Linear Inequations
ICSE Class 10 · Mathematics · Chapter 4
What this chapter covers
- Inequalities and the four signs
- Linear inequations in one variable
- The replacement set: x ∈ N, W, Z, R
- Rules for solving linear inequations
- Solving algebraically and writing the solution in set notation form
- Combined inequations
- Representation of the solution on the number line
- Common errors and how to avoid them
Key points
- An inequation uses one of the signs <, >, ≤ or ≥; a linear inequation in one variable has one unknown, to the first power only.
- The replacement set (N, W, Z or R) states which values x may take; the solution set is the part of it that makes the inequation true.
- The same inequation can have different solution sets for different replacement sets.
- Adding or subtracting any number, or multiplying or dividing by a positive number, keeps the sign; multiplying or dividing by a negative number reverses it.
- Never multiply or divide both sides by an expression containing x, or by zero.
- Write finite or listable answers in roster form; write real-number answers in set-builder form, \x : x in R, ldots\.
- An empty solution set, φ, is a valid answer.
- In a combined inequation, apply each step to all three parts; if x appears in more than one part, split it into two inequations and take the common values.
- On the number line, N, W and Z answers are separate solid dots; R answers are a thick line.
- For real numbers, a hollow circle marks a boundary that is excluded (<, >) and a solid circle a boundary that is included (≤, ≥); an arrow-head shows that the solutions continue without end.
- Check every answer with one value inside the solution set and one just outside it.
Questions answered
Define a linear inequation in one variable and give one example.
A linear inequation in one variable is an inequality containing one unknown that appears only to the first power, for example 4x - 1 ≥ 7.
Define the solution set of an inequation.
The solution set is the set of all values from the replacement set that make the inequation true.
Write each statement as an inequation: (a) the number of passengers p in an auto-rickshaw is at most 3; (b) the cost c of a cricket bat is more than ₹1200.
(a) p ≤ 3; (b) c > 1200.
Solve 2x + 7 ≤ 15, x in W.
2x ≤ 8, so x ≤ 4. Solution set = \0, 1, 2, 3, 4\.
Solve 7 - 3x > 22, x in Z.
-3x > 15, so x < -5 (sign reversed on dividing by -3). Solution set = \ldots, -8, -7, -6\.
Solve (x + 1)/(2) - (x)/(3) ≥ 1, x in R, and write the solution set.
Multiplying by 6: 3x + 3 - 2x ≥ 6, so x ≥ 3. Solution set = \x : x in R, x ≥ 3\.
Solve -5 ≤ 3x + 1 < 10, x in Z.
-6 ≤ 3x < 9, so -2 ≤ x < 3. Solution set = \-2, -1, 0, 1, 2\.
Solve -2 < 4 - 3x ≤ 10, x in R, and describe its graph.
Subtract 4: -6 < -3x ≤ 6. Divide by -3 and reverse both signs: 2 > x ≥ -2, that is, -2 ≤ x < 2. Graph: a solid circle at -2, a hollow circle at 2, joined by a thick line.
Past board exam questions
ICSE 2027 sample paper · 3 marks
Solve the following inequation, write the solution set and represent it on the real number line. -6 + (12x)/(5) ≤ (7x)/(5) - 3 < (2)/(5) + 2x, x in R.
Show answer
-6 + (12x)/(5) ≤ (7x)/(5) - 3 ⇒ (12x)/(5) - (7x)/(5) ≤ 3 ⇒ x ≤ 3
(7x)/(5) - 3 < (2)/(5) + 2x ⇒ (7x)/(5) - 2x < (17)/(5) ⇒ -(3x)/(5) < (17)/(5) ⇒ x > -(17)/(3) = -5(2)/(3)
Solution set: \x : -5(2)/(3) < x ≤ 3, x in R\
On the number line: an open (hollow) circle at -5(2)/(3), a closed (filled) circle at 3, and the segment between them darkened.
ICSE 2026 board paper · 3 marks
Solve the following inequation, write the solution set and represent it on the real number line. -1 < (2x - 3)/(3) - (x)/(5) ≤ 1, x in R
Show answer
Multiply throughout by 15: -15 < 5(2x - 3) - 3x ≤ 15 ⇒ -15 < 7x - 15 ≤ 15
Add 15: 0 < 7x ≤ 30 ⇒ 0 < x ≤ (30)/(7), i.e. 0 < x ≤ 4(2)/(7).
Solution set = \x : 0 < x ≤ 4(2)/(7), x in R\.
On the real number line: an open (hollow) circle at 0, a filled (dark) circle at 4(2)/(7), and the segment between them darkened.
ICSE 2025 compartment paper · 1 mark
If 2x - 15 > 4x + 9, then:
- x < -12
- x < 12
- x > -12
- x > 12
Show answer
x < -12
ICSE 2025 sample paper · 1 mark
Given, x + 2 ≤ (x)/(3) + 3 and x is a prime number. The solution set for x is:
- ∅
- {0}
- {1}
- {0, 1}
Show answer
∅
ICSE 2025 board paper · 1 mark
The solution set for 0 < -(x)/(3) < 2, x in Z is:
- {– 5, – 4, – 3, – 2, – 1}
- {– 6, – 5, – 4, – 3, – 2, – 1}
- {– 5, – 4, – 3, – 2, – 1, 0}
- {– 6, – 5, – 4, – 3, – 2, – 1, 0}
Show answer
{– 5, – 4, – 3, – 2, – 1}
ICSE 2025 sample paper · 3 marks
Solve the following inequation, write the solution set and represent it on the real number line. 5x - 21 < (5x)/(7) - 6 ≤ -3(3)/(7) + x, x in R.
Show answer
Left part: 5x - 21 < (5x)/(7) - 6 ⇒ 5x - (5x)/(7) < 15 ⇒ (30x)/(7) < 15 ⇒ 30x < 105 ⇒ x < 3.5.
Right part: (5x)/(7) - 6 ≤ -(24)/(7) + x ⇒ (5x)/(7) - x ≤ 6 - (24)/(7) ⇒ -(2x)/(7) ≤ (18)/(7) ⇒ -2x ≤ 18 ⇒ x ≥ -9.
Solution set: \x : -9 ≤ x < (7)/(2), x in R\.
Number line: a thick line from −9 to 3.5, with a filled (solid) circle at −9 and an open (hollow) circle at 3.5.
ICSE 2025 board paper · 3 marks
Solve the following inequation, write the solution set and represent it on the real number line. 2x - (5)/(3) < (3x)/(5) + 10 ≤ (4x)/(5) + 11; x in R
Show answer
2x - (5)/(3) < (3x)/(5) + 10 ⇒ 30x - 25 < 9x + 150 ⇒ 21x < 175 ⇒ x < (25)/(3), i.e. x < 8(1)/(3).
(3x)/(5) + 10 ≤ (4x)/(5) + 11 ⇒ -1 ≤ (x)/(5) ⇒ x ≥ -5.
Solution set = \x : -5 ≤ x < 8(1)/(3), x in R\.
On the number line: a filled dot at –5, a hollow circle at 8(1)/(3), and the segment between them darkened.
ICSE 2025 compartment paper · 3 marks
Solve the inequation, write down the solution set and represent it on a real number line: 3x - 16 < (2x)/(5) - 3 ≤ -(3)/(5) + 2x ; x in R
Show answer
3x - 16 < (2x)/(5) - 3 ⇒ 15x - 80 < 2x - 15 ⇒ 13x < 65 ⇒ x < 5.
(2x)/(5) - 3 ≤ -(3)/(5) + 2x ⇒ 2x - 15 ≤ -3 + 10x ⇒ -12 ≤ 8x ⇒ x ≥ -1.5.
Solution set: \x : -1.5 ≤ x < 5, x in R\.
On the real number line: a solid (filled) dot at –1.5, a hollow (open) circle at 5, and the segment between them darkened.