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Class 11 Mathematics Chapter 6 of 16

Chapter 6 — Linear Inequalities

Overview

This chapter introduces linear inequalities — algebraic statements that compare expressions using <, >, ≤, ≥. Beginning with inequalities in one variable, students learn rules for manipulating inequalities (including the crucial sign reversal when multiplying or dividing by a negative number), represent solution sets on the number line and in interval notation, and solve compound inequalities. The chapter then extends to linear inequalities in two variables: representing linear equations as boundary lines, identifying the half-plane that satisfies an inequality, and graphing solution regions. It also covers solving systems of linear inequalities graphically to find common solution regions and includes straightforward applications (word problems) that translate real-life constraints into inequalities. Emphasis is placed on logical reasoning, accurate graphing, interpreting boundary inclusion (strict vs non-strict), and checking solutions. Understanding these topics builds foundations for higher algebra (quadratic inequalities, absolute-value inequalities) and for optimization (linear programming).

Learning Objectives

  • Define a linear inequality and describe its solution set in one variable
  • Explain the difference between strict (<, >) and non-strict (≤, ≥) inequalities and their boundary representations
  • Solve linear inequalities in one variable algebraically, including those containing parentheses and fractions
  • Apply the properties of inequalities for addition, subtraction, multiplication and division, and account for direction reversal when multiplying or dividing by a negative
  • Represent solution sets of one-variable inequalities on the number line and express them in interval notation
  • Solve and graph compound inequalities (conjunctions and disjunctions) and interpret their combined solution sets
  • Translate verbal problems into linear inequalities and solve them to obtain meaningful numeric answers
  • Define a linear inequality in two variables and identify its solution region as a half-plane

Topics in this chapter

12 topics · tap a topic title to jump straight to it.

🔢1

Introduction to Inequalities

What is an inequality? An inequality is a mathematical statement that compares two quantities and shows that they are not necessarily equal. Common inequality symbols are:

  • < (less than)
  • > (greater than)
  • ≤ (less than or equal to)
  • ≥ (greater than or equal to)

For example, x < 5 means x is any number strictly smaller than 5; x ≤ 5 means x may be 5 or any number less than 5.

Types: Strict inequalities (<, >) exclude the boundary point; non-strict (≤, ≥) include it.

Solution set: The set of all values that satisfy an inequality is its solution set. We often express it as a set, on a number line, or with interval notation.

Basic idea for linear inequalities (one variable): An inequality of the form ax + b < 0 (or >, ≤, ≥) is solved much like a linear equation: isolate x by adding/subtracting and dividing. Key difference: when you multiply or divide both sides by a negative number, the direction of the inequality reverses.

Key properties and operations:

  • Addition/Subtraction: if a < b then a + c < b + c (for any real c).
  • Multiplication/Division by positive: if a < b and k > 0 then ka < kb.
  • Multiplication/Division by negative: if a < b and k < 0 then ka > kb (inequality reverses).
  • Transitivity: if a < b and b < c then a < c.

Representation: On the number line use an open dot for strict inequalities and a closed dot for non-strict, shading to the left for < or ≤ and to the right for > or ≥. Interval notation: (a, b), [a, b], (a, b], [a, b) according to open/closed endpoints.

Extension to two variables: A linear inequality in x and y, e.g. ax + by + c < 0, represents a half-plane. The boundary line ax + by + c = 0 is drawn dashed for strict inequalities and solid for non-strict ones; then shade the side that satisfies the inequality (test one point to decide).

Why inequalities matter: They model limits, constraints and feasible ranges in real life: budgets, speed limits, tolerances, thresholds, optimization constraints, etc.

📌 Examples
  • 1) Solve 3x - 5 < 10. Solution: 3x < 15 => x < 5. Interval: (-∞, 5). On number line: open dot at 5, shade left.
  • 2) Solve -2x + 4 ≥ 10. Solution: -2x ≥ 6 => divide by -2 (reverse sign) x ≤ -3. Interval: (-∞, -3].
  • 3) Compound inequality: 1 < 2x + 1 ≤ 7. Solve stepwise: subtract 1: 0 < 2x ≤ 6 => divide by 2: 0 < x ≤ 3. Interval: (0, 3].
  • 4) Real-life: Budget constraint. If you have at most Rs. 1500 to spend and each item costs Rs. 120, the number n of items satisfies 120n ≤ 1500 => n ≤ 12.5 so n ≤ 12 (integer items).
  • 5) Two-variable inequality: For x + 2y ≤ 4. Boundary line x + 2y = 4 (solid). Test (0,0): 0 ≤ 4 true, so shade the side containing (0,0).
🧮 Formulas
  1. Addition: If a < b then a + c < b + c (for any c).
  2. Subtraction: If a < b then a - c < b - c (for any c).
  3. Multiplication/Division (positive): If a < b and k > 0 then ka < kb and a/k < b/k.
  4. Multiplication/Division (negative): If a < b and k < 0 then ka > kb and a/k > b/k (inequality reverses).
  5. Transitive property: If a < b and b < c then a < c.
  6. Solve ax + b < 0 => x < -b/a (if a > 0). If a < 0, reverse inequality when dividing.
📊 Visual ideas
Number-line for x < 5: draw an open circle at 5 and shade leftwards to -∞. Label arrow at ends.
Number-line for x ≤ -3: draw a filled/solid dot at -3 and shade leftwards.
Compound inequality 0 < x ≤ 3: open dot at 0, closed at 3, shade between them.
Boundary and half-plane for ax + by + c < 0: draw the line ax + by + c = 0 as dashed (since strict), then pick a test point (e.g., (0,0)) to decide which side to shade; if test point satisfies inequality, shade that side.
🔢2

Properties of Inequalities

An inequality compares two expressions using symbols >, <, ≥, ≤. The properties of inequalities are rules that let us transform inequalities while keeping track of whether the relation stays the same or reverses. These rules are essential to solve and manipulate linear inequalities in one or two variables.

  • Transitive property: If a > b and b > c, then a > c. (Similarly for ≥.)
  • Add/subtract the same number: For any real c, if a > b then a + c > b + c and a - c > b - c. Adding or subtracting preserves the inequality.
  • Multiply/divide by a positive number: If c > 0 and a > b then ac > bc and a/c > b/c. Multiplying or dividing by a positive number preserves the direction.
  • Multiply/divide by a negative number (sign flip): If c < 0 and a > b then ac < bc and a/c < b/c. Multiplying or dividing by a negative number reverses the inequality sign.
  • Addition of inequalities: If a > b and c > d then (a + c) > (b + d).
  • Reciprocals (positive numbers): If 0 < a < b then 1/a > 1/b (the inequality reverses because the reciprocal function is decreasing on (0, ∞)). For negative a, b, similar care with sign applies.
  • Multiplication of positive intervals: If 0 < a < b and 0 < c < d then ac < bd.
  • Monotone functions: If f is increasing, a < b ⇒ f(a) < f(b). If f is decreasing, a < b ⇒ f(a) > f(b). This explains effects like squaring (increasing on [0,∞)) and reciprocals (decreasing on (0,∞)).
  • Care with non‑known sign: When multiplying or dividing by an expression whose sign is unknown, you must determine its sign first; otherwise the inequality direction may be incorrectly handled.

These properties apply both to one-variable inequalities and, with geometric interpretation, to linear inequalities in two variables (which represent half-planes). They are the backbone of solving, combining, and graphing inequalities.

📌 Examples
  • Simple addition: From 5 &gt; 3, add 2 to both sides to get 7 &gt; 5.
  • Multiplying by positive: From 4 &gt; 1, multiply by 3 to get 12 &gt; 3.
  • Multiplying by negative (sign flips): From 5 &gt; 2, multiply by -1 to get -5 &lt; -2.
  • Reciprocal (positive numbers): From 2 &gt; 1, take reciprocals to get 1/2 &lt; 1/1.
  • System with no solution: x + 2 &gt; 5 ⇒ x &gt; 3 and 3x &lt; 9 ⇒ x &lt; 3. Intersection is empty (no solution).
  • Inequality in two variables: 2x + 3y &lt; 6 is graphed as the half-plane below the line 2x + 3y = 6; use a dashed boundary and test a point (0,0) to decide shading.
🧮 Formulas
  1. Transitive: (a &gt; b and b &gt; c) ⇒ a &gt; c
  2. Addition/Subtraction: a &gt; b ⇒ a ± c &gt; b ± c (for any real c)
  3. Multiply/Divide (positive): c &gt; 0 and a &gt; b ⇒ ac &gt; bc, a/c &gt; b/c
  4. Multiply/Divide (negative): c &lt; 0 and a &gt; b ⇒ ac &lt; bc, a/c &lt; b/c (sign reverses)
  5. Add inequalities: (a &gt; b and c &gt; d) ⇒ a + c &gt; b + d
  6. Reciprocal (for positives): 0 &lt; a &lt; b ⇒ 1/a &gt; 1/b
📊 Visual ideas
Number line for single-variable inequality: draw an open circle at the endpoint for strict inequalities (>, <) and a closed dot for (≥, ≤); shade right for x &gt; a, left for x &lt; a.
Interval intersections/unions: show overlapping shaded regions for systems; intersection = common shaded part, union = any shaded part combined.
Linear inequalities in two variables: draw the boundary line ax + by = c. Use a dashed line for strict inequalities and solid line for inclusive ones. Choose a test point (e.g., (0,0)) to decide which side to shade (half-plane).
Visual of sign flip: plot y = x and y = -x or show multiplying by a negative flips the order by reflecting points about 0 on the number line.
🔢3

Linear Inequalities in One Variable

Definition: A linear inequality in one variable is an inequality that can be written in the form ax + b < 0, ax + b > 0, ax + b <= 0 or ax + b >= 0, where a and b are real numbers and a ≠ 0. The solution is the set of all real numbers x that satisfy the inequality.

General method to solve:

  • Isolate the variable on one side by using addition/subtraction and multiplication/division.
  • When you add or subtract the same number from both sides, the inequality sign does not change.
  • When you multiply or divide both sides by a positive number, the inequality sign does not change. When you multiply or divide by a negative number, the inequality sign reverses direction.
  • Express the solution on a number line and/or in interval notation. For strict inequalities (< or >) use open endpoints; for inclusive (<= or >=) use closed endpoints.

Compound inequalities: You may see expressions like a <= bx + c <= d. Treat them like two inequalities joined by AND (intersection). OR situations lead to unions of solution sets.

Verification: Check at least one test value from the solution set and one from its complement to ensure correctness.

Graphical idea: A linear inequality in one variable is represented on the real number line. Solutions form rays or intervals: shade to the left or right of a boundary point; use an open circle for < or > and a filled circle for <= or >=.

📌 Examples
  • Example 1: Solve 3x - 5 < 7. Steps: 3x < 12 ⇒ x < 4. Solution set: (-∞, 4). Number-line: open circle at 4, shade left.
  • Example 2: Solve -2x + 3 ≥ 7. Steps: -2x ≥ 4 ⇒ divide by -2 (reverse sign) ⇒ x ≤ -2. Solution set: (-∞, -2]. Number-line: closed dot at -2, shade left.
  • Example 3 (compound): Solve 1 < 2x + 3 ≤ 7. Steps: subtract 3 ⇒ -2 < 2x ≤ 4 ⇒ divide by 2 ⇒ -1 < x ≤ 2. Solution set: (-1, 2]. On the number line use an open circle at -1 and a filled circle at 2, shade between them.
  • Example 4 (real life): A commuter has at most ₹200. A taxi charges a base fare of ₹50 plus ₹12 per km. How many whole kilometres can the commuter travel? Inequality: 50 + 12x ≤ 200 ⇒ 12x ≤ 150 ⇒ x ≤ 12.5. For whole kilometres x ≤ 12. So at most 12 km. Represent: closed dot at 12.5 (or at 12 if integer constraint) and shade left.
🧮 Formulas
  1. Basic linear form: ax + b < 0, ax + b ≤ 0, ax + b > 0, ax + b ≥ 0 (a ≠ 0).
  2. Boundary point: ax + b = 0 ⇒ x = -b/a.
  3. Addition/subtraction rule: If A < B then A + c < B + c (sign unchanged).
  4. Multiplication/division rule: If A < B and k > 0 then kA < kB (sign unchanged). If k < 0 then kA > kB (sign reverses).
  5. Transitive property: If A < B and B < C then A < C.
  6. Interval notation: x < a ⇒ (-∞, a); x ≤ a ⇒ (-∞, a]; x > a ⇒ (a, ∞); x ≥ a ⇒ [a, ∞); a < x ≤ b ⇒ (a, b].
📊 Visual ideas
Single strict inequality x < 4: draw a number line, place an open circle at 4 and shade the line to the left (toward -∞).
Single non-strict inequality x ≤ -2: draw a number line, place a filled (closed) dot at -2 and shade the line to the left.
Compound inequality -1 < x ≤ 2: draw a number line, open circle at -1, closed circle at 2, shade the segment between -1 and 2.
Real-life (taxi) example 50 + 12x ≤ 200: boundary x = 12.5. On number line place closed dot at 12.5 and shade to the left to show all allowed distances up to 12.5 km. If integer solution required, mark integer points 0,1,...,12 as valid.
⚗️4

Compound Inequalities (One Variable)

What is a compound inequality? A compound inequality joins two or more simple inequalities connected by the words 'and' or 'or'. It describes a set of values of one variable that satisfy both (AND) or at least one (OR) of the inequalities.

Types

  • Conjunction (AND): both conditions must hold. Example: a < x ≤ b means x > a and x ≤ b. The solution is the intersection of the solution sets.
  • Disjunction (OR): at least one condition must hold. Example: x < c or x ≥ d. The solution is the union of the solution sets.
  • Chained inequalities: a < b < c may be used to write simultaneous relations, e.g. a < x ≤ b is shorthand for (x > a) and (x ≤ b).

How to solve

  1. For a chained inequality like A < expression ≤ B, treat it as two inequalities linked by AND and perform the same algebraic operations on all three parts simultaneously (add/subtract, multiply/divide). Remember to flip the inequality sign if you multiply or divide by a negative number.
  2. For an OR compound inequality, solve each inequality separately, then take the union of the solution intervals.
  3. Always express final answers in interval notation and/or with a number-line graph, using open circles for strict inequalities (< or >) and closed dots for inclusive (≤ or ≥).

Important rules

  • If you add or subtract the same number from all parts of an inequality, the inequality relations do not change.
  • If you multiply or divide all parts by a positive number, the inequality relations do not change. If you multiply or divide by a negative number, you must reverse all inequality signs.
  • Chained inequality a < x ≤ b is equivalent to (x > a) ∩ (x ≤ b).

Representation: Use interval notation (a, b), [a, b], (a, b], [a, b) and number-line graphs. For OR, write unions: (-∞, p] ∪ (q, ∞), etc.

📌 Examples
  • Example 1 (AND, chained): Solve 2 < 3x + 1 ≤ 10. Step 1: Treat as two inequalities: 2 < 3x + 1 and 3x + 1 ≤ 10. Step 2: Subtract 1 from all parts: 1 < 3x ≤ 9. Step 3: Divide all parts by 3 (positive, no flip): 1/3 < x ≤ 3. Answer: x ∈ (1/3, 3]. Graph: number line with an open circle at 1/3, closed circle at 3, shade between them.
  • Example 2 (OR): Solve 4x - 5 > 3 or 2x + 1 ≤ -1. Solve each: 4x - 5 > 3 → 4x > 8 → x > 2. 2x + 1 ≤ -1 → 2x ≤ -2 → x ≤ -1. Answer: x ∈ (-∞, -1] ∪ (2, ∞). Graph: two shaded regions: left with closed dot at -1, and right with open dot at 2.
  • Example 3 (multiplying by negative in chained inequality): Solve -2 ≤ 5 - 3x < 7. Subtract 5: -7 ≤ -3x < 2. Divide by -3 (negative → flip inequalities): 7/3 ≥ x > -2/3. Rewrite in increasing order: -2/3 < x ≤ 7/3. Answer: x ∈ (-2/3, 7/3]. Graph: open circle at -2/3, closed at 7/3, shade between.
  • Example 4 (real-life): A medicine is effective when blood concentration C satisfies 50 ≤ C < 100 mg/L. This is a compound inequality. Express: C ∈ [50, 100). On a number line use closed dot at 50, open at 100, shade in between.
🧮 Formulas
  1. Addition/Subtraction: If a < x and c is real, then a + c < x + c (inequalities unchanged).
  2. Multiplication/Division: If k > 0, then a < b ⇒ ka < kb; if k < 0, a < b ⇒ ka > kb (sign flips when multiplying/dividing by negative).
  3. Chained equivalence: a < x ≤ b ⇔ (x > a) AND (x ≤ b) ⇔ x ∈ (a, b].
  4. Union/Intersection: Solution(OR) = Solution(ineq1) ∪ Solution(ineq2); Solution(AND) = Solution(ineq1) ∩ Solution(ineq2).
  5. Transitive property: If a < b and b < c then a < c (useful when combining steps).
📊 Visual ideas
Number-line shading for AND (connected interval): draw a line, mark endpoints a and b. Use an open circle at endpoint for strict inequality and a filled dot for inclusive. Shade the region between. Example: 1/3 < x ≤ 3 → open at 1/3, filled at 3, shade between.
Number-line shading for OR (union of intervals): draw separate shaded regions. Example: x ≤ -1 or x > 2 → shade left of -1 including -1 (filled dot) and shade right of 2 excluding 2 (open circle).
Chained with negative multiplication: show three-part operation on line. For -7 ≤ -3x < 2, illustrate how dividing by -3 flips arrows: annotate the flip step visibly so students see sign reversal.
Overlay method: to show AND as intersection, draw shading for each inequality in different translucent colors (e.g., blue and yellow); the overlapping (green) region is the final solution. For OR, the union is the total area covered by either color.
🔢5

Interval and Set-Builder Notation

What is being described: Interval and set‑builder notations are compact ways to describe sets of real numbers, especially solutions of linear inequalities.

Set‑builder notation: Write a set by a rule. General form: {x ∈ R : condition on x}. Example: {x ∈ R : x > 2} means all real numbers x such that x > 2.

Interval notation: Describes all numbers between endpoints. Brackets [ ] mean endpoint included (≤ or ≥). Parentheses ( ) mean endpoint excluded (< or >). Examples:

  • (a, b) = {x ∈ R : a < x < b}
  • [a, b] = {x ∈ R : a ≤ x ≤ b}
  • [a, b) = {x ∈ R : a ≤ x < b}
  • (-∞, b), (a, ∞), (-∞, ∞) for unbounded intervals. Note: ∞ and -∞ are never included, so always use parentheses with them.

How inequalities correspond: Use interval or set‑builder form to represent solution sets of inequalities:

  • x > a ↔ (a, ∞) ↔ {x ∈ R : x > a}
  • x ≥ a ↔ [a, ∞) ↔ {x ∈ R : x ≥ a}
  • a < x ≤ b ↔ (a, b] ↔ {x ∈ R : a < x ≤ b}

Operations on intervals: Intersection and union are used to combine conditions:

  • Intersection: numbers satisfying both conditions. Example: [a,b] ∩ [c,d] = [max(a,c), min(b,d)] if the result is nonempty.
  • Union: numbers satisfying at least one condition. Example: disjoint intervals (-∞, -1) ∪ (1, ∞).
  • Complement: R \ (a,b) = (-∞, a] ∪ [b, ∞).

Endpoints and open/closed circles on number line: On a number line draw a filled (solid) dot for included endpoints (bracket) and an open circle for excluded endpoints (parenthesis). Shade between endpoints or to infinity as required.

Why it matters: Interval and set‑builder notations make writing and combining solution sets of linear inequalities precise and compact. They are used in real problems such as tolerance ranges, eligibility ranges, and constraints.

📌 Examples
  • Inequality x > 3: set-builder {x ∈ R : x > 3}, interval notation (3, ∞).
  • Inequality -1 ≤ x < 4: set-builder {x ∈ R : -1 ≤ x < 4}, interval notation [-1, 4).
  • Solution of x^2 < 9: set-builder {x ∈ R : x^2 < 9}, interval (-3, 3).
  • Temperature comfort zone 18°C to 25°C inclusive: interval [18, 25].
  • Age eligibility 'age must be at least 18': interval [18, ∞).
  • Manufacturing tolerance diameter 5.00 ± 0.02 mm (inclusive): [4.98, 5.02].
🧮 Formulas
  1. a < x < b ↔ (a, b); a ≤ x ≤ b ↔ [a, b]; x > a ↔ (a, ∞); x ≥ a ↔ [a, ∞)
  2. (-∞, ∞) = R (the set of all real numbers); ∅ denotes the empty set
  3. Intersection: [a,b] ∩ [c,d] = [max(a,c), min(b,d)] if max(a,c) ≤ min(b,d), otherwise ∅
  4. Union (overlapping): [a,b] ∪ [c,d] = [min(a,c), max(b,d)] if intervals overlap or touch; if disjoint keep as union of pieces
  5. Complement: R \ (a,b) = (-∞, a] ∪ [b, ∞)
  6. Set-builder logical form: {x ∈ R : condition}, e.g. {x ∈ R : x^2 < 4} = (-2, 2)
📊 Visual ideas
Number line for x > a: draw open circle at a and shade to the right with an arrow to indicate (a, ∞). Use a filled dot for x ≥ a.
Number line for an interval [a, b]: draw filled dots at a and b and shade the segment between them. For (a, b) use open circles at endpoints.
Union example: draw two disjoint shaded segments, e.g. (-∞, -1) ∪ (1, ∞) with open circles at -1 and 1 and arrows pointing outward.
Intersection example: draw two overlapping intervals and highlight only the overlapping part (the intersection) using a different color or thicker line.
🔢6

Graphical Representation on Number Line

What it means: Graphical representation on a number line shows the set of real numbers that satisfy a given linear inequality. Each solution set is drawn as a region on a horizontal line with marked boundary points.

Basic rules

  • Represent each boundary value (solution of equality) as a point on the line.
  • Use a closed dot (●) at a boundary if the inequality includes equality (≤ or ≥).
  • Use an open dot (○) at a boundary if the inequality is strict (< or >).
  • Shade the line to the right of a boundary for > or ≥ (larger numbers); shade to the left for < or ≤ (smaller numbers).
  • If the solution extends indefinitely, draw an arrow (→ or ←) to indicate extension toward +∞ or −∞.

Steps to graph a linear inequality

  1. Algebraically solve the inequality for x (isolate x). If you multiply or divide by a negative number, reverse the inequality sign.
  2. Mark the boundary point(s) on the number line.
  3. Decide open/closed dot depending on strictness.
  4. Test one sample point (optional) to confirm which side to shade, then shade the correct region; add arrows for infinite directions.

Compound inequalities and set operations: Use intersection (AND) when both conditions must hold (e.g., 2 ≤ x < 5 is the overlap: closed dot at 2, open at 5, shade between). Use union (OR) when either condition suffices (e.g., x < −1 or x ≥ 4 gives two disjoint shaded regions).

Interval notation corresponds to graphing: (a,b) for a < x < b (open ends), [a,b] for a ≤ x ≤ b (closed ends), (−∞, b], [a, ∞), etc.

📌 Examples
  • x > 3 — mark an open circle at 3 and shade the line to the right with an arrow toward +∞.
  • x ≤ −2 — place a closed dot at −2 and shade the line to the left toward −∞.
  • 2 ≤ x < 5 — closed dot at 2, open dot at 5, shade the segment between 2 and 5 (interval [2,5)).
  • -3 < 2x + 1 ≤ 5 — solve: subtract 1: -4 < 2x ≤ 4; divide by 2 (positive): -2 < x ≤ 2. Graph: open dot at -2, closed dot at 2, shade between.
  • x < −1 or x ≥ 4 — graph two parts: open dot at −1 with left shading, closed dot at 4 with right shading (union of two regions).
  • Real-life: A car speed must be at least 40 km/h but less than 80 km/h. If s denotes speed, 40 ≤ s < 80. Graph shows acceptable speed interval [40,80).
🧮 Formulas
  1. If a < b, then a + c < b + c (addition/subtraction preserves inequality).
  2. If k > 0, multiply: ka < kb (inequality preserved); if k < 0, multiply: ka > kb (inequality sign reverses).
  3. Transitive: if a < b and b < c, then a < c.
  4. Reciprocal (for positive numbers): if 0 < a < b, then 1/a > 1/b.
  5. Interval notation equivalents: (a,b) = {x | a < x < b}, [a,b] = {x | a ≤ x ≤ b}, (−∞, b], [a, ∞).
  6. To solve ax + b < c: isolate x → ax < c − b → x < (c − b)/a (remember to reverse sign if a < 0).
📊 Visual ideas
Simple: a horizontal number line with tick marks, an open circle at 3 and a shaded ray to the right ending in an arrow (represents x > 3).
Closed-left: closed dot at −2 and shaded ray to the left with arrow (represents x ≤ −2).
Segment: closed dot at 2, open dot at 5, shaded straight line between them (represents 2 ≤ x < 5).
Compound OR: two separate shaded regions — open dot at −1 with left shade, closed dot at 4 with right shade (represents x < −1 or x ≥ 4).
🔢7

Linear Inequalities in Two Variables

Definition: A linear inequality in two variables x and y is an inequality of the form ax + by + c < 0, ax + by + c > 0, ax + by + c ≤ 0 or ax + by + c ≥ 0 where a and b are not both zero. Its solution set is a region (half-plane) in the xy‑plane, not just points.

Boundary and solution region: Replace the inequality by the corresponding equality ax + by + c = 0. This straight line is the boundary of the solution region. If the inequality is strict (< or >) the boundary is not included (draw a dashed line). If it is non‑strict (≤ or ≥) the boundary is included (draw a solid line). The solution set is the half‑plane on one side of the boundary line.

Slope‑intercept form and intercepts: The boundary line can be written as y = (-a/b)x + (-c/b) (when b ≠ 0). The slope is m = -a/b and the y‑intercept is -c/b. x‑intercept is -c/a when a ≠ 0.

Graphing method (step by step):

  • 1. Write the boundary: ax + by + c = 0.
  • 2. Draw the boundary line (solid for ≤, ≥; dashed for <, >).
  • 3. Choose a test point not on the line (commonly (0,0) if it is not on the boundary). Substitute into the inequality.
  • 4. If the test point satisfies the inequality, shade the side of the boundary containing that point; otherwise shade the opposite side.

Systems of linear inequalities: The solution to a system of linear inequalities is the intersection of the individual solution regions (often a polygonal region). This is the basic feasible region used in linear programming.

Notes: Always label axes, show intercepts or two points used to draw the boundary, and indicate whether the boundary is included. For strict inequalities use dashed lines; for nonstrict use solid lines.

📌 Examples
  • Example 1: x + y ≤ 4. Boundary: x + y = 4 (solid line). Intercepts: (4,0) and (0,4). Test (0,0): 0 + 0 ≤ 4 true, so shade the region containing (0,0) (the portion below and left of the line).
  • Example 2: 2x - y > 2. Rewrite as y < 2x - 2. Boundary: y = 2x - 2 (dashed line). Test (0,0): 0 > 2 is false, so shade the side opposite (0,0) — i.e., the region below the line y = 2x - 2 is the solution.
  • Example 3 (system): x ≥ 0, y ≥ 0, x + 2y ≤ 8. Boundaries: x = 0 (y‑axis), y = 0 (x‑axis), and x + 2y = 8. Feasible region is the triangle with vertices (0,0), (8,0), and (0,4).
  • Real‑life example: Budget constraint — if a product A costs ₹3 and product B costs ₹5 and you have at most ₹60, then 3x + 5y ≤ 60 (x,y ≥ 0). The feasible integer/non‑integer pairs (x,y) are all combinations you can buy within the budget; graphically it is the half‑plane under the line 3x + 5y = 60 in the first quadrant.
🧮 Formulas
  1. General linear inequality: ax + by + c < 0, ax + by + c ≤ 0, ax + by + c > 0, ax + by + c ≥ 0
  2. Boundary (line): ax + by + c = 0
  3. Slope (if b ≠ 0): y = (-a/b)x + (-c/b) ⇒ slope m = -a/b
  4. y‑intercept: (0, -c/b) (if b ≠ 0); x‑intercept: (-c/a, 0) (if a ≠ 0)
  5. Graphing rule: use dashed line for '<' or '>'; solid line for '≤' or '≥'. Use a test point to decide which half‑plane to shade.
  6. System solution: intersection of half‑planes (use to form feasible region in linear programming)
📊 Visual ideas
Graph suggestion 1 — Single inequality: For x + y ≤ 4, plot points (4,0) and (0,4), draw a solid line through them and shade the region containing (0,0). Mark the line as included (solid).
Graph suggestion 2 — Strict inequality: For y > 2x + 1, draw the line y = 2x + 1 as dashed. Test (0,0): 0 > 1 false, so shade the region above the line (where y is larger than 2x + 1).
Graph suggestion 3 — Using intercepts: To draw ax + by + c = 0, find x‑intercept (-c/a) and y‑intercept (-c/b), draw the line through them, then test a point to choose the shading side.
Graph suggestion 4 — System (feasible region): For x ≥ 0, y ≥ 0, x + 2y ≤ 8, draw x=0 and y=0 (axes) as boundaries and x + 2y = 8 as a solid line, then shade the triangular intersection. Label vertices and, if needed, compute their coordinates for optimization problems.
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Graphical Solution in the Plane

What it is: A graphical solution of a linear inequality (or a system of linear inequalities) in two variables finds the set of all points (x, y) in the plane that satisfy the inequality. Each linear inequality divides the plane into two half-planes; the solution is the half-plane (or intersection of half-planes) that satisfies the inequality.

General form: ax + by + c < 0, ax + by + c <= 0, ax + by + c > 0 or ax + by + c >= 0 (where a and b are not both zero).

Steps to graph a single linear inequality:

  • 1. Replace the inequality by the equality ax + by + c = 0 and draw this straight line (the boundary).
  • 2. If the inequality is < or > (strict), draw the boundary as a dashed line. If it is ≤ or ≥ (non-strict), draw it as a solid line.
  • 3. Pick a test point that is not on the boundary (commonly (0,0) if it is not on the line). Substitute into the inequality. If the inequality is true at the test point, shade the side of the line containing the test point; otherwise shade the opposite side.

System of linear inequalities: Graph each inequality as above. The solution of the system is the intersection (common shaded region) of all individual solution regions. This common region can be:

  • empty (no solution),
  • a bounded polygon (when inequalities form a closed region), or
  • an unbounded region extending to infinity.

Useful remarks:

  • To convert to slope-intercept form: if b ≠ 0, ax + by + c ≤ 0 ⟹ y ≤ (-a/b)x - c/b. This makes plotting easier (y-intercept and slope).
  • Intersection points (vertices) of boundary lines are found by solving the corresponding equalities (two linear equations). Solve by substitution, elimination, or determinants (Cramer's rule).
  • For optimization problems (linear programming), the optimal value occurs at a vertex (corner point) of the feasible polygonal region.
  • Special cases: 0·x + 0·y ≤ k: if k ≥ 0 the solution is the whole plane; if k < 0 the solution is empty.

How to check intersection algebraically: For two boundary lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, their intersection exists and is unique if the determinant D = a1b2 - a2b1 ≠ 0. If D = 0, lines are parallel (either coincident or distinct).

📌 Examples
  • Example 1 (single inequality): Graph y &lt; 2x + 1. Boundary: y = 2x + 1 (dashed). Test point (0,0): 0 &lt; 2(0)+1 → 0 &lt; 1 true, so shade the region below the line that contains (0,0).
  • Example 2 (system, bounded region): Solve graphically x ≥ 0, y ≥ 0, x + 2y ≤ 8. Graph x=0 (y-axis, solid), y=0 (x-axis, solid), and x + 2y = 8 (solid). The feasible region is the triangle with vertices (0,0), (8,0), (0,4).
  • Example 3 (no solution): Graph x + y &lt; 1 and x + y &gt; 3. Boundaries are parallel lines x+y=1 and x+y=3. The shaded half-planes do not overlap, so the system has no solution.
  • Example 4 (whole plane or empty): Inequality 0x + 0y ≤ 5 is true for every point (whole plane). Inequality 0x + 0y &gt; 0 has no solution (empty set).
🧮 Formulas
  1. General linear inequality: ax + by + c ≤ 0 (or ≥, <, >).
  2. Slope-intercept form (if b ≠ 0): y ≤ (-a/b)x - c/b.
  3. Boundary line: ax + by + c = 0 (use dashed for strict inequalities, solid for non-strict).
  4. Test-point method: substitute a point not on the boundary (often (0,0)); if inequality holds, shade that side.
  5. Intersection of two lines (Cramer's rule): For a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, solve x = (b1*c2 - b2*c1)/D, y = (a2*c1 - a1*c2)/D where D = a1*b2 - a2*b1 (if D ≠ 0).
📊 Visual ideas
Single inequality visual: Plot the line y = 2x + 1 as dashed. Mark test point (0,0). Since 0 &lt; 1, shade the half-plane containing (0,0). Label the shaded side 'y < 2x + 1'.
System producing a triangle: Plot x = 0 (y-axis), y = 0 (x-axis) and x + 2y = 8. Mark and join points (0,0), (8,0), (0,4). Shade the triangular region (including boundaries because inequalities are ≥ or ≤).
Parallel boundaries (no solution): Draw lines x + y = 1 and x + y = 3. Shade the region below the first and above the second to show non-overlapping regions—highlight empty intersection.
Strict vs non-strict comparison: Draw the same line twice—once dashed for y &lt; x + 1 and once solid for y ≤ x + 1—to show inclusion/exclusion of the boundary.
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Systems of Linear Inequalities (Two Variables)

What it is: A system of linear inequalities in two variables is a set of two or more inequalities each of the form ax + by < c, ax + by > c, ax + by ≤ c or ax + by ≥ c (a and b not both zero). The solution of the system is the set of all ordered pairs (x, y) that satisfy every inequality in the system simultaneously.

How to understand it:

  • Each inequality divides the plane into two half-planes: the points that satisfy the inequality and the points that do not.
  • The boundary of each half-plane is the straight line given by replacing the inequality with equality (ax + by = c). If the inequality is ≤ or ≥, the boundary line is included (drawn solid); if it is < or >, the boundary line is not included (drawn dashed).
  • The solution of the system is the intersection of the individual half-planes — i.e., the region (possibly empty) common to all shaded regions.

Graphical method (summary): 1) Rewrite the boundary line in slope-intercept form if convenient: y = (-a/b)x + c/b (when b ≠ 0). 2) Draw the boundary line (solid for ≤ or ≥, dashed for < or >). 3) Pick a test point (often (0,0) if not on the boundary). If the test point satisfies the inequality, shade the side containing it; otherwise shade the opposite side. 4) Repeat for each inequality and identify the overlapping shaded region as the solution set.

Key properties:

  • The solution set of a system of linear inequalities in two variables is a convex region — an intersection of half-planes (could be empty, a polygonal region, a ray, a half-plane, or an unbounded region).
  • If you need extreme points (vertices) of a bounded feasible region, they occur at intersections of the boundary lines. Solve the corresponding equalities simultaneously to find vertex coordinates.
  • This topic is the basis for linear programming in two variables, where a linear objective function is optimized over the feasible (common) region; optimal values occur at vertex points when the feasible region is bounded.
📌 Examples
  • Example 1 (Production constraints): Suppose a factory makes two products x and y. Each unit of x uses 1 hour of machine time and 2 hours of labor; each unit of y uses 2 hours of machine time and 1 hour of labor. Machine hours available = 8, labor hours available = 8. Inequalities: x + 2y ≤ 8 (labor), 2x + y ≤ 8 (machine), and x ≥ 0, y ≥ 0. Graph each inequality, shade feasible side and find the common region; vertices are (0,0), (0,4), (8/3,8/3), (4,0).
  • Example 2 (Feasibility/infeasibility): System: x + y ≤ 1 and x + y ≥ 3. The boundaries are parallel lines; their shaded half-planes do not overlap, so the system has no solution (inconsistent).
  • Example 3 (Intersection region unbounded): System: y ≥ x - 1 and y ≥ -x + 2. Graph both boundary lines (solid). The common region is the unbounded wedge above both lines; its vertex is intersection of the lines at solving x - 1 = -x + 2 ⇒ 2x = 3, x = 1.5 giving y = 0.5. The feasible region extends to infinity from that vertex.
🧮 Formulas
  1. General linear inequality: ax + by < c, ax + by ≤ c, ax + by > c, ax + by ≥ c (a and b not both 0).
  2. Boundary line (if b ≠ 0): y = (-a/b)x + c/b. If b = 0, boundary is vertical line x = c/a.
  3. Test-point method: Substitute a test point (x0, y0) into ax + by ? c; if true, shade the side containing (x0, y0).
  4. Intersection (vertex) of two boundary lines: solve the system of equalities ax + by = c and a'x + b'y = c' simultaneously (use substitution or elimination).
  5. Property: Intersection of half-planes is convex; for bounded feasible region, extrema of a linear objective occur at vertices.
📊 Visual ideas
Graphing steps: 1) For each inequality, draw the boundary line ax + by = c. Use a solid line for ≤ or ≥, dashed for < or >. 2) Choose an easy test point (often (0,0) unless it lies on the boundary) and check the inequality; shade the side that satisfies it. 3) Repeat for all inequalities and mark the overlapping shaded region as the solution set. 4) Label intersection points (vertices) where boundaries meet, and indicate if region is bounded or unbounded (draw arrows if unbounded).
Visual suggestions: use different colors or hatch patterns for each inequality's shaded half-plane so the common region is clear. Emphasize boundary type by line style (solid/dashed) and mark included boundary points with a solid line or filled dot, excluded with dashed line or open circle at key points.
Software/tools: Use graph paper or tools like GeoGebra/Desmos. Enter boundary lines as equalities, set inequality directions (or shade manually). For example, in Desmos enter 'y <= 8 - x/2' (converted form) or 'y >= x - 1' to see shaded half-planes and their intersection.
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Strict vs Non-Strict Inequalities and Boundary Inclusion

In inequalities we compare expressions using one of four relation signs: the strict inequalities < and >, and the non-strict (or inclusive) inequalities ≤ and ≥. The essential difference is whether the boundary value where the two sides are equal is allowed as part of the solution set.

One-variable view (number line):

  • x < a or x > a are strict: the value x = a is not included in the solution. On a number line show an open circle at a and shade to the left for x < a or to the right for x > a.
  • x ≤ a or x ≥ a are non-strict (inclusive): the value x = a is included. On a number line show a closed (filled) circle at a and shade accordingly.

Two-variable (linear inequality) / geometric view:

  • A linear inequality ax + by + c < 0 or ax + by + c > 0 defines an open half-plane. Its boundary is the straight line ax + by + c = 0. For the strict case (< or >) the boundary line is not included (draw as dashed).
  • For ax + by + c ≤ 0 or ax + by + c ≥ 0 the boundary line is included (draw as solid). The solution is the half-plane on the side where the inequality holds.
  • To decide which side to shade, pick any convenient test point (commonly (0,0) if it is not on the boundary). Substitute into ax + by + c; if the inequality is true use the side containing the test point, otherwise shade the opposite side.

Key operational rules that affect boundary inclusion:

  • Add/subtract the same number from both sides: inequality direction unchanged.
  • Multiply/divide both sides by a positive number: direction unchanged.
  • Multiply/divide both sides by a negative number: inequality direction reverses (this is important when solving; boundary inclusion depends only on <, >, ≤, ≥, not on sign changes).

Why boundary matters: In many problems (real constraints, safety limits, legal thresholds) allowing equality changes whether a candidate value is acceptable. Graphically, this is the difference between taking the boundary line/point as part of the shaded solution set (inclusive) or leaving it out (exclusive).

📌 Examples
  • Speed limit: If the speed limit is “at most 60 km/h” we model v ≤ 60 (non-strict). If a regulation says “must be strictly under 60 km/h”, use v < 60 (strict).
  • Elevator capacity: If a lift’s limit is 10 people, n ≤ 10 (non-strict) — 10 people are allowed.
  • Age-restricted entry: If only people aged 18 or older may enter, age ≥ 18 (non-strict); someone exactly 18 is allowed.
  • Clearance requirement: A bridge clearance must be greater than 4.0 m to pass safely, h > 4.0 (strict) — exactly 4.0 m would not clear.
  • Exam pass mark: If pass mark is 40%, score ≥ 40% means a 40% score passes (non-strict).
🧮 Formulas
  1. One-variable solution intervals: x < a → (-∞, a), x ≤ a → (-∞, a], x > a → (a, ∞), x ≥ a → [a, ∞).
  2. Linear inequality boundary: ax + by + c = 0 is the boundary line. ax + by + c < 0 and ax + by + c ≤ 0 are the half-planes on one side; ax + by + c > 0 and ax + by + c ≥ 0 are the half-planes on the other side.
  3. Test-point method: If P(x0,y0) is any point not on the boundary, evaluate S = a x0 + b y0 + c. If S satisfies the inequality, shade the half-plane containing P; otherwise shade the opposite half-plane.
  4. Operations rules: For real k>0, (if) a < b ⇒ ka < kb. For k<0, a < b ⇒ ka > kb (inequality reverses when multiplying/dividing by negative).
  5. Intersection/union: For systems of inequalities, the solution is the intersection of half-planes (common shaded region).
📊 Visual ideas
Number line pictures: show an open circle at a for x &lt; a or x &gt; a; show a closed (filled) circle at a for x ≤ a or x ≥ a. Shade left for &lt; and ≤, shade right for &gt; and ≥.
2D half-plane: draw the boundary line ax + by + c = 0. For ax + by + c &lt; 0 use a dashed line and shade the side where ax + by + c evaluates negative. For ax + by + c ≤ 0 use a solid line and shade the same side.
Example sketch: line x + y = 2. For x + y &lt; 2 draw the line dashed and shade the region below-left (points with sum less than 2). For x + y ≥ 2 draw the line solid and shade the opposite region (sum at least 2).
Show test-point annotation: mark (0,0) and compute substitution; annotate whether (0,0) is in the shaded area or not. This clarifies boundary-choice visually.
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Techniques and Problem Solving

Overview: Techniques and Problem Solving for linear inequalities covers methods to solve inequalities in one variable, compound inequalities, rational inequalities, inequalities with absolute values, and systems of linear inequalities in two variables. The main goal is to transform the inequality into a form where the solution set can be read easily (interval form or graph on a number line/plane).

Basic techniques (one variable)

  1. Isolate the variable: use addition/subtraction to bring variable terms to one side.
  2. Clear coefficients: multiply or divide by a nonzero number. Remember: if you multiply or divide by a negative number, reverse the inequality sign.
  3. Combine like terms and simplify to get x < a, x > b, x ≤ c or x ≥ d.
  4. Represent solutions on a number line and in interval notation.

Handling compound inequalities: For chained inequalities like a ≤ bx + c < d, perform the same operation on all three segments simultaneously (e.g., subtract c from all parts), keeping track of sign reversals if you multiply/divide by a negative.

Rational inequalities and sign charts: For expressions of the form P(x)/Q(x) > 0 or < 0,

  1. bring all terms to one side and write as single rational expression;
  2. find critical points: zeros of numerator and denominator (they partition the number line); denominator zeros are excluded from solution;
  3. use a sign chart or test points on each interval to determine where the expression is positive or negative;
  4. include zeros of numerator if inequality is ≤ or ≥, exclude denominator zeros.

Absolute-value inequalities: Solve |A(x)| < k as -k < A(x) < k. Solve |A(x)| > k as A(x) < -k or A(x) > k. Then solve resulting linear inequalities.

Systems of linear inequalities (two variables): Convert each inequality to standard form (e.g., y < mx + c or ax + by ≤ c), draw the boundary line (dashed for > or <, solid for ≥ or ≤), pick a test point (commonly (0,0) unless on boundary) to decide which side to shade, and the solution is the intersection (common shaded region) for a system.

Problem-solving tips:

  • Always perform the same operation on every part of a chained inequality.
  • Keep track of strict (<, >) vs non-strict (≤, ≥) to decide open vs closed endpoints on graphs.
  • When clearing denominators, multiply both sides by the square of denominators is sometimes used to avoid sign issues, but easier is to bring to one side and use sign charts.
  • Check boundary points and special values (where expressions are undefined).
  • For word problems, translate constraints into inequalities, solve, and interpret in context (rounding or integer constraints may apply).
📌 Examples
  • Solve 3x - 5 < 10. Solution: Add 5 to both sides: 3x < 15. Divide by 3 (positive): x < 5. Interval: (-∞, 5). On number line: open circle at 5, shade left.
  • Solve -2 ≤ 3x + 1 < 10. Subtract 1 from all parts: -3 ≤ 3x < 9. Divide by 3: -1 ≤ x < 3. Interval: [-1, 3). Number line: closed dot at -1, open at 3, shade between.
  • Solve the rational inequality (x - 2)/(x + 3) > 0. Critical points: numerator zero at x = 2, denominator zero at x = -3 (exclude). Intervals: (-∞, -3), (-3, 2), (2, ∞). Test points: x = -4 gives ( - / - ) = + so true; x = 0 gives ( - / + ) = - false; x = 3 gives ( + / + ) = + true. Solution: (-∞, -3) ∪ (2, ∞). Exclude x = -3; x = 2 not included because strict > 0.
  • System in two variables: x + 2y ≤ 6 and x - y ≥ 1. For x + 2y = 6 draw a solid line; for x - y = 1 draw a solid line. Test (0,0): for first, 0 ≤ 6 true, so shade side containing (0,0); for second, 0 ≥ 1 false, so shade opposite side. Intersection of shaded regions is the solution region (a polygonal area).
  • Word problem: A student has at most Rs.600 to spend on pens (Rs.50 each) and notebooks (Rs.30 each). If x = pens and y = notebooks, constraint is 50x + 30y ≤ 600. This is a linear inequality region: draw the line 50x + 30y = 600, use (0,0) to test (0 ≤ 600 true) so shade the region including the origin. Also include x ≥ 0, y ≥ 0 for nonnegative quantities. Integer solutions are lattice points in the shaded region.
🧮 Formulas
  1. Transposition: If a < b then a + c < b + c (for any real c).
  2. Multiplication/Division rule: If a < b and k > 0 then ka < kb; if k < 0 then ka > kb (inequality sign reverses).
  3. Transitivity: If a < b and b < c then a < c.
  4. Compound inequality operations: Apply same operation to all parts: if a ≤ bx + c < d, subtract or divide on all three sides carefully tracking sign flips.
  5. Rational inequality approach: Solve P(x)/Q(x) &gt; 0 by finding zeros of P and Q, making sign chart. Zeros of Q are excluded.
  6. Absolute value: |A(x)| &lt; k ⇔ -k &lt; A(x) &lt; k; |A(x)| &gt; k ⇔ A(x) &lt; -k or A(x) &gt; k (for k ≥ 0).
📊 Visual ideas
Number line graph for simple inequality x < 5: open circle at 5 with arrow/shading to the left. For x ≤ 3: closed dot at 3 with shading left.
Compound inequality [-1, 3): closed dot at -1, open at 3, shade between. Label endpoints and show interval notation.
Sign chart for rational inequality (x-2)/(x+3)>0: draw number line, mark -3 and 2, list signs of numerator/denominator on each interval and product sign, then shade intervals where product is positive.
Half-plane graph for y < 2x + 1: draw the line y = 2x + 1 dashed (strict), pick (0,0) to test (0 < 1 true), shade the side containing (0,0). For y ≥ -x + 2: draw solid line and shade appropriate side. Intersection of shaded regions shows solution to the system.
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Connections and Applications

Overview
A linear inequality is like a linear equation but describes a range of values (a region) rather than a single line. The topic "Connections and Applications" emphasizes how linear inequalities link to other parts of mathematics (intervals, absolute values, systems of linear equations, linear programming) and how they model real-life constraints.

Key ideas

  • One-variable inequalities (e.g. ax + b < 0) describe intervals on the number line. Use algebraic operations to isolate x, remembering to reverse the inequality when multiplying/dividing by a negative number.
  • Compound inequalities use AND (intersection) or OR (union). For example, 2 < x + 1 < 5 is an AND that yields an interval.
  • Absolute-value inequalities like |x − a| < r become double inequalities a − r < x < a + r; they describe distances from a point.
  • Two-variable linear inequalities (ax + by ≤ c) represent half-planes in the plane. The boundary ax + by = c is a straight line; the inequality picks one side.
  • Systems of linear inequalities produce feasible regions given by the intersection of half-planes — often a polygon. This is the geometric base for linear programming (optimize a linear objective subject to linear constraints).

How to graph and interpret a two-variable inequality

  1. Convert to the boundary line ax + by = c. Draw it (solid if ≤ or ≥, dashed if < or >).
  2. Choose a test point (often (0,0) if not on the line). Substitute into the inequality.
  3. If the test point satisfies the inequality, shade the side containing that point; otherwise shade the opposite side.

Connections

  • With intervals: solutions of one-variable inequalities are intervals; endpoints depend on strictness (open/closed).
  • With absolute value: rewrite absolute inequalities as double inequalities (distance interpretation).
  • With systems of linear equations: equalities form the boundary lines for inequalities; intersection points (vertices) of these boundaries are critical in optimization problems.
  • With linear programming: constraints are linear inequalities; objective functions attain extremes at vertices of the feasible polygon.

Practical interpretation
Linear inequalities model constraints: limits on resources, thresholds, safety bounds, tolerances. The feasible region shows all possible choices meeting all constraints; checking vertices can give best/worst values for linear objectives.

📌 Examples
  • Budgeting: If a student has at most Rs. 1200 to spend on books (x) and pens (y) with prices Rs. 200 per book and Rs. 50 per pen, constraint: 200x + 50y ≤ 1200. Graph the half-plane and find feasible integer pairs (x,y).
  • Travel time: If one must arrive within 2 hours and travel time t satisfies t = d/v + 0.25 (stop time), then inequality d/v + 0.25 ≤ 2 gives v ≥ d / 1.75 (minimum speed needed).
  • Diet/nutrition: A food mix must contain at least 30% protein and at most 15% fat. If x and y are quantities of two ingredients with known protein/fat percentages, inequalities model permissible mixtures.
  • Safety tolerance: A machine part diameter x must satisfy 9.98 mm ≤ x ≤ 10.02 mm. This is |x − 10| ≤ 0.02, i.e. 9.98 ≤ x ≤ 10.02.
  • Manufacturing constraints (linear programming): Producing two products P1 and P2 requires limited labor and material. If each P1 needs 2h labor and each P2 needs 3h, and 120 labor-hours available: 2x + 3y ≤ 120. Combined with other constraints, feasible region gives possible production plans.
  • Temperature control: If safe operating temperature T must be above 15°C and below 35°C: 15 < T < 35 modeled as a compound inequality.
🧮 Formulas
  1. If a &gt; b, then a + c &gt; b + c (addition/subtraction preserves inequality).
  2. If a &gt; b and c &gt; 0, then ac &gt; bc; if c &lt; 0 then ac &lt; bc (multiplication by negative reverses inequality).
  3. Transitivity: if a &gt; b and b &gt; c then a &gt; c.
  4. Solve linear inequality ax + b &lt; 0 → x &lt; −b/a (if a &gt; 0). If a &lt; 0 then inequality sign reverses when dividing by a.
  5. Absolute-value: |x − a| &lt; r ⇔ a − r &lt; x &lt; a + r; and |x − a| ≤ r ⇔ a − r ≤ x ≤ a + r.
  6. Interval notation: (a, b) for a &lt; x &lt; b (open ends), [a, b] for a ≤ x ≤ b (closed), (−∞, c) and (d, ∞) for unbounded sets.
📊 Visual ideas
Number-line graph for x: show boundary points as open circles for strict inequalities (x &lt; a) and filled circles for non-strict (x ≥ a); shade left for &lt; and right for &gt;.
Graph of ax + by &lt; c: draw the boundary line ax + by = c. Use dashed line for &lt; or &gt;, solid for ≤ or ≥. Pick a test point (0,0) if not on the line; shade the side that satisfies the inequality.
Intersection of two half-planes: sketch two boundary lines and shade the overlapping region (feasible region). Mark intersection points (vertices) — these are candidate points for optimization.
Feasible polygon with labeled vertices: show objective-level lines (px + qy = k) sliding parallel until the last contact with feasible region — the contact vertex gives maximum/minimum.

Key Concepts

Inequality
A mathematical statement that compares two expressions using symbols <, >, ≤, ≥.
Strict inequality
An inequality using < or >, which does not allow equality of the two sides.
Non-strict inequality
An inequality using ≤ or ≥, which allows equality of the two sides.
Linear inequality in one variable
An inequality of the form ax + b < 0, ax + b ≤ 0, ax + b > 0 or ax + b ≥ 0 where a ≠ 0.
Solution set
The set of all values of the variable(s) that satisfy the inequality.
Interval notation
A compact way to denote sets of real numbers using parentheses and brackets, e.g. (a,b), [a,b].
Open interval
An interval (a, b) that contains all real numbers between a and b but not the endpoints a and b.
Closed interval
An interval [a, b] that includes its endpoints a and b along with all numbers between them.
Boundary point
A value at which the corresponding linear expression becomes equal (equality case) and which separates regions satisfying or not satisfying the inequality.
Compound inequality
A combination of two inequalities joined by 'and' (intersection) or 'or' (union).
Equivalent inequalities
Inequalities that have the same solution set after performing allowed algebraic operations.
Linear inequality in two variables
An inequality of the form ax + by + c < 0, ≤ 0, > 0 or ≥ 0 where a and b are not both zero; solutions form a region in the plane.
Half-plane
The set of points on one side of a straight line (including or excluding the line) in the plane that satisfy a linear inequality in two variables.
Boundary line
The straight line given by replacing the inequality sign with equality (ax + by + c = 0); it divides the plane into two half-planes.
Graph of an inequality
A depiction of all points that satisfy the inequality: the boundary (solid for ≤/≥, dashed for < />) and the shaded solution region.
Test point method
A method to determine which side of the boundary line is the solution region by substituting a convenient point not on the line.
Feasible region
The common intersection of solution regions of a system of inequalities; all points that satisfy every inequality in the system.
Multiplication by negative number rule
When both sides of an inequality are multiplied or divided by a negative number, the inequality sign reverses direction.
Intersection of solution sets
The set of values that satisfy all of several inequalities (logical AND).
Union of solution sets
The set of values that satisfy at least one of several inequalities (logical OR).

End-of-Chapter Trial Paper & Test Questions

Topic-wise questions to test your understanding of every concept in this chapter.

  1. State the rule that distinguishes solving an inequality from solving an equation, and explain why it is needed. / असमिका हल करने को समीकरण हल करने से अलग करने वाला नियम बताइए और स्पष्ट कीजिए कि इसकी आवश्यकता क्यों है।
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    When both sides are multiplied or divided by a negative number, the inequality sign must be reversed; this is needed because multiplying by a negative reverses the order of numbers on the number line. / जब दोनों पक्षों को किसी ऋणात्मक संख्या से गुणा या भाग किया जाता है, तो असमिका का चिह्न उलट देना चाहिए; यह इसलिए आवश्यक है क्योंकि ऋणात्मक से गुणा करने पर संख्या रेखा पर संख्याओं का क्रम उलट जाता है।

  2. Solve −2x + 4 ≥ 10 and represent the solution in interval notation. / −2x + 4 ≥ 10 को हल कीजिए और हल को अंतराल संकेतन में लिखिए।
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    −2x ≥ 6; dividing by −2 reverses the sign: x ≤ −3, so the solution is (−∞, −3]. / −2x ≥ 6; −2 से भाग देने पर चिह्न उलटता है: x ≤ −3, अतः हल (−∞, −3] है।

  3. Solve the compound inequality 1 < 2x + 1 ≤ 7 and describe its number-line graph. / संयुक्त असमिका 1 < 2x + 1 ≤ 7 को हल कीजिए और इसके संख्या-रेखा आलेख का वर्णन कीजिए।
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    Subtract 1: 0 < 2x ≤ 6; divide by 2: 0 < x ≤ 3, i.e. (0, 3]; graph has an open circle at 0, a closed dot at 3, shaded between. / 1 घटाएँ: 0 < 2x ≤ 6; 2 से भाग दें: 0 < x ≤ 3, अर्थात् (0, 3]; आलेख में 0 पर खुला वृत्त, 3 पर भरा बिंदु, बीच में छायांकित होगा।

  4. Explain the difference between strict and non-strict inequalities in terms of boundary inclusion on a graph. / आलेख पर सीमा सम्मिलन के संदर्भ में दृढ़ और अदृढ़ असमिकाओं के बीच अंतर समझाइए।
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    Strict inequalities (<, >) exclude the boundary, shown by an open circle on a number line or a dashed boundary line in a plane; non-strict (≤, ≥) include the boundary, shown by a closed dot or a solid line. / दृढ़ असमिकाएँ (<, >) सीमा को बाहर रखती हैं, जो संख्या रेखा पर खुले वृत्त या तल में बिंदुदार रेखा से दर्शाई जाती हैं; अदृढ़ (≤, ≥) सीमा को सम्मिलित करती हैं, जो भरे बिंदु या ठोस रेखा से दर्शाई जाती हैं।

  5. A commuter has at most ₹200; a taxi charges ₹50 base plus ₹12 per km. Form an inequality and find the maximum whole kilometres travelled. / एक यात्री के पास अधिकतम ₹200 हैं; टैक्सी ₹50 आधार शुल्क और ₹12 प्रति किमी लेती है। एक असमिका बनाइए और तय की जाने वाली अधिकतम पूर्ण किलोमीटर ज्ञात कीजिए।
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    50 + 12x ≤ 200 ⇒ 12x ≤ 150 ⇒ x ≤ 12.5; since kilometres must be whole, the maximum is 12 km. / 50 + 12x ≤ 200 ⇒ 12x ≤ 150 ⇒ x ≤ 12.5; क्योंकि किलोमीटर पूर्ण होने चाहिए, अधिकतम 12 किमी है।

  6. Describe how to graph the two-variable inequality x + 2y ≤ 4 in the plane. / दो-चर असमिका x + 2y ≤ 4 को तल में आलेखित करने की विधि बताइए।
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    Draw the boundary line x + 2y = 4 as a solid line (non-strict); test (0,0): 0 ≤ 4 is true, so shade the half-plane containing the origin. / सीमा रेखा x + 2y = 4 को ठोस रेखा (अदृढ़) के रूप में खींचें; (0,0) का परीक्षण करें: 0 ≤ 4 सत्य है, अतः मूल बिंदु वाले अर्ध-तल को छायांकित करें।

  7. Solve the chained inequality −2 ≤ 5 − 3x < 7, being careful with the sign. / श्रृंखलित असमिका −2 ≤ 5 − 3x < 7 को चिह्न का ध्यान रखते हुए हल कीजिए।
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    Subtract 5: −7 ≤ −3x < 2; divide by −3 and reverse signs: 7/3 ≥ x > −2/3, i.e. x ∈ (−2/3, 7/3]. / 5 घटाएँ: −7 ≤ −3x < 2; −3 से भाग दें और चिह्न उलटें: 7/3 ≥ x > −2/3, अर्थात् x ∈ (−2/3, 7/3]।

  8. Why does the system x + 2 > 5 and 3x < 9 have no solution? / निकाय x + 2 > 5 और 3x < 9 का कोई हल क्यों नहीं है?
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    The first gives x > 3 and the second gives x < 3; their intersection is empty because no number is both greater than 3 and less than 3. / पहली से x > 3 और दूसरी से x < 3 मिलता है; इनका प्रतिच्छेदन रिक्त है क्योंकि कोई संख्या 3 से बड़ी और 3 से छोटी दोनों नहीं हो सकती।

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