Linear Inequalities Class 11 Notes

• Add or subtract the same quantity from both sides without changing the inequality direction.
• Multiply or divide both sides by a positive number — direction stays the same.
• Multiply or divide both sides by a negative number — the inequality sign reverses.
• The solution is represented on the real number line using open circles (strict) or closed circles (non-strict).

• Transitive property: if a > b and b > c, then a > c.
• Addition property: if a > b, then a + c > b + c for any real c.
• Multiplication by positive: if a > b and c > 0, then ac > bc.
• Multiplication by negative: if a > b and c < 0, then ac < bc (sign reverses).
• Combine like terms first, then isolate the variable.

A system such as $2x-3>1$ and $3x+1\le 10$ is solved by finding the solution set of each inequality separately and then taking their intersection. The final answer is the set of x-values that satisfy ALL inequalities simultaneously.

• |x| < a represents all points within distance a of the origin — an open interval.
• |x| > a represents points farther than a from the origin — two disjoint rays.
• For |ax + b| < c, rewrite as −c < ax + b < c and solve the compound inequality.
• For |ax + b| > c, split into two separate inequalities and take their union.

• The solution region (half-plane) is determined by testing any one point not on the boundary — typically the origin (0, 0).
• If the test point satisfies the inequality, its half-plane is shaded; otherwise the opposite half-plane is shaded.
• For a strict inequality (< or >), the boundary line is dashed (excluded). For ≤ or ≥, the boundary line is solid (included).

To solve a system of linear inequalities in two variables graphically, graph each inequality's half-plane on the same axes. The feasible region is the intersection of all half-planes — the set of points satisfying every inequality in the system.

• Graph each boundary line, decide solid or dashed, and shade the correct half-plane.
• The feasible region is where all shaded regions overlap.
• If the feasible region is unbounded, the system has infinitely many solutions in that direction.
• If the feasible region is empty, the system has no solution.

Word problems on linear inequalities typically involve constraints on quantities — budget limits, mixing ratios, speed–time limits, or ticket pricing. Translate the verbal conditions into inequalities, solve algebraically, and interpret the solution in context.

• Identify the unknown and assign a variable.
• Translate each condition into an inequality.
• Solve the system and interpret the result — the answer must make physical sense (e.g., positive quantities).
• Check boundary cases: does the solution include the endpoint?

Example: Solve $3x-7<5x+1$ and represent the solution on a number line.

Solution: Subtract 3x from both sides: $-7<2x+1$. Subtract 1: $-8<2x$. Divide by 2 (positive, no reversal): $x>-4$. The solution is the open interval (−4, ∞), shown on the number line as an open circle at −4 with shading to the right.

Example: Solve $|2x-5|\le 9$.

Solution: Rewrite as $-9\le 2x-5\le 9$. Add 5: $-4\le 2x\le 14$. Divide by 2: $-2\le x\le 7$. The solution is the closed interval [−2, 7].

Exercises (NCERT Q1–Q26)

• 1) Solve 24x < 100, when (i) x is a natural number, (ii) x is an integer.
• 2) Solve −12x > 30, when (i) x is a natural number, (ii) x is an integer.
• 3) Solve 5x − 3 < 7, when (i) x is an integer, (ii) x is a real number.
• 4) Solve 3x + 8 > 2, when (i) x is an integer, (ii) x is a real number.
• 5) Solve the inequality 4x + 3 < 5x + 7 for real x.
• 6) Solve the inequality 3x − 7 > 5x − 1 for real x.
• 7) Solve the inequality 3(x − 1) ≤ 2(x − 3) for real x.
• 8) Solve the inequality 3(2 − x) ≥ 2(1 − x) for real x.
• 9) Solve the inequality x/2 + x/3 < 11 for real x.
• 10) Solve the inequality x/3 > x/2 + 1 for real x.
• 11) Solve the inequality 3(x − 2)/5 ≤ 5(2 − x)/3 for real x.
• 12) Solve the inequality (1/2)(3x/5 + 4) ≥ (1/3)(x − 6) for real x.
• 13) Solve the inequality 2(2x + 3) − 10 < 6(x − 2) for real x.
• 14) Solve the inequality 37 − (3x + 5) ≥ 9x − 8(x − 3) for real x.
• 15) Solve the inequality x/4 < (5x − 2)/3 − (7x − 3)/5 for real x.
• 16) Solve the inequality (2x − 1)/3 ≥ (3x − 2)/4 − (2 − x)/5 for real x.
• 17) Solve 3x − 2 < 2x + 1 and show the graph of the solution on the number line.
• 18) Solve 5x − 3 ≥ 3x − 5 and show the graph of the solution on the number line.
• 19) Solve 3(1 − x) < 2(x + 4) and show the graph of the solution on the number line.
• 20) Solve x/2 ≥ (5x − 2)/3 − (7x − 3)/5 and show the graph of the solution on the number line.
• 21) Ravi obtained 70 and 75 marks in the first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
• 22) To receive Grade A in a course, one must obtain an average of 90 marks or more in five examinations, each of 100 marks. If Sunita's marks in the first four examinations are 87, 92, 94 and 95, find the minimum marks she must obtain in the fifth examination to get Grade A.
• 23) Find all pairs of consecutive odd positive integers, both of which are smaller than 10, such that their sum is more than 11.
• 24) Find all pairs of consecutive even positive integers, both of which are larger than 5, such that their sum is less than 23.
• 25) The longest side of a triangle is 3 times the shortest side, and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
• 26) A man wants to cut three lengths from a single piece of board of length 91 cm. The second length is 3 cm longer than the shortest and the third is twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5 cm longer than the second?

Answers at a glance

1. 1
   Q1 → (i) {1, 2, 3, 4}, (ii) {…, −1, 0, 1, 2, 3, 4}.
2. 2
   Q2 → (i) No solution, (ii) {…, −5, −4, −3}.
3. 3
   Q3 → (i) {…, −1, 0, 1}, (ii) (−∞, 2).
4. 4
   Q4 → (i) {−1, 0, 1, 2, …}, (ii) (−2, ∞).
5. 5
   Q5 → x ∈ (−4, ∞).
6. 6
   Q6 → x ∈ (−∞, −3).
7. 7
   Q7 → x ∈ (−∞, −3].
8. 8
   Q8 → x ∈ (−∞, 4].
9. 9
   Q9 → x ∈ (−∞, 66/5).
10. 10
    Q10 → x ∈ (−∞, −6).
11. 11
    Q11 → x ∈ (−∞, 2].
12. 12
    Q12 → x ∈ (−∞, 120].
13. 13
    Q13 → x ∈ (4, ∞).
14. 14
    Q14 → x ∈ (−∞, 2].
15. 15
    Q15 → x ∈ (4, ∞).
16. 16
    Q16 → x ∈ (−∞, 2].
17. 17
    Q17 → x ∈ (−∞, 3). Graph: open circle at 3, arrow to the left.
18. 18
    Q18 → x ∈ [−1, ∞). Graph: filled circle at −1, arrow to the right.
19. 19
    Q19 → x ∈ (−1, ∞). Graph: open circle at −1, arrow right.
20. 20
    Q20 → x ∈ [−2/7, ∞). Graph: filled circle at −2/7, arrow right.
21. 21
    Q21 → At least 35 marks.
22. 22
    Q22 → At least 82 marks.
23. 23
    Q23 → (5, 7) and (7, 9).
24. 24
    Q24 → (6, 8), (8, 10) and (10, 12).
25. 25
    Q25 → Minimum shortest side = 9 cm.
26. 26
    Q26 → Shortest board between 8 cm and 22 cm.

MCQ Questions

1 Mark Questions

1. 1
   Write the solution of x − 1 > 0. (Ans: x > 1)
2. 2
   State the rule when an inequality is multiplied by a negative number.
3. 3
   Find the solution set of 3x < 12 for natural x. (Ans: {1, 2, 3})
4. 4
   Write the interval (−∞, 4) as an inequality. (Ans: x < 4)

2 Mark Questions

1. 1
   Solve 3x + 2 ≤ 17 and represent on the number line. (Ans: x ≤ 5)
2. 2
   Solve 4(2 − x) ≥ 8. (Ans: x ≤ 0)
3. 3
   Find the integer solutions of −3 < x < 4. (Ans: −2, −1, 0, 1, 2, 3)

3 Mark Questions

1. 1
   Solve (2x − 1)/3 ≥ (3x − 2)/4 and show the solution on the number line. (Ans: 8x − 4 ≥ 9x − 6 ⇒ x ≤ 2)
2. 2
   A man rides 30 km in at most... Formulate: he cycles x km at 15 km/h and y km at 20 km/h with at most 2 hours — write the inequalities. (Ans: x/15 + y/20 ≤ 2, x, y ≥ 0)
3. 3
   Solve the double inequality 1 ≤ 3x − 2 ≤ 7. (Ans: 1 ≤ x ≤ 3)

4 Mark Questions

1. 1
   Solve the system 2x + 3y ≤ 12, x ≥ 0, y ≥ 0 graphically: shade the feasible region, state the corner points and verify a point inside it satisfies all constraints. (Ans: region in first quadrant under 2x + 3y = 12)
2. 2
   A student needs at least 10 pens whose total cost does not exceed ₹120, with a pen costing ₹15 and a pencil ₹10. If he buys x pens and y pencils, form the inequalities and draw the region. (Ans: x + y ≥ 10, 15x + 10y ≤ 120, x, y ≥ 0)