In this page we have **Class 8 Maths Chapter 2 Exercise 2.2 Solutions**.This exercise on question on equation which can be reduced to form of linear equation 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

- Class 8 Maths Chapter 2 Exercise 2.2 Solutions has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also

Download NCERT Solutions Exercise 2.2 as pdf - This chapter 2 has total 2 Exercise 2.1,2.2. Exercise 2.3,2.4,2.5.2.6 has been deleted as per latest syllabus.This is the last exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below

**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

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