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Quantitative Aptitude · Linear Inequalities

Solving Inequalities in One Variable (CA Foundation Quantitative Aptitude)

Updated 1 October 2026

A linear inequality in one variable compares two expressions using <, >, ≤ or ≥, such as 3x − 5 < 7. To solve it, isolate x using the same steps as an equation, but reverse the sign when you multiply or divide by a negative number. Then show the solution set on a number line.

Understand Solving Inequalities in One Variable

An equation says two sides are equal. An inequality says one side is smaller or larger. The symbols are < (less than), > (greater than), ≤ (less than or equal to) and ≥ (greater than or equal to).

An equation like x + 2 = 5 has one answer, x = 3. An inequality like x + 2 < 5 has many answers. Any x less than 3 works: 2, 0, −7, 2.9 and so on. The set of all values that satisfy it is called the solution set.

You solve it almost like an equation. You add or subtract the same number on both sides, or multiply or divide both sides by the same number. There is one big change. If you multiply or divide by a negative number, the inequality sign flips. Check with numbers: 2 < 5 is true, but −2 and −5 give −2 > −5. The order reverses.

The solution set is shown on a number line. A hollow (open) circle means the end point is not included (for < and >). A filled (closed) circle means it is included (for ≤ and ≥). Then you shade the part of the line that holds the solutions. The same set can be written as an interval: (a, b) is open, [a, b] is closed. Unless the question says otherwise, x is a real number. If x must be a natural number or integer, you list only those values.

Key formulas to remember

Adding or subtracting
If a < b, then a + c < b + c and a − c < b − c
Holds for any real c. The sign does not change.
Multiplying or dividing by a positive number
If a < b and c > 0, then ac < bc and a ÷ c < b ÷ c
The sign stays the same.
Multiplying or dividing by a negative number
If a < b and c < 0, then ac > bc and a ÷ c > b ÷ c
The sign reverses. This is the most tested rule.
Compound inequality
a < x ≤ b means x > a and x ≤ b together
Do the same operation on all three parts. Flip all signs if you multiply by a negative number.
Number line and interval notation
x > a: (a, ∞) | x ≥ a: [a, ∞) | x < b: (−∞, b) | x ≤ b: (−∞, b]
Hollow circle with round bracket, filled circle with square bracket. Infinity always takes a round bracket.

How to solve Solving Inequalities in One Variable questions

Use this method for any single-variable linear inequality. Treat it like an equation, with one extra check for sign reversal.

  1. 1Remove fractions by multiplying every term by the LCM of the denominators. The LCM is positive, so the sign does not change.
  2. 2Expand brackets and combine like terms on each side.
  3. 3Move all x terms to one side and constants to the other. Adding or subtracting never changes the sign.
  4. 4Divide both sides by the coefficient of x. If it is negative, reverse the inequality sign.
  5. 5Write the solution as x < a, x ≥ a and so on, or as an interval.
  6. 6Draw the number line. Use a hollow circle for < or >, a filled circle for ≤ or ≥, and shade in the direction of the sign.
  7. 7If x must be a natural number or integer, list only the allowed values from the solution set.

Quickest way: Option testing and sign check

When to use it: Use when the MCQ gives four solution sets or four intervals. It saves time and avoids sign errors.

  1. Solve in one or two lines, and note only the final boundary value and direction.
  2. Look at the options. Two or more usually differ only in direction or in open versus closed ends.
  3. If unsure, pick a test value inside your answer and one outside. Put both in the original inequality.
  4. Substitute the boundary value into the original inequality. If it makes the statement true (original sign ≤ or ≥), the end is closed. If it gives equality while the original sign is < or >, the statement is false, so the end is open.
  5. If the question asks for integer or natural solutions, count only those inside the interval. Do not forget that 0 is not a natural number.
  6. If a question needs more than three minutes, skip it. Wrong answers cost 0.25 marks.

Common mistakes in Solving Inequalities in One Variable

  • Not reversing the sign when dividing by a negative number

    Students are used to equations, where nothing changes on division.

    Fix: Every time the coefficient of x is negative, circle it and flip the sign immediately. Check with x = 0 in the original.

  • Using the wrong circle on the number line

    Students mix up strict and non-strict signs.

    Fix: Remember: equal sign present (≤, ≥) means filled circle. No equal sign (<, >) means hollow circle.

  • Shading in the wrong direction

    Students shade by habit instead of reading the final sign.

    Fix: x > a means shade to the right of a. x < a means shade to the left. Test one point from the shaded part.

  • Flipping the sign when moving a term across

    Students confuse the sign of the term with the inequality sign.

    Fix: Moving a term changes the sign of the term only. The inequality sign changes only when multiplying or dividing by a negative.

  • Ignoring the type of x in the question

    Students give the full real-number interval when the question asks for natural numbers or integers.

    Fix: Read the domain first. For natural numbers list 1, 2, 3 and so on. List only the values inside the solution set.

  • Multiplying by an unknown-sign expression

    Students cross-multiply when x is in a denominator without checking the sign.

    Fix: At this level, questions keep x in the numerator. If x is in a denominator, do not multiply by it unless its sign is known.

Worked examples

Example 1

The solution set of 3x − 5 < 7 for real x is: (a) x < 4 (b) x > 4 (c) x < 12 (d) x > 2

Show the solution
  1. Add 5 to both sides: 3x < 12.
  2. Divide both sides by 3, which is positive, so the sign stays: x < 4.
  3. Check x = 0: 3(0) − 5 = −5 < 7, which is true. Check x = 4: 12 − 5 = 7, not less than 7, so 4 is excluded.

Answer: (a) x < 4. On the number line: hollow circle at 4, shade to the left.

Example 2

If 5 − 2x ≥ 11, then x lies in: (a) x ≥ −3 (b) x ≤ −3 (c) x ≤ 3 (d) x ≥ 3

Show the solution
  1. Subtract 5 from both sides: −2x ≥ 6.
  2. Divide both sides by −2, a negative number, so reverse the sign: x ≤ −3.
  3. Check x = −3: 5 − 2(−3) = 11 ≥ 11, true. So −3 is included. Check x = 0: 5 ≥ 11 is false, so right-side values are not solutions.

Answer: (b) x ≤ −3. On the number line: filled circle at −3, shade to the left.

Example 3

How many natural numbers satisfy 2x + 3 ≤ 13? (a) 4 (b) 5 (c) 6 (d) 7

Show the solution
  1. Subtract 3 from both sides: 2x ≤ 10.
  2. Divide by 2: x ≤ 5.
  3. Natural numbers are 1, 2, 3, 4, 5.
  4. Count them: there are 5.

Answer: (b) 5. Option (c) 6 would wrongly include 0.

Exam tips

  • Always test one value in the original inequality before you mark. It takes ten seconds and catches sign errors.
  • Read the domain of x (real, integer or natural) before solving. Many options differ only on this point.
  • In option-based questions, check the end point. Options often differ only by round versus square bracket.
  • Fractions and brackets cause most slips. Multiply out by the LCM first and then solve.
  • If a question looks long and you cannot see the first step in half a minute, skip it and return later.

Practice questions from Linear Inequalities

Solving Inequalities in One Variable: frequently asked questions

What is the solution set of a linear inequality in one variable?

It is the set of all values of the variable that make the inequality true. It is usually an interval, such as x < 4 or −2 ≤ x < 5. You can write it in interval form or show it on a number line.

When do I reverse the inequality sign?

Reverse the sign only when you multiply or divide both sides by a negative number. Adding or subtracting any number never changes the sign. Multiplying or dividing by a positive number also leaves it unchanged.

How do I show an inequality on a number line?

Mark the boundary value. Use a hollow circle for < or > and a filled circle for ≤ or ≥. Then shade to the right for greater than and to the left for less than.

How is the solution different for natural numbers and real numbers?

For real numbers the solution is the whole interval. For natural numbers you take only the whole numbers from 1 upward that lie in that interval. The answer is then a short list of values, or a count.