Linear Inequalities

2.1 Core concepts

• An inequality is formed when two real numbers or two algebraic expressions are related by <, >, ≤, or ≥; statements like 30x < 200 and 40x + 20y ≤ 120 are inequalities (NCERT §5.2, Definition 1, p. 90). They differ from equations in that the solution is typically an interval rather than a single value.
• Inequalities are classified as numerical (e.g., 3 < 5, 7 > 5) or literal (e.g., x < 5, y > 2); double inequalities like 3 < 5 < 7 or 3 ≤ x < 5 express two relations at once (NCERT §5.2, p. 90).
• Inequalities with < or > are called strict; those with ≤ or ≥ are called slack (NCERT §5.2, p. 90). Slack inequalities allow equality; strict ones do not.
• ax + b < 0, ax + b > 0, ax + b ≤ 0, ax + b ≥ 0 (with a ≠ 0) are linear inequalities in one variable; ax + by < c, ax + by > c, ax + by ≤ c, ax + by ≥ c (with a ≠ 0, b ≠ 0) are linear inequalities in two variables (NCERT §5.2, p. 90).
• Inequalities involving ax² + bx + c are not linear; they are quadratic inequalities in one variable (NCERT §5.2, p. 90). Quadratic inequalities have a different solution structure (typically a union of intervals).
• A solution of an inequality in one variable is any value of the variable that makes the inequality a true statement; the set of all such values is the solution set (NCERT §5.3, p. 91). The solution set may be empty, finite, or an infinite interval.
• Rule 1: Equal numbers may be added to (or subtracted from) both sides of an inequality without affecting the sign of inequality (NCERT §5.3, p. 92). This is the basis for transposing terms.
• Rule 2: Both sides of an inequality may be multiplied (or divided) by the same positive number; but when both sides are multiplied (or divided) by a negative number, the sign of inequality is reversed (NCERT §5.3, p. 92). Illustration: 3 > 2, yet –3 < –2; –8 < –7, yet (–8)(–2) > (–7)(–2), i.e., 16 > 14 (NCERT §5.3, p. 91). The sign-flip rule is the single most error-prone step.
• Solutions may be sought in N, Z, or R; unless stated otherwise NCERT solves inequalities in R (NCERT §5.3, p. 92, after Example 2). The same inequality 30x < 200 has different solution sets in N (1, 2, …, 6) and R ((−∞, 20/3)).
• Worked algebraic technique: shift like terms, apply Rule 1 then Rule 2, remembering the sign flip; e.g., 5x – 3 < 3x + 1 ⇒ 2x < 4 ⇒ x < 2, giving solution set x ∈ (–∞, 2) (NCERT §5.3, Example 2, p. 92).
• Graphical representation on the number line: for x < 3 (from 7x + 3 < 5x + 9), the solution is shown on the number line in Fig 5.1 (open/hollow circle at 3 and the line darkened to the left of 3) (NCERT §5.3, Example 5, p. 93). For x ≥ 1 (from (3x – 4)/2 ≥ (x + 1)/4 – 1), Fig 5.2 shows a dark/filled circle at 1 and the line darkened to the right (NCERT §5.3, Example 6, p. 93–94). The Summary on p. 99 codifies: open circle for < / >, dark circle for ≤ / ≥.
• Double inequalities are solved by operating on all three parts simultaneously: –8 ≤ 5x – 3 < 7 ⇒ –5 ≤ 5x < 10 ⇒ –1 ≤ x < 2 (Misc. Example 9, p. 96). The chain notation is shorthand for a system of two inequalities joined by AND.
• System of inequalities (one variable): the solution is the intersection of the individual solution sets; e.g., 3x – 7 < 5 + x gives x < 6, and 11 – 5x ≤ 1 gives x ≥ 2, so the common solution is 2 ≤ x < 6, shown as the bold portion of the number line in Fig 5.3 (Misc. Example 11, p. 96–97).
• Word-problem applications: minimum marks for a target average (Example 7, p. 94: (62 + 48 + x)/3 ≥ 60 ⇒ x ≥ 70); consecutive odd natural numbers larger than 10 with sum less than 40 give pairs (11,13), (13,15), (15,17), (17,19) (Example 8, p. 94); a temperature kept between 30°C and 35°C corresponds to 86°F < F < 95°F (Misc. Example 12, p. 97); acid-mixture problem yields 120 < x < 300 litres of 30% acid solution to be added (Misc. Example 13, p. 97–98).
• Two-variable graphical work is outside the current reprint, but its algebra mirrors the one-variable techniques. Two-variable inequalities reappear in Linear Programming (Class XII Ch. 12).

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

• Fig 5.1 (p. 93): Number-line graph of x < 3 (solution of 7x + 3 < 5x + 9) — open circle at 3, line darkened to the left.
• Fig 5.2 (p. 94): Number-line graph of x ≥ 1 (solution of (3x – 4)/2 ≥ (x + 1)/4 – 1) — dark/filled circle at 1, line darkened to the right.
• Fig 5.3 (p. 97): Number-line graph of the system x < 6 and x ≥ 2 — bold segment from 2 (dark circle) to 6 (open circle), representing the common solution 2 ≤ x < 6.
• Procedure to solve a linear inequality in one variable (§5.3, p. 91–92): (1) transpose like terms using Rule 1, (2) collect the variable on one side, (3) divide by the coefficient using Rule 2, flipping the sign if the coefficient is negative, (4) write the solution set in interval form, (5) mark on the number line using open/dark circles per the convention on p. 99.
• Procedure to solve a system of inequalities: (1) solve each inequality separately, (2) write each solution set as an interval, (3) intersect the intervals — graphically, take the overlap of the shaded portions on the number line.
• Procedure to solve a double inequality: treat as two inequalities joined by AND and operate on all three "parts" simultaneously.
• Word-problem translation pattern: define the variable, translate "at least" / "at most" / "less than" / "more than" into ≥, ≤, <, >, set up the inequality, solve, and check the answer makes physical sense (e.g., x > 0 for litres).
• Process — temperature/unit conversion: substitute the formula into the inequality, multiply/divide each part by the conversion factor, simplify. Direction is preserved because the conversion factor (e.g., 9/5) is positive.

2.4 Common confusions / NTA trap points

• Forgetting the sign flip when dividing/multiplying by a negative number — e.g., –5x ≤ –40 ⇒ x ≥ 8 (Example 4, p. 93), not x ≤ 8. NTA distractors often offer the un-flipped option.
• Open vs dark circle on the number line: strict </> uses an open (hollow) circle; slack ≤/≥ uses a filled (dark) circle (Summary, p. 99). Mixing these up is a classic trap.
• Interval notation: x < 2 corresponds to (–∞, 2) (open at 2); x ≥ 8 corresponds to [8, ∞) (closed at 8) (Examples 2 and 4, p. 92–93). Square brackets for slack, parentheses for strict.
• Domain of the variable matters: solving 30x < 200 for natural numbers gives {1,2,3,4,5,6}, but for integers it also includes 0 and all negative integers (Example 1, p. 92).
• Solving double inequalities: every operation must be applied to all three members of the chain; e.g., –8 ≤ 5x – 3 < 7 becomes –5 ≤ 5x < 10, not just one side (Misc. Example 9, p. 96).
• Quadratic inequalities are not linear: ax² + bx + c ≤ 0 is excluded from this chapter (NCERT §5.2, p. 90); a distractor that asks students to classify such an expression as "linear" is a known NTA trap.
• Forgetting to multiply through to clear fractions. When the inequality has denominators, multiplying by the LCM of denominators clears them — but check the sign of the multiplier (always positive if denominators are positive constants).
• Misreading "between A and B" as inclusive. "Between" in everyday English is ambiguous, but NCERT word problems usually mean strictly between unless "at least"/"at most" is specified.
• Reversing the inequality when squaring. Squaring both sides is not a valid operation on inequalities unless both sides are known to be non-negative; the NCERT chapter avoids squaring entirely.
• Forgetting that an inequality with variables on both sides reduces to a single-side form first. E.g., 5x − 3 < 3x + 1 becomes 2x < 4 before dividing.
• Treating "or" as intersection. A system joined by "or" is a union of solution sets, not an intersection.
• Mis-flipping with negative variables. If −x > 0, then x < 0; the flip happens because we multiply by −1.

2.5 Key formulas & theorems

2.6 Solved examples (NCERT-grounded)

Example A (NCERT Example 2, p. 92). Solve 5x − 3 < 3x + 1 (real x).

Step 1 — transpose: 5x − 3x < 1 + 3 ⇒ 2x < 4. Step 2 — divide by positive 2: x < 2. Step 3 — write in interval form: x ∈ (−∞, 2).

Example B (NCERT Example 4, p. 93). Solve (5 − 2x)/3 ≤ x/6 − 5.

Step 1 — multiply by 6 (positive): 2(5 − 2x) ≤ x − 30 ⇒ 10 − 4x ≤ x − 30. Step 2 — transpose: −5x ≤ −40. Step 3 — divide by −5 (flip): x ≥ 8. Answer: x ∈ [8, ∞).

Example C (NCERT Misc. Example 9, p. 96). Solve −8 ≤ 5x − 3 < 7.

Step 1 — add 3 throughout: −5 ≤ 5x < 10. Step 2 — divide by 5 (positive): −1 ≤ x < 2. Step 3 — write: x ∈ [−1, 2).

Example D (NCERT Example 7, p. 94). Find minimum marks Ravi must obtain in the third unit test (62 and 48 in the first two) to maintain an average ≥ 60.

Step 1 — set up: (62 + 48 + x)/3 ≥ 60. Step 2 — multiply by 3: 110 + x ≥ 180 ⇒ x ≥ 70. Step 3 — interpret: Ravi must score at least 70 marks.

Example E (NCERT Misc. Example 12, p. 97). Convert "temperature between 30°C and 35°C" to °F using C = (5/9)(F − 32).

Step 1 — write: 30 < (5/9)(F − 32) < 35. Step 2 — multiply each part by 9/5: 54 < F − 32 < 63. Step 3 — add 32: 86 < F < 95 °F.