## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.1

- Class 8 Maths Linear Equations in One Variable Exercise 2.1
- Class 8 Maths Linear Equations in One Variable Exercise 2.2
- Class 8 Maths Linear Equations in One Variable Exercise 2.3
- Class 8 Maths Linear Equations in One Variable Exercise 2.4
- Class 8 Maths Linear Equations in One Variable Exercise 2.5
- Class 8 Maths Linear Equations in One Variable Exercise 2.6
- Linear Equations in One Variable Class 8 Extra Questions

**NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.1**

Ex 2.1 Class 8 Maths Question 1.

Solve the equation: x – 2 = 7.

Solution:

Given: x – 2 = 7

⇒ x – 2 + 2 = 7 + 2 (adding 2 on both sides)

⇒ x = 9 (Required solution)

Ex 2.1 Class 8 Maths Question 2.

Solve the equation: y + 3 = 10.

Given: y + 3 = 10

⇒ y + 3 – 3 = 10 – 3 (subtracting 3 from each side)

⇒ y = 7 (Required solution)

Ex 2.1 Class 8 Maths Question 3.

Solve the equation: 6 = z + 2

Solution:

We have 6 = z + 2

⇒ 6 – 2 = z + 2 – 2 (subtracting 2 from each side)

⇒ 4 = z

Thus, z = 4 is the required solution.

Ex 2.1 Class 8 Maths Question 4.

Solve the equations: \(\frac { 3 }{ 7 }\) + x = \(\frac { 17 }{ 7 }\)

Solution:

Ex 2.1 Class 8 Maths Question 5.

Solve the equation 6x = 12.

Solution:

We have 6x = 12

⇒ 6x ÷ 6 = 12 ÷ 6 (dividing each side by 6)

⇒ x = 2

Thus, x = 2 is the required solution.

Ex 2.1 Class 8 Maths Question 6.

Solve the equation \(\frac { t }{ 5 }\) = 10.

Solution:

Given \(\frac { t }{ 5 }\) = 10

⇒ \(\frac { t }{ 5 }\) × 5 = 10 × 5 (multiplying both sides by 5)

⇒ t = 50

Thus, t = 50 is the required solution.

Ex 2.1 Class 8 Maths Question 7.

Solve the equation \(\frac { 2x }{ 3 }\) = 18.

Solution:

We have \(\frac { 2x }{ 3 }\) = 18

⇒ \(\frac { 2x }{ 3 }\) × 3 = 18 × 3 (multiplying both sides by 3)

⇒ 2x = 54

⇒ 2x ÷ 2 = 54 ÷ 2 (dividing both sides by 2)

⇒ x = 27

Thus, x = 27 is the required solution.

Ex 2.1 Class 8 Maths Question 8.

Solve the equation 1.6 = \(\frac { y }{ 1.5 }\)

Solution:

Given: 1.6 = \(\frac { y }{ 1.5 }\)

⇒ 1.6 × 1.5 = \(\frac { y }{ 1.5 }\) × 1.5 (multiplying both sides by 1.5)

⇒ 2.40 = y

Thus, y = 2.40 is the required solution.

Ex 2.1 Class 8 Maths Question 9.

Solve the equation 7x – 9 = 16.

Solution:

We have 7x – 9 = 16

⇒ 7x – 9 + 9 = 16 + 9 (adding 9 to both sides)

⇒ 7x = 25

⇒ 7x ÷ 7 = 25 ÷ 7 (dividing both sides by 7)

⇒ x = \(\frac { 25 }{ 7 }\)

Thus, x = \(\frac { 25 }{ 7 }\) is the required solution.

Ex 2.1 Class 8 Maths Question 10.

Solve the equation 14y – 8 = 13.

Solution:

We have 14y – 8 = 13

⇒ 14y – 8 + 8 = 13 + 8 (adding 8 to both sides)

⇒ 14y = 21

⇒ 14y ÷ 14 = 21 ÷ 14 (dividing both sides by 14)

⇒ y = \(\frac { 21 }{ 14 }\)

⇒ y = \(\frac { 3 }{ 2 }\)

Thus, y = \(\frac { 3 }{ 2 }\) is the required solution.

Ex 2.1 Class 8 Maths Question 11.

Solve the equation 17 + 6p = 9.

Solution:

We have, 17 + 6p = 9

⇒ 17 – 17 + 6p = 9 – 17 (subtracting 17 from both sides)

⇒ 6p = -8

⇒ 6p ÷ 6 = -8 ÷ 6 (dividing both sides by 6)

⇒ p = \(\frac { -8 }{ 6 }\)

⇒ p = \(\frac { -4 }{ 3 }\)

Thus, p = \(\frac { -4 }{ 3 }\) is the required solution.

Ex 2.1 Class 8 Maths Question 12.

Solve the equation \(\frac { x }{ 3 }\) + 1 = \(\frac { 7 }{ 15 }\)

Solution:

#### More CBSE Class 8 Study Material

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