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
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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 equations in One variable Exercise 2.1
- NCERT Solution Linear equations in One variable Exercise 2.2
- NCERT Solution Linear equations in One variable Exercise 2.3
- NCERT Solution Linear equations in One variable Exercise 2.4
- NCERT Solution Linear equations in One variable Exercise 2.5
- NCERT Solution Linear equations in One variable Exercise 2.6

Class 8 Maths Class 8 Science