How do you solve a system of linear inequalities in two variables?

Solve each inequality separately and graph the solution region on the coordinate plane. The solution of the system is the intersection (overlap) of the individual solution regions. Shade the common region to represent the solution set.

What is the difference between a strict and a non-strict inequality?

A strict inequality uses (the boundary is not included, shown with a dashed line on graph). A non-strict inequality uses ≤ or ≥ (the boundary is included, shown with a solid line on graph).

Linear Inequalities — NCERT Class 11 Maths

Q: What happens when you multiply both sides of an inequality by a negative number?

The direction of the inequality sign reverses. For example, if x > 3 and you multiply by −2, you get −2x < −6. This is the most common mistake students make — always flip the sign when multiplying or dividing by a negative number.

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

View Solutions →

How to Use This Page

Solve each NCERT exercise on your own first, then compare with the step-by-step solutions below. Every question from Chapter 5 (Linear Inequalities) is fully worked — isolate the variable, handle sign reversals, and verify your answer in the original inequality.

Key Concepts (Quick Recap)

Inequality rules: Adding/subtracting same number on both sides — inequality unchanged
Multiplying/dividing by positive number — inequality unchanged
Multiplying/dividing by negative number — inequality sign reverses
Graphical solution: Line divides plane into two half-planes; test a point to pick the correct one
System of inequalities: Solution is the intersection of all individual solution regions
Interval notation: (a, b) = open; [a, b] = closed; (a, ∞) or (−∞, b) = unbounded

Exercise 5.1

Q1. Solve 24x < 100 when: (i) x is a natural number (ii) x is an integer.

Solution

24x < 100 ⟹ x < 100/24 = 25/6 ≈ 4.17

(i) Natural numbers less than 25/6: {1, 2, 3, 4} Solution set = {1, 2, 3, 4}

(ii) Integers less than 25/6: {..., −3, −2, −1, 0, 1, 2, 3, 4} Solution set = {..., −2, −1, 0, 1, 2, 3, 4}

Q2. Solve −12x > 30 when: (i) x is a natural number (ii) x is an integer.

Solution

−12x > 30 ⟹ x < 30/(−12) (sign reverses when dividing by −12) x < −5/2 = −2.5

(i) No natural number is less than −2.5. Solution set = ∅ (empty set)

(ii) Integers less than −2.5: {..., −5, −4, −3} Solution set = {..., −5, −4, −3}

Q3. Solve 5x − 3 < 7 when x is an integer.

Solution

5x − 3 < 7 ⟹ 5x < 10 ⟹ x < 2

Integers less than 2: {..., −2, −1, 0, 1} Solution set = {..., −1, 0, 1}

Q4. Solve 3x + 8 > 2 when x is a natural number.

Solution

3x + 8 > 2 ⟹ 3x > −6 ⟹ x > −2

Natural numbers greater than −2: {1, 2, 3, ...} Solution set = {1, 2, 3, 4, ...} = N

Q5. Solve 4x + 3 < 5x + 7.

Solution

4x + 3 < 5x + 7 3 − 7 < 5x − 4x −4 < x x > −4, or in interval notation: (−4, ∞)

Q6. Solve 3x − 7 > 5x − 1.

Solution

3x − 7 > 5x − 1 −7 + 1 > 5x − 3x −6 > 2x x < −3 Solution: (−∞, −3)

Q7. Solve 3(x − 2) ≤ 5x − 8.

Solution

3x − 6 ≤ 5x − 8 −6 + 8 ≤ 5x − 3x 2 ≤ 2x 1 ≤ x Solution: [1, ∞)

Q8. Solve 3(2 − x) ≥ 2(1 − x).

Solution

6 − 3x ≥ 2 − 2x 6 − 2 ≥ −2x + 3x 4 ≥ x x ≤ 4, Solution: (−∞, 4]

Q9. Solve x + 3/2 > 2(2x − 1).

Solution

x + 3/2 > 4x − 2 3/2 + 2 > 4x − x 7/2 > 3x 7/6 > x x < 7/6, Solution: (−∞, 7/6)

Q10. Solve 3x − 2 < 2x + 1.

Solution

3x − 2x < 1 + 2 x < 3 Solution: (−∞, 3)

Q11. Solve 5x − 3 > 3x − 5.

Solution

5x − 3x > −5 + 3 2x > −2 x > −1 Solution: (−1, ∞)

Q12. Solve 2(2x + 1) < 3(x − 2) + 4.

Solution

4x + 2 < 3x − 6 + 4 4x + 2 < 3x − 2 4x − 3x < −2 − 2 x < −4 Solution: (−∞, −4)

Exercise 5.2

Q1. Solve the inequality 2 ≤ 3x − 4 ≤ 5.

Solution

2 ≤ 3x − 4 ≤ 5 Adding 4: 6 ≤ 3x ≤ 9 Dividing by 3: 2 ≤ x ≤ 3 Solution: [2, 3]

Q2. Solve the inequality 6 ≤ −3(2x − 4) < 12.

Solution

6 ≤ −3(2x − 4) < 12 Dividing by −3 (reversing both inequalities): −2 ≥ 2x − 4 > −4 Adding 4: 2 ≥ 2x > 0 Dividing by 2: 1 ≥ x > 0 Solution: (0, 1]

Q3. Solve the inequality −3 ≤ 4 − 7x/2 ≤ 18.

Solution

−3 ≤ 4 − 7x/2 ≤ 18 Subtracting 4: −7 ≤ −7x/2 ≤ 14 Multiplying by −2/7 (reversing): 2 ≥ x ≥ −4 i.e., −4 ≤ x ≤ 2 Solution: [−4, 2]

Q4. Solve the inequality −15 < 3(x − 2)/5 ≤ 0.

Solution

−15 < 3(x − 2)/5 ≤ 0 Multiplying by 5: −75 < 3(x − 2) ≤ 0 Dividing by 3: −25 < x − 2 ≤ 0 Adding 2: −23 < x ≤ 2 Solution: (−23, 2]

Q5. Solve the inequality −12 < 4 − 3x/5 ≤ 2.

Solution

−12 < 4 − 3x/5 ≤ 2 Subtracting 4: −16 < −3x/5 ≤ −2 Multiplying by −5/3 (reversing): 80/3 > x ≥ 10/3 i.e., 10/3 ≤ x < 80/3 Solution: [10/3, 80/3)

Q6. Solve the inequality 7 ≤ (3x + 11)/2 ≤ 11.

Solution

7 ≤ (3x + 11)/2 ≤ 11 Multiplying by 2: 14 ≤ 3x + 11 ≤ 22 Subtracting 11: 3 ≤ 3x ≤ 11 Dividing by 3: 1 ≤ x ≤ 11/3 Solution: [1, 11/3]

Q7. Solve: (5x/2) − (3x − 2)/3 ≥ (x − 1)/2 + 1.

Solution

LCM of denominators 2, 3, 2 = 6. Multiply throughout by 6: 15x − 2(3x − 2) ≥ 3(x − 1) + 6 15x − 6x + 4 ≥ 3x − 3 + 6 9x + 4 ≥ 3x + 3 6x ≥ −1 x ≥ −1/6 Solution: [−1/6, ∞)

Q8. Solve: (2x − 1)/3 ≥ (3x − 2)/4 − (2 − x)/5.

Solution

LCM of 3, 4, 5 = 60. Multiply throughout by 60: 20(2x − 1) ≥ 15(3x − 2) − 12(2 − x) 40x − 20 ≥ 45x − 30 − 24 + 12x 40x − 20 ≥ 57x − 54 −20 + 54 ≥ 57x − 40x 34 ≥ 17x 2 ≥ x x ≤ 2, Solution: (−∞, 2]

Q9. Solve: 3(2 − x) ≥ 2(1 − x).

Solution

6 − 3x ≥ 2 − 2x 6 − 2 ≥ 3x − 2x 4 ≥ x x ≤ 4, Solution: (−∞, 4]

Q10. Solve: 3x − 2 < 2x + 1 and 3x + 4 > 2x + 3.

Solution

From first inequality: x < 3 From second inequality: x > −1 Both must hold: −1 < x < 3 Solution: (−1, 3)

Exercise 5.3

Q1. Solve the following system of inequalities graphically: x ≥ 0, y ≥ 0.

Solution

x ≥ 0 represents the region to the right of the y-axis (including the y-axis). y ≥ 0 represents the region above the x-axis (including the x-axis). The solution is the intersection — the first quadrant (including the axes).

Q2. Solve graphically: x + y ≤ 9, y > x, x ≥ 0.

Solution

x + y ≤ 9: Region on and below the line x + y = 9 (passes through (9,0) and (0,9))
y > x: Region above the line y = x (not including the line)
x ≥ 0: Right of y-axis

Solution region**: The bounded area in the first quadrant between y = x and x + y = 9, above the line y = x (not including y = x).

Corner points: (0, 0), (0, 9), (4.5, 4.5) — but y > x strictly, so the region is the interior triangle with vertices approaching (0,0), (0,9), and (4.5, 4.5) from above y = x.

Q3. Solve graphically: 2x + y ≥ 6, 3x + 4y ≤ 12.

Solution

2x + y = 6: passes through (3, 0) and (0, 6). Region: above this line.
3x + 4y = 12: passes through (4, 0) and (0, 3). Region: below this line.

Intersection point: 2x + y = 6 and 3x + 4y = 12 From first: y = 6 − 2x. Substituting: 3x + 4(6 − 2x) = 12 ⟹ 3x + 24 − 8x = 12 ⟹ −5x = −12 ⟹ x = 12/5 y = 6 − 24/5 = 6/5

SolutionThe bounded region with vertices at (12/5, 6/5), and where the lines intersect the axes.

Q4. Solve graphically: x + y ≤ 9, y > x, x ≥ 0.

(This is the same as Q2.)

Q5. Solve graphically: 2x − y > 1, x − 2y < −1.

Solution

2x − y = 1: passes through (1/2, 0) and (0, −1). Region: 2x − y > 1 means y < 2x − 1 (below the line)
x − 2y = −1: passes through (−1, 0) and (0, 1/2). Region: x − 2y < −1 means y > (x + 1)/2 (above the line)

Intersection point: 2x − y = 1 and x − 2y = −1 From first: y = 2x − 1. Substituting: x − 2(2x − 1) = −1 ⟹ x − 4x + 2 = −1 ⟹ −3x = −3 ⟹ x = 1, y = 1

The lines intersect at (1, 1). The solution region is the area below 2x − y = 1 and above x − 2y = −1.

Q6. Solve graphically: x + y ≤ 6, x + y ≥ 4.

Solution

x + y = 6: line through (6,0) and (0,6). Region: on or below the line.
x + y = 4: line through (4,0) and (0,4). Region: on or above the line.

SolutionThe strip between the two parallel lines, including both lines. This is a bounded region between x + y = 4 and x + y = 6.

Q7. Solve graphically: 2x + y ≥ 8, x + 2y ≥ 10.

Solution

2x + y = 8: passes through (4, 0) and (0, 8). Region: above the line.
x + 2y = 10: passes through (10, 0) and (0, 5). Region: above the line.

Intersection point: 2x + y = 8, x + 2y = 10 From first: y = 8 − 2x. Substituting: x + 2(8 − 2x) = 10 ⟹ x + 16 − 4x = 10 ⟹ −3x = −6 ⟹ x = 2, y = 4

SolutionThe unbounded region above both lines, with the intersection at (2, 4).

Q8. Solve graphically: x + y ≤ 6, x ≥ 0, y ≥ 0.

Solution

x + y ≤ 6: Region on and below line x + y = 6
x ≥ 0: Right of y-axis
y ≥ 0: Above x-axis

SolutionThe right triangle in the first quadrant with vertices (0,0), (6,0), (0,6), bounded by x + y = 6 and the axes.

Q9. Ravi obtained 70 and 75 marks in two unit tests. What is the minimum marks he should get in the third test to have an average of at least 60 marks?

Solution

Let marks in third test = x. Average = (70 + 75 + x)/3 ≥ 60 145 + x ≥ 180 x ≥ 35

Minimum marks needed = 35

Q10. To receive Grade A in a course, one must obtain an average of 90 marks or more in five examinations. If marks in first four are 87, 92, 94, 95, find the minimum marks in the fifth to get grade A.

Solution

Let marks in fifth = x. (87 + 92 + 94 + 95 + x)/5 ≥ 90 368 + x ≥ 450 x ≥ 82

Minimum marks needed = 82

Miscellaneous Exercise

Q1. Solve: 2(x − 1) < 3(x − 3) + 4.

Solution

2x − 2 < 3x − 9 + 4 2x − 2 < 3x − 5 −2 + 5 < 3x − 2x 3 < x x > 3, Solution: (3, ∞)

Q2. Solve: 4(2x − 1) ≤ 3(x + 2).

Solution

8x − 4 ≤ 3x + 6 5x ≤ 10 x ≤ 2 Solution: (−∞, 2]

Q3. Solve: 3x − 2 ≥ 2x + 1 and 5x − 4 ≤ 4x + 2.

Solution

First: 3x − 2x ≥ 1 + 2 ⟹ x ≥ 3 Second: 5x − 4x ≤ 2 + 4 ⟹ x ≤ 6

Both: 3 ≤ x ≤ 6 Solution: [3, 6]

Q4. Solve: 3x − 7 > 2x − 5 and 2x + 5 > x + 6.

Solution

First: x > 2 Second: x > 1

Both must hold: x > 2 (since if x > 2, then x > 1 automatically) Solution: (2, ∞)

Q5. A solution is to be kept between 68°F and 77°F. What is the range of temperature in Celsius if the conversion formula is C = 5(F − 32)/9?

Solution

68 < F < 77 C = 5(F − 32)/9

When F = 68: C = 5(36)/9 = 20 When F = 77: C = 5(45)/9 = 25

Since the conversion is linear and increasing: 20°C < C < 25°C, or (20, 25) in °C

Q6. A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution. The resultant mixture must have more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how much of the 2% solution must be added?

Solution

Let x litres of 2% solution be added. Total mixture = (640 + x) litres Boric acid in mixture = 8% × 640 + 2% × x = 51.2 + 0.02x

Condition: 4% < concentration < 6% 0.04(640 + x) < 51.2 + 0.02x < 0.06(640 + x)

Lower bound: 25.6 + 0.04x < 51.2 + 0.02x 0.02x < 25.6 x < 1280

Upper bound: 51.2 + 0.02x < 38.4 + 0.06x 12.8 < 0.04x x > 320

320 litres < x < 1280 litres

Internal Links

[Chapter 5 Guide — Linear Inequalities](/school/cbse/class/11/maths/cbse-linear-inequalities)
[All Class 11 Maths Solutions](/school/cbse/class/11/maths/solutions)
[Class 11 Maths Subject Guide](/school/cbse/class/11/maths)

Frequently Asked Questions

The direction of the inequality sign reverses. For example, if x > 3 and you multiply by −2, you get −2x < −6. This is the most common mistake students make — always flip the sign when multiplying or dividing by a negative number.