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

**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.

Solution:

⇒ 30x – 12 = 20x + 15

⇒ 30x – 20x = 15 + 12 (Transposing 20x to LHS and 12 to RHS)

⇒ 10x = 27

⇒ x = 27/10

Ex 2.5 Class 8 Maths Question 2.

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.

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:

⇒ (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.

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.

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 = 7/5 (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

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.