Chapter 5 of 14

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Telangana (TSBIE)Class 11 Mathematics★ Unit II · part of the 25-mark Algebra block shared with Complex Numbers and Quadratic Equations, Permutations and Combinations, Binomial Theorem, and Sequences and Series — CBSE states no chapter-wise split

Linear Inequalities

Not every real relationship is an equality — Ravi's ₹200 rice budget and Reshma's stationery shopping both translate into inequalities, not equations. This chapter builds the two rules for solving them, chief among them the sign-flip rule. The entire two-variable graphical half-plane method most coaching material drills for this chapter — plotting ax+by+c>0 as a shaded region — is graded formatively only and does not appear in the current book at all. CBSE Class 11 Mathematics Chapter 5.

⏱ 8 min readEasy difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Distinguish an inequality from an equation, and classify inequalities as numerical, literal, strict, or slack
2Apply the two solving rules correctly, especially reversing the sign when multiplying or dividing by a negative number
3Represent a solution set using set-builder notation, interval notation, and a number line
4Translate a real-world constraint into a linear inequality and solve it
5Solve a compound inequality as a single chain, and a system of two separate inequalities by intersecting their individual solutions

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Why this chapter matters

This chapter builds the two rules for solving inequalities in one variable, chief among them the sign-flip rule when multiplying or dividing by a negative number. It also trains a specific translation skill — turning a real-world constraint (a budget, a temperature range, an acid mixture) into an inequality — that recurs throughout word-problem-heavy exam sections.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Solving linear equations in one variable, since the same two rules apply here with one added exception

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Basic set notation and interval notation, used throughout to write solution sets

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Plotting points and reading a number line, needed for the graphical representation of a solution

Linear Inequalities

1. Check this before you revise anything

The single biggest chunk of "standard" content for this chapter is not in the current book.

Topic	NCERT 2026-27 chapter	CBSE 2026-27
Linear inequalities in two variables — plotting $a x + b y + c > 0$ as a half-plane, the test-point/shading method, systems of two-variable inequalities and feasible regions	Absent — the chapter has exactly one exercise (5.1) plus a Miscellaneous Exercise, both purely one-variable	Formative-only — CBSE's own note excludes this from summative assessment
Linear inequalities in one variable — the two solving rules, number-line and interval representation, real-world word problems	Present in full, with 13 worked examples	Listed and summatively assessed

This is a bigger removal than usual. Graphing a half-plane and shading the feasible region is the traditional lead-in to Linear Programming (Class 12), and many coaching sheets still treat it as this chapter's centrepiece. The current NCERT text confines itself entirely to inequalities in one variable — solved algebraically and shown on a number line, never a coordinate plane.

Section 5.2 does mention two-variable inequalities exist, purely to classify them ( $a x + b y + c > 0$ is linear in two variables, $a x^{2} + b x + c > 0$ is quadratic) — but the chapter never returns to teach solving or graphing them.

2. What this chapter covers

Textbook section	Topic
5.2	What an inequality is; numerical, literal, and double inequalities; classifying by variable count and degree
5.3	Algebraic solutions of linear inequalities in one variable, and their number-line representation

3. What an inequality actually is

Two real numbers or algebraic expressions joined by $<$ , $>$ , $\leq$ , or $\geq$ form an inequality. $3 < 5$ is a numerical inequality; $x < 5$ is a literal one; $3 < x < 5$ is a double (compound) inequality, read " $x$ is greater than 3 and less than 5."

Worked, mirroring the textbook's own motivating example. Ravi has ₹200 to spend on rice at ₹30 per packet. If $x$ is the number of packets, he spends $30 x$ rupees — and since he can't spend more than he has, $30 x < 200$ , not $30 x = 200$ . This is genuinely not an equation: there's no reason the amount spent must hit ₹200 exactly.

Inequalities are classified by variable count and degree, though this chapter only solves the first kind:

Form	Variables	Degree	Example
$a x + b ≶ 0$	One	Linear	$2 x - 3 > 0$
$a x + b y ≶ c$	Two	Linear	$2 x + 3 y \leq 12$
$a x^{2} + b x + c ≶ 0$	One	Quadratic	$x^{2} - 4 > 0$

$<$ and $>$ are strict; $\leq$ and $\geq$ are slack (non-strict).

4. The two rules for solving inequalities

Solving an inequality means finding every value of the variable that makes it a true statement — its solution set. Two rules do all the work, and they match the rules for equations with exactly one difference:

Rule	Statement
Rule 1	Adding or subtracting the same number from both sides never changes the inequality sign
Rule 2	Multiplying or dividing both sides by the same positive number never changes the sign — but by a negative number, the sign reverses

Rule 2's second half is the entire reason this chapter exists as a separate topic from equation-solving, and it's easy to see why it must be true: $3 > 2$ , but multiplying both sides by $- 1$ gives $- 3 < - 2$ — the inequality has to flip, or the resulting statement would simply be false.

Worked, mirroring the textbook's own Example 2. Solve $5 x - 3 < 3 x + 1$ . Adding 3 to both sides: $5 x < 3 x + 4$ . Subtracting $3 x$ : $2 x < 4$ , so $x < 2$ — no sign flip needed, since both operations used addition/subtraction or division by a positive number.

Worked, mirroring the textbook's own Example 4. Solve $\frac{5 - 2 x}{3} \leq \frac{x}{6} - 5$ . Clearing denominators: $2 (5 - 2 x) \leq x - 30$ , so $10 - 4 x \leq x - 30$ , giving $- 5 x \leq - 40$ . Dividing by $- 5$ flips the sign: $x \geq 8$ .

5. Representing the solution

Form	Example	Meaning
Set-builder	${x : x < 3, x \in R}$	Explicit condition
Interval	$(- \infty, 3)$	Round bracket = endpoint excluded
Number line	Open circle at 3, shaded left	Visual — open circle for strict, filled circle for $\leq / \geq$

Worked, mirroring the textbook's own Example 5. Solve $7 x + 3 < 5 x + 9$ and graph it. Subtracting $5 x$ and 3: $2 x < 6$ , so $x < 3$ — an open circle at 3 with the number line shaded to the left, since 3 itself doesn't satisfy the strict inequality.

6. Turning word problems into inequalities

Worked, mirroring the textbook's own Example 7. A student scored 62 and 48 in the first two exams. What's the minimum third score for an average of at least 60? Let $x$ be the third score: $\frac{62 + 48 + x}{3} \geq 60$ , so $110 + x \geq 180$ , giving $x \geq 70$ — a minimum of 70, not an exact target.

Worked, mirroring the textbook's own Example 8. Find pairs of consecutive odd natural numbers, both greater than 10, summing to less than 40. With $x$ the smaller number: $x > 10$ and $x + (x + 2) < 40$ , so $x < 19$ . Combining: $10 < x < 19$ , and since $x$ must be odd, $x \in {11, 13, 15, 17}$ , giving the pairs $(11, 13), (13, 15), (15, 17), (17, 19)$ .

This pattern — one inequality from a lower bound, one from an upper bound, then intersect — is the standard shape for every word problem in this chapter's exercises.

7. Compound inequalities, and mixture/range problems

A double inequality like $- 8 \leq 5 x - 3 < 7$ is solved as one single chain, performing the same operation on all three parts at once — not split into two separate inequalities and intersected afterward.

Worked, mirroring the textbook's own Example 9. Solve $- 8 \leq 5 x - 3 < 7$ . Add 3 across all three parts: $- 5 \leq 5 x < 10$ . Divide by 5: $- 1 \leq x < 2$ .

A system of two separately-written inequalities is a different shape from a single chain, and needs a different approach: solve each one on its own, then intersect. Worked, mirroring the textbook's own Example 11: solve $3 x - 7 < 5 + x$ and $11 - 5 x \leq 1$ together. The first gives $x < 6$ ; the second gives $x \geq 2$ . The values satisfying both — the intersection — are $2 \leq x < 6$ .

Worked, mirroring the textbook's own Example 12. A solution must be kept between $30°$ and $35°$ Celsius; find the Fahrenheit range, given $C = \frac{5}{9} (F - 32)$ . From $30 < C < 35$ : $30 < \frac{5}{9} (F - 32) < 35$ . Multiply through by $\frac{9}{5}$ : $54 < F - 32 < 63$ , so $86 < F < 95$ .

Worked, mirroring the textbook's own Example 13. A manufacturer has 600 L of 12% acid solution. How many litres of 30% solution, $x$ , must be added so the mixture is between 15% and 18% acid? The acid content gives two inequalities at once: $30 x + 7200 > 15 (x + 600)$ and $30 x + 7200 < 18 (x + 600)$ . Solving each: $x > 120$ and $x < 300$ , so $120 < x < 300$ litres.

Summary

An inequality relates two expressions with $<, >, \leq, \geq$ ; solving one means finding its full solution set, not a single value.
Rule 1: adding/subtracting the same number never flips the sign. Rule 2: multiplying/dividing by a positive number never flips it, but a negative number always does.
Solutions are written in set-builder form, interval notation, or on a number line — open circle for strict ( $<, >$ ), filled circle for slack ( $\leq, \geq$ ).
Word problems typically produce two inequalities — one lower bound, one upper bound — solved and then intersected, exactly like the mixture and temperature-range examples.
A compound inequality such as $a \leq f (x) < b$ is solved as one chain, applying the same operation to all three parts simultaneously.
Linear inequalities in two variables — the half-plane, test-point, and shaded-region method most commonly associated with this chapter's name — are formative-only under the 2026-27 CBSE syllabus and do not appear anywhere in the current NCERT text.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Rule 1 — addition and subtraction

Adding or subtracting the same number from both sides never changes the inequality sign

Identical to the corresponding rule for equations

Rule 2 — multiplication and division

Multiplying/dividing both sides by a positive number: sign unchanged. By a negative number: sign reverses

The one genuinely new rule in this chapter, and its most common source of errors

Interval notation

x > a means (a, infinity); x >= a means [a, infinity); x < b means (-infinity, b); a < x < b means (a,b); a <= x <= b means [a,b]

Round bracket = endpoint excluded (strict); square bracket = endpoint included (slack)

Compound inequality (single chain)

For a <= f(x) < b, apply the same operation to all three parts at once

Never split a single chained inequality into two and solve separately — perform each step across all three parts together

System of two inequalities

Solve each inequality separately, then take the intersection of both solution sets

Different from a compound chain — used when two inequalities are given as separate statements, as in the textbook's Example 11

Word-problem pattern

A lower-bound condition and an upper-bound condition, solved separately then intersected

The standard shape for mixture, average-marks, and range word problems in this chapter

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Forgetting to flip the inequality sign when multiplying or dividing by a negative number

✓ -5x <= -40 divided by -5 gives x >= 8, not x <= 8. This single rule accounts for the large majority of errors in this chapter.

WATCH OUT

✗ Expecting the two-variable graphical method (plotting ax+by+c>0 as a shaded half-plane, testing a point) as this chapter's main skill

✓ It is graded formatively only under the 2026-27 CBSE syllabus and does not appear anywhere in the current textbook — the chapter's actual content is entirely one-variable, solved algebraically and shown on a number line.

WATCH OUT

✗ Splitting a single chained inequality like -8 <= 5x-3 < 7 into two inequalities and solving them independently

✓ Apply each operation to all three parts of the chain simultaneously: add 3 to get -5 <= 5x < 10, then divide by 5 to get -1 <= x < 2 — one continuous chain throughout.

WATCH OUT

✗ Treating a 'system of two separate inequalities' (e.g. 3x-7<5+x and 11-5x<=1, given as two statements) the same way as a single compound chain

✓ A system needs each inequality solved on its own first, then the two solution sets intersected — there is no single three-part chain to manipulate at once.

WATCH OUT

✗ Using an open circle on a number line for <= or >=

✓ An open circle means the endpoint is excluded (strict inequality, < or >). A filled/dark circle means the endpoint is included (<= or >=).

WATCH OUT

✗ Stopping at an algebraic inequality without translating back into the word problem's units or context

✓ A result like x >= 70 in an 'average marks' problem must be reported as 'a minimum of 70 marks', not left as a bare inequality — exam marks are awarded for the interpretation, not just the algebra.

WATCH OUT

✗ In a mixture or range word problem, writing only one of the two required inequalities (just the lower bound, or just the upper bound)

✓ Phrases like 'more than 15% but less than 18%' always produce two separate inequalities that must both be solved and then intersected — one alone gives an incomplete answer.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 5.1

Exercise 5.1

Solving one-variable linear inequalities algebraically (Q1-16), with number-line graphs (Q17-20), and real-world word problems on averages, consecutive integers, and geometric constraints (Q21-26)

Questions

Miscellaneous Ex

Miscellaneous Exercise

Compound inequalities and systems of two inequalities (Q1-10), plus word problems on temperature conversion, acid-solution mixtures, and the IQ formula (Q11-14)

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Linear Inequalities?

9 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

9 questions~6 min worth ~25 marks in Telangana (TSBIE) exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Adding/subtracting the same number: inequality sign unchanged
•Multiplying/dividing by a positive number: sign unchanged. By a negative number: sign reverses — the one rule this whole chapter is built around
•Open circle on a number line = strict inequality excluded; filled circle = non-strict inequality included
•Interval (a,b) = both ends excluded; [a,b] = both ends included; mixed brackets match mixed strictness
•A compound chain a <= f(x) < b is solved by applying each step to all three parts at once
•A system of two separately-stated inequalities is solved by finding each solution set separately, then intersecting them
•Word problems (averages, mixtures, temperature ranges) typically produce two inequalities — a lower and an upper bound — solved together
•Linear inequalities in two variables (half-plane graphing, test-point shading, feasible regions) are formative-only under CBSE 2026-27 and absent from this chapter's own exercises

Telangana (TSBIE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Part of Unit II's 25-mark Algebra block (no chapter-wise split, per CBSE)

Question type	Marks each	Typical count	What it tests
One variable inequalities and the sign flip rule	2-3	1	Solving a linear inequality algebraically, representing the solution in interval notation or on a number line
Compound inequalities and systems of inequalities	3-4	1	Solving a chained double inequality, or a system of two separately-stated inequalities by intersection
Word problem translation and range	4-5	0-1	Translating a real-world constraint (average, mixture, temperature range) into one or two inequalities and solving

Prep strategy

Drill the sign-flip rule until it is automatic — it is the single most tested rule in this entire chapter
For word problems, identify whether the situation gives one bound (minimum/maximum) or two (a range) before writing the inequality
Distinguish a compound chain (solve all three parts together) from a system of two inequalities (solve separately, then intersect) before starting

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Budget constraints

Any 'you have at most X to spend' situation, like Ravi's rice budget or Reshma's stationery shopping in this chapter's own opening examples, is a direct linear inequality rather than an equation.

Acceptable ranges in science and manufacturing

Keeping a chemical solution's temperature or concentration within a safe band (this chapter's own acid-mixture and temperature-conversion examples) is exactly a compound inequality — one lower bound, one upper bound, solved together.

Grading and eligibility thresholds

Minimum-average requirements ('score at least 60 overall') and grade cutoffs translate directly into a one-variable inequality, exactly as in this chapter's own average-marks word problems.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Write the inequality sign explicitly at every intermediate step — dropping it and just tracking numbers is how sign-flip errors go unnoticed

For word problems, translate each sentence into its own inequality first, then combine — do not try to write one inequality straight from the full paragraph

For a compound chain, perform every operation identically across all three parts in the same line of working

State which number system (natural, integer, or real) the final solution set belongs to — the same algebra gives different final answers depending on this

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

AM-GM inequality: for positive reals a, b, the arithmetic mean (a+b)/2 is always at least the geometric mean sqrt(ab) — the foundational inequality behind most olympiad inequality problems

STRETCH

Solving a quadratic inequality ax^2+bx+c>0 using the roots and a sign chart is a natural next step beyond this chapter's purely linear scope, and a common JEE Main question type

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainSolving a fractional linear inequalitySign-flip application

Solve $\frac{2 x - 1}{3} \geq \frac{3 x - 2}{4} - \frac{2 - x}{5}$ for real $x$ .

Stuck? Show the approach

Clear all three denominators at once by multiplying through by their LCM (60), then solve the resulting linear inequality using the two standard rules.

Show the full solution

Multiplying every term by 60: $20 (2 x - 1) \geq 15 (3 x - 2) - 12 (2 - x)$ . Expanding: $40 x - 20 \geq 45 x - 30 - 24 + 12 x$ , i.e. $40 x - 20 \geq 57 x - 54$ . Rearranging: $34 \geq 17 x$ , so $x \leq 2$ .

Answer: x ≤ 2, i.e. the interval (−∞, 2]

The trap

Multiplying only the fractional terms by the LCM and forgetting to scale the whole-number terms by the same factor is the most common slip in a multi-denominator inequality like this one.

JEE MainSystem of two inequalities from a single problemIntersection of solution sets

Find all real $x$ satisfying both $3 x - 2 < 2 x + 1$ and $2 x + 3 > x - 4$ .

Stuck? Show the approach

Solve each inequality independently first, then find the overlap of the two resulting intervals — there is no way to combine them into one chain since they involve different-sized coefficients.

Show the full solution

First: $3 x - 2 x < 1 + 2$ , so $x < 3$ . Second: $2 x - x > - 4 - 3$ , so $x > - 7$ . Intersecting: $- 7 < x < 3$ .

Answer: −7 < x < 3, i.e. the interval (−7, 3)

The trap

Assuming the two inequalities must share the same variable coefficient before they can be intersected is unnecessary — any two solved one-variable inequalities can always be intersected on the same number line.

JEE AdvancedInequality with a variable that must also satisfy a domain restrictionCombining an algebraic condition with a natural constraint

A rectangle's length is 3 cm more than its width, and its perimeter is at least 26 cm but the width itself is a positive integer less than 8. Find every possible width.

Stuck? Show the approach

Translate the perimeter condition into an inequality first, then intersect it with the explicitly stated bounds on the width (positive integer, less than 8) — a word problem with three separate conditions, not just one.

Show the full solution

Let $w$ be the width, so the length is $w + 3$ . Perimeter $= 2 (w + (w + 3)) = 4 w + 6 \geq 26$ , giving $w \geq 5$ . Combined with $0 < w < 8$ and $w$ a positive integer: $w \in {5, 6, 7}$ .

Answer: w = 5, 6, or 7

The trap

Solving only the perimeter inequality and reporting w ≥ 5 as the final answer misses that the problem separately restricts w to a positive integer under 8 — every stated condition in a word problem must be intersected, not just the one requiring algebra.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 BoardMedium

JEE MainMedium

NDA MathematicsHigh

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

No. CBSE's 2026-27 syllabus note marks it formative-only, and it does not appear anywhere in the current NCERT chapter — no exercise asks for a graph on the coordinate plane. The chapter is entirely about one-variable inequalities, shown on a number line.

Consider 3 > 2. Multiplying both sides by -1 gives -3 and -2 — but on the number line, -3 is to the left of -2, so -3 < -2. The relative order reverses under a sign change, so the inequality symbol must flip to stay true.

A compound inequality like -8 <= 5x-3 < 7 is one continuous chain, solved by operating on all three parts simultaneously. A system, like '3x-7 < 5+x and 11-5x <= 1', is two separate statements — solve each individually, then intersect the two solution sets.

Yes. The same inequality can have a different-looking solution set depending on this: solving 24x < 100 for natural numbers gives {1,2,...,4}, for integers it includes all negative integers too, and for real numbers it's the interval (-infinity, 25/6) — always check what type of number the problem allows before finalising the answer.

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