Class 8 Maths Chapter 2 – Linear Equations in One Variable Exercise 2.3

Class 8 Maths Chapter 2 – Linear Equations in One Variable Exercise 2.3 :-

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Solve the following equations and check your results.

1. 3x = 2x + 18

Solution:

3x = 2x + 18

⇒ 3x – 2x = 18

⇒ x = 18

Putting the value of x in RHS and LHS we get, 3 × 18 = (2 × 18) +18

⇒ 54 = 54

⇒ LHS = RHS

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2. 5t – 3 = 3t – 5

Solution:

5t – 3 = 3t – 5

⇒ 5t – 3t = -5 + 3

⇒ 2t = -2

⇒ t = -1

Putting the value of t in RHS and LHS we get, 5× (-1) – 3 = 3× (-1) – 5

⇒ -5 – 3 = -3 – 5

⇒ -8 = -8

⇒ LHS = RHS

3. 5x + 9 = 5 + 3x

Solution:

5x + 9 = 5 + 3x

⇒ 5x – 3x = 5 – 9

⇒ 2x = -4

⇒ x = -2

Putting the value of x in RHS and LHS we get, 5× (-2) + 9 = 5 + 3× (-2)

⇒ -10 + 9 = 5 + (-6)

⇒ -1 = -1

⇒ LHS = RHS

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4. 4z + 3 = 6 + 2z

Solution:

4z + 3 = 6 + 2z

⇒ 4z – 2z = 6 – 3

⇒ 2z = 3

⇒ z = 3/2

Putting the value of z in RHS and LHS we get,

(4 × 3/2) + 3 = 6 + (2 × 3/2)

⇒ 6 + 3 = 6 + 3

⇒ 9 = 9

⇒ LHS = RHS

5. 2x – 1 = 14 – x

Solution:

2x – 1 = 14 – x

⇒ 2x + x = 14 + 1

⇒ 3x = 15

⇒ x = 5

Putting the value of x in RHS and LHS we get, (2×5) – 1 = 14 – 5

⇒ 10 – 1 = 9

⇒ 9 = 9

⇒ LHS = RHS

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6. 8x + 4 = 3 (x – 1) + 7

Solution:

8x + 4 = 3 (x – 1) + 7

⇒ 8x + 4 = 3x – 3 + 7

⇒ 8x + 4 = 3x + 4

⇒ 8x – 3x = 4 – 4

⇒ 5x = 0

⇒ x = 0

Putting the value of x in RHS and LHS we get, (8×0) + 4 = 3 (0 – 1) + 7

⇒ 0 + 4 = 0 – 3 + 7

⇒ 4 = 4

⇒ LHS = RHS

7. x = 4/5 (x + 10)

Solution:

x = 4/5 (x + 10)

⇒ x = 4x/5 + 40/5

⇒ x – (4x/5) = 8

⇒ (5x – 4x)/5 = 8

⇒ x = 8 × 5

⇒ x = 40

Putting the value of x in RHS and LHS we get,

40 = 4/5 (40 + 10)

⇒ 40 = 4/5 × 50

⇒ 40 = 200/5

⇒ 40 = 40

⇒ LHS = RHS

8. 2x/3 + 1 = 7x/15 + 3

Solution:

2x/3 + 1 = 7x/15 + 3

⇒ 2x/3 – 7x/15 = 3 – 1

⇒ (10x – 7x)/15 = 2

⇒ 3x = 2 × 15

⇒ 3x = 30

⇒ x = 30/3

⇒ x = 10

Putting the value of x in RHS and LHS we get,

9. 2y + 5/3 = 26/3 – y

Solution:

2y + 5/3 = 26/3 – y

⇒ 2y + y = 26/3 – 5/3

⇒ 3y = (26 – 5)/3

⇒ 3y = 21/3

⇒ 3y = 7

⇒ y = 7/3

Putting the value of y in RHS and LHS we get,

⇒ (2 × 7/3) + 5/3 = 26/3 – 7/3

⇒ 14/3 + 5/3 = 26/3 – 7/3

⇒ (14 + 5)/3 = (26 – 7)/3

⇒ 19/3 = 19/3

⇒ LHS = RHS

10. 3m = 5m – 8/5

Solution:

3m = 5m – 8/5

⇒ 5m – 3m = 8/5

⇒ 2m = 8/5

⇒ 2m × 5 = 8

⇒ 10m = 8

⇒ m = 8/10

⇒ m = 4/5

Putting the value of m in RHS and LHS we get,

⇒ 3 × (4/5) = (5 × 4/5) – 8/5

⇒ 12/5 = 4 – (8/5)

⇒ 12/5 = (20 – 8)/5

⇒ 12/5 = 12/5

⇒ LHS = RHS

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