Courses

# NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev

## Class 8 Mathematics by VP Classes

Created by: Vp Classes

## Class 8 : NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev

The document NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev is a part of the Class 8 Course Class 8 Mathematics by VP Classes.
All you need of Class 8 at this link: Class 8

EXERCISE 2.3

Ques: Solve the following equations and check your results.
1. 3x = 2x + 18
2. 5t â€“ 3 = 3t â€“ 5
3. 5x + 9 = 5 + 3x
4. 4z + 3 = 6 + 2z
5. 2x â€“ 1 = 14 â€“ x
6. 8x + 4 = 3(x â€“ 1) + 7
7.
8.
9.
10. 3m = 5m - 8/5

Ans:
1. 3x = 2x + 18
Transposing 2x from RHS to LHS, we have

3x â€“ 2x = 18
â‡’ x = 18

2. 5t â€“ 3 = 3t â€“ 5
Transposing (â€“3) to RHS, we have
5t = 3t â€“ 5 + 3
â‡’ 5t = 3t â€“ 2
Transposing 3t to LHS, we have
5t â€“ 3t = â€“2
â‡’ 2t = â€“2
Diving both sides by 2, we have
t = â€“1

3. 5x + 9 = 5 + 3x

Transposing 9 to RHS, we have
5x = 5 + 3x â€“ 9
â‡’ 5x = â€“4 + 3x
Transposing 3x to LHS, we have
5x â€“ 3x = â€“4
â‡’ 2x = â€“4
Dividing both sides by 2, we have
x = -4/2 = -2
x =  -2

4. 4z + 3 = 6 + 2z

Transposing 3 to RHS, we have

4z = 6 â€“ 3 + 2z
â‡’ 4z = 3 + 2z

Transposing 2z to LHS, we have

4z â€“ 2z = 3
â‡’ 2z = 3

Dividing both sides by 2, we have
z = 3/2

5. 2x â€“ 1 = 14 â€“ x

Transposing (â€“1) to RHS, we have

2x = 14 â€“ x + 1
â‡’ 2x = 15 â€“ x

Transposing (â€“x) to LHS, we have
2x + x = 15
â‡’ 3x = 15
Dividing both sides by 3, we have

x = 15/3 = 5
âˆ´ x = 5

6.  8x + 4 = 3(x â€“ 1) + 7
8x + 4 = 3x â€“ 3 + 7
â‡’ 8x + 4 = 3x + 4
Transposing 4 to RHS, we have
8x = 3x + 4 â€“ 4
Transposing 3x to LHS, we have
8x â€“ 3x = 0
â‡’ 5x = 0
âˆ´  x = 0

7.
Using distributive property,

â‡’

Transposing  x to LHS, we have

â‡’
â‡’ x/5 = 8

Multiplying both sides by 5, we have
x = 8 Ã— 5 = 40
âˆ´ x = 40

8.
Transposing 1 to RHS, we have

â‡’

Transposing 7x/15 to LHS, we have

â‡’
â‡’ 3x/15 = 2

Multiplying both sides by 15, we have

3x = 2 Ã— 15 = 30

Dividing both sides by 3, we have

âˆ´ x = 10

9.

Transposing 5/3  to RHS, we have

Transposing (â€“y) to LHS, we have

â‡’ 3y = 7

Dividing both sides by 3, we have

3y = 3 = 7/3

âˆ´ y = 7/3

10. 3m = 5m

Transposing 5m to LHS, we have

3m - 5m =

â‡’ -2m =

Dividing both sides by (â€“2), we have

âˆ´ m = 4/5

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

93 docs|16 tests

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;