To solve a linear inequality, use the same steps as for an equation, with one extra rule: when you multiply or divide both sides by a negative number, reverse the inequality sign. Forget the rule and your answer describes exactly the wrong set of numbers.
This skill opens the inequalities and feasible regions module. It returns when you draw boundaries and shade regions, because a flipped sign puts the shading on the wrong side.
Why does the sign reverse?
Start with a true statement: 2 < 5. Multiply both sides by −1 and you get −2 and −5.
On the number line, −5 sits to the left of −2, so −2 > −5. The order reversed.
Multiplying by a negative reflects the whole number line across zero. Numbers that were on the right end up on the left. Dividing by a negative does the same, so the symbol must turn round to keep the statement true.
What changes the sign and what does not?
| Operation on both sides | Sign |
|---|---|
| Add or subtract any number | stays the same |
| Multiply or divide by a positive number | stays the same |
| Multiply or divide by a negative number | reverses |
A useful habit is to circle the coefficient of x the moment you are about to divide. If the circled number has a minus sign, the flip happens on that very line.
Worked example
Solve 5 − 3x ≥ 14.
Step 1, subtract 5 from both sides: −3x ≥ 9. The sign stays the same, because we only subtracted.
Step 2, divide both sides by −3: this is a negative, so the sign reverses. x ≤ −3.
Step 3, check with two test values. Inside the answer, try x = −4: 5 − 3(−4) = 17, and 17 ≥ 14 is true. Outside the answer, try x = 0: 5 − 0 = 5, and 5 ≥ 14 is false. Both tests agree.
Answer: x ≤ −3
A second route avoids dividing by a negative. Start again from 5 − 3x ≥ 14 and add 3x to both sides: 5 ≥ 14 + 3x.
Subtract 14: −9 ≥ 3x. Divide by 3, which is positive: −3 ≥ x.
Read it backwards and it says x ≤ −3. Choose whichever route feels safer.
The mistake to watch for
Mistaken working: 5 − 3x ≥ 14, so −3x ≥ 9, so x ≥ −3.
The student divided by −3 and kept the ≥ sign.
The test value shows the problem. Take x = 0, which satisfies x ≥ −3.
Put it into the original: 5 ≥ 14 is false. So the answer x ≥ −3 cannot be right.
The correction is to spot the negative divisor and turn the sign, giving x ≤ −3. Making a two-number test part of every inequality costs ten seconds and catches this slip each time.
Check yourself
Try these on paper first, then open each answer.
1. Solve −2x < 10.
Show answer
Divide both sides by −2 and reverse the sign: x > −5.
Check: x = −4 gives −2(−4) = 8, and 8 < 10 is true. x = −6 gives 12, and 12 < 10 is false.
x > −5
2. Solve 7 − 4x ≤ 19.
Show answer
Subtract 7: −4x ≤ 12. Divide by −4 and reverse the sign: x ≥ −3.
Check: x = −3 gives 7 + 12 = 19, and 19 ≤ 19 is true. x = −4 gives 23, and 23 ≤ 19 is false.
x ≥ −3
3. Solve x/(−2) > 3.
Show answer
Multiply both sides by −2 and reverse the sign: x < −6.
Check: x = −8 gives −8/(−2) = 4, and 4 > 3 is true. x = −4 gives 2, and 2 > 3 is false.
x < −6
Where this leads next
Once the flip is automatic, go to representing an interval on a number line to show the answer as a picture. The non-calculator working trainer is handy for checking arithmetic steps. You can also review equation-solving habits in the equations and formulas module.
Some students follow each step here but slip when the negative sits inside a longer question. A teacher on online one-to-one Mathematics tuition can trace which step your habit changes without you noticing.