Solving Inequalities

The method, step by step

1. 1

   Solve as though it were an equation

   3x + 4 < 19 → 3x < 15 → x < 5

   Subtract, divide, and treat the inequality sign exactly as you would an equals sign. Nothing about the process changes while you are adding, subtracting, or dividing by positive numbers.

2. 2

   Flip the sign when multiplying or dividing by a negative

   −2x > 6 → x < −3

   Dividing by −2 reverses the direction. Check it with numbers: 5 > 3 is true, but multiplying both by −1 gives −5 > −3, which is false. The reversal is what keeps the statement true.

3. 3

   Show the answer on a number line

   x < 5 → open circle at 5, arrow pointing left

   An open circle excludes the endpoint and a filled circle includes it, so < and > get open circles while ≤ and ≥ get filled ones. This distinction is worth a mark on its own and is lost more often than the algebra.

4. 4

   Handle double inequalities on both sides at once

   −3 ≤ 2x + 1 < 7 → −4 ≤ 2x < 6 → −2 ≤ x < 3

   Whatever you do, do it to all three parts. The variable stays in the middle throughout, and the two outer bounds are processed simultaneously rather than as two separate inequalities.

5. 5

   List integer solutions carefully

   −2 ≤ x < 3 → x = −2, −1, 0, 1, 2

   The −2 is included because the sign is ≤; the 3 is excluded because the sign is <. Getting the endpoints right is usually the whole of the mark, and it is decided entirely by which symbols the inequality used.

Why the sign flips

Multiplying by a negative reflects every number across zero, and reflection reverses order. Five is greater than three, but −5 is less than −3, because they have swapped positions on the number line. The inequality sign has to flip to keep describing the same relationship.

Students taught this as a rule apply it inconsistently — often flipping when they subtract a negative, which does not require it. Students who have seen the reflection can check any case in two seconds by testing with real numbers.

• Add and subtract freely — no flip
• Multiply or divide by a positive — no flip
• Multiply or divide by a negative — flip the sign
• < and > → open circle. ≤ and ≥ → filled circle
• Double inequality → operate on all three parts

You can usually avoid the negative entirely

For −2x > 6, adding 2x to both sides and subtracting 6 gives −6 > 2x, then −3 > x, which is the same as x < −3. No flip was needed because nothing was divided by a negative. Keeping the variable on whichever side makes its coefficient positive sidesteps the rule altogether.

This is worth teaching as a genuine strategy rather than a dodge. The flip is the topic's main source of error, and a method that never triggers it is a method that never gets it wrong.

The answer is a set, not a number

x < 5 is not a value; it is every number below five. Students accustomed to equations having one answer often want to name a particular number, and questions asking for the largest integer satisfying an inequality exploit exactly that discomfort.

Reading x < 5 aloud as 'x is any number less than five' rather than 'x equals less than five' helps more than it sounds like it should. The second phrasing keeps the idea of a single answer alive.

How we teach inequalities

We test the flip with numbers rather than asserting it. A student who has checked that 5 > 3 becomes −5 < −3 owns the rule, and can rebuild it in an exam when memory is unreliable.

We also mark the number line notation strictly from the start. Open and filled circles are worth marks, they take no extra thought once the habit is formed, and students who are allowed to be casual about them early stay casual about them.