NCERT Solution Class 8th Maths Chapter – 2 Linear Equations in One Variable Exercise – 2.2

Solve the following linear equations.

1. x/2 – 1/5 = x/3 + ¼

Solution:
x/2 – 1/5 = x/3 + ¼
⇒ x/2 – x/3 = ¼+ 1/5
⇒ (3x – 2x)/6 = (5 + 4)/20
⇒ 3x – 2x = 9/20 × 6
⇒ x = 54/20
⇒ x = 27/10

2. n/2 – 3n/4 + 5n/6 = 21

Solution:
n/2 – 3n/4 + 5n/6 = 21
⇒ (6n – 9n + 10n)/12 = 21
⇒ 7n/12 = 21
⇒ 7n = 21 × 12
⇒ n = 252/7
⇒ n = 36

3. x + 7 – 8x/3 = 17/6 – 5x/2

Solution:
x + 7 – 8x/3 = 17/6 – 5x/2
⇒ x – 8x/3 + 5x/2 = 17/6 – 7
⇒ (6x – 16x + 15x)/6 = (17 – 42)/6
⇒ 5x/6 = – 25/6
⇒ 5x = – 25
⇒ x = – 5

4. (x – 5)/3 = (x – 3)/5

Solution:
(x – 5)/3 = (x – 3)/5
⇒ 5(x – 5) = 3(x – 3)
⇒ 5x – 25 = 3x – 9
⇒ 5x – 3x = -9 + 25
⇒ 2x = 16
⇒ x = 8

5. (3t – 2)/4 – (2t + 3)/3 = 2/3 – t

Solution:
(3t – 2)/4 – (2t + 3)/3 = 2/3 – t
⇒ ((3t – 2)/4) × 12 – ((2t + 3)/3) × 12
⇒ (3t – 2) × 3 – (2t + 3) × 4 = 2 × 4 – 12t
⇒ 9t – 6 – 8t – 12 = 8 – 12t
⇒ 9t – 6 – 8t – 12 = 8 – 12t
⇒ t – 18 = 8 – 12t
⇒ t + 12t = 8 + 18
⇒ 13t = 26
⇒ t = 2

6. m – (m – 1)/2 = 1 – (m – 2)/3

Solution:
m – (m – 1)/2 = 1 – (m – 2)/3
⇒ m – m/2 – 1/2 = 1 – (m/3 – 2/3)
⇒ m – m/2 + ½ = 1 – m/3 + 2/3
⇒ m – m/2 + m/3 = 1 + 2/3 – ½
⇒ m/2 + m/3 = ½ + 2/3
⇒ (3m + 2m)/6 = (3 + 4)/6
⇒ 5m/6 = 7/6
⇒ m = 7/6 × 6/5
⇒ m = 7/5

Simplify and solve the following linear equations.

7. 3(t – 3) = 5(2t + 1)

Solution:
3(t – 3) = 5(2t + 1)
⇒ 3t – 9 = 10t + 5
⇒ 3t – 10t = 5 + 9
⇒ -7t = 14
⇒ t = 14/-7
⇒ t = -2

8. 8. 15(y – 4) –2(y – 9) + 5(y + 6) = 0

Solution:
15(y – 4) –2(y – 9) + 5(y + 6) = 0
⇒ 15y – 60 -2y + 18 + 5y + 30 = 0
⇒ 15y – 2y + 5y = 60 – 18 – 30
⇒ 18y = 12
⇒ y = 12/18
⇒ y = 2/3

9. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

Solution:
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
⇒ 15z – 21 – 18z + 22 = 32z – 52 – 17
⇒ 15z – 18z – 32z = -52 – 17 + 21 – 22
⇒ -35z = -70
⇒ z = -70/-35
⇒ z = 2

10. 0.25(4f – 3) = 0.05(10f – 9)

Solution:
0.25(4f – 3) = 0.05(10f – 9)
⇒ f – 0.75 = 0.5f – 0.45
⇒ f – 0.5f = -0.45 + 0.75
⇒ 0.5f = 0.30
⇒ f = 0.30/0.5
⇒ f = 3/5
⇒ f = 0.6