NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in one Variable in English and Hindi Medium updated for academic session 2022-2023.
8th Maths Chapter 2 Solutions in Hindi Medium
Class 8 Maths Chapter 2 all Exercises Solution
NCERT Solutions for Class 8 Maths Chapter 2
Class VIII Maths based on latest NCERT Books for new session. Download Prashnavali 2.1, Prashnavali 2.2, Prashnavali 2.3, Prashnavali 2.4, Prashnavali 2.5 and Prashnavali 2.6 in Hindi Medium or Exercise 2.1, Exercise 2.2, Exercise 2.3, Exercise 2.4, Exercise 2.5 and Exercise 2.6 in English Medium free to use online or download in PDF to use offline. NCERT Solutions 2022-23 for other subjects are also given as downloadable format. All the solutions are in updated format based on latest NCERT Books.
|Chapter 2:||Linear Equations in One Variable|
|Mode:||Text and Videos Solutions|
|Medium:||Hindi and English Medium|
Class 8 Maths Chapter 2 Solutions
NCERT Solutions of Class 8 Mathematics Chapter 2 Linear Equations in One Variable all Exercises are in English as well as in Hindi Medium to use online. NCERT Solutions Offline Apps 2022-23 are also given to free download based on latest NCERT Books.
More about Class 8 Maths Chapter 2
In Chapter 2, we are concerned about algebraic equations in an equality involving variables. It means the value of expression on the both sides are equal. The solution of the equation may be any Rational Number. During the solutions, the variables can be transported from one side to other side of the equation. In order to solve the equation, first make it simplified and convert it into simple linear equation and then shift all the variables one side and constants on the other side.
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Class 8 Maths Chapter 2 Practice questions with Solution
A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Let another number be x.
Then positive number = 5x
According to the question,
5x + 21 = 2 (x + 21)
⇒ 5x + 21 = 2x + 42
⇒ 5x + 21 = 2x + 42 – 21
⇒5x = 21
⇒ x = 21/3 = 7
Hence another number = 7 and positive number = 5 x 7 =35
Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present age?
Let Shobo’s present age be x years.
And Shobo’s mother’s present age = 6x years
According to the question,
x + 5 = (1/3) × 6x
⇒ x + 5 = 2x
⇒ 2x = x + 5
⇒ 2x – x = 5
⇒ x = 5 years.
Hence, Shobo’s present age = 5 years and Shobo’s mother’s present age = 6 x 5 = 30 years
Application of Linear Eqations
Convert multiple situations of linear equations into simple equations first and then solve them. In this lesson, questions based on numbers, questions on age and distances, questions based on perimeters and currency have to be converted first into simple equations and then solved.
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