MATHS CBSE VIII Chapter 2 - Linear Equations in One Variable | Exercise 2.5 Page 34 question 6

Question 6:

Solve the linear equation

Answer

L.C.M. of the denominators, 2 and 3, is 6.

Multiplying both sides by 6, we obtain

6m − 3(m − 1) = 6 − 2(m − 2)

⇒ 6m − 3m + 3 = 6 − 2m + 4 (Opening the brackets)

⇒ 6m − 3m + 2m = 6 + 4 − 3

⇒ 5m = 7

MATHS CBSE VIII Chapter 2 - Linear Equations in One Variable | Exercise 2.5 Page 34 question 5

Question 5:

Solve the linear equation

Answer

L.C.M. of the denominators, 3 and 4, is 12.

Multiplying both sides by 12, we obtain

3(3t − 2) − 4(2t + 3) = 8 − 12t

⇒ 9t − 6 − 8t − 12 = 8 − 12t (Opening the brackets)

⇒ 9t − 8t + 12t = 8 + 6 + 12

⇒ 13t = 26

MATHS CBSE VIII Chapter 2 - Linear Equations in One Variable | Exercise 2.5 Page 34 question 4

Question 4:

Solve the linear equation

Answer

L.C.M. of the denominators, 3 and 5, is 15.

Multiplying both sides by 15, we obtain

5(x − 5) = 3(x − 3)

⇒ 5x − 25 = 3x − 9 (Opening the brackets)

⇒ 5x − 3x = 25 − 9

⇒ 2x = 16

MATHS CBSE VIII Chapter 2 - Linear Equations in One Variable | Exercise 2.5 Page 33 question 3

Question 3:

Solve the linear equation

 

Answer

L.C.M. of the denominators, 2, 3, and 6, is 6.

Multiplying both sides by 6, we obtain

6x + 42 − 16x = 17 − 15x

⇒ 6x − 16x + 15x = 17 − 42

⇒ 5x = −25

MATHS CBSE VIII Chapter 2 - Linear Equations in One Variable | Exercise 2.5 Page 33 question 2

Question 2:

Solve the linear equation

Answer

L.C.M. of the denominators, 2, 4, and 6, is 12.

Multiplying both sides by 12, we obtain

6n − 9n + 10n = 252

⇒ 7n = 252

MATHS CBSE VIII Chapter 2 - Linear Equations in One Variable | Exercise 2.5 Page 33 question 1

 
Question 10:

Simplify and solve the linear equation

Answer

0.25(4f − 3) = 0.05(10f − 9)

Multiplying both sides by 20, we obtain

5(4f − 3) = 10f − 9

⇒ 20f − 15 = 10f − 9 (Opening the brackets)

⇒ 20f − 10f = − 9 + 15

⇒ 10f = 6