In this page we have **NCERT book Solutions for Class 8th Maths:Linear equations Chapter 2 ** for
Exercise 2.5 . Hope you like them and do not forget to like , social share
and comment at the end of the page.

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

$30x- 12 = 20x+ 15$

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

$30x- 20x= 15 + 12$

10x= 27

X=27/10=2.7

Solve the linear equation

L.C.M. of the denominators, 2, 4, and 6, is 12.

Multiplying both sides by 12, we obtain

$6n- 9n+ 10n= 252$

7n= 252

Dividing by 7 on both the sides

n=36

Solve the linear equation

L.C.M. of the denominators, 2, 3, and 6, is 6.

Multiplying both sides by 6, we obtain

6x+ 42 - 16x= 17 - 15x

Transposing 15x to RHS and 42 to LHS

6x- 16x+ 15x= 17 - 42

5x= -25

Dividing by 5 on both the sides

x=-5

Solve the linear equation

L.C.M. of the denominators, 3 and 5, is 15.

Multiplying both sides by 15, we obtain

$5(x- 5) = 3(x- 3)$

$5x- 25 = 3x- 9$

Transposing 3x to RHS and 25 to LHS

5x- 3x= 25 - 9

2x= 16

Dividing both the sides by 2

x=8

Solve the linear equation

L.C.M. of the denominators, 3 and 4, is 12.

Multiplying both sides by 12, we obtain

$3(3t- 2) - 4(2t+ 3) = 8 - 12t$

$9t- 6 - 8t- 12 = 8 - 12t$

Transposing 12t to RHS and 6 and 12 to LHS

9t- 8t+ 12t= 8 + 6 + 12

13t= 26

Dividing by 13 on both the sides

t=2

Solve the linear equation

L.C.M. of the denominators, 2 and 3, is 6.

Multiplying both sides by 6, we obtain

$6m- 3(m- 1) = 6 - 2(m- 2)$

$6m- 3m+ 3 = 6 - 2m+ 4$

Transposing 2m to RHS and 3 to LHS

6m- 3m+ 2m= 6 + 4 - 3

5m= 7

Dividing by 5 on both the sides

m=7/5

Simplify and solve the linear equation $3(t- 3) = 5(2t+ 1)$

$3(t- 3) = 5(2t+ 1)$

3t- 9 = 10t+ 5

-9 - 5 = 10t- 3t

-14 = 7t

t=-2

Simplify and solve the linear equation $15(y- 4) - 2(y- 9) + 5(y+ 6) = 0$

$15(y- 4) - 2(y- 9) + 5(y+ 6) = 0$

15y- 60 - 2y+ 18 + 5y+ 30 = 0

18y- 12 = 0

18y= 12 y=12/18=2/3

Simplify and solve the linear equation

$3(5z- 7) - 2(9z- 11) = 4(8z- 13)-17$

$3(5z- 7) - 2(9z- 11) = 4(8z- 13)-17$

15z- 21 - 18z+ 22 = 32z- 52 - 17

-3z+ 1 = 32z- 69

-3z- 32z= -69 - 1

-35z= -70

z=2

Simplify and solve the linear equation $0.25(4f- 3) = 0.05(10f- 9)$

$0.25(4f- 3) = 0.05(10f- 9)$

Multiplying both sides by 20, we obtain

$5(4f- 3) = 10f- 9$

20f- 15 = 10f- 9

20f- 10f= - 9 + 15

10f= 6

f==6/10=3/5

Download NCERT Solutions Exercise 2.5 as pdf

**Notes****Worksheets****Ncert Solutions**- NCERT Solution Linear equation in One variable Exercise 2.1
- NCERT Solution Linear equation in One variable Exercise 2.2
- NCERT Solution Linear equation in One variable Exercise 2.3
- NCERT Solution Linear equation in One variable Exercise 2.4
- NCERT Solution Linear equation in One variable Exercise 2.5
- NCERT Solution Linear equation in One variable Exercise 2.6

Class 8 Maths Class 8 Science

Given below are the links of some of the reference books for class 8 Math.

- Mathematics Foundation Course for JEE/Olympiad : Class 8 This book can take students maths skills further. Only buy if child is interested in Olympiad/JEE foundation courses.
- Mathematics for Class 8 by R S Aggarwal Detailed Mathematics book to clear basics and concepts. I would say it is a must have book for class 8 student.
- Pearson Foundation Series (IIT -JEE / NEET) Physics, Chemistry, Maths & Biology for Class 8 (Main Books) | PCMB Combo : These set of books could help your child if he aims to get extra knowledge of science and maths. These would be helpful if child wants to prepare for competitive exams like JEE/NEET. Only buy if you can provide help to the child while studying.
- Reasoning Olympiad Workbook - Class 8 :- Reasoning helps sharpen the mind of child. I would recommend students practicing reasoning even though they are not appearing for Olympiad.

You can use above books for extra knowledge and practicing different questions.

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