**NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.1 , has been designed by the NCERT to test the knowledge of the student on the following topics : **

- Introduction
- Solving Equations which have Linear Expressions on one Side and Numbers on the other Side

### NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.1 – Linear Equation in one variable

**NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.1 – Linear Equation in one variable**

**1. Solve x – 2 = 7**

Transposing -2 to RHS

x = 7 + 2

x = 9

**2. Solve y + 3 = 10**

Transposing +3 to R.H.S

y = 10 – 3

y = 7

**3. Solve 6 = z + 2**

Transposing +2 to L.H.S

6 – 2 = z

4 = z

z = 4

**4. Solve (3/7) + x = (17/7)**

Transposing + (3/7) to R.H.S

x = (17/7) – (3/7)

x = (17 – 3/7)

x = 14/7

x = 2

**5. Solve 6x = 12**

Dividing both sides by 6

6x/6 = 12/6

x = 2

**6. Solve (t/5) = 10**

Multiply 5 on both sides

(t/5) x 5 = 10 x 5

t = 50

**7. Solve (2x/3) = 18**

Multiply both sides with (3/2) [Multiplicative inverse of (2/3)]

(2x/3) x (3/2) = 18 x (3/2)

x = 27

**8. Solve 1.6 = (y/1.5)**

Multiply both sides with 1.5

1.6 x 1.5 = 1.5 x (y/1.5)

y = 2.4

**9. Solve 7x – 9 = 16**

Transposing -9 to RHS

7x = 16 + 9

7x = 25

Dividing both sides by 7

(7x/7) = (25/7)

x = (25/7)

**10. Solve 14y – 8 = 13**

Transposing -8 to RHS

14y = 13 + 8

14y = 21

Dividing both sides by 14

(14y/14) = (21/14)

y = 21/14

**11. Solve 17 + 6p = 9**

Transposing +17 to R.H.S

6p = 9 – 17

6p = -8

Dividing both sides by 6

(6p/6) = -(8/6)

p = -(4/3)

**12. Solve (x/3) + 1 = (7/15)**

Transposing +1 to R.H.S

(x/3) = (7/15) – 1

(x/3) = (-8/15)

Multiply both sides with 3

(x/3) x 3 = (-8/15) x 3

x = (-24/15)

x = (-8/5)

**The next Exercise for** **NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.2 – Linear Equation in two Variables can be accessed by clicking here **

**Download NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.1 – Linear Equation in one Variable**