MathsICSE Class 10 · Linear Inequations

Solving Linear Inequations: Rules, Replacement Sets and Solved Questions (ICSE Class 10)

Q: When does the inequality sign change?

Only when you multiply or divide both sides by a negative number. Adding, subtracting, or multiplying by a positive number keeps the sign.

Q: What is the difference between a replacement set and a solution set?

The replacement set is the set of values x is allowed to take, such as N, W, Z or R. The solution set is the part of it that makes the inequation true.

Q: Is 0 a natural number in ICSE?

No. Natural numbers start at 1. Whole numbers include 0.

Q: How do I write the answer when x is real?

Use set-builder form, such as {x : x ≥ 3, x ∈ R}, and show it on a number line with a filled or hollow circle at the end-point.

Updated 15 October 2026 · Fact-checked

Solve a linear inequation like an equation: add or subtract the same number on both sides and multiply or divide by the same positive number. The one new rule: multiplying or dividing both sides by a negative number reverses the sign. Then list only those values of the replacement set that satisfy the result.

Multiplying by −1 mirrors numbers about 0, so their order flips: 2 < 5 becomes −2 > −5, hence reverse the sign.

Illustration drawn for QuestionBuddy.

Understand Solving Linear Inequations

A linear inequation in one variable is a statement such as $3 x - 5 < 7$ or $\frac{x}{2} + 1 \geq x - 3$ , using one of $<, >, \leq, \geq$ . Unlike an equation, it is usually satisfied by many values.

Replacement set and solution set. The replacement set is the set from which $x$ may be chosen; the CISCE syllabus uses $N$ (natural numbers $1, 2, 3, \dots$ ), $W$ (whole numbers $0, 1, 2, \dots$ ), $Z$ or $I$ (integers) and $R$ (real numbers). The solution set is the set of members of the replacement set that make the inequation true. The same inequation can have different solution sets for different replacement sets: $x < 3$ gives ${1, 2}$ for $x \in N$ , ${0, 1, 2}$ for $x \in W$ , ${\dots, - 1, 0, 1, 2}$ for $x \in Z$ , and ${x : x < 3, x \in R}$ for real $x$ .

The rules. For any numbers $a, b$ with $a < b$ :

1. $a + c < b + c$ and $a - c < b - c$ : you may add or subtract (transpose) freely.
2. If $c > 0$ : $a c < b c$ and $\frac{a}{c} < \frac{b}{c}$ : multiplying or dividing by a positive number keeps the sign.
3. If $c < 0$ : $a c > b c$ and $\frac{a}{c} > \frac{b}{c}$ : multiplying or dividing by a negative number reverses the sign.

Why rule 3? $2 < 5$ , but $- 2 > - 5$ : multiplying by $- 1$ flips the order of numbers on the number line. In 2025 Q1(xiv), $0 < - \frac{x}{3} < 2$ needed this rule; the examiners reported that candidates did not reverse the sign.

Fractions. Multiply every term by the LCM of the denominators (a positive number, so the sign stays). Then collect the $x$ terms on the side where they will be positive, which avoids dividing by a negative.

Writing the answer. For $N, W, Z$ , list the members in roster form, using '…' if the set is infinite. For $R$ , use set-builder form, for example ${x : x \geq \frac{22}{7}, x \in R}$ . Board questions usually also ask for the number line, covered in the related topic on number lines.

Checking. Test one value inside the solution set and one just outside it in the original inequation.

Solving Linear Inequations: formulas to remember

Adding or subtracting: $a < b \Rightarrow a \pm c < b \pm c$; Transposing a term changes its sign; the inequality sign stays.
Multiplying or dividing by a positive number: $a < b, c > 0 \Rightarrow a c < b c, \frac{a}{c} < \frac{b}{c}$; Sign stays.
Multiplying or dividing by a negative number: $a < b, c < 0 \Rightarrow a c > b c, \frac{a}{c} > \frac{b}{c}$; Sign reverses.
Replacement sets: $N = {1, 2, \dots}, W = {0, 1, 2, \dots}, Z = {\dots, - 1, 0, 1, \dots}$; $R$ is all real numbers.
Solution set in set-builder form: ${x : x \geq a, x \in R}$; Used when the replacement set is $R$ .

How to solve Solving Linear Inequations questions

Treat the inequation like an equation, watching only for negative multipliers.

1Note the replacement set ( $N$ , $W$ , $Z$ or $R$ ) and any extra condition on $x$ .
2Clear brackets, then multiply every term by the LCM of the denominators.
3Transpose so that the $x$ terms are on one side and numbers on the other.
4Divide by the coefficient of $x$ ; if it is negative, reverse the sign.
5Write the solution set from the replacement set: roster form for $N, W, Z$ ; set-builder form for $R$ .
6Check one value inside and one just outside the answer in the original inequation.

Solving Linear Inequations in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

An upper limit: at most

Suppose a goods lift can carry at most 600 kg. Its operator weighs 70 kg and each box to be loaded weighs 45 kg.

Form an inequation and find the greatest number of boxes the lift can carry with the operator.

Try it first, then show the solutionSolution

Let $x$ be the number of boxes ( $x \in W$ ): $70 + 45 x \leq 600$ .
$45 x \leq 530 \Rightarrow x \leq 11. \overline{7}$ .
Whole numbers only, so $x \leq 11$ .

Answer: At most 11 boxes.

Memory hook: 'At most' means ≤; round down to stay safe.

A lower limit: at least

Imagine you need at least ₹500 to buy a gift. You have ₹140 now and can save ₹45 a week.

Find the least number of whole weeks you must save.

Try it first, then show the solutionSolution

$140 + 45 w \geq 500 \Rightarrow 45 w \geq 360$ .
$w \geq 8$ .

Answer: 8 weeks.

Memory hook: 'At least' means ≥; the smallest member of the solution set is the answer.

Tricks to remember Solving Linear Inequations

Short memory aids. Say each one aloud once, then use it on a question the same day.

Negative flips the see-saw
Multiply or divide by a negative and the see-saw flips: < becomes >, ≤ becomes ≥.
Use it when: The one rule that differs from equations.
Keep x positive
Collect the $x$ terms on the side where their coefficient is positive; then you never divide by a negative.
Use it when: Avoiding sign-reversal slips.
N starts at 1, W at 0
Natural numbers are the counting numbers (no zero); Whole numbers add the zero.
Use it when: Writing the correct roster.
Inside and outside test
Plug one value inside the answer and one outside into the original; the first must work, the second must fail.
Use it when: Checking every solution.

Solving Linear Inequations: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Dividing by a negative number · 1 mark · MCQBoard level

The solution set of $5 - 2 x > 11$ , $x \in Z$ , is:
(a) ${\dots, - 5, - 4}$
(b) ${- 3, - 2, - 1, \dots}$
(c) ${\dots, - 4, - 3}$
(d) ${- 2, - 1, 0, \dots}$

Try it first, then show the solutionSolution

$- 2 x > 6$ .
Divide by $- 2$ and reverse: $x < - 3$ , so ${\dots, - 5, - 4}$ .

Answer: (a)

Example 2: Fractions and a finite replacement set · 3 marks · Short answerBoard level

Solve $\frac{x}{3} + 2 \leq \frac{x}{2} + 1$ , where $x \in {x : x \in N, x < 10}$ .

Try it first, then show the solutionSolution

Multiply by 6: $2 x + 12 \leq 3 x + 6$ .
$12 - 6 \leq 3 x - 2 x \Rightarrow x \geq 6$ .
From ${1, 2, \dots, 9}$ : solution set ${6, 7, 8, 9}$ .

Answer: ${6, 7, 8, 9}$

Example 3: Brackets, whole numbers · 3 marks · Short answerHarder board level

Solve $3 (x - 2) - 4 < 2 (x + 1)$ , $x \in W$ , and write the solution set.

Try it first, then show the solutionSolution

$3 x - 6 - 4 < 2 x + 2 \Rightarrow 3 x - 10 < 2 x + 2$ .
$x < 12$ .
Whole numbers: ${0, 1, 2, \dots, 11}$ .

Answer: ${0, 1, 2, \dots, 11}$

Example 4: Three fractions, real numbers · 4 marks · Short answerHarder board level

Solve $\frac{2 x - 1}{3} - \frac{x + 2}{4} \geq 1 - \frac{x}{6}$ , $x \in R$ . Hence write the smallest integer that satisfies it.

Try it first, then show the solutionSolution

Multiply by 12: $4 (2 x - 1) - 3 (x + 2) \geq 12 - 2 x$ .
$8 x - 4 - 3 x - 6 \geq 12 - 2 x \Rightarrow 5 x - 10 \geq 12 - 2 x$ .
$7 x \geq 22 \Rightarrow x \geq \frac{22}{7} = 3 \frac{1}{7}$ .
Solution set ${x : x \geq \frac{22}{7}, x \in R}$ ; the smallest integer is 4.

Answer: ${x : x \geq \frac{22}{7}, x \in R}$ ; smallest integer 4.

Example 5: Greatest integer satisfying an inequation · 3 marks · Short answerHOTS / competency

Find the greatest integer $x$ for which $3 - \frac{x + 5}{2} > \frac{x - 1}{3}$ .

Try it first, then show the solutionSolution

Multiply by 6: $18 - 3 (x + 5) > 2 (x - 1)$ .
$18 - 3 x - 15 > 2 x - 2 \Rightarrow 3 - 3 x > 2 x - 2$ .
$5 > 5 x \Rightarrow x < 1$ .
Greatest integer less than 1 is 0. Check $x = 0$ : $3 - 2.5 = 0.5 > - \frac{1}{3}$ ✓; $x = 1$ : $0 > 0$ is false.

Answer: 0

Example 6: Case study: the marks needed · 4 marks · Case-basedHOTS / competency

Anya scored 62 and 74 (out of 100) in two tests. She wants an average of at least 70 over three tests. Marks are whole numbers.
(a) Form an inequation for her third score $x$ .
(b) Find the solution set.
(c) Could she reach an average above 80? Explain.

Try it first, then show the solutionSolution

(a) $\frac{62 + 74 + x}{3} \geq 70$ , with $x \in W$ , $x \leq 100$ .
(b) $136 + x \geq 210 \Rightarrow x \geq 74$ ; so ${74, 75, \dots, 100}$ .
(c) Average above 80 needs $136 + x > 240$ , so $x > 104$ , which is impossible since $x \leq 100$ .

Answer: (a) $\frac{136 + x}{3} \geq 70$ (b) ${74, 75, \dots, 100}$ (c) No: it would need more than 104 marks.

Where students lose marks in Solving Linear Inequations

The slips examiners see most often, and what to write instead.

Mistake: Not reversing the sign when dividing by a negative
Multiplying by a negative number reverses the order of numbers.
Wrong working
$- \frac{x}{3} < 2 \Rightarrow x < - 6$ .
Right working
$- \frac{x}{3} < 2 \Rightarrow x > - 6$ .
Do this: Circle any negative divisor and flip the sign in the same line.
Usually costs: The examiners named this for 2025 Q1(xiv).
Mistake: Transposing errors
A term changes sign when it crosses to the other side.
Wrong working
$- 3 + x \leq \frac{7 x}{2} + 2 \Rightarrow x \leq - 2$ .
Right working
$- 10 \leq 5 x \Rightarrow x \geq - 2$ .
Do this: Collect $x$ terms on the side where they stay positive.
Usually costs: Reported for 2024 Q7(ii).
Mistake: Including 0 for natural numbers
$N$ starts at 1.
Wrong working
$x \leq 4$ , $x \in N$ : ${0, 1, 2, 3, 4}$ .
Right working
${1, 2, 3, 4}$ .
Do this: Write the first few members of the replacement set before choosing.
Usually costs: The answer mark.
Mistake: Writing a real solution set as a list of integers
For $x \in R$ every real number in the interval is a solution.
Wrong working
$x \geq \frac{22}{7}$ , $x \in R$ : ${4, 5, 6, \dots}$ .
Right working
${x : x \geq \frac{22}{7}, x \in R}$ .
Do this: Use set-builder form for $R$ .
Usually costs: Reported for 2025 Q4(i).

Exam tips

Write the replacement set at the start of your answer and the solution set at the end, in braces.
In 2025 Q1(xiv) the examiners reported candidates who did not reverse the sign on dividing by a negative number.
In 2024 Q7(ii) the examiners reported transposing errors that gave $x \leq - 2$ instead of $x \geq - 2$ .
Show the LCM step for fractions; it earns the first method mark.

Practice questions on Solving Linear Inequations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If $- 3 x \geq 12$ , then:
(a) $x \geq - 4$
(b) $x \leq - 4$
(c) $x \geq 4$
(d) $x \leq 4$
Hint
Dividing by −3 reverses the sign.
Answer
(b) $x \leq - 4$
Q2 · 3 marks · Short answerBoard level
Solve $2 x + 7 \leq 15$ , $x \in N$ .
Hint
$2 x \leq 8$ .
Answer
$x \leq 4$ ; solution set ${1, 2, 3, 4}$ .
Q3 · 3 marks · Short answerHarder board level
Solve $\frac{x}{2} - 5 \geq \frac{x}{3} - 4$ , where $x \in W$ and $x \leq 10$ .
Hint
Multiply by 6.
Answer
$3 x - 30 \geq 2 x - 24 \Rightarrow x \geq 6$ ; solution set ${6, 7, 8, 9, 10}$ .
Q4 · 1 mark · MCQHarder board level
The solution of $4 - x < 1$ , $x \in R$ , is:
(a) $x < 3$
(b) $x > 3$
(c) $x < - 3$
(d) $x > - 3$
Hint
$- x < - 3$ .
Answer
(b) $x > 3$
Q5 · 3 marks · Short answerHarder board level
Solve $5 x - 3 (2 x - 7) > 3 x - 1$ , where $x \in Z$ and $x > - 2$ .
Hint
Expand carefully: $- 3 \times - 7 = + 21$ .
Answer
$- x + 21 > 3 x - 1 \Rightarrow 22 > 4 x \Rightarrow x < 5.5$ . With $x > - 2$ : ${- 1, 0, 1, 2, 3, 4, 5}$ .
Q6 · 3 marks · Short answerHarder board level
A taxi charges ₹50 plus ₹12 per km. You have ₹290. Form an inequation and find the greatest whole number of kilometres you can travel.
Hint
Total fare must not exceed ₹290.
Answer
$50 + 12 x \leq 290 \Rightarrow x \leq 20$ ; at most 20 km.
Q7 · 4 marks · Short answerHOTS / competency
For which values of the integer $k$ does $x = 3$ NOT satisfy $2 x - k > 1$ , while $x = 4$ does satisfy it?
Hint
Write one inequality in $k$ for each value of $x$ .
Answer
$x = 3$ fails: $6 - k \leq 1 \Rightarrow k \geq 5$ . $x = 4$ works: $8 - k > 1 \Rightarrow k < 7$ . So $k \in {5, 6}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Solve $- 2 x < 8$ .Show answer
$x > - 4$ .
2.Solution set of $x < 3$ , $x \in W$ ?Show answer
${0, 1, 2}$ .
3.Solve $3 x - 4 \geq 11$ , $x \in N$ .Show answer
$x \geq 5$ : ${5, 6, 7, \dots}$ .
4.Does multiplying both sides by 4 change the sign?Show answer
No, 4 is positive.
5.Greatest integer satisfying $x < 2.5$ ?Show answer
2.

Solving Linear Inequations: frequently asked questions

When does the inequality sign change?

Only when you multiply or divide both sides by a negative number. Adding, subtracting, or multiplying by a positive number keeps the sign.

What is the difference between a replacement set and a solution set?

The replacement set is the set of values x is allowed to take, such as N, W, Z or R. The solution set is the part of it that makes the inequation true.

Is 0 a natural number in ICSE?

No. Natural numbers start at 1. Whole numbers include 0.

How do I write the answer when x is real?

Use set-builder form, such as {x : x ≥ 3, x ∈ R}, and show it on a number line with a filled or hollow circle at the end-point.