- 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

In this page we have *NCERT Solutions for Class 8 Maths Chapter 2 : Linear equations in One Variable * for
Exercise 1.2. Hope you like them and do not forget to like , social share
and comment at the end of the page.

Transposing 2 to R.H.S, we obtain

Transposing 3 to R.H.S, we obtain

Or

z+2=6

Transposing 2 to R.H.S, we obtain

z=6 -2=4

Transposing 3 to R.H.S, we obtain

7x=17-3=14

7x=14

Dividing both the sides by 7

x=2

6x=12

6x=12

Dividing both the sides by 6

x=2

Multiplying 5 on both sides

t=50

Multiplying both the sides by 3

2x=54

Dividing both the sides by 2

x=27

Multiplying both the sides by 1.5

1.6 × 1.5=y

y=2.4

Solve:

7x-9=16

7x-9=16

Transposing 9 to R.H.S, we obtain

7x=16+9

7x=25

Dividing by 7 on both the sides

x=25/7

Solve:

14y-8=13

14y-8=13

Transposing 8 to R.H.S, we obtain

14y=13+8

14y=21

Dividing 14 on both the sides

y=21/14=3/2

Solve:

17+6p=9

17+6p=9

Transposing 17 to R.H.S, we obtain

6p=9-17

6p=-8

Dividing 6 on both the sides

p=-8/6=-4/3

The LCM of denominator is 15, So multiplying both sides by 15

Transposing 15 to R.H.S, we obtain

5x=-8

Dividing 5 on both the sides

x=-8/5

Download NCERT solution linear equation Exercise 1.2 as pdf

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