Inequalities

A range of values

An equation such as x = 4 has one solution. An inequality such as x > 4, read "x is greater than 4", has a whole range of them: 5 works, 4.1 works, 100 works, and so does every number in between. 4 itself does not work, because 4 is not greater than 4.

On a number line the solutions make a ray: a shaded stretch that starts at 4 and runs on to the right without end. The circle at 4 is hollow, to show that 4 is the boundary but is not one of the solutions.

Every value to the right of 4 is a solution. The hollow circle leaves 4 out.

Including the boundary

The sign $\ge$≥ means "greater than or equal to". $x\ge 4$x ≥ 4 has every solution that x > 4 has, and one more: 4 itself, because 4 = 4. The line under the sign is the "or equal to" part, and on the number line it fills the circle.

There are four signs to know. < is less than, and > is greater than: these are strict, so they leave the boundary out and draw a hollow circle. $\le$≤ is less than or equal to, and $\ge$≥ is greater than or equal to: these let the boundary in and draw a filled circle. With x written first, > and $\ge$≥ shade to the right, where the larger numbers are, and < and $\le$≤ shade to the left.

The same ray, with the circle at 4 filled in: now 4 is a solution too.

Solve it like an equation

To solve 2x > 8, do what you would do to 2x = 8: divide both sides by 2. That gives x > 4, the same ray as before.

This works because the move keeps the larger side larger. If 2x is more than 8, then half of 2x is more than half of 8. The same is true when you add the same number to both sides, subtract it from both sides, or multiply or divide both sides by the same positive number. None of these moves changes the sign.

Check the answer with one value on each side of the boundary. x = 5 gives 2 × 5 = 10, and 10 > 8 is true. x = 3 gives 2 × 3 = 6, and 6 > 8 is false. So the boundary is at 4 and the solutions are on the side of 5.

Multiplying by a negative number

One move behaves differently. Start with something true: 2 < 3. Multiply both numbers by −1 to get −2 and −3. On the number line, −2 is to the right of −3, so −2 > −3. The smaller number has become the larger one, and the sign has to turn around for the statement to stay true.

Multiplying by a negative number reflects every number across zero, and the reflection reverses their order. So whenever you multiply or divide both sides of an inequality by a negative number, reverse the sign: < becomes >, > becomes <, $\le$≤ becomes $\ge$≥, and $\ge$≥ becomes $\le$≤.

2 is to the left of 3, so 2 < 3. Their negatives swap sides: −2 is to the right of −3, so −2 > −3.

a = 1 and b = 3, so a < b. The handle sets a number k and moves the two points to ka and kb. While k is positive the gold arrow points right and ka < kb. Drag k below zero and the arrow turns around: ka > kb.

Solving with a negative

Solve −3x < 12. Divide both sides by −3, and because −3 is negative, reverse the sign: x > −4.

Check with x = 0, which is on the side of the answer: −3 × 0 = 0, and 0 < 12 is true. Check with x = −5, on the other side: −3 × (−5) = 15, and 15 < 12 is false.

You can also avoid dividing by a negative. Add 3x to both sides of −3x < 12 to get 0 < 12 + 3x. Subtract 12 from both sides: −12 < 3x. Divide by 3, which is positive: −4 < x. That says the same thing as x > −4.

The solutions of −3x < 12 are every number to the right of −4.

The usual mistakes

Forgetting to reverse the sign. Dividing −3x < 12 by −3 and writing x < −4 gives numbers such as −5, and −3 × (−5) = 15 is not less than 12.

Reversing the sign because a number is negative. In 2x > −8, you divide by 2, which is positive, so the sign stays: x > −4. Only multiplying or dividing by a negative number reverses the sign. A negative number on one side does not.

Flipping the number instead of the sign. In −2x > −8, dividing by −2 gives x < 4, not x < −4: −8 ÷ (−2) = 4, and it is the inequality sign that turns around.

Two moves, and a whole-number answer

Some inequalities need two moves. In $8+3n\le 40$8 + 3n ≤ 40, n is multiplied by 3 and then 8 is added. Undo them in the reverse order, as you would for an equation: subtract 8 from both sides to get $3n\le 32$3n ≤ 32, then divide both sides by 3 to get $n\le \frac{32}{3}$n ≤ 32/3, which is $10\frac{2}{3}$10 2/3. Neither move multiplies or divides by a negative number, so the sign stays as $\le$≤.

When n counts something that only comes in whole numbers, such as rides, the answer is the greatest whole number that satisfies the inequality. $n\le 10\frac{2}{3}$n ≤ 10 2/3 allows 10 but not 11, because 3 × 11 = 33 is more than 32. So the answer is 10: round down, even though $10\frac{2}{3}$10 2/3 is nearer to 11.