Quantitative Aptitude · Linear Inequalities

Rules for Solving Linear Inequalities (CA Foundation Quantitative Aptitude)

Q: Why does the inequality sign change when multiplying by a negative number?

Multiplying by a negative number flips the number line, so the order of two numbers reverses. For example, 2 -5. So the sign must reverse to keep the statement true.

Updated 1 October 2026 · Fact-checked

A linear inequality compares two expressions using <, >, ≤ or ≥. You solve it like an equation: add or subtract the same number on both sides, or multiply or divide by the same positive number. If you multiply or divide by a negative number, reverse the inequality sign.

Understand Rules for Solving Linear Inequalities

An inequality says one quantity is smaller or larger than another. An equation such as x + 2 = 5 has one answer. An inequality such as x + 2 < 5 has many answers: every x less than 3 works.

Think of it as a weighing scale. If the left pan is heavier, adding the same weight to both pans keeps it heavier. Removing the same weight from both keeps it heavier. Doubling both weights also keeps it heavier. So adding, subtracting, and multiplying or dividing by a positive number never change the sign.

Negative numbers flip the number line. Take 2 < 5, which is true. Multiply both sides by -1 and you get -2 and -5. But -2 is to the right of -5, so -2 > -5. The order has reversed. That is why the sign must reverse whenever you multiply or divide both sides by a negative number.

The answer is usually a range, not one number. You show it as x > 3, as an interval such as (3, ∞), or on a number line. A round bracket means the end value is not included. A square bracket means it is included. Strict signs (< and >) use round brackets. Signs with equality (≤ and ≥) use square brackets.

Key formulas to remember

Addition rule: If a < b, then a + c < b + c; True for any real number c, positive or negative. The sign does not change.
Subtraction rule: If a < b, then a - c < b - c; True for any real number c. The sign does not change.
Multiplication or division by a positive number: If a < b and c > 0, then ac < bc and a ÷ c < b ÷ c; The sign stays the same.
Multiplication or division by a negative number: If a < b and c < 0, then ac > bc and a ÷ c > b ÷ c; The sign reverses. < becomes >, ≤ becomes ≥, and the other way round.
Transposing a term: x + a < b ⇒ x < b - a; Moving a term across the sign changes its sign. The inequality sign stays the same.
Interval notation: x > a is (a, ∞); x ≥ a is [a, ∞); x < a is (-∞, a); x ≤ a is (-∞, a]; Infinity always takes a round bracket.

How to solve Rules for Solving Linear Inequalities questions

Use this method for any linear inequality in one variable. The aim is to get x alone on one side, watching the sign at each step.

1Remove brackets and fractions. To clear fractions, multiply every term by the positive LCM of the denominators. The sign stays the same.
2Collect the x terms on one side and the constant terms on the other. Change the sign of a term when it crosses the inequality sign.
3Simplify both sides to the form ax < b (or >, ≤, ≥).
4Divide both sides by the coefficient a. If a is positive, keep the sign. If a is negative, reverse the sign.
5Write the solution as x compared with a number, then as an interval if the question asks.
6Check with one test value from your solution. Put it in the original inequality. It should make the statement true.
7Match your answer with the options. Look at the bracket type and the direction of the sign.

Quickest way: Keep x positive and test one value

When to use it: Use in MCQs where the options differ in direction of the sign or in bracket type.

Move the x terms to the side where the x coefficient will be positive. Then you never divide by a negative number and cannot forget the reversal.
Solve to get x < k or x > k.
Pick a test value from your answer, such as k - 1 for x < k. Put it in the original inequality.
If it fails, you made a sign error. Reverse the sign.
Pick a value that is just at the boundary. If the original sign is strict, the boundary must fail. If it has equality, the boundary must work. This settles round versus square brackets.
If you cannot decide in about a minute, skip the question. A wrong answer costs 0.25 marks.

Common mistakes in Rules for Solving Linear Inequalities

Not reversing the sign when dividing by a negative number, for example -3x < 12 ⇒ x < -4.
Students treat the inequality like an equation, where the sign never changes.
Fix: Circle any negative coefficient before dividing. Here -3x < 12 gives x > -4.
Reversing the sign when adding or subtracting a negative number.
Students confuse the negative number with the multiplication rule.
Fix: Only multiplication and division by a negative number reverse the sign. Adding or subtracting never does.
Multiplying only some terms when clearing fractions.
Students rush and multiply the fraction terms but leave the whole number terms alone.
Fix: Multiply every term on both sides by the LCM. Write each product on a new line.
Using the wrong bracket or ignoring the equality part of ≤ and ≥.
Students focus on the number and forget what the sign means.
Fix: Strict sign means the end value is excluded, so use ( ). Signs with equality include it, so use [ ].
Reversing the sign when the variable is on the right, such as 5 < 2x ⇒ 2x < 5.
Students mix up reading the inequality with rewriting it.
Fix: Reading from the other side reverses the sign: 5 < 2x is the same as 2x > 5. Then divide by 2 to get x > 5/2.

Worked examples

Example 1

The solution of -3x + 7 > 22 is: (a) x > -5 (b) x < -5 (c) x > 5 (d) x < 5

Show the solution

Subtract 7 from both sides: -3x > 15.
Divide both sides by -3. This is negative, so reverse the sign: x < -5.
Check with x = -6: -3(-6) + 7 = 25, and 25 > 22. True.

Answer: (b) x < -5

Example 2

Solve (2x - 1)/3 ≤ (x + 4)/2. The solution is: (a) x ≥ 14 (b) x ≤ 14 (c) x ≤ 2 (d) x ≥ 2

Show the solution

The LCM of 3 and 2 is 6. Multiply both sides by 6, which is positive: 2(2x - 1) ≤ 3(x + 4).
Expand: 4x - 2 ≤ 3x + 12.
Subtract 3x from both sides: x - 2 ≤ 12.
Add 2 to both sides: x ≤ 14.
Check with x = 14: left side is 27/3 = 9, right side is 18/2 = 9. 9 ≤ 9 is true, so 14 is included.

Answer: (b) x ≤ 14

Example 3

If 5 - 2x ≥ 3x - 10, the greatest integer value of x is: (a) 2 (b) 3 (c) 4 (d) 5

Show the solution

Move the x terms to the right side so the coefficient stays positive: 5 + 10 ≥ 3x + 2x.
Simplify: 15 ≥ 5x.
Divide by 5, which is positive, so the sign stays: 3 ≥ x, which means x ≤ 3.
The greatest integer satisfying x ≤ 3 is 3.
Check x = 3: left side is 5 - 6 = -1, right side is 9 - 10 = -1. -1 ≥ -1 is true.

Answer: (b) 3

Exam tips

Questions often hide a negative coefficient. Look for it first, because the options usually include both directions of the sign.
Check the brackets in interval options. Two options may have the same numbers but one includes the end value and one does not.
For questions asking for the greatest or least integer, solve the inequality first, then pick the integer. Check whether the boundary value is allowed.
Test one value from your answer in the original inequality. It takes about ten seconds and catches most sign errors.
If a question is long and you are not sure, skip it and return later. Wrong answers cost 0.25 marks each.

Practice questions from Linear Inequalities

Rules for Solving Linear Inequalities: frequently asked questions

Why does the inequality sign change when multiplying by a negative number?

Multiplying by a negative number flips the number line, so the order of two numbers reverses. For example, 2 < 5, but -2 > -5. So the sign must reverse to keep the statement true.

Does the sign change when I add or subtract a negative number?

No. Adding or subtracting any number, positive or negative, keeps the sign the same. Only multiplying or dividing both sides by a negative number reverses it.

What happens if I multiply by zero?

Do not multiply or divide an inequality by zero. Multiplying by zero turns both sides into 0, which loses the original information. Division by zero is not defined.

How do I solve a linear inequality in one variable?

Clear fractions, collect x terms on one side and constants on the other, then divide by the coefficient of x. Reverse the sign only if that coefficient is negative. Then check with a test value.