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5.1 Introduction

🎓 Class 11 Mathematics CBSE Theory Ch 5 — Linear Inequalities ⏱ ~18 min
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Targeting Class 11 level in Algebra, with Advanced difficulty.

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5.1 Introduction

In earlier classes, we have studied equations in one variable and two variables and also solved some statement problems by translating them into equations. Now, a natural question arises: in real life, do we always translate a given real-world situation into an equation? Consider the example — Ravi goes to a market to buy a notebook costing at most ₹200?. If the price of a notebook is denoted by \(x\), then the statement "price is less than or equal to 200" is captured by \(x \leq 200\). This is not an equation; it is an inequality.

Statements involving symbols \(<,>,\leq,\geq\) are called inequalities. In this chapter we learn the algebraic and graphical solutions of linear inequalities in one and two variables, along with applications from the fields of science, mathematics, statistics, economics and psychology.

5.2 Inequalities

Let us consider the following situations.

(i) Ravi goes to market with ₹200 to buy rice which is available in packets of 1 kg. The price of one packet of rice is ₹30. If \(x\) denotes the number of packets, then the total amount spent is \(30x\). Since Ravi cannot spend more than ₹200, we need

\(30x < 200\)   ...(1)

Clearly the statement (i) is not an equation as it does not involve the sign of equality.

(ii) Reshma has ₹120 and wants to buy some registers and pens. The cost of one register is ₹40 and that of a pen is ₹20. In this case, if \(x\) denotes the number of registers and \(y\) the number of pens which Reshma buys, then

\(40x + 20y \leq 120\)   ...(2)

Since the total amount she spends must be at most ₹120. Statement (1) is an inequality in one variable \(x\) while statement (2) is an inequality in two variables \(x\) and \(y\).

Definition 1
Two real numbers or two algebraic expressions related by the symbols \(<,>,\leq\) or \(\geq\) form an inequality.

Statements such as (1), (2), (3) and (4) are examples of numerical inequalities and literal inequalities:

  • \(3 < 5\), \(7 > 5\) — numerical inequalities
  • \(x < 5\), \(y > 2\), \(x \geq 3\), \(y \leq 4\) — literal inequalities
  • \(3 < 5 < 7\) (read: 5 is greater than 3 and less than 7), \(3 < x < 5\), \(2 < y < 4\) — double inequalities

Some more examples:

Classification of inequalities
\(ax + b < 0\) ...(5), \(ax + b > 0\) ...(6), \(ax + b \leq 0\) ...(7), \(ax + b \geq 0\) ...(8) — linear inequalities in one variable \(x\) when \(a \neq 0\).
\(ax + by < c\) ...(9), \(ax + by > c\) ...(10), \(ax + by \leq c\) ...(11), \(ax + by \geq c\) ...(12) — linear inequalities in two variables \(x\) and \(y\) when \(a \neq 0, b \neq 0\).
\(ax^2 + bx + c \leq 0\) ...(13), \(ax^2 + bx + c > 0\) ...(14) — quadratic inequalities when \(a \neq 0\).

Inequalities (5), (6), (9), (10) and (14) are strict inequalities while (7), (8), (11), (12) and (13) are called slack inequalities. Inequalities from (5) to (8) are inequalities in one variable \(x\) while inequalities from (9) to (12) are in two variables \(x\) and \(y\). In this chapter, we shall confine ourselves to the study of linear inequalities in one and two variables only.

5.3 Algebraic Solutions of Linear Inequalities in One Variable & Graphical Representation

Let us consider the inequality \(30x < 200\). Note that here \(x\) denotes the number of packets of rice.

Obviously, \(x\) cannot be a negative integer or a fraction. Left hand side (L.H.S.) of this inequality is \(30x\) and right hand side (RHS) is 200. Therefore, we have

  • For \(x=0\), L.H.S. \(= 30(0)=0 < 200\) (R.H.S.), which is true.
  • For \(x=1\), L.H.S. \(=30(1)=30 < 200\) (R.H.S.), which is true.
  • For \(x=2\), L.H.S. \(=30(2)=60 < 200\), true.
  • For \(x=3\), \(90 < 200\), true. For \(x=4\), \(120 < 200\), true. For \(x=5\), \(150 < 200\), true. For \(x=6\), \(180 < 200\), true.
  • For \(x=7\), L.H.S. \(=30(7)=210 < 200\) is false.

In the above situation, we find that the values of \(x\), which makes the above inequality a true statement, are \(0,1,2,3,4,5,6\). These values are called solutions of the inequality and the set \(\{0,1,2,3,4,5,6\}\) is called its solution set.

Solution of an inequality
The value(s) of the variable(s) which make an inequality a true statement, are called its solutions. The set of all solutions is called the solution set of the inequality.

For example, \(-1\) is a solution of inequality \(x+10>0\) because, when \(x=-1\), \(-1+10=9>0\), which is true. Note that all real numbers greater than \(-10\) are the solutions.

We have found solutions of the inequality \(30x < 200\) by trial and error method, which is not always convenient. Obviously, this method is time consuming and sometimes not feasible. We must have some better or systematic techniques for solving inequalities. Before that, we should go through some more properties of numerical inequalities and follow them as rules.

Rules for solving inequalities
Rule 1: Equal numbers may be added to (or subtracted from) both sides of an inequality without affecting the sign of inequality.
Rule 2: Both sides of an inequality can be multiplied (or divided) by the same positive number. But when both sides are multiplied or divided by a negative number, then the sign of inequality is reversed.

Worked Examples

Example 1. Solve \(30x < 200\) when (i) \(x\) is a natural number, (ii) \(x\) is an integer.
Dividing both sides by 30 (a positive number): \(x < \frac{200}{30}\) i.e. \(x < \frac{20}{3}\).
(i) When \(x\) is a natural number, the solutions are \(\{1,2,3,4,5,6\}\).
(ii) When \(x\) is an integer, solutions are \(\{..., -3,-2,-1,0,1,2,3,4,5,6\}\).
Example 2. Solve \(5x-3 < 3x+1\) when (i) \(x\) is an integer, (ii) \(x\) is a real number.
\(5x - 3 < 3x+1 \Rightarrow 5x-3x < 1+3 \Rightarrow 2x < 4 \Rightarrow x < 2\).
(i) Integer solutions: \(\{...,-2,-1,0,1\}\).
(ii) Real solutions: all real numbers \(x\) such that \(x < 2\); solution set \((-\infty, 2)\).
Example 3. Solve \(4x+3 < 6x+7\).
\(4x+3 < 6x+7 \Rightarrow 4x - 6x < 7 - 3 \Rightarrow -2x < 4 \Rightarrow x > -2\) (dividing by \(-2\) reverses sign). Solution set: \((-2, \infty)\).
Example 4. Solve \(\dfrac{5-2x}{3} \leq \dfrac{x}{6} - 5\).
Multiply both sides by 6: \(2(5-2x) \leq x - 30 \Rightarrow 10 - 4x \leq x - 30 \Rightarrow -5x \leq -40 \Rightarrow x \geq 8\). Solution set: \([8, \infty)\).

Graphical Solution on the Number Line

Example 5. Solve \(7x+3 < 5x+9\). Show the graph of the solutions on number line.
\(7x-5x < 9-3 \Rightarrow 2x < 6 \Rightarrow x < 3\). Graph: open circle at 3, shaded line to the left (see Fig 5.1).
-4 -3 -2 -1 0 1 2 3 4 5 6
Fig 5.1 — Graphical solution of \(7x+3 < 5x+9\): \(x<3\), i.e. \((-\infty,3)\). Open circle at 3 indicates 3 is not included.
Example 6. Solve \(\dfrac{3x-4}{2} \geq \dfrac{x+1}{4} - 1\). Show the graph of the solutions on number line.
Multiply by 4: \(2(3x-4) \geq (x+1)-4 \Rightarrow 6x - 8 \geq x - 3 \Rightarrow 5x \geq 5 \Rightarrow x \geq 1\). Graph: closed dot at 1, shaded line to the right (Fig 5.2).
-4 -3 -2 -1 0 1 2 3 4 5 6
Fig 5.2 — Graph of \(x \geq 1\) i.e. \([1,\infty)\). Closed dot at 1 means 1 is included.
Example 7. The marks obtained by a student of Class XI in first and second terminal examinations are 62 and 48, respectively. Find the minimum marks he should get in the annual examination to have an average of at least 60 marks.
Let \(x\) be the marks in the annual examination. Then \(\frac{62+48+x}{3} \geq 60 \Rightarrow 110 + x \geq 180 \Rightarrow x \geq 70\). He must obtain a minimum of 70 marks.
Example 8. Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 40.
Let \(x\) be the smaller of two consecutive odd natural numbers; the other is \(x+2\). Then \(x > 10\) ...(1) and \(x+(x+2) < 40 \Rightarrow 2x < 38 \Rightarrow x < 19\) ...(2). Combining: \(10 < x < 19\). Since \(x\) is odd: \(x \in \{11,13,15,17\}\). Pairs: (11,13), (13,15), (15,17), (17,19).
Activity 5.1 — Feel the Sign-Flip Rule
Predict: If we multiply the true inequality \(3 < 5\) by \(-2\), will the new statement still be true without flipping the inequality?
  1. Write the inequality \(3 < 5\). Clearly it is true.
  2. Multiply both sides by a positive number, say 4: \(12 < 20\). Still true.
  3. Now multiply both sides of \(3 < 5\) by \(-2\) keeping the sign: \(-6 < -10\). Check on the number line — is \(-6\) really less than \(-10\)?
  4. Reverse the inequality: \(-6 > -10\). True!
  5. Repeat with \(-2 < 7\) multiplied by \(-3\).
Observation: Multiplying by a negative number mirrors numbers about 0 on the number line — so the order (left/right) flips. This is why Rule 2 states the inequality sign must be reversed when multiplying or dividing both sides by a negative number.

Figure it Out — Exercise 5.1 (Selected)

Q1. Solve \(24x < 100\) when (i) \(x\) is a natural number, (ii) \(x\) is an integer.
\(x < \frac{100}{24} = \frac{25}{6} \approx 4.17\). (i) \(\{1,2,3,4\}\). (ii) \(\{...,-2,-1,0,1,2,3,4\}\).
Q2. Solve \(-12x > 30\) when (i) \(x\) is a natural number, (ii) \(x\) is an integer.
Divide by \(-12\) (flip): \(x < -\frac{30}{12} = -\frac{5}{2}\). (i) No natural-number solutions. (ii) \(\{...,-5,-4,-3\}\).
Q3. Solve \(5x - 3 < 7\) when (i) \(x\) is an integer, (ii) \(x\) is a real number.
\(5x < 10 \Rightarrow x < 2\). (i) \(\{...,-1,0,1\}\). (ii) \((-\infty, 2)\).
Q4. Solve \(3x + 8 > 2\) when (i) \(x\) is an integer, (ii) \(x\) is a real number.
\(3x > -6 \Rightarrow x > -2\). (i) \(\{-1,0,1,2,...\}\). (ii) \((-2, \infty)\).
Q5–Q10. Solve: (5) \(4x+3 < 5x+7\); (6) \(3x - 7 > 5x - 1\); (7) \(3(x-1) \leq 2(x-3)\); (8) \(3(2-x) \geq 2(1-x)\); (9) \(x + \frac{x}{2} + \frac{x}{3} < 11\); (10) \(\frac{x}{3} > \frac{x}{2} + 1\).
(5) \(-x < 4 \Rightarrow x > -4\), i.e. \((-4,\infty)\). (6) \(-2x > 6 \Rightarrow x < -3\), \((-\infty,-3)\). (7) \(3x - 3 \leq 2x - 6 \Rightarrow x \leq -3\), \((-\infty,-3]\). (8) \(6 - 3x \geq 2 - 2x \Rightarrow -x \geq -4 \Rightarrow x \leq 4\), \((-\infty,4]\). (9) \(\frac{11x}{6} < 11 \Rightarrow x < 6\), \((-\infty,6)\). (10) \(\frac{x}{3} - \frac{x}{2} > 1 \Rightarrow -\frac{x}{6} > 1 \Rightarrow x < -6\), \((-\infty,-6)\).
Competency-Based Questions — Inequalities in Real Life
Anika's school is organising a stationery drive. She has a budget of ₹500 to buy notebooks that cost ₹45 each and pens that cost ₹20 each. Let \(x\) = number of notebooks and \(y\) = number of pens she buys.
Q1. Which inequality best models Anika's budget constraint?
L3 Apply
  • A) \(45x + 20y = 500\)
  • B) \(45x + 20y \leq 500\)
  • C) \(45x + 20y \geq 500\)
  • D) \(45x - 20y < 500\)
Answer: B. "At most ₹500" translates to \(\leq\). It is a slack linear inequality in two variables.
Q2. If she buys 6 notebooks, what is the maximum whole number of pens she can buy?
L3 Apply
\(45(6) + 20y \leq 500 \Rightarrow 270 + 20y \leq 500 \Rightarrow y \leq \frac{230}{20} = 11.5\). So maximum 11 pens.
Q3. In first two papers a student scored 70 and 75 marks. What minimum score in the third paper ensures an average greater than or equal to 80?
L4 Analyse
\(\frac{70+75+x}{3} \geq 80 \Rightarrow 145 + x \geq 240 \Rightarrow x \geq 95\). Minimum = 95 marks.
Q4. Design: Reshma wants a set of two consecutive even natural numbers, both greater than 5, whose sum is less than 23. Find all such pairs and justify why no other pair works.
L6 Create
Let \(x\) be smaller even number. \(x > 5\) and \(x + (x+2) < 23 \Rightarrow 2x < 21 \Rightarrow x < 10.5\). Even \(x\) in \((5,10.5]\): \(x = 6,8,10\). Pairs: (6,8), (8,10), (10,12).
Assertion–Reason Questions
Assertion (A): The solution of \(-3x > 9\) is \(x < -3\).
Reason (R): When both sides of an inequality are divided by a negative number, the inequality sign is reversed.
A) Both A and R true; R correct explanation of A
B) Both true; R not correct explanation
C) A true, R false
D) A false, R true
Answer: A. Dividing by \(-3\) flips the sign: \(x < -3\). Rule 2 directly justifies it.
Assertion (A): An open circle on the number line indicates that the endpoint is a solution of the inequality.
Reason (R): Strict inequalities \(<\) and \(>\) exclude the boundary value.
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: D. An open circle indicates the endpoint is not a solution (so A is false). R is true — it explains the use of open circles.
Assertion (A): The inequality \(0 \cdot x + 5 > 0\) is a linear inequality.
Reason (R): Every inequality of the form \(ax + b > 0\) is linear provided \(a \neq 0\).
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: D. A is false because the coefficient of \(x\) is zero, violating the linearity requirement. R correctly states the definition.

Frequently Asked Questions — Linear Inequalities

What is Part 1 — Linear Inequalities: Introduction & Algebraic Solutions | Class 11 Maths | MyAiSchool in NCERT Class 11 Mathematics?

Part 1 — Linear Inequalities: Introduction & Algebraic Solutions | Class 11 Maths | MyAiSchool is a key concept covered in NCERT Class 11 Mathematics, Chapter 5: Linear Inequalities. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Part 1 — Linear Inequalities: Introduction & Algebraic Solutions | Class 11 Maths | MyAiSchool step by step?

To solve problems on Part 1 — Linear Inequalities: Introduction & Algebraic Solutions | Class 11 Maths | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 11 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 5: Linear Inequalities?

The essential formulas of Chapter 5 (Linear Inequalities) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Part 1 — Linear Inequalities: Introduction & Algebraic Solutions | Class 11 Maths | MyAiSchool important for the Class 11 board exam?

Part 1 — Linear Inequalities: Introduction & Algebraic Solutions | Class 11 Maths | MyAiSchool is part of the NCERT Class 11 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Part 1 — Linear Inequalities: Introduction & Algebraic Solutions | Class 11 Maths | MyAiSchool?

Common mistakes in Part 1 — Linear Inequalities: Introduction & Algebraic Solutions | Class 11 Maths | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Part 1 — Linear Inequalities: Introduction & Algebraic Solutions | Class 11 Maths | MyAiSchool?

End-of-chapter NCERT exercises for Part 1 — Linear Inequalities: Introduction & Algebraic Solutions | Class 11 Maths | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 5, and solve at least one previous-year board paper to consolidate your understanding.

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