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Chapter 5: Linear Inequalities

Class 11 • Mathematics • Chapter 5

Linear Inequalities

Comparing quantities that are not equal, and finding every value that fits.

CBSE / NCERT Aligned

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Chapter Roadmap

Inequalities • Rules for Solving • One Variable • On the Number Line • Two Variables • Graphical Solution • Systems of Inequalities • Applications

Why Linear Inequalities Matter

In real life, things are rarely exactly equal. A train must reach the station in at most two hours. A budget must stay within 5,000 rupees. A bridge must carry at least a certain load. Each of these is a comparison, not an equation, and that is exactly what an inequality describes.

So far you have solved equations, where two sides are equal and the answer is usually a single value. An inequality compares two sides that are not equal, using the signs less than, greater than, less than or equal to, and greater than or equal to. Its answer is not one number but a whole range of numbers. In this chapter you will solve linear inequalities in one variable and show the answer on a number line, then move to two variables and shade the region of the plane that works, and finally handle several inequalities at once as a system.

Key idea

An equation has an equals sign and usually one solution. An inequality uses <, >, ≤ or ≥ and has a whole set of solutions. The one rule that is easy to forget: when you multiply or divide both sides by a negative number, the inequality sign reverses.

Key Terms You Must Know

Term

Meaning

Example

Inequality

A statement that two quantities are not equal, using <, >, ≤ or ≥.

x < 5

Strict inequality

Uses < or >; the boundary value is not included.

x > 2

Slack inequality

Uses ≤ or ≥; the boundary value is included.

x ≤ 7

Solution

Any value of the variable that makes the inequality true.

x = 1 satisfies x < 5

Solution set

The set of all values that satisfy the inequality.

{x : x < 5}

Number line

A line used to picture the solution as a ray or a segment.

An open ray for x > 2

Half-plane

The region on one side of a line; the solution of a two-variable inequality.

x + y ≤ 4

Core Concepts, Step by Step

1. Reading an Inequality

An inequality compares two quantities. The sign < means ‘less than’, > means ‘greater than’, ≤ means ‘less than or equal to’, and ≥ means ‘greater than or equal to’. The signs < and > are called strict, because the boundary value is left out. The signs ≤ and ≥ are called slack, because the boundary value is allowed. A linear inequality is one where the variable appears only to the first power, such as 3x + 2 < 11, with no x² or higher.

2. The Rules for Solving

Solving an inequality is almost the same as solving an equation. You may add or subtract the same number on both sides, and the sign stays the same. You may multiply or divide both sides by the same positive number, and the sign stays the same. But there is one special case: if you multiply or divide both sides by a negative number, you must reverse the inequality sign. The table below collects the rules at a glance.

What each operation does to a < b

Operation

Result

Sign

Add c to both sides

a + c < b + c

stays the same

Subtract c from both sides

a − c < b − c

stays the same

Multiply by k (k > 0)

ka < kb

stays the same

Divide by k (k > 0)

a/k < b/k

stays the same

Multiply by k (k < 0)

ka > kb

reverses

Divide by k (k < 0)

a/k > b/k

reverses

3. Solving in One Variable and Showing It on a Number Line

To solve a linear inequality in one variable, collect the variable terms on one side and the numbers on the other, then make the coefficient of the variable equal to 1, remembering the sign rule. The answer is a range of values, which we picture on a number line. We use an open (hollow) circle for < and >, because the boundary value is excluded, and a closed (solid) circle for ≤ and ≥, because it is included. The figure below shows the idea.

Showing a solution on the number line

Solution of x greater than 2 on a number line

4. Inequalities in Two Variables

An inequality such as 2x + 3y ≤ 6 involves two variables, so its solution is not a set of points on a line but a whole region of the plane. The matching equation 2x + 3y = 6 is a straight line that splits the plane into two halves, called half-planes. One half-plane satisfies the inequality and the other does not. We draw the boundary line solid for ≤ or ≥ (the line is included) and dashed for < or > (the line is excluded).

5. The Graphical Method

To solve a two-variable inequality, draw the boundary line, then choose any test point not on the line (the origin (0, 0) is easiest when the line does not pass through it). Put the test point into the inequality. If it makes the inequality true, shade the side that contains the test point; if it is false, shade the other side. The shaded region is the complete solution. The figure below shows this for x + y ≤ 4.

Shading the correct half-plane

Shaded half-plane for x + y at most 4

6. Systems of Linear Inequalities

Often several conditions must hold at the same time, for example x ≥ 0, y ≥ 0 and x + y ≤ 5. Each inequality shades its own region, and the solution of the whole system is the region where all of them overlap. This common region is the foundation of linear programming, which you will meet in Class 12 for finding the best possible value of a quantity under several constraints.

Key Results & Proofs

Two rules sit underneath everything in this chapter, and both can be proved from one simple idea: a < b is just another way of saying that b − a is positive.

Result 1: Adding the Same Number Keeps the Inequality

Statement. If a < b, then a + c < b + c for every real number c.

Proof

Start from what a < b actually means.

a < b means b − a > 0. – the meaning of less-than

Now (b + c) − (a + c) = b − a, which is still > 0. – the c cancels out

A positive difference means a + c < b + c. – same order is kept

The same reasoning works for subtracting a number, and for the signs >, ≤ and ≥.

Result 2: Multiplying by a Negative Reverses the Sign

Statement. If a < b and k > 0, then ka < kb; but if a < b and k < 0, then ka > kb.

Proof

Again begin from b − a > 0, and watch the sign of the product.

a < b means b − a > 0. – starting point

Multiply by k > 0: k(b − a) > 0, so kb − ka > 0, giving ka < kb. – positive times positive stays positive

Multiply instead by k < 0: k(b − a) < 0, so kb − ka < 0, giving ka > kb. – positive times negative becomes negative

This is the one rule students most often forget: dividing or multiplying by a negative number flips the inequality sign.

Worked Examples

Example 1

Question: Solve 3x − 5 < 7, where x is a real number.

▶ Show full working

Treat it like an equation: get x by itself. We only add and divide by a positive number, so the sign stays the same.

3x − 5

<

7

given

3x

<

12

add 5 to both sides

x

<

4

divide both sides by 3 (positive, so sign stays)

Answer: x < 4, that is, every real number less than 4.

Example 2

Question: Solve −2x + 1 ≥ 7.

▶ Show full working

Watch the final step, where we divide by a negative number and the sign reverses.

−2x + 1

≥

7

given

−2x

≥

6

subtract 1 from both sides

x

≤

−3

divide by −2 (negative, so ≥ becomes ≤)

Answer: x ≤ −3.

Example 3

Question: Solve 2(x − 1) < x + 3.

▶ Show full working

Open the bracket first, then gather x on one side and numbers on the other.

2(x − 1)

<

x + 3

given

2x − 2

<

x + 3

open the bracket

2x − x

<

3 + 2

collect terms

x

<

5

simplify

Answer: x < 5.

Example 4

Question: Solve the double inequality 3 ≤ 2x − 1 < 9.

▶ Show full working

Do the same operation to all three parts at once.

3 ≤ 2x − 1 < 9 – given

3 + 1 ≤ 2x < 9 + 1 – add 1 to all three parts

4 ≤ 2x < 10 – simplify

2 ≤ x < 5 – divide all three parts by 2

Answer: 2 ≤ x < 5.

Example 5

Question: Solve x/3 > (x − 2)/2.

▶ Show full working

Clear the fractions by multiplying both sides by a positive number, then finish carefully.

x/3

>

(x − 2)/2

given

6 · x/3

>

6 · (x − 2)/2

multiply both sides by 6 (positive)

2x

>

3(x − 2)

simplify

2x

>

3x − 6

open the bracket

−x

>

−6

collect terms

x

<

6

divide by −1 (negative, so reverse the sign)

Answer: x < 6.

Example 6

Question: Solve 5x − 3 < 3x + 7 and describe the solution on a number line.

▶ Show full working

Collect terms, solve, then read off the picture.

5x − 3

<

3x + 7

given

5x − 3x

<

7 + 3

collect terms

2x

<

10

simplify

x

<

5

divide by 2

Answer: x < 5, shown as an open ray pointing left from 5 (5 not included).

Example 7

Question: Solve 24x < 100 when x is a natural number.

▶ Show full working

Solve as usual, then keep only the natural numbers that fit.

24x

<

100

given

x

<

100/24

divide by 24

x

<

4.17 (approx.)

simplify

The natural numbers below 4.17 are 1, 2, 3 and 4. – list the values that fit

Answer: x ∈ {1, 2, 3, 4}.

Example 8

Question: Is the point (1, 1) a solution of 2x + 3y ≤ 6?

▶ Show full working

Put the point into the left side and compare with the right side.

2x + 3y

=

2(1) + 3(1)

substitute x = 1, y = 1

This equals 5, and 5 ≤ 6 is true. – compare with the right side

Answer: Yes, (1, 1) satisfies 2x + 3y ≤ 6.

Example 9

Question: Solve x + y ≤ 4 graphically.

▶ Show full working

Draw the boundary line, test the origin, then shade.

Boundary line x + y = 4 passes through (4, 0) and (0, 4); draw it solid (because of ≤). – draw the line

Test the origin (0, 0): 0 + 0 = 0, and 0 ≤ 4 is true. – pick a test point

So shade the side of the line that contains the origin. – choose the side

Answer: The solution is the half-plane on the origin side of x + y = 4, including the line itself.

Example 10

Question: Solve the system x ≥ 0, y ≥ 0, x + y ≤ 5.

▶ Show full working

Each inequality gives a region; the answer is where they all overlap.

x ≥ 0 keeps us to the right of the y-axis.

y ≥ 0 keeps us above the x-axis.

x + y ≤ 5 keeps us below that line.

The common region is a triangle with corners (0, 0), (5, 0) and (0, 5).

Answer: The overlapping triangle with corners (0, 0), (5, 0) and (0, 5).

Example 11

Question: Solve 7x + 3 < 5x + 9.

▶ Show full working

Collect terms and divide.

7x + 3

<

5x + 9

given

7x − 5x

<

9 − 3

collect terms

2x

<

6

simplify

x

<

3

divide by 2

Answer: x < 3.

Example 12

Question: The longest side of a triangle is twice the shortest side, and the third side is 2 cm shorter than the longest. If the perimeter is at least 166 cm, find the smallest the shortest side can be.

▶ Show full working

Name the shortest side, write the other sides in terms of it, then form an inequality from the perimeter.

Let the shortest side be x cm. Then the longest is 2x and the third is 2x − 2. – set up the sides

x + 2x + (2x − 2)

≥

166

perimeter is at least 166

5x − 2

≥

166

add the sides

5x

≥

168

add 2

x

≥

33.6

divide by 5

Answer: The shortest side must be at least 33.6 cm.

Where You Meet This in Real Life

Budgets and shopping

Staying within a fixed amount of money is a less-than-or-equal-to condition; inequalities tell you how many items you can afford.

Marks and cut-offs

A pass mark, a distinction, or a cut-off for admission is an at-least condition written with ≥.

Speed and safety limits

Speed limits, weight limits on lifts and bridges, and dosage limits in medicine are all inequalities that keep us safe.

Business and profit

A business makes a profit only when revenue is greater than cost, which is a simple inequality between two expressions.

Linear programming

Factories decide how much of each product to make under limits on time, material and labour; the allowed plans form a region defined by many inequalities.

Practice Sets A to D

Practice Set A – Basics

A1. Solve x + 5 < 9.

▶ Reveal full working

Move the number across.

x + 5

<

9

given

x

<

4

subtract 5

Answer: x < 4.

A2. Solve 4x ≥ 20.

▶ Reveal full working

Divide by a positive number.

4x

≥

20

given

x

≥

5

divide by 4

Answer: x ≥ 5.

A3. Solve −3x < 12.

▶ Reveal full working

Divide by a negative number and reverse the sign.

−3x

<

12

given

x

>

−4

divide by −3, reverse the sign

Answer: x > −4.

A4. Write the solution of x ≤ 3 in natural numbers.

▶ Reveal full working

List the natural numbers that fit.

Natural numbers that are 3 or less are 1, 2 and 3.

Answer: x ∈ {1, 2, 3}.

Practice Set B – Conceptual

B1. Why does the inequality sign reverse when we multiply by a negative number?

▶ Reveal full working

Test it on a true statement.

Take a true statement, say 2 < 5.

Multiply both sides by −1 to get −2 and −5.

On the number line −2 lies to the right of −5, so −2 > −5.

The order flipped, so the sign must reverse.

Answer: Because multiplying by a negative reverses order on the number line.

B2. What is the difference between an open circle and a closed circle on a number line?

▶ Reveal full working

Think about whether the boundary is included.

An open (hollow) circle means the boundary value is not included, used for < and >.

A closed (solid) circle means the boundary value is included, used for ≤ and ≥.

Answer: Open means excluded (strict); closed means included.

B3. When is the boundary line of a two-variable inequality solid, and when is it dashed?

▶ Reveal full working

Decide whether points on the line count.

Solid line for ≤ or ≥, because points on the line are part of the answer.

Dashed line for < or >, because points on the line are not part of the answer.

Answer: Solid for ≤ and ≥; dashed for < and >.

B4. What does the solution of a system of inequalities look like?

▶ Reveal full working

Combine the separate regions.

Each inequality shades one region.

The system is satisfied only where all those regions overlap.

Answer: The common overlapping region of all the inequalities.

Practice Set C – Application / Numerical

C1. Solve 2(2x + 3) − 10 < 6(x − 2).

▶ Reveal full working

Open the brackets, collect, and finish (watch the negative).

2(2x + 3) − 10

<

6(x − 2)

given

4x + 6 − 10

<

6x − 12

open the brackets

4x − 4

<

6x − 12

simplify the left side

4x − 6x

<

−12 + 4

collect terms

−2x

<

−8

simplify

x

>

4

divide by −2, reverse the sign

Answer: x > 4.

C2. Solve the double inequality −8 ≤ 5x − 3 < 7.

▶ Reveal full working

Apply each step to all three parts.

−8 ≤ 5x − 3 < 7 – given

−8 + 3 ≤ 5x < 7 + 3 – add 3 to all parts

−5 ≤ 5x < 10 – simplify

−1 ≤ x < 2 – divide all parts by 5

Answer: −1 ≤ x < 2.

C3. Find all integers x satisfying 5x − 7 > 2x + 5.

▶ Reveal full working

Solve, then list the integers.

5x − 7

>

2x + 5

given

3x

>

12

collect and simplify

x

>

4

divide by 3

Integers greater than 4 are 5, 6, 7, and so on.

Answer: x ∈ {5, 6, 7, …}.

C4. Solve the system x ≥ 0, y ≥ 0, 2x + y ≤ 6 and name the corner points of the region.

▶ Reveal full working

Each inequality gives a region; find the overlap.

x ≥ 0 and y ≥ 0 keep us in the first quadrant.

The line 2x + y = 6 passes through (3, 0) and (0, 6).

Test the origin: 0 ≤ 6 is true, so shade towards the origin.

The region is the triangle with corners (0, 0), (3, 0) and (0, 6).

Answer: Triangle with corners (0, 0), (3, 0) and (0, 6).

Practice Set D – HOTS / Word Problems

D1. A small business sells x units. Its revenue is 5x and its cost is 2x + 90. For how many units does it make a profit?

▶ Reveal full working

Profit means revenue greater than cost.

5x

>

2x + 90

revenue greater than cost

3x

>

90

collect terms

x

>

30

divide by 3

Answer: More than 30 units (x > 30).

D2. Solve (2x + 1)/3 > (x − 1)/2.

▶ Reveal full working

Clear the fractions first.

(2x + 1)/3

>

(x − 1)/2

given

6 · (2x + 1)/3

>

6 · (x − 1)/2

multiply both sides by 6

2(2x + 1)

>

3(x − 1)

simplify

4x + 2

>

3x − 3

open the brackets

x

>

−5

collect terms and simplify

Answer: x > −5.

D3. Find the values of x for which both 3x − 7 > 2 and 2x + 1 < 11 hold.

▶ Reveal full working

Solve each, then take the overlap.

First: 3x − 7 > 2 gives 3x > 9, so x > 3.

Second: 2x + 1 < 11 gives 2x < 10, so x < 5.

Both hold together when 3 < x < 5.

Answer: 3 < x < 5.

D4. A student scores 70 and 75 in the first two tests, each out of 100. What is the least she must score in the third test so that the average of the three is at least 80?

▶ Reveal full working

Write the average as an inequality.

(70 + 75 + x)/3

≥

80

average is at least 80

145 + x

≥

240

multiply both sides by 3

x

≥

95

subtract 145

Answer: At least 95 marks.

Chapter Summary

Everything in One Glance

Inequality

Compares two unequal sides with <, >, ≤ or ≥; the answer is a range.

Sign Rule

Adding, subtracting, and multiplying or dividing by a positive keep the sign; a negative reverses it.

One Variable

Solve like an equation, then show the answer on a number line (open or closed circle).

Two Variables

The solution is a half-plane; the boundary is solid for ≤/≥ and dashed for </>.

Graphical Method

Draw the line, test a point such as the origin, and shade the correct side.

Systems

The solution is the region where all the separate regions overlap.

Are You Exam-Ready?

8-Point Exam Quick-Check

Solve 2x + 5 < 13.

Solve −4x ≥ 16 and state how the sign changes.

Show the solution of x ≥ −1 on a number line: open or closed circle?

Is (2, 0) a solution of x + 2y ≤ 3?

When is the boundary line of a two-variable inequality drawn dashed?

Solve the double inequality 1 ≤ 3x − 2 < 7.

Find all natural numbers satisfying 3x < 14.

What does the solution of a system of inequalities represent?

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Class 11 Mathematics Chapter 5: Linear Inequalities, Complete Notes and Practice

These free Class 11 Maths Chapter 5 Linear Inequalities notes follow the latest NCERT 2026-27 syllabus and give complete, exam-ready coverage for board exams and competitive foundations. Includes worked examples, step-by-step proofs and graded practice, free on SchoolRevise.com.