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: 

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

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

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

2.0CBSE 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, 

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

3.0Key Features - Linear Equations in One Variable

1. 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.
2. 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.

4.0How Do You Find Solutions To Equations? 

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

1. Let us assume that the two sides of the equation are balanced. 
2. Perform the same mathematical operations on both sides of the given equation so that the balance is not disturbed.
3. 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: 

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

$4x+7=-6$

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

$4x=-13$     (Divide by 4)

$x=-3.25$

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

Solution: 

Multiply both sides by 9, 

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

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

$27x+3=x-5$

Subtract x on both sides,

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

$26x+3=-5$

Subtract 3 on both sides, 

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

$26x=-8$

$x=-4/13$

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

Solution:

First, let’s open the bracket 

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

$=5x-4x+14$

$=x+14$

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

$=6x-2+7/2$

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

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

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

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

Now, solve the equation 

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

Multiply by 2 on both sides, 

$2x+28=12x+3$

Subtract by 2x on both sides

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

$28=10x+3$

Subtract by 3 on both sides, 

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

$10x=25$

Divide by 10 on both sides

$x=2.5$

5.0Key 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.