Class 8 Maths Exercise 2.5 Easy Solutions NCERT

Class 8 Maths Exercise 2.5 Easy Solutions NCERT focuses on solving linear equations with more complex steps and variables. This exercise is designed to strengthen your understanding of balancing equations and applying algebraic operations. With clear, easy-to-follow solutions based on the NCERT textbook, students can build their confidence and master the skills needed to solve equations efficiently. Each step-by-step solution helps break down the process for better clarity and learning.

Class 8 Maths Exercise 2.5 Easy Solutions NCERT

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.5

Solve the following linear equations.
Ex 2.5 Class 8 Maths Question 1.
NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Q1
Solution:

Linearity training

NCERT Books
⇒ 30x – 12 = 20x + 15
⇒ 30x – 20x = 15 + 12 (Transposing 20x to LHS and 12 to RHS)
⇒ 10x = 27
⇒ x = 2710

Ex 2.5 Class 8 Maths Question 2.
NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Q2
Solution:
LCM of 2, 4 and 6 = 12

(Multiplying both sides by 12)
⇒ 6n – 9n + 10n = 252
⇒ 7n = 252
⇒ n = 252 ÷ 7
⇒ n = 36

Ex 2.5 Class 8 Maths Question 3.
NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Q3
Solution:

⇒ -10x + 42 = 17 – 15x
⇒ -10x + 15x = 17 – 42 [Transposing 15x to LHS and 42 to RHS]
⇒ 5x = -25
⇒ x = -25 ÷ 5 [Transposing 5 to RHS]
⇒ x = -5

Ex 2.5 Class 8 Maths Question 4.

Solution:
NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Q4.1
⇒ (x – 5) × 5 = (x – 3) × 3
⇒ 5x – 25 = 3x – 9 (Solving the brackets)
⇒ 5x – 3x = 25 – 9 (Transposing 3x to LHS and 25 to RHS)
⇒ 2x = 16
⇒ x = 16 ÷ 2 = 8 (Transposing 2 to RHS)
⇒ x = 8

Ex 2.5 Class 8 Maths Question 5.
NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Q5
Solution:

⇒ (3t – 2) × 3 – (2t + 3) × 4 = 2 × 4 – 12t
⇒ 9t – 6 – 8t – 12 = 8 – 12t (Solving the brackets)
⇒ t – 18 = 8 – 12t
⇒ t + 12t = 8 + 18 (Transposing 12t to LHS and 18 to RHS)
⇒ 13t = 26
⇒ t = 2 (Transposing 13 to RHS)
Hence t = 2 is the required solution.

Ex 2.5 Class 8 Maths Question 6.
NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Q6
Solution.

⇒ 6m – (m – 1) × 3 = 6 – (m – 2) × 2
⇒ 6m – 3m + 3 = 6 – 2m + 4 (Solving the brackets)
⇒ 3m + 3 = 10 – 2m
⇒ 3m + 2m = 10 – 3 (Transposing 2m to LHS and 3 to RHS)
⇒ 5m = 7
⇒ m = 75 (Transposing 5 to RHS)

Simplify and solve the following linear equations.
Ex 2.5 Class 8 Maths Question 7.
3(t – 3) = 5(21 + 1)
Solution:
We have
3(t – 3) = 5(2t + 1)
⇒ 3t – 9 = 10t + 5 (Solving the brackets)
⇒ 3t – 10t = 9 + 5 (Transposing 10t to LHS and 9 to RHS)
⇒ -7t = 14
⇒ t = -2(Transposing -7 to RHS)
Hence, t = -2 is the required solution.

Ex 2.5 Class 8 Maths Question 8.
15(y – 4) – 2(y – 9) + 5(y + 6) = 0
Solution:
We have 15(y – 4) – 2(y – 9) + 5(y + 6) = 0
⇒ 15y – 60 – 2y + 18 + 5y + 30 = 0 (Solving the brackets)
⇒ 8y – 12 = 0
⇒ 8y = 12 (Transposing 12 to RHS)
⇒ y = 23
Hence, y = 23 is the required solution.

Ex 2.5 Class 8 Maths Question 9.
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Solution:
We have
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
⇒ 15z – 21 – 18z + 22 = 32z – 52 – 17 (Solving the bracket)
⇒ -3z + 1 = 32z – 69
⇒ -3z – 32z = – 69 – 1 (Transposing 322 to LHS and 1 to RHS)
⇒ -35z = -70
⇒ z = 2
Hence, z = 2 is the required solution.

Ex 2.5 Class 8 Maths Question 10.
0.25(4f – 3) = 0.05(10f – 9)
Solution:
We have
0.25(4f – 3) = 0.05(10f – 9)
⇒ 0.25 × 4f – 3 × 0.25 = 0.05 × 10f – 9 × 0.05 (Solving the brackets)
⇒ 1.00f – 0.75 = 0.5f – 0.45
⇒ f – 0.5f = -0.45 + 0.75 (Transposing 0.5 to LHS and 0.75 to RHS)
⇒ 0.5f = 0.30
⇒ f = 0.6
Hence, f = 0.6 is the required solution.