Linear Inequalities — Class 11 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Linear Inequalities" — 7 important questions with detailed answers for CBSE board exam preparation.

TL;DR: Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Linear Inequalities" — 7 important questions with detailed answers for CBSE board…

By Syllab.in · Updated Jun 14, 2026

Key Questions Covered:

Solve the linear inequality 3x - 5 > 7 and represent the solution on a number…
Solve the compound inequality -3 ≤ 2x + 1 < 5.
Solve the absolute value inequality |x - 3| ≤ 2.
Solve the system of linear inequalities: 2x + y ≤ 8 x - y ≥ -2 x ≥ 0, y ≥ 0
Solve the quadratic inequality x² - 5x + 6 > 0.
Solve the inequality (x + 1)/(x - 2) > 0.
+ 1 more questions in the full chapter

Solutions Summary:

Question	Status
Solve the linear inequality 3x - 5 > 7 and represent the …	✓ Solved
Solve the compound inequality -3 ≤ 2x + 1 < 5.	✓ Solved
Solve the absolute value inequality \|x - 3\| ≤ 2.	✓ Solved
Solve the system of linear inequalities: 2x + y ≤ 8 x - y…	✓ Solved
Solve the quadratic inequality x² - 5x + 6 > 0.	✓ Solved
Solve the inequality (x + 1)/(x - 2) > 0.	✓ Solved

Showing 6 of 7 questions

Q1: Solve the linear inequality 3x - 5 > 7 and represent the solution on a number line.

Step 1: Isolate the variable term. 3x - 5 > 7 3x > 7 + 5 3x > 12 Step 2: Divide both sides by 3. x > 12/3 x > 4 Step 3: Write the solution set. Solution = {x ∈ ℝ : x > 4} or in interval notation: (4, ∞) Step 4: Represent on a number line. ←────○────────────→ 0 4 5 6 7 Open circle at 4, arrow pointing right to indicate all numbers greater than 4. Step 5: Verify with test points. Test x = 5: 3(5) - 5 = 15 - 5 = 10 > 7 ✓ Test x = 4: 3(4) - 5 = 12 -...

Q2: Solve the compound inequality -3 ≤ 2x + 1 < 5.

Step 1: Break into two separate inequalities. -3 ≤ 2x + 1 AND 2x + 1 < 5 Step 2: Solve the left inequality -3 ≤ 2x + 1. -3 ≤ 2x + 1 -3 - 1 ≤ 2x -4 ≤ 2x -2 ≤ x Or: x ≥ -2 Step 3: Solve the right inequality 2x + 1 < 5. 2x + 1 < 5 2x < 5 - 1 2x < 4 x < 2 Step 4: Combine the solutions. x ≥ -2 AND x < 2 Solution: -2 ≤ x < 2 Step 5: Write in interval notation. [-2, 2) Step 6: Verify with test points. Test x = -2: -3 ≤ 2(-2) + 1 = -3 ≤ -3 ✓ and -3 < 5 ✓ Test x = 0: -3 ≤ ...

Q3: Solve the absolute value inequality |x - 3| ≤ 2.

Step 1: Recall the definition of absolute value inequality. |A| ≤ B (where B ≥ 0) is equivalent to -B ≤ A ≤ B Step 2: Apply the definition. |x - 3| ≤ 2 -2 ≤ x - 3 ≤ 2 Step 3: Add 3 to all parts. -2 + 3 ≤ x - 3 + 3 ≤ 2 + 3 1 ≤ x ≤ 5 Step 4: Write the solution set. Solution: [1, 5] Step 5: Verify with test points. Test x = 1: |1 - 3| = |-2| = 2 ≤ 2 ✓ Test x = 3: |3 - 3| = 0 ≤ 2 ✓ Test x = 5: |5 - 3| = 2 ≤ 2 ✓ Test x = 0: |0 - 3| = 3 ≤ 2 ✗ Test x = 6: |6 - 3| = 3 ≤ 2 ✗ Final Answer: 1 ≤ x ≤ 5 ...

Q4: Solve the system of linear inequalities: 2x + y ≤ 8 x - y ≥ -2 x ≥ 0, y ≥ 0

Step 1: Identify the inequalities. (1) 2x + y ≤ 8 (2) x - y ≥ -2, or x ≥ y - 2 (3) x ≥ 0 (non-negative x) (4) y ≥ 0 (non-negative y) Step 2: Find boundary lines by converting to equations. Line 1: 2x + y = 8 ⟹ y = 8 - 2x Line 2: x - y = -2 ⟹ y = x + 2 Line 3: x = 0 (y-axis) Line 4: y = 0 (x-axis) Step 3: Find key intersection points. Intersection of 2x + y = 8 and x - y = -2: 2x + y = 8 x - y = -2 Adding: 3x = 6 ⟹ x = 2, y = 4 Point: (2, 4) Intersection of 2x + y = 8 and x = 0: 2(0) + y = 8 ⟹...

Q5: Solve the quadratic inequality x² - 5x + 6 > 0.

Step 1: Factor the quadratic expression. x² - 5x + 6 = (x - 2)(x - 3) Step 2: Find the zeros (critical points). (x - 2)(x - 3) = 0 x = 2 or x = 3 Step 3: Determine the sign of (x - 2)(x - 3) in each interval. Test intervals: (-∞, 2), (2, 3), (3, ∞) Test x = 0 (from interval (-∞, 2)): (0 - 2)(0 - 3) = (-2)(-3) = 6 > 0 ✓ Test x = 2.5 (from interval (2, 3)): (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25 < 0 ✗ Test x = 4 (from interval (3, ∞)): (4 - 2)(4 - 3) = (2)(1) = 2 > 0 ✓ Step 4: Wri...

Q6: Solve the inequality (x + 1)/(x - 2) > 0.

Step 1: Find critical points where the expression equals zero or is undefined. Numerator: x + 1 = 0 ⟹ x = -1 Denominator: x - 2 = 0 ⟹ x = 2 (undefined, asymptote) Step 2: Create a sign chart with intervals. Intervals: (-∞, -1), (-1, 2), (2, ∞) Step 3: Test the sign in each interval. Test x = -2 (from (-∞, -1)): (-2 + 1) / (-2 - 2) = (-1) / (-4) = 1/4 > 0 ✓ Test x = 0 (from (-1, 2)): (0 + 1) / (0 - 2) = 1 / (-2) = -1/2 < 0 ✗ Test x = 3 (from (2, ∞)): (3 + 1) / (3 - 2) = 4 / 1 = 4 > 0...

Showing 6 of 7 questions. Visit the full page for complete solutions.