In this page we will explain the topics for the chapter 8 of Linear equations in One variable Class 8 Maths.We have given quality Linear equations in One variable Class 8 Notes and video to explain various things so that students can benefits from it and learn maths in a fun and easy manner, Hope you like them and do not forget to like , social share
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- What is algebraic expression
- What is algebraic equation
- What is Linear equation in one Variable
- Solving Equations which have Linear Expressions on one Side and Numbers on the other Side
- Word Problem
- Solving Equations having the Variable on both Sides
- Reducing Equations to Simpler Form
- Equations Reducible to the Linear Form

$10x$

$2x -3$

$3x + y$

$2xy + 11$

$x^2 +2$

An equation is a condition on a variable.

A

An algebraic equation is an equality involving variables. It says that the value of the expression on one side of the equality sign is equal to the value of the expression on the other side.

$5x = 25$

$2x - 3 = 9$

$11xy+1 =98$

$x^2 +1=y^2$

- These all above equation contains the equality (=) sign.

- The above equation can have more than one variable

- The above equation can have highest power of variable > 1

- The expression on the left of the equality sign is the Left Hand Side (LHS). The expression
on the right of the equality sign is the Right Hand Side (RHS)

- In an equation the values of the expressions on the LHS and RHS are equal. This
happens to be true only for certain values of the variable. These values are the solutions of the equation.

- We assume that the two sides of the equation are balanced. We perform the same mathematical operations on both sides of the equation, so that the balance is not disturbed. We get the solution after generally performing few steps

a) algebraic equation in one variable

b) variable will have power 1 only

$5x = 25$

$2x - 3 = 9$

We will focus on Linear equation in One variable in this chapter.

$2x -3= 5$

$3x-11= 22$

$3x-11= 22$

- Transpose (changing the side of the number) the Numbers to the side where all number are present. We know the sign of the number changes when we transpose it to other side

- Now you will have a equation have variable on one side and number on other side. Add/subtract on both the side to get single term

- Now divide or multiply on both the side to get the value of the variable

$2x -3= 5$

Transposing 3 to other side

$2x=5+3$

$2x=8$

Dividing both the sides by 2

x=4

- First read the problem carefully. Write down the unknown and known

- Assume one of the unknown to x and find the other unknown in term of that

- Create the linear equation based on the condition given

- Solve them by using the above method

The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?

The unknown are length and breadth.

Let the breadth be x m.

Then as per question the length will be (2x + 2) m.

Perimeter of swimming pool = 2(l + b) = 154 m

$2(2x + 2 + x) = 154$

$2(3x + 2) = 154$

Dividing both sides by 2

$3x + 2 = 77$

Transposing 2 to R.H.S, we obtain

$3x = 77 - 2$

$3x = 75$

Dividing 3 on both the sides

$x = 25$

So

Breath is 25 m

Length =2x + 2 = 2 × 25 + 2 = 52m

Hence, the breadth and length of the pool are 25 m and 52 m respectively.

$2x -3= 6-x$

$3x-11= 4x$

- Here we Transpose (changing the side of the number) both the variable and Numbers to the side so that one side contains only the number and other side contains only the variable. We know the sign of the number changes when we transpose it to other side.Same is the case with Variable

- Now you will have a equation have variable on one side and number on other side. Add/subtract on both the side to get single term

- Now divide or multiply on both the side to get the value of the variable

$2x -3= 6-x$

Transposing 3 to RHS and x to LHS

$2x+x=6+3$

$3x=9$

Dividing both the sides by 3

$x=3$

- Take the LCM of the denominator of both the LHS and RHS

- Multiple the LCM on both the sides, this will reduce the number without denominator and we can solve using the method described above

Solve the linear equation

L.C.M. of the denominators, 2, 3, 4, and 5, is 60.

So Multiplying both sides by 60, we obtain

$30x - 12 = 20x + 15 +60$

Transposing 20x to R.H.S and 12 to L.H.S, we obtain

$30x- 20x = 15 + 12+60$

$10x= 87$

Dividing both the sides by 10

$x=\frac {[87}{10}=8.7$

The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.

Let the common ratio between their ages be x. Therefore, Hari’s age and Harry’s age will be 5x years and 7x years respectively and four years later, their ages will be (5x + 4) years and (7x + 4) years respectively.

According to the situation given in the question,

Multiplying both the sides by (7x+4)

$(5x+4) =\frac {3(7x+4)}{4}$

or

$20x+16=21x+12$

Transposing 20x to RHS and 12 to LHS

4=x

x=4

Hari’s age = 5x years = (5 × 4) years = 20 years

Harry’s age = 7x years = (7 × 4) years = 28 years

Therefore, Hari’s age and Harry’s age are 20 years and 28 years respectively.

- $6 x + 3 = 3 x + 4$
- $11 x + 4 = 4 x + 10$
- $13 x + 1 = x + 9$
- $3 x + 1 = x + 3$
- $15 x + 5 = 5 x + 5$
- $12 x + 9 = 9 x + 7$
- $7 x + 1 = x + 4$
- $19 x + 10 = 10 x + 4$
- $5 x + 4 = 4 x + 6$
- $9 x + 5 = 5 x + 4$
- $11 x + 1 = x + 8$
- $14 x + 10 = 10 x + 6$

- $x = \frac{1}{3}$
- $x = \frac{6}{7}$
- $x = \frac{2}{3}$
- $x = 1$
- $x = 0$
- $x = - \frac{2}{3}$
- $x = \frac{1}{2}$
- $x = - \frac{2}{3}$
- $x = 2$
- $x = - \frac{1}{4}$
- $x = \frac{7}{10}$
- $x = -1$

In an equation ,The expression on the left of the equality sign is the Left Hand Side (LHS). The expression on the right of the equality sign is the Right Hand Side (RHS)

The general form of the linear equations in one variable is Ax + B = 0 where x is the variable and A is the coefficient of x and b is the constant term. Example 2x +3=0

Properties of Linear equation (i) Variable will have Power 1 only (ii) It cannot be denominator of the fraction

- Linear equations in One Variable is an algebriac equation where we have one variable and it has power 1 only.
- The solution of the linear equation can be integers or fractions
- We can transpose both the constant and linear expression in the linear equations to solve the equation

**Notes****Worksheets****Ncert Solutions**- class 8 maths chapter 2 exercise 2.1 solutions
- class 8 maths chapter 2 exercise 2.2 solutions
- NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.3
- NCERT Solutions for Linear equations Class 8 maths Chapter 2 Exercise 2.4
- class 8 maths chapter 2 exercise 2.5 solutions
- class 8 maths chapter 2 exercise 2.6 solutions

Class 8 Maths Class 8 Science