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

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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.
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EXERCISE 2.3 

Ques: Solve the following equations and check your results.
1. 3x = 2x + 18
2. 5t – 3 = 3t – 5
3. 5x + 9 = 5 + 3x
4. 4z + 3 = 6 + 2z
5. 2x – 1 = 14 – x
6. 8x + 4 = 3(x – 1) + 7
7. NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev
8.NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev
9.NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev
10. 3m = 5m - 8/5

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

3x – 2x = 18
x = 18

2. 5t – 3 = 3t – 5 
Transposing (–3) to RHS, we have
5t = 3t – 5 + 3
5t = 3t – 2 
Transposing 3t to LHS, we have
5t – 3t = –2
2t = –2
Diving both sides by 2, we have
t = –1

3. 5x + 9 = 5 + 3x

Transposing 9 to RHS, we have
5x = 5 + 3x – 9
 5x = –4 + 3x
Transposing 3x to LHS, we have
5x – 3x = –4
2x = –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 – 3 + 2z
4z = 3 + 2z 

Transposing 2z to LHS, we have

4z – 2z = 3
2z = 3 

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

5. 2x – 1 = 14 – x 

Transposing (–1) to RHS, we have 

2x = 14 – x + 1
2x = 15 – x 

Transposing (–x) to LHS, we have
2x + x = 15
 3x = 15 
Dividing both sides by 3, we have

x = 15/3 = 5
∴ x = 5

6.  8x + 4 = 3(x – 1) + 7
8x + 4 = 3x – 3 + 7
8x + 4 = 3x + 4 
Transposing 4 to RHS, we have 
8x = 3x + 4 – 4 
Transposing 3x to LHS, we have
8x – 3x = 0
5x = 0
∴  x = 0

7. NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev
Using distributive property,
 NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev
 NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev

Transposing NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev x to LHS, we have 

NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev
  NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev
 x/5 = 8

Multiplying both sides by 5, we have
x = 8 × 5 = 40
∴ x = 40

8. NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev 
Transposing 1 to RHS, we have
NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev
 NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev

Transposing 7x/15 to LHS, we have

NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev
 NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev
3x/15 = 2

Multiplying both sides by 15, we have

3x = 2 × 15 = 30

Dividing both sides by 3, we have

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

∴ x = 10

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

Transposing 5/3  to RHS, we have

NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev
Transposing (–y) to LHS, we have 

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

 3y = 7

Dividing both sides by 3, we have

3y = 3 = 7/3

∴ y = 7/3

10. 3m = 5m NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev 

Transposing 5m to LHS, we have

3m - 5m = NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev

-2m = NCERT Solutions(Part- 3)- Linear Equations in One Variable Class 8 Notes | EduRev

Dividing both sides by (–2), we have

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

∴ m = 4/5


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