Linear Inequalities
Strict vs slack rules, the negative sign-flip law, ℕ/ℤ/ℝ solution sets, number line dots, double intervals & word problems.
Topic 5.2: Strict vs Slack Inequalities & Everyday Constraint Modelling
₹200 in your pocket and each pazhampori costs ₹25. You don't HAVE to spend every single rupee — you only need 25x ≤ 200! Real-world rules rarely use exact equals (=); they use limits like "less than", "at most", "more than", and "at least" — that's Linear Inequalities!x < 5 forbids 5!)≤, "at least" = ≥!)ax + b < 0, ≤ 0, > 0, or ≥ 0 (with a ≠ 0)≥ 60); "At most 200" means 200 or LESS (≤ 200)!h ≥ 120). If it says "Age under 12", a 12-year-old cannot ride — that's strict (a < 12)!| Everyday Phrase | Symbol | Type | Example |
|---|---|---|---|
| Strictly less than / below | < | Strict | x < 10 (10 fails!) |
| Strictly greater than / above | > | Strict | x > 5 (5 fails!) |
| At most / not exceeding / max | ≤ | Slack | 25x ≤ 200 (200 passes!) |
| At least / minimum / not less than | ≥ | Slack | marks ≥ 35 (35 passes!) |
>, or reading "at most" and writing <! Both "at least" (≥) and "at most" (≤) are SLACK inequalities that INCLUDE the boundary number!Topic 5.3: Golden Rules of Inequality (Negative Sign Flip!)
2 < 5 (clearly 2 is smaller than 5!). Now multiply BOTH sides by −1: you get −2 and −5! Wait — which is larger on a thermometer: −2°C or −5°C? −2 > −5! Multiplying or dividing by a NEGATIVE number spins the whole number line backward — so the inequality sign MUST FLIP!< ↔ >)!< to >, and ≤ to ≥!−1 flip the sign? Think of debt! Having ₹2 is less than having ₹5 (2 < 5). Flipping signs turns money into debt: owing ₹2 (−2) is BETTER than owing ₹5 (−5), so −2 > −5!- 1Safe Moves (No Flip): Transposing
+5to the other side as−5, or dividing by+4never flips the sign. - 2Danger Move (MUST Flip!): Dividing
−4x ≤ 20by−4flips≤into≥→x ≥ −5! - 3Or Dodge the Flip: Move
−4xto the other side so its coefficient stays positive:−20 ≤ 4x ⇒ −5 ≤ x(same answer!).
−3x < 12 and writing x < −4! Wait — test x = 0: −3(0) = 0 < 12 (TRUE!), yet 0 < −4 is FALSE! Dividing by −3 MUST flip < to >, giving x > −4! Also NEVER cross-multiply by an unknown variable x whose sign might be negative!Topic 5.3: Algebraic Solving in One Variable (ℕ, ℤ & ℝ Solution Sets)
30x < 200 tells you the number of notebooks you can buy is x < 6.67. Can you ask a shopkeeper for −3 notebooks or 4.25 notebooks? NO — when x is a Natural Number (ℕ), the solution set is ONLY {1, 2, 3, 4, 5, 6}! Always check which number family x belongs to!{1, 2, 3, ...} (roster list!)0 & negative integers: {..., −2, −1, 0, 1, ...}x (flip if dividing by negative!). Step 3: Filter answer through ℕ, ℤ, or ℝ!−12x > 30 ⇒ x < −2.5 give completely different answers depending on the question's universe!• If
x ∈ ℕ: No solution (∅) (no natural number is less than −2.5!)• If
x ∈ ℤ: {..., −5, −4, −3} (notice −2 is NOT less than −2.5!)3(x − 2)5 ≤ 5(2 − x)3 for real x:1) Multiply both sides by positive LCM
15 (NO flip!): 9(x − 2) ≤ 25(2 − x).2) Expand:
9x − 18 ≤ 50 − 25x.3) Group
x: 34x ≤ 68 ⇒ x ≤ 2 → Interval: (−∞, 2]!
x < −2.5 with x ∈ ℤ, students write {..., −4, −3, −2} — WRONG! Check a number line: −2 is GREATER than −2.5, so the largest integer below −2.5 is −3! (2) Leaving answer as interval (−∞, 6.67) when question specifies x ∈ ℕ!Topic 5.3: Open Circles vs Solid Dots on the Real Number Line
≤ or ≥), the boundary point joins the party with a solid filled dot (●). When it is strict (< or >), the boundary stays locked out with a hollow open circle (○)!< or > → endpoint is EXCLUDED≤ or ≥ → endpoint is INCLUDED>, ≥; LEFT for <, ≤!≤ and ≥ have an extra bar underneath, so colour the dot SOLID (●)! No bar (<, >) = hollow circle (○)!x on the LEFT side before choosing your arrow direction! If you have x ≥ 3, the arrow > literally points RIGHT! (Whereas 3 ≤ x points left visually even though x goes right!)- 1Put
xon the left: e.g.5 ≥ xbecomesx ≤ 5. - 2Mark boundary
5with a Solid Dot (●) for≤, ≥or Open Circle (○) for<, >. - 3Shade thickly along the number line in the arrow's direction (left toward
−∞or right toward+∞).
x sits on the right (reading 3 < x and shading left of 3 instead of right!). Always flip it so x stands on the LEFT: x > 3! (2) Drawing a solid dot for strict < or forgetting to make the dot hollow ○!Topic 5.3: Double Inequalities & Interval Notation [a, b] vs (a, b)
T > 40°C) AND not too hot (T ≤ 65°C) — the safe zone is 40 < T ≤ 65, written sleekly as (40, 65]! Solving two inequalities together means finding the shared overlap (intersection) where BOTH conditions hold true at the exact same time!−8 ≤ 5x − 3 < 7 → do the SAME operation on all 3 parts![ ] = Solid dot (≤, ≥) | ( ) = Open dot (<, >)−∞ and +∞ ALWAYS use round ( ) — NEVER [ ]!S₁ ∩ S₂)! If they don't overlap, answer is ∅ (No Solution)!x ≥ 3 is ALWAYS written [3, ∞) (closed on 3, round open on ∞!).−3 ≤ 4 − 7x2 ≤ 18:1) Subtract 4 from all 3 parts:
−7 ≤ −7x2 ≤ 14.2) Multiply all 3 parts by
−2/7 (NEGATIVE → flip BOTH signs!):(−7)(−2/7) ≥ x ≥ (14)(−2/7) → 2 ≥ x ≥ −4, i.e., −4 ≤ x ≤ 2 → [−4, 2]!
[2, ∞] with a square bracket on ∞ — ALWAYS wrong, infinity takes )! (2) Flipping only ONE sign when multiplying a double inequality a < −x ≤ b by −1 — BOTH signs flip to −b ≤ x < −a! (3) Writing [5, 2] with larger number first — intervals ALWAYS run (smaller, larger)!Topic 5.3: Real-Life Word Problems: Marks Average, Temperature & Acid Mixture
68°F to 77°F? How much 30% acid should a lab pour into 600 L of 12% acid? Every single one of these real-life Kerala DHSE 3–4 mark problems turns straight into a 1-variable inequality!m₁ + m₂ + x3 ≥ TargetF = 95C + 32 | C = 59(F − 32)12%(600) + 30%(x) out of TOTAL (600 + x)!x and x + 1! Odd/Even numbers jump by 2: always pick x and x + 2!x litres of 30% solution increases BOTH the pure acid AND the total tank volume to (600 + x)!x L of 30% acid to 600 L of 12% acid so resulting mixture lies strictly between 15% and 18%:15%(600 + x) < 12%(600) + 30%(x) < 18%(600 + x).Multiply by 100:
• Left part:
9000 + 15x < 7200 + 30x ⇒ 1800 < 15x ⇒ x > 120.• Right part:
7200 + 30x < 10800 + 18x ⇒ 12x < 3600 ⇒ x < 300.Answer: Add strictly between
120 and 300 litres!
15%(600) instead of 15%(600 + x) (forgetting that adding x litres expands total tank volume!). (2) In consecutive odd/even integers, writing x and x + 1 — consecutive odd (e.g. 5, 7) or even (e.g. 4, 6) integers ALWAYS differ by 2, so use x and x + 2!Solved Previous Year Questions (Kerala DHSE & NCERT)
Q1 (Kerala HSE March • 2 Marks): Solve 24x < 100 when (i) x is a natural number, (ii) x is an integer.
x < 10024 = 256 ≈ 4.17.• (i) For
x ∈ ℕ: {1, 2, 3, 4}.• (ii) For
x ∈ ℤ: {..., −2, −1, 0, 1, 2, 3, 4}.Q2 (Kerala HSE Say • 2 Marks): Solve −12x > 30 when (i) x is a natural number, (ii) x is an integer.
−12 and FLIP sign: x < −3012 = −2.5.• (i) For
x ∈ ℕ: ∅ (No natural number is < −2.5).• (ii) For
x ∈ ℤ: {..., −5, −4, −3}.Q3 (Kerala HSE March • 2 Marks): Solve 5x − 3 < 3x + 1 when (i) x is an integer, (ii) x is a real number.
5x − 3x < 1 + 3 ⇒ 2x < 4 ⇒ x < 2.• (i) Integer solution:
{..., −2, −1, 0, 1}.• (ii) Real solution:
(−∞, 2).Q4 (Kerala HSE March • 3 Marks): Solve the inequality x/4 < (5x − 2)/3 − (7x − 3)/5 for real x and show the interval.
60:15x < 20(5x − 2) − 12(7x − 3)15x < 100x − 40 − 84x + 36 = 16x − 4−x < −4 ⇒ x > 4 → Interval: (4, ∞) (open circle at 4, shade right).Q5 (Kerala HSE Model • 3 Marks): Solve −15 < (3(x − 2))/5 ≤ 0.
5/3 (positive, no flip):−25 < x − 2 ≤ 0.Add 2 everywhere:
−23 < x ≤ 2 → (−23, 2].Q6 (Kerala HSE March • 3 Marks): Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 40.
x and x + 2 (jump by 2!).Conditions:
x > 10 AND x + (x + 2) < 40 ⇒ 2x < 38 ⇒ x < 19.Odd integers in
(10, 19): x ∈ {11, 13, 15, 17}.Pairs:
(11, 13), (13, 15), (15, 17), (17, 19).Q7 (Kerala HSE Say • 3 Marks): Ravi obtained 70 and 75 marks in first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
x.70 + 75 + x3 ≥ 60 ⇒ 145 + x ≥ 180 ⇒ x ≥ 35.Minimum marks required =
35.Q8 (NCERT / DHSE Model • 4 Marks): How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?
x litres of pure water (0% acid) gives total volume (1125 + x).Pure acid stays
0.45 × 1125:0.25(1125 + x) < 0.45(1125) < 0.30(1125 + x).• Right:
0.45(1125) < 0.30(1125) + 0.30x ⇒ 0.15(1125) < 0.30x ⇒ x > 562.5 L.• Left:
0.25x < 0.20(1125) ⇒ x < 900 L.Add between
562.5 L and 900 L of water!