).
- |x − 2| ≤ 3 means −1 ≤ x ≤ 5 (within 3 of 2).
Inequalities that are always true (absolute inequalities)
Some inequalities hold for every allowed value. They are called absolute (or identical) inequalities. Proving one usually means turning it into 'a square is never negative'.
- x² ≥ 0 for every real x. So (a − b)² ≥ 0, which gives a² + b² ≥ 2ab.
- AM–GM inequality: for a, b ≥ 0, (a + b)/2 ≥ √(ab). The average is never less than the geometric mean. Equal only when a = b.
- For x > 0: x + 1/x ≥ 2.
Proof of AM–GM: (√a − √b)² ≥ 0 → a − 2√(ab) + b ≥ 0 → (a + b)/2 ≥ √(ab).
Use: of all rectangles with perimeter 20 cm, the square (5 × 5) has the biggest area, 25 cm².
Try it at home
Draw a number line from −6 to 6 on paper. Put a coin on 1 and a button on 3. Now 'multiply by −1': move each to the opposite side of 0. Which one is on the left now? Write both statements with < or >. You have just seen why the sign flips.
In the 3D free play, pick |x| < a and set a = 0. Predict first: how many numbers turn green?
Key formulas and definitions
- If a < b then a + c < b + c
- If a < b and c > 0 then ac < bc
- If a < b and c < 0 then ac > bc (flip)
- If 0 < a < b then 1/a > 1/b
- |x| < a ⇔ −a < x < a (a > 0)
- |x| > a ⇔ x < −a or x > a
- (a + b)/2 ≥ √(ab) for a, b ≥ 0 (AM–GM)
- [ ] included, ( ) not included; ∞ always ( )
Worked examples
1. Solve 3x + 4 < 19 and write the interval.
3x < 15 (subtract 4). x < 5 (divide by 3, positive, sign stays). Interval (−∞, 5); hollow dot at 5.
2. Solve 7 − 2x ≥ 1.
−2x ≥ −6 (subtract 7). Divide by −2 and flip: x ≤ 3. Interval (−∞, 3]. Check x = 0: 7 ≥ 1 ✓.
3. Solve −3 < 2x + 1 ≤ 9.
Subtract 1 from all parts: −4 < 2x ≤ 8. Divide by 2: −2 < x ≤ 4. Interval (−2, 4].
4. Solve the system x + 2 > 0 and 2x − 5 ≤ 3.
First: x > −2. Second: 2x ≤ 8, x ≤ 4. Both: −2 < x ≤ 4, interval (−2, 4].
5. Solve x² − 5x + 4 ≤ 0.
(x − 1)(x − 4) ≤ 0. Roots 1 and 4. The parabola opens upward, so it is ≤ 0 between the roots, ends included: 1 ≤ x ≤ 4, interval [1, 4].
6. Solve |2x − 1| < 5.
−5 < 2x − 1 < 5. Add 1: −4 < 2x < 6. Divide by 2: −2 < x < 3. Interval (−2, 3).
7. Prove that x + 1/x ≥ 2 for every x > 0.
Since x > 0, multiply by x (sign stays): x² + 1 ≥ 2x ⇔ x² − 2x + 1 ≥ 0 ⇔ (x − 1)² ≥ 0, which is always true. Equality when x = 1.
Common mistakes
- Forgetting to flip the sign after dividing by a negative: −2x > 6 gives x < −3, not x > −3.
- Using a filled dot or square bracket for < or >; those ends are NOT included. And writing ∞] — infinity always takes a round bracket.
- Multiplying both sides by an expression like (x + 4) whose sign is unknown. Use a sign chart instead.
- Solving x² < 9 as x < ±3. The correct answer is −3 < x < 3; and x² > 9 is x < −3 or x > 3.