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Linear Inequalities

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Chapter Notes

Overview
This chapter introduces the concept of inequalities, which are algebraic expressions related by symbols like '<', '>', '≤', or '≥', as opposed to equations which use '='. It explains how to solve linear inequalities in one and two variables, both algebraically and graphically, and applies these to real-world problems like determining minimum marks or mixture concentrations. The one big idea is that solving inequalities follows the same rules as equations, except that multiplying or dividing by a negative number reverses the inequality sign.
Key Concepts (11)
Inequality
Two real numbers or two algebraic expressions related by the symbol '<', '>', '≤' or '≥' form an inequality.
e.g. 30x < 200, where x is the number of rice packets Ravi can buy with ₹200.
Numerical inequality
An inequality involving only numbers, without variables.
e.g. 3 < 5; 7 > 5.
Literal inequality
An inequality involving variables (letters).
e.g. x < 5; y > 2; x ≥ 3; y ≤ 4.
Double inequality
A statement that combines two inequalities, indicating a range for a variable.
e.g. 3 < x < 5 (read as x is greater than 3 and less than 5); 2 < y < 4.
Strict inequality
An inequality using the symbols '<' or '>', which does not include the boundary value.
e.g. ax + b < 0; ax + b > 0; ax + by < c; ax + by > c.
Slack inequality
An inequality using the symbols '≤' or '≥', which includes the boundary value.
e.g. ax + b ≤ 0; ax + b ≥ 0; ax + by ≤ c; ax + by ≥ c.
Linear inequality in one variable
An inequality that can be written in the form ax + b < 0, ax + b > 0, ax + b ≤ 0, or ax + b ≥ 0, where a ≠ 0.
e.g. 30x < 200; 5x – 3 < 3x + 1.
Linear inequality in two variables
An inequality that can be written in the form ax + by < c, ax + by > c, ax + by ≤ c, or ax + by ≥ c, where a ≠ 0 and b ≠ 0.
e.g. 40x + 20y ≤ 120, where x is the number of registers and y is the number of pens Reshma buys.
Solution of an inequality
A value of the variable which makes the inequality a true statement.
e.g. For 30x < 200, x = 0,1,2,3,4,5,6 are solutions because they make the statement true.
Solution set
The set of all solutions of an inequality.
e.g. For 30x < 200 when x is a natural number, the solution set is {1,2,3,4,5,6}.
Quadratic inequality
An inequality of the form ax² + bx + c ≤ 0 or ax² + bx + c > 0, where a ≠ 0. (Not covered in detail in this chapter.)
e.g. ax² + bx + c ≤ 0; ax² + bx + c > 0.
Formulas & Equations (4)
30x < 200
Reads as: 30 times x is less than 200
Variables: x = number of packets of rice (unitless integer), 30 = cost per packet (₹), 200 = total money (₹)
Use when: Solving word problems involving a maximum or minimum limit, like budget constraints.
⚠ Forgetting that x must be a natural number (or integer) in context, leading to fractional solutions like x = 6.67.
40x + 20y ≤ 120
Reads as: 40 times x plus 20 times y is less than or equal to 120
Variables: x = number of registers, y = number of pens, 40 = cost per register (₹), 20 = cost per pen (₹), 120 = total money (₹)
Use when: Solving two-variable inequality word problems, e.g., buying multiple items within a budget.
⚠ Treating the inequality as an equation and finding only one pair (x, y) instead of the entire solution region.
C = (5/9)(F – 32)
Reads as: C equals five-ninths times (F minus 32)
Variables: C = temperature in degrees Celsius, F = temperature in degrees Fahrenheit
Use when: Converting temperature scales in inequality word problems, e.g., 'keep between 30°C and 35°C'.
⚠ Incorrectly multiplying or dividing when isolating F; e.g., writing 30 < (5/9)(F – 32) < 35 and then multiplying only the middle term by 9/5.
IQ = (MA / CA) × 100
Reads as: IQ equals mental age divided by chronological age, times 100
Variables: IQ = intelligence quotient (unitless), MA = mental age (years), CA = chronological age (years)
Use when: Solving problems involving IQ ranges, e.g., '80 ≤ IQ ≤ 140 for 12-year-olds'.
⚠ Forgetting to multiply by 100, or confusing which age is in the numerator.
How It Works
Solving a linear inequality in one variable (algebraic method)
  1. Step 1: Simplify both sides of the inequality by removing brackets and combining like terms.
  2. Step 2: Use Rule 1: Add or subtract the same number from both sides to isolate the variable term on one side.
  3. Step 3: Use Rule 2: Multiply or divide both sides by the same positive number to solve for the variable. If multiplying or dividing by a negative number, reverse the inequality sign.
  4. Step 4: Write the solution set, considering the domain (natural numbers, integers, or real numbers) if specified.
  5. Step 5: Represent the solution on a number line if required: use an open circle for '<' or '>' and a closed (dark) circle for '≤' or '≥'.
💡 The most critical step is reversing the inequality sign when multiplying or dividing by a negative number. Forgetting this is the #1 mistake in board exams.
Solving a system of linear inequalities (simultaneous inequalities)
  1. Step 1: Solve each inequality separately to find its solution set.
  2. Step 2: Find the intersection (common values) of the solution sets.
  3. Step 3: Represent the common solution on a number line, showing the overlapping region.
  4. Step 4: Write the final solution as a compound inequality (e.g., 2 ≤ x < 6).
💡 Always check whether the endpoints are included (≤ or ≥) or excluded (< or >) in the final answer.
Solving double inequalities (e.g., –8 ≤ 5x – 3 < 7)
  1. Step 1: Treat the double inequality as two separate inequalities: –8 ≤ 5x – 3 and 5x – 3 < 7.
  2. Step 2: Solve both simultaneously by performing the same operation on all three parts (left, middle, right).
  3. Step 3: Isolate the variable in the middle part.
  4. Step 4: Write the final solution as a single compound inequality (e.g., –1 ≤ x < 2).
💡 When multiplying or dividing all three parts by a negative number, reverse both inequality signs. For example, –15 ≤ –3x ≤ 11 becomes 5 ≥ x ≥ –11/3.
Solving word problems involving inequalities
  1. Step 1: Read the problem carefully and identify the unknown quantity. Assign a variable (e.g., x).
  2. Step 2: Translate the given conditions into one or more inequalities using keywords: 'at least' (≥), 'at most' (≤), 'more than' (>), 'less than' (<), 'between' (double inequality).
  3. Step 3: Solve the inequality(ies) algebraically.
  4. Step 4: Interpret the solution in the context of the problem (e.g., minimum marks, number of items, range of temperature).
  5. Step 5: Write the final answer as a complete sentence.
💡 Always consider the practical domain (e.g., number of items must be a whole number, marks cannot exceed 100). Many students lose marks by giving a fractional answer when only integers make sense.
Exam Traps — Don't Make These Mistakes
⚠ Forgetting to reverse the inequality sign when multiplying/dividing by a negative number
✗ –2x < 4 ⇒ x < –2
→ When dividing both sides by –2, the inequality sign must be reversed. The correct step is –2x < 4 ⇒ x > –2.
✓ –2x < 4 ⇒ x > –2
⚠ Mishandling double inequalities when multiplying by a negative
✗ –5 ≤ (5 – 3x)/2 ≤ 8 ⇒ –10 ≤ 5 – 3x ≤ 16 ⇒ –15 ≤ –3x ≤ 11 ⇒ 5 ≤ x ≤ –11/3
→ When dividing all three parts by –3, both inequality signs must be reversed. The correct step is –15 ≤ –3x ≤ 11 ⇒ 5 ≥ x ≥ –11/3, which is written as –11/3 ≤ x ≤ 5.
✓ –15 ≤ –3x ≤ 11 ⇒ 5 ≥ x ≥ –11/3 ⇒ –11/3 ≤ x ≤ 5
⚠ Confusing open and closed circles on the number line
✗ For x < 3, they draw a dark circle at 3 and shade to the left.
→ A dark (closed) circle means the point is included (≤ or ≥). For '<' or '>', an open circle must be used to show the point is not included.
✓ For x < 3, draw an open circle at 3 and shade the number line to the left.
⚠ Not considering the domain (natural numbers, integers) in word problems
✗ For 30x < 200, they write the solution set as x < 20/3, i.e., all real numbers less than 6.67.
→ The problem specifies x is a natural number (or integer), so the solution set must list only those values: {1,2,3,4,5,6} for natural numbers.
✓ When x is a natural number, the solution set is {1,2,3,4,5,6}.
⚠ Treating '≤' as '<' in the final answer
✗ For x ≥ 1, they write the solution set as (1, ∞) instead of [1, ∞).
→ The bracket notation matters: (1, ∞) excludes 1, while [1, ∞) includes 1. Since x ≥ 1, the correct notation is [1, ∞).
✓ x ∈ [1, ∞)
Compare & Contrast
Relationship between expressions
Equation
Inequality
An equation uses the '=' sign to show equality, while an inequality uses '<', '>', '≤', or '≥' to show a range of possible values.
Inclusion of boundary value
Strict inequality (< or >)
Slack inequality (≤ or ≥)
A strict inequality does not include the boundary value (e.g., x < 3 means x can be 2.999 but not 3), while a slack inequality includes the boundary value (e.g., x ≤ 3 includes 3).
Exam Focus
Marks pattern: 1 MCQ (1 mark), 1 Very Short Answer (1 mark), 1 Short Answer (3 marks), 1 Long Answer (5 marks) — total 10 marks typically.
Most asked: Solving linear inequalities in one variable (algebraic and graphical representation on number line) · Solving systems of linear inequalities (simultaneous inequalities) · Word problems involving inequalities (e.g., marks, mixtures, temperature conversion, consecutive numbers)
Don't skip: Word problems on inequalities (e.g., finding minimum marks, range of mixture percentages, consecutive odd/even numbers) — these carry 4-5 marks and require careful translation of conditions into inequalities.
One-Line Revision (read aloud before the exam)
  1. An inequality relates two expressions using <, >, ≤, or ≥.
  2. Numerical inequalities involve only numbers (e.g., 3 < 5), while literal inequalities involve variables (e.g., x < 5).
  3. Double inequalities combine two inequalities into one statement (e.g., 3 < x < 5).
  4. Strict inequalities use < or >; slack inequalities use ≤ or ≥.
  5. Linear inequalities in one variable are of the form ax + b < 0, etc., with a ≠ 0.
  6. Linear inequalities in two variables are of the form ax + by < c, etc., with a ≠ 0, b ≠ 0.
  7. A solution of an inequality is a value that makes it true; the set of all solutions is the solution set.
  8. Rule 1: Adding or subtracting the same number from both sides does not change the inequality sign.
  9. Rule 2: Multiplying or dividing both sides by a positive number does not change the sign; by a negative number reverses the sign.
  10. On a number line, use an open circle for < or >, and a closed (dark) circle for ≤ or ≥.
  11. For word problems, translate 'at least' as ≥, 'at most' as ≤, 'more than' as >, and 'less than' as <.
  12. When solving double inequalities, perform the same operation on all three parts simultaneously.
  13. The conversion formula between Celsius and Fahrenheit is C = (5/9)(F – 32).
  14. IQ is calculated as (Mental Age / Chronological Age) × 100.
  15. Always consider the practical domain (natural numbers, integers) when solving real-world inequality problems.

MCQs — 10 Questions with Answers

Q1. Which of the following is an example of a slack inequality?

  • A. 3x + 2 < 5
  • B. 5y − 1 ≤ 9 ✓
  • C. 2x > 4
  • D. 7 − x < 10

Answer: B — Slack inequalities use ≤ or ≥; option B uses ≤, making it a slack inequality.

Q2. Solve the inequality 3x − 5 > 7 for real x.

  • A. x < 4
  • B. x > 4 ✓
  • C. x ≥ 4
  • D. x ≤ 4

Answer: B — 3x − 5 > 7 → 3x > 12 → x > 4 (dividing by positive 3 keeps the sign).

Q3. What is the solution set of 2x + 3 ≤ 11 in interval notation?

  • A. [4, ∞)
  • B. (−∞, 4] ✓
  • C. (−∞, 4)
  • D. [−4, ∞)

Answer: B — 2x + 3 ≤ 11 → 2x ≤ 8 → x ≤ 4, so solution set is (−∞, 4] in interval notation.

Q4. When solving −4x > 12, what happens to the inequality sign?

  • A. The sign remains the same
  • B. The sign reverses to < ✓
  • C. The sign becomes ≤
  • D. No change, division preserves strict inequalities

Answer: B — Dividing both sides by −4 (a negative number) reverses the inequality sign: −4x > 12 → x < −3.

Q5. If 5x − 2 ≤ 3x + 6, then the solution is:

  • A. x ≥ 4
  • B. x ≤ 4 ✓
  • C. x < 4
  • D. x > −4

Answer: B — 5x − 2 ≤ 3x + 6 → 2x ≤ 8 → x ≤ 4 (dividing by positive 2 keeps the sign).

Q6. Which statement is NOT correct about solving inequalities?

  • A. The same number can be subtracted from both sides
  • B. Both sides can be divided by any non-zero number without changing the sign ✓
  • C. Multiplying by a negative number reverses the inequality sign
  • D. Adding the same number to both sides preserves the inequality sign

Answer: B — Option B is incorrect because dividing by a negative number reverses the inequality sign, contradicting 'without changing the sign'.

Q7. Consider the statements: (I) 3 < x means x > −3 only. (II) The inequality x ≤ 5 includes x = 5. Which is true?

  • A. Both I and II are true
  • B. Only I is true
  • C. Only II is true ✓
  • D. Neither I nor II is true

Answer: C — Statement I is false because 3 < x means x > 3, not x > −3; Statement II is true because ≤ includes the boundary value.

Q8. Solve (x − 2)/3 ≤ (x + 4)/5 and determine the solution set.

  • A. x ≥ 11
  • B. x ≤ 11 ✓
  • C. x < 11
  • D. x > −11

Answer: B — Multiply by 15: 5(x − 2) ≤ 3(x + 4) → 5x − 10 ≤ 3x + 12 → 2x ≤ 22 → x ≤ 11.

Q9. A linear inequality in two variables has the form:

  • A. ax² + bx + c < 0
  • B. ax + by < c where a ≠ 0 and b ≠ 0 ✓
  • C. ax + b > 0 where a ≠ 0
  • D. ax < by where a = b

Answer: B — Linear inequalities in two variables have the form ax + by with two distinct variables and both coefficients non-zero.

Q10. If Ravi solves 30x < 200 and x must be a natural number, what is the solution set?

  • A. {0, 1, 2, 3, 4, 5, 6}
  • B. {1, 2, 3, 4, 5, 6} ✓
  • C. x < 20/3 as real numbers
  • D. {2, 3, 4, 5, 6, 7}

Answer: B — 30x < 200 → x < 20/3 ≈ 6.67; natural numbers satisfying this are {1, 2, 3, 4, 5, 6} (natural numbers start at 1, not 0).

Flashcards

What is the definition of an inequality?

Two real numbers or algebraic expressions related by <, >, ≤, or ≥ symbols form an inequality.

Name the two types of inequalities based on the symbol used.

Strict inequalities use < or >, while slack inequalities use ≤ or ≥.

State Rule 1 for solving inequalities.

Equal numbers may be added to or subtracted from both sides of an inequality without changing the inequality sign.

State Rule 2 for solving inequalities.

Both sides can be multiplied or divided by the same positive number; if multiplied/divided by a negative number, the inequality sign reverses.

What is a linear inequality in one variable?

An inequality of the form ax + b < c, ax + b > c, ax + b ≤ c, or ax + b ≥ c where a ≠ 0.

What is a linear inequality in two variables?

An inequality of the form ax + by < c, ax + by > c, ax + by ≤ c, or ax + by ≥ c where a ≠ 0 and b ≠ 0.

What happens to the inequality sign when you multiply both sides by a negative number?

The inequality sign reverses: < becomes >, ≤ becomes ≥, and vice versa.

What is the solution set of x < 2 in interval notation?

The solution set is (−∞, 2), representing all real numbers less than 2.

Solve 5x − 3 < 3x + 1 and express the solution in interval notation.

Simplifying gives 2x < 4, so x < 2, with solution set (−∞, 2).

What is a double inequality and give one example.

A double inequality relates three quantities using two inequality symbols; for example, 3 < x < 5 means x is greater than 3 and less than 5.

Important Board Questions

Define a linear inequality in one variable. Give two examples and classify them as strict or slack inequalities. [2 marks]

A linear inequality in one variable has form ax + b < c (or >, ≤, ≥) where a ≠ 0. Strict uses < or >, slack uses ≤ or ≥. Provide specific examples like 2x + 3 < 7 and 5x − 1 ≥ 4.

Solve the inequality (3x − 2)/4 ≥ (2x + 3)/3 for real x and express the solution in interval notation. Show all working steps. [5 marks]

Clear the fractions by multiplying by LCM (12), simplify the linear inequality step-by-step using Rules 1 and 2, then convert to interval notation like [a, ∞) or (−∞, b].

Reshma has ₹120 and wants to buy registers costing ₹40 each and pens costing ₹20 each. (a) Form a linear inequality in two variables representing her spending. (b) Explain whether she can buy 2 registers and 2 pens. (c) Find one possible combination of registers and pens she can buy within her budget. Show complete working. [6 marks]

Set up 40x + 20y ≤ 120 where x is registers and y is pens. Check the constraint by substituting values, then verify at least one solution pair satisfies the inequality. Use Rule 1 to simplify if needed.

Assertion-Reason Questions for Linear Inequalities

Q1. Assertion (A):

The inequality 3x - 5 < 7 has solution set x < 4 when x is a real number.

Reason (R):

When we add 5 to both sides and then divide by 3 (a positive number), the inequality sign remains unchanged.

Answer: A — Both A and R are true, and R is the correct explanation of A — The assertion is true: 3x - 5 < 7 gives 3x < 12, so x < 4. The reason is also true and correctly explains WHY: adding to both sides doesn't change the sign (Rule 1), and dividing by positive 3 doesn't change the sign (Rule 2). The reason is the correct explanation of the assertion.

Q2. Assertion (A):

If -2x ≥ 8, then x ≤ -4.

Reason (R):

Multiplying or dividing both sides of an inequality by a negative number reverses the inequality sign.

Answer: A — Both A and R are true, and R is the correct explanation of A — The assertion is true: -2x ≥ 8, divide by -2 (negative), sign reverses to x ≤ -4. The reason is true and correctly explains why: this is exactly Rule 2 for negative divisors. Both statements are true and the reason is the correct explanation.

Q3. Assertion (A):

Double inequalities like 3 < x < 7 mean x is greater than 3 AND less than 7 simultaneously.

Reason (R):

The inequality 3 < x < 7 can be written as two separate inequalities 3 < x OR x < 7.

Answer: C — A is true but R is false — The assertion is true: 3 < x < 7 means both 3 < x AND x < 7 must be satisfied at the same time (conjunction). The reason is false: it uses OR when it should use AND. In logic, a double inequality is a single statement requiring BOTH conditions simultaneously, not either/or. The assertion is correct, but the reason gives wrong logic.

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Case-Based Questions (CBQ) for Linear Inequalities

Case Study:

A school cafeteria purchases milk packets for breakfast. Each milk packet costs ₹15. The school has a budget of ₹450 for milk procurement. Additionally, the school policy states that if the total expenditure exceeds ₹300, they receive a 10% discount on the purchase. Let x denote the number of milk packets purchased.

Q1. Which inequality represents the number of packets that can be bought within the budget of ₹450?

  • A. 15x ≤ 450 ✓
  • B. 15x < 450
  • C. 15x ≥ 450
  • D. 15x > 450

Answer: A

Q2. Solve the inequality to find the maximum number of milk packets that can be purchased.

  • A. x ≤ 30 ✓
  • B. x < 30
  • C. x ≥ 30
  • D. x > 30

Answer: A

Q3. For the 10% discount to apply, the school must spend more than ₹300. How many packets must be bought?

  • A. x ≥ 20
  • B. x > 20 ✓
  • C. x ≤ 20
  • D. x < 20

Answer: B

Q4. What is the range of packets that allows both budget compliance AND discount eligibility?

  • A. x ≤ 30
  • B. 20 ≤ x < 30
  • C. x > 20
  • D. 20 < x ≤ 30 ✓

Answer: D

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