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