Can a linear equation have more than one solution?

A linear equation in one variable usually has exactly one solution. However, if the equation reduces to a true statement (such as 0 = 0), it has infinitely many solutions (an identity). If the equation reduces to a false statement, such as 5 = 3, it has no solution.

What is an identity in linear equations?

An identity is an equation that holds for all values of the variable. For instance, if the equation is 3x + 2 = 3x + 2, then both sides are the same no matter what value of x might be chosen, so it is an identity.

Can a linear equation have no solution?

Yes, a linear equation in maths may have no solution; this occurs when simplification of the equation results in a contradiction, such as 0 = 5. Consider the equation 2x + 3 = 2x + 5. It reduces to 3 = 5, which is false, so it has no solution.

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CBSE Notes

Class 8

Maths

Chapter 2 Linear Equations In One Variable

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CBSE Class 8 Syllabus

CBSE syllabus outlines the topics and subtopics that will be covered in the exam, helping students focus on what’s important and avoid wasting time on irrelevant material.

CBSE Class 8 Notes

CBSE Class 8 Notes are important for students who want to prepare effectively for the exam. They are also helpful in...

CBSE Class 8: Complete Overview

The Central Board of Education (CBSE) administers the class 8 exam, a prerequisite for all NCERT-based schools to be affiliate...

NCERT Solutions Class 8

The NCERT Solutions for Class 8 provide systematically well-structured and very explanatory answers to all the questions included in NCERT Class 8 textbooks. These...

CBSE Exam

CBSE is a national-level board of education in India, responsible for public and private school education, and is managed by the Gover...

Important CBSE Maths Topics

Mathematics is the study of patterns, structure, and relationships, rooted in fundamental practices like counting...

CBSE Notes

Class 8

Maths

Chapter 2 Linear Equations In One Variable

CBSE Notes Class 8 Maths Chapter 2 Linear Equation in One Variable

1.0Introduction

One of the very basic concepts in algebra is the linear equations in one variable. They apply to model situations where an unknown quantity exists and should be found under some conditions. The graph, when plotted, gives the shape of a line, hence the name “Linear equation”. The "one variable" part indicates that the equation involves only one unknown quantity, sometimes represented by x. Here are some examples of one variable:

$5 x, 6 x + 1, 9 y - y, 2 x + 2$

The following are not examples of linear equations because the highest power in all the equations is more than “1”.

$6 x^{2}, x^{2} + 4 x - 90, y^{2}$

2.0Download CBSE Notes Class 8 Maths Chapter 2 Linear Equation in One Variable: Free PDF

Download the free PDF of CBSE Notes for Class 8 Maths Chapter 2 Linear Equation in One Variable to simplify your exam prep. These notes cover key concepts, formulas, and solved examples, following the latest CBSE syllabus. The PDF helps you understand and solve linear equations easily.

Class 8 Maths Chapter 2 Revision Notes:

3.0CBSE Class 8 Chapter-2 Linear Equation in One Variable - Revision Notes

What are Linear Equations in one variable?

A linear equation in one variable is an equation that includes only one variable in maths. It can be written as:

Ax + B = 0

Here,

A and B are the constants.
A can not be equal to 0 as there will be no point in the equation then.
x is the variable that changes.

4.0Key Features - Linear Equations in One Variable

An algebraic equation is an equality equation that contains an equality sign(=) and variables. The expression on the left side of the equality sign is the LHS, and on the right side is RHS.
The values of the LHS and RHS of an equation are equal. This is true only for specific values of the variable. These values are called the solutions of the equation.

5.0How Do You Find Solutions To Equations?

To get the solutions of linear equations in one variable in maths:

Let us assume that the two sides of the equation are balanced.
Perform the same mathematical operations on both sides of the given equation so that the balance is not disturbed.
Repeat Step 2 until you get the solution to the equation.

Let us try an example to get a better understanding of how to get solutions of equation:

Example 1: Solve 6x + 7 = 2x - 6 for x.

Solution:

Subtract 2x on both sides of the equation:

$6 x + 7 - 2 x = 2 x - 6 - 2 x$

$4 x + 7 = - 6$

$4 x + 7 - 7 = - 6 - 7$ (Subtract 7 on both sides)

$4 x = - 13$ (Divide by 4)

$x = - 3.25$

Example 2: Solve $\frac{9 x + 1}{3} = \frac{x - 5}{9}$

Solution:

Multiply both sides by 9,

$\frac{9 ( 9 x + 1 )}{3} = \frac{9 ( x - 5 )}{9}$

$3 (9 x + 1) = (x - 5)$

$27 x + 3 = x - 5$

Subtract x on both sides,

$27 x + 3 - x = x - 5 - x$

$26 x + 3 = - 5$

Subtract 3 on both sides,

$26 x + 3 - 3 = - 5 - 3$

$26 x = - 8$

$x = - 4/13$

Example 3: Solve $5 x -2 (2 x -7) = 2 (3 x -1) + 7/2$

Solution:

First, let’s open the bracket

LHS $= 5 x - 2 (2 x - 7)$

$= 5 x - 4 x + 14$

$= x + 14$

RHS = $2 (3 x - 1) + 7/2$

$= 6 x - 2 + 7/2$

For $6 x - \frac{2}{1} + \frac{7}{2}$

Take LCM of Denominators of $- \frac{2}{1}$ and $\frac{7}{2}$

RHS $= 6 x + \frac{- 4 + 7}{2}$

$= 6 x + \frac{3}{2}$

Now, solve the equation

$x + 14 = 6 x + \frac{3}{2}$

Multiply by 2 on both sides,

$2 x + 28 = 12 x + 3$

Subtract by 2x on both sides

$2 x - 2 x + 28 = 12 x + 3 - 2 x$

$28 = 10 x + 3$

Subtract by 3 on both sides,

$28 - 3 = 10 x + 3 - 3$

$10 x = 25$

Divide by 10 on both sides

$x = 2.5$

6.0Key Features of CBSE Math Notes for Class 8 Chapter 2

The notes provide a step-by-step guide along with examples to solve the questions.
The notes are updated with the latest CBSE curriculum.
It covers all essential topics in a well-organised manner and is easy to understand.
Notes are tailored to align with the exam patterns and syllabus.

Frequently Asked Questions

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

CBSE Class 8 Syllabus

CBSE syllabus outlines the topics and subtopics that will be covered in the exam, helping students focus on what’s important and avoid wasting time on irrelevant material.

CBSE Class 8 Notes

CBSE Class 8 Notes are important for students who want to prepare effectively for the exam. They are also helpful in...

CBSE Class 8: Complete Overview

The Central Board of Education (CBSE) administers the class 8 exam, a prerequisite for all NCERT-based schools to be affiliate...

NCERT Solutions Class 8

The NCERT Solutions for Class 8 provide systematically well-structured and very explanatory answers to all the questions included in NCERT Class 8 textbooks. These...

CBSE Exam

CBSE is a national-level board of education in India, responsible for public and private school education, and is managed by the Gover...

Important CBSE Maths Topics

Mathematics is the study of patterns, structure, and relationships, rooted in fundamental practices like counting...

CBSE Notes Class 8 Maths Chapter 2 Linear Equation in One Variable

1.0Introduction

One of the very basic concepts in algebra is the linear equations in one variable. They apply to model situations where an unknown quantity exists and should be found under some conditions. The graph, when plotted, gives the shape of a line, hence the name “Linear equation”. The "one variable" part indicates that the equation involves only one unknown quantity, sometimes represented by x. Here are some examples of one variable:

$5 x, 6 x + 1, 9 y - y, 2 x + 2$

The following are not examples of linear equations because the highest power in all the equations is more than “1”.

$6 x^{2}, x^{2} + 4 x - 90, y^{2}$

2.0Download CBSE Notes Class 8 Maths Chapter 2 Linear Equation in One Variable: Free PDF

Download the free PDF of CBSE Notes for Class 8 Maths Chapter 2 Linear Equation in One Variable to simplify your exam prep. These notes cover key concepts, formulas, and solved examples, following the latest CBSE syllabus. The PDF helps you understand and solve linear equations easily.

Class 8 Maths Chapter 2 Revision Notes:

3.0CBSE Class 8 Chapter-2 Linear Equation in One Variable - Revision Notes

What are Linear Equations in one variable?

A linear equation in one variable is an equation that includes only one variable in maths. It can be written as:

Ax + B = 0

Here,

A and B are the constants.
A can not be equal to 0 as there will be no point in the equation then.
x is the variable that changes.

4.0Key Features - Linear Equations in One Variable

An algebraic equation is an equality equation that contains an equality sign(=) and variables. The expression on the left side of the equality sign is the LHS, and on the right side is RHS.
The values of the LHS and RHS of an equation are equal. This is true only for specific values of the variable. These values are called the solutions of the equation.

5.0How Do You Find Solutions To Equations?

To get the solutions of linear equations in one variable in maths:

Let us assume that the two sides of the equation are balanced.
Perform the same mathematical operations on both sides of the given equation so that the balance is not disturbed.
Repeat Step 2 until you get the solution to the equation.

Let us try an example to get a better understanding of how to get solutions of equation:

Example 1: Solve 6x + 7 = 2x - 6 for x.

Solution:

Subtract 2x on both sides of the equation:

$6 x + 7 - 2 x = 2 x - 6 - 2 x$

$4 x + 7 = - 6$

$4 x + 7 - 7 = - 6 - 7$ (Subtract 7 on both sides)

$4 x = - 13$ (Divide by 4)

$x = - 3.25$

Example 2: Solve $\frac{9 x + 1}{3} = \frac{x - 5}{9}$

Solution:

Multiply both sides by 9,

$\frac{9 ( 9 x + 1 )}{3} = \frac{9 ( x - 5 )}{9}$

$3 (9 x + 1) = (x - 5)$

$27 x + 3 = x - 5$

Subtract x on both sides,

$27 x + 3 - x = x - 5 - x$

$26 x + 3 = - 5$

Subtract 3 on both sides,

$26 x + 3 - 3 = - 5 - 3$

$26 x = - 8$

$x = - 4/13$

Example 3: Solve $5 x -2 (2 x -7) = 2 (3 x -1) + 7/2$

Solution:

First, let’s open the bracket

LHS $= 5 x - 2 (2 x - 7)$

$= 5 x - 4 x + 14$

$= x + 14$

RHS = $2 (3 x - 1) + 7/2$

$= 6 x - 2 + 7/2$

For $6 x - \frac{2}{1} + \frac{7}{2}$

Take LCM of Denominators of $- \frac{2}{1}$ and $\frac{7}{2}$

RHS $= 6 x + \frac{- 4 + 7}{2}$

$= 6 x + \frac{3}{2}$

Now, solve the equation

$x + 14 = 6 x + \frac{3}{2}$

Multiply by 2 on both sides,

$2 x + 28 = 12 x + 3$

Subtract by 2x on both sides

$2 x - 2 x + 28 = 12 x + 3 - 2 x$

$28 = 10 x + 3$

Subtract by 3 on both sides,

$28 - 3 = 10 x + 3 - 3$

$10 x = 25$

Divide by 10 on both sides

$x = 2.5$

6.0Key Features of CBSE Math Notes for Class 8 Chapter 2

The notes provide a step-by-step guide along with examples to solve the questions.
The notes are updated with the latest CBSE curriculum.
It covers all essential topics in a well-organised manner and is easy to understand.
Notes are tailored to align with the exam patterns and syllabus.

Chapter-wise CBSE Notes for Class 8 Maths:

Class 8 Maths Chapter 1 - Rational Numbers Notes

Class 8 Maths Chapter 2 - Linear Equations In One Variable Notes

Class 8 Maths Chapter 3 - Understanding Quadrilaterals Notes

Class 8 Maths Chapter 4 - Data Handling Notes

Class 8 Maths Chapter 5 - Squares and Square Roots Notes

Class 8 Maths Chapter 6 - Cubes and Cube Roots Notes

Class 8 Maths Chapter 7 - Comparing Quantities Notes

Class 8 Maths Chapter 8 - Algebraic Expressions and Identities Notes

Class 8 Maths Chapter 9 - Mensuration Notes

Class 8 Maths Chapter 10 - Exponents and Powers Notes

Class 8 Maths Chapter 11 - Direct and Inverse Proportions Notes