MathsISC Class 11 · Linear and Quadratic Inequalities

Linear Inequalities in One Variable: Rules, Double Inequalities and Modulus

Q: Why does the inequality flip on multiplying by a negative number?

Multiplication by a negative number reflects the number line, reversing the order: 2 −5.

Updated 17 October 2026 · Fact-checked

Add or subtract the same number on both sides of an inequality freely; multiply or divide by a positive number keeps the sign, but multiplying or dividing by a negative number reverses it. Solutions are intervals shown on the number line; for simultaneous inequalities take the intersection; |x − a| < b means a − b < x < a + b.

Points within distance 3 of 2 give −1 < x < 5; points farther than 3 give two rays; hollow dots show −1 and 5 are not included.

Illustration drawn for QuestionBuddy.

Understand Linear Inequalities in One Variable and Modulus

Rules. (1) Adding or subtracting the same quantity to both sides keeps the inequality. (2) Multiplying or dividing by a positive number keeps it. (3) Multiplying or dividing by a negative number reverses it: $- 2 x < 6$ gives $x > - 3$ . (4) Taking reciprocals of positive numbers reverses it. A solution is a set of values; the solution set is usually an interval.

One inequality. Solve $a x + b < c$ like an equation, watching the sign when dividing by a negative number. Show the answer on a number line (a hollow dot for < and >, a filled dot for ≤ and ≥).

Double inequality. $c < a x + b < d$ is solved by performing the same operation on all three parts. If a is negative, the direction of both signs reverses.

Simultaneous inequalities. Solve each inequality separately and take the intersection of the solution sets (the values satisfying all). If the intersection is empty there is no solution.

Modulus inequalities. For $b > 0$ : $∣ x - a ∣ < b ⟺ a - b < x < a + b$ ; $∣ x - a ∣ \leq b ⟺ a - b \leq x \leq a + b$ ; $∣ x - a ∣ > b ⟺ x < a - b$ or $x > a + b$ . If $b < 0$ , $∣ x - a ∣ < b$ has no solution and $∣ x - a ∣ > b$ holds for all x.

Word problems. Define the variable, write the condition as an inequality, solve it, and interpret the result in context (for example only whole numbers of items).

Linear Inequalities in One Variable and Modulus: formulas to remember

Rules: $M u l t i pl y or d i v i d e b y an e g a t i v e n u mb er : r e v er se t h es i g n$; Adding or subtracting never changes the sign.
Modulus (b > 0): $$|x - a| < b ⟺ a - b < x < a + b$$; And $∣ x - a ∣ > b ⟺ x < a - b$ or $x > a + b$ .
Double inequality: $$c < ax + b < d$$; Operate on all three parts.
Simultaneous: $S o l u t i o n se t = in t er sec t i o n o f t h e in d i v i d u a l so l u t i o n se t s$; Empty intersection means no solution.

Linear Inequalities in One Variable and Modulus 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.

A plumber's bill

Suppose a plumber charges Rs 300 as a call-out fee plus Rs 150 per hour, and a customer wants the total to be between Rs 1000 and Rs 1600 inclusive.

Find the possible number of hours and the largest whole number of hours.

Try it first, then show the solutionSolution

$1000 \leq 300 + 150 h \leq 1600$ .
$700 \leq 150 h \leq 1300$ , so $\frac{14}{3} \leq h \leq \frac{26}{3}$ , that is $4.67 \leq h \leq 8.67$ .
The largest whole number of hours is 8 (cost Rs 1500).

Answer: 4.67 ≤ h ≤ 8.67; at most 8 whole hours.

Memory hook: Apply each operation to all three parts of the inequality.

Tricks to remember Linear Inequalities in One Variable and Modulus

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

Negative flips
Multiplying or dividing by a negative number flips the sign.
Use it when: The most common error.
Modulus less than: sandwiched
$∣ x - a ∣ < b$ means x lies within b of a.
Use it when: Modulus inequalities.

Linear Inequalities in One Variable and Modulus: solved examples

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

Example 1: Solving a linear inequality · 1 mark · MCQBoard level

The solution of $3 x - 7 > 5$ is:
(a) $x > 4$
(b) $x > 2$
(c) $x < 4$
(d) $x < 12$

Try it first, then show the solutionSolution

$3 x > 12$ , so $x > 4$ .

Answer: (a) $x > 4$

Example 2: A double inequality · 3 marks · Short answerBoard level

Solve $- 5 \leq \frac{2 - 3 x}{4} \leq 9$ and write the solution in interval notation.

Try it first, then show the solutionSolution

Multiply by 4: $- 20 \leq 2 - 3 x \leq 36$ . (1 mark)
Subtract 2: $- 22 \leq - 3 x \leq 34$ . (0.5 mark)
Divide by −3 (reverse): $\frac{22}{3} \geq x \geq - \frac{34}{3}$ . (1 mark)
Solution $[- \frac{34}{3}, \frac{22}{3}]$ . (0.5 mark)

Answer: [−34/3, 22/3]

Example 3: Simultaneous inequalities · 3 marks · Short answerHarder board level

Solve $3 x - 2 < 2 x + 1$ and $4 x + 3 \geq 2 x - 5$ together, and mark the solution on a number line.

Try it first, then show the solutionSolution

First: $x < 3$ . (1 mark)
Second: $2 x \geq - 8$ , so $x \geq - 4$ . (1 mark)
Intersection: $- 4 \leq x < 3$ , i.e. $[- 4, 3)$ (filled dot at −4, hollow dot at 3). (1 mark)

Answer: [−4, 3)

Example 4: Modulus inequalities · 4 marks · Short answerHarder board level

Solve (i) $∣2 x - 3∣ \leq 5$ and (ii) $∣ x - 2∣ > 3$ .

Try it first, then show the solutionSolution

(i) $- 5 \leq 2 x - 3 \leq 5$ gives $- 2 \leq 2 x \leq 8$ and $- 1 \leq x \leq 4$ . (2 marks)
(ii) $x - 2 > 3$ or $x - 2 < - 3$ , so $x > 5$ or $x < - 1$ . (1.5 marks)
Interval notation: $[- 1, 4]$ and $(- \infty, - 1) \cup (5, \infty)$ . (0.5 mark)

Answer: [−1, 4]; (−∞, −1) ∪ (5, ∞).

Example 5: A word problem · 5 marks · Long answerHOTS / competency

A student has scored 62 and 71 in two tests. What score in a third test will give an average of at least 70 but less than 80? If the maximum is 100 marks, what is the range of possible scores?

Try it first, then show the solutionSolution

Let the third score be x. Average $= \frac{62 + 71 + x}{3} = \frac{133 + x}{3}$ . (1 mark)
$70 \leq \frac{133 + x}{3} < 80$ gives $210 \leq 133 + x < 240$ . (1.5 marks)
Subtract 133: $77 \leq x < 107$ . (1.5 marks)
The maximum is 100, so $77 \leq x \leq 100$ . (1 mark)

Answer: 77 ≤ x ≤ 100.

Where students lose marks in Linear Inequalities in One Variable and Modulus

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

Mistake: Not reversing the sign
Dividing by −3 without flipping.
Do this: Whenever you multiply or divide by a negative number, reverse the inequality sign.
Usually costs: 2 marks.
Mistake: Wrong union/intersection
Taking the union for simultaneous inequalities.
Do this: Simultaneous means both must hold: intersection.
Usually costs: 1 to 2 marks.
Mistake: Open and closed end points confused
Writing [ for <.
Do this: Strict inequality: round bracket; ≤ or ≥: square bracket.
Usually costs: 1 mark.

Exam tips

Show each algebraic step.
Draw the number line.
Check one value from the solution set.
Interpret word problems: whole numbers and limits.

Practice questions on Linear Inequalities in One Variable and Modulus

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 $- 2 x > 6$ then:
(a) $x > - 3$
(b) $x < - 3$
(c) $x > 3$
(d) $x < 3$
Hint
Dividing by a negative reverses the sign.
Answer
(b) $x < - 3$
Q2 · 2 marks · Short answerBoard level
Solve $5 (x - 1) + 2 \geq 3 x - 7$ .
Hint
Expand first.
Answer
$5 x - 3 \geq 3 x - 7$ gives $2 x \geq - 4$ , so $x \geq - 2$ .
Q3 · 3 marks · Short answerHarder board level
Solve $- 4 < \frac{3 x + 1}{2} \leq 8$ .
Hint
Multiply by 2 first.
Answer
$- 8 < 3 x + 1 \leq 16$ gives $- 9 < 3 x \leq 15$ , so $- 3 < x \leq 5$ .
Q4 · 3 marks · Short answerHarder board level
Solve $∣3 x + 1∣ < 4$ and $∣ x ∣ \geq 2$ together.
Hint
Take the intersection.
Answer
$∣3 x + 1∣ < 4$ : $- 5 < 3 x < 3$ gives $- \frac{5}{3} < x < 1$ . $∣ x ∣ \geq 2$ : $x \leq - 2$ or $x \geq 2$ . The intersection is empty.
Q5 · 3 marks · Short answerHarder board level
The perimeter of a rectangle of length 12 cm is to be less than 40 cm and its area at least 36 cm². Find the possible widths w.
Hint
Two conditions on w.
Answer
Perimeter $2 (12 + w) < 40$ gives $w < 8$ . Area $12 w \geq 36$ gives $w \geq 3$ . So $3 \leq w < 8$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a mobile plan A costs Rs 200 plus Rs 2 per minute, and plan B costs a flat Rs 500.
(i) Write the inequality for plan A to be cheaper. (1 mark)
(ii) Solve it. (1 mark)
(iii) State the largest whole number of minutes for which A is cheaper. (1 mark)
(iv) For what number of minutes is plan B cheaper? (1 mark)
Hint
Compare 200 + 2m with 500.
Answer
(i) $200 + 2 m < 500$ . (ii) $m < 150$ . (iii) 149 minutes. (iv) $m > 150$ minutes (equal at 150).

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.Multiply by −1?Show answer
Reverse the inequality.
2.|x − a| < b?Show answer
a − b < x < a + b.
3.Simultaneous solution?Show answer
Intersection.
4.Hollow dot means?Show answer
The end point is excluded.

Linear Inequalities in One Variable and Modulus: frequently asked questions

Why does the inequality flip on multiplying by a negative number?

Multiplication by a negative number reflects the number line, reversing the order: 2 < 5 but −2 > −5.

What if the solution sets do not overlap?

Then there is no solution; the solution set is empty.

Can I square both sides?

Only if both sides are non-negative; otherwise squaring may change the inequality.