# RD Sharma Solutions For Class 7 Maths Exercise 8.2 Chapter 8 Linear Equations in One Variable

Students can view and download the PDF of RD Sharma Solutions for Class 7 Maths Exercise 8.2 of Chapter 8 Linear Equations in One Variable are available here. The BYJUâ€™S experts in Maths formulate the questions present in this exercise. This exercise explains about solving linear equations. One of the methods is the systematic method which includes:

• Equation involving subtraction
• Equation involving multiplication
• Equation involving division
• Equation solvable by using more than one rule or operation

By practising RD Sharma Solutions for Class 7, students will able to solve problems with Math skills.

## Download the PDF of RD Sharma Solutions For Class 7 Maths Chapter 8 – Linear Equations in One variable Exercise 8.2

### Access answers to Maths RD Sharma Solutions For Class 7 Chapter 8 – Linear Equation in One Variable Exercise 8.2

1. x â€“ 3 = 5

Solution:

Given x â€“ 3 = 5

Adding 3 to both sides we get,

x â€“ 3 + 3 = 5 + 3

x = 8

Verification:

Substituting x = 8 in LHS, we get

LHS = x â€“ 3 and RHS = 5

LHS = 8 â€“ 3 = 5 and RHS = 5

LHS = RHS

Hence, verified.

2. x + 9 = 13

Solution:

Given x + 9 = 13

Subtracting 9 from both sides i.e. LHS and RHS, we get

x + 9 â€“ 9 = 13 â€“ 9

x = 4

Verification:

Substituting x = 4 on LHS, we get

LHS = 4 + 9 = 13 = RHS

LHS = RHS

Hence, verified.

3. x â€“ (3/5) = (7/5)

Solution:

Given x â€“ (3/5) = (7/5)

Add (3/5) to both sides, we get

x â€“ (3/5) + (3/5) = (7/5) + (3/5)

x = (7/5) + (3/5)

x = (10/5)

x = 2

Verification:

Substitute x = 2 in LHS of given equation, then we get

2 â€“ (3/5) = (7/5)

(10 â€“ 3)/5 = (7/5)

(7/5) = (7/5)

LHS = RHS

Hence, verified

4. 3x = 0

Solution:

Given 3x = 0

On dividing both sides by 3 we get,

(3x/3) = (0/3)

x = 0

Verification:

Substituting x = 0 in LHS we get

3 (0) = 0

And RHS = 0

Therefore LHS = RHS

Hence, verified.

5. (x/2) = 0

Solution:

Given x/2Â = 0

Multiplying both sides by 2, we get

(x/2) Ã— 2 = 0 Ã— 2

x = 0

Verification:

Substituting x = 0 in LHS, we get

LHS = 0/2 = 0 and RHS = 0

LHS = 0 and RHS = 0

Therefore LHS = RHS

Hence, verified.

6. x â€“ (1/3) = (2/3)

Solution:

Given x â€“ (1/3) = (2/3)

Adding (1/3) to both sides, we get

x â€“ (1/3) + (1/3) = (2/3) + (1/3)

x = (2 + 1)/3

x = (3/3)

x =1

Verification:

Substituting x = 1 in LHS, we get

1 â€“ (1/3) = (2/3)

(3 â€“ 1)/3 = (2/3)

(2/3) = (2/3)

Therefore LHS = RHS

Hence, verified.

7. x + (1/2) = (7/2)

Solution:

Given x + (1/2) = (7/2)

Subtracting (1/2) from both sides, we get

x + (1/2) â€“ (1/2) = (7/2) â€“ (1/2)

x = (7 â€“ 1)/2

x = (6/2)

x = 3

Verification:

Substituting x = 3 in LHS we get

3 + (1/2) = (7/2)

(6 + 1)/2 = (7/2)

(7/2) = (7/2)

Therefore LHS = RHS

Hence, verified.

8. 10 â€“ y = 6

Solution:

Given 10 â€“ y = 6

Subtracting 10 from both sides, we get

10 â€“ y â€“ 10 = 6 â€“ 10

-y = -4

Multiplying both sides by -1, we get

-yÂ Ã—Â -1 = – 4Â Ã—Â – 1

y = 4

Verification:

Substituting y = 4 in LHS, we get

10 â€“ y = 10 â€“ 4 = 6 and RHS = 6

Therefore LHS = RHS

Hence, verified.

9. 7 + 4y = -5

Solution:

Given 7 + 4y = -5

Subtracting 7 from both sides, we get

7 + 4y â€“ 7 = -5 -7

4y = -12

Dividing both sides by 4, we get

y = -12/ 4

y = -3

Verification:

Substituting y = -3 in LHS, we get

7 + 4y = 7 + 4(-3) = 7 â€“ 12 = -5, and RHS = -5

Therefore LHS = RHS

Hence, verified.

10. (4/5) â€“ x = (3/5)

Solution:

Given (4/5) â€“ x = (3/5)

Subtracting (4/5) from both sides, we get

(4/5) â€“ x â€“ (4/5) = (3/5) â€“ (4/5)

– x = (3 -4)/5

– x = (-1/5)

x = (1/5)

Verification:

Substituting x = (1/5) in LHS we get

(4/5) â€“ (1/5) = (3/5)

(4 -1)/5 = (3/5)

(3/5) = (3/5)

Therefore LHS =RHS

Hence, verified.

11. 2y â€“ (1/2) = (-1/3)

Solution:

Given 2y â€“ (1/2) = (-1/3)

Adding (1/2) from both the sides, we get

2y â€“ (1/2) + (1/2) = (-1/3) + (1/2)

2y = (-1/3) + (1/2)

2y = (-2 + 3)/6 [LCM of 3 and 2 is 6]

2y = (1/6)

Now divide both the side by 2, we get

y = (1/12)

Verification:

Substituting y = (1/12) in LHS we get

2 (1/12) â€“ (1/2) = (-1/3)

(1/6) â€“ (1/2) = (-1/3)

(2 â€“ 6)/12 = (-1/3) [LCM of 6 and 2 is 12]

(-4/12) = (-1/3)

(-1/3) = (-1/3)

Therefore LHS = RHS

Hence, verified.

12. 14 = (7x/10) â€“ 8

Solution:

Given 14 = (7x/10) â€“ 8

Adding 8 to both sides we get,

14 + 8 = (7x/10) â€“ 8 + 8

22 = (7x/10)

Multiply both sides by 10 we get,

220 = 7x

x = (220/7)

Verification:

Substituting x = (220/7) in RHS we get,

14 = (7/10) Ã— (220/7) â€“ 8

14 = 22 -8

14 = 14

Therefore LHS = RHS.

Hence, verified.

13. 3 (x + 2) = 15

Solution:

Given 3 (x + 2) = 15

Dividing both sides by 3 we get,

3 (x + 2)/3 = (15/3)

(x + 2) = 5

Now subtracting 2 by both sides, we get

x + 2 -2 = 5 -2

x = 3

Verification:

Substituting x =3 in LHS we get,

3 (3 + 2) = 15

3 (5) = 15

15 = 15

Therefore LHS = RHS

Hence, verified.

14. (x/4) = (7/8)

Solution:

Given (x/4) = (7/8)

Multiply both sides by 4 we get,

(x/4) Ã— 4 = (7/8) Ã— 4

x = (7/2)

Verification:

Substituting x = (7/2) in LHS we get,

(7/2)/4 = (7/8)

(7/8) = (7/8)

Therefore LHS = RHS

Hence, verified.

15. (1/3) â€“ 2x = 0

Solution:

Given (1/3) â€“ 2x = 0

Subtract (1/3) from both sides we get,

(1/3) â€“ 2x â€“ (1/3) = 0 â€“ (1/3)

– 2x = – (1/3)

2x = (1/3)

Divide both side by 2 we get,

2x/2 = (1/3)/2

x = (1/6)

Verification:

Substituting x = (1/6) in LHS we get,

(1/3) â€“ 2 (1/6) = 0

(1/3) â€“ (1/3) = 0

0 = 0

Therefore LHS = RHS

Hence, verified.

16. 3 (x + 6) = 24

Solution:

Given 3 (x + 6) = 24

Divide both the sides by 3 we get,

3 (x + 6)/3 = (24/3)

(x + 6) = 8

Now subtract 6 from both sides we get,

x + 6 â€“ 6 = 8 â€“ 6

x = 2

Verification:

Substituting x = 2 in LHS we get,

3 (2 + 6) = 24

3 (8) =24

24 = 24

Therefore LHS =RHS

Hence, verified.

17. 3 (x + 2) â€“ 2 (x â€“ 1) = 7

Solution:

Given 3 (x + 2) â€“ 2 (x â€“ 1) = 7

On simplifying the brackets, we get

3 Ã—Â x + 3Â Ã—Â 2 â€“ 2Â Ã—Â x + 2Â Ã—Â 1 = 7

3x + 6 â€“ 2x + 2 = 7

3x â€“ 2x + 6 + 2 = 7

x + 8 = 7

Subtracting 8 from both sides, we get

x + 8 â€“ 8 = 7 â€“ 8

x = -1

Verification:

Substituting x = -1 in LHS, we get

3 (x + 2) -2(x -1) = 7

3 (-1 + 2) -2(-1-1) = 7

(3Ã—1) â€“ (2Ã—-2) = 7

3 + 4 = 7

Therefore LHS = RHS

Hence, verified.

18. 8 (2x â€“ 5) â€“ 6(3x â€“ 7) = 1

Solution:

Given 8 (2x â€“ 5) â€“ 6(3x â€“ 7) = 1

On simplifying the brackets, we get

(8Â Ã—Â 2x) â€“ (8Â Ã—Â 5) â€“ (6Â Ã—Â 3x) + (-6)Â Ã—Â (-7) = 1

16x â€“ 40 â€“ 18x + 42 = 1

16x â€“ 18x + 42 â€“ 40 = 1

-2x + 2 = 1

Subtracting 2 from both sides, we get

-2x+ 2 â€“ 2 = 1 -2

-2x = -1

Multiplying both sides by -1, we get

-2xÂ Ã—Â (-1) = -1Ã—Â (-1)

2x = 1

Dividing both sides by 2, we get

2x/2 = (1/2)

x = (1/2)

Verification:

Substituting x = (1/2) in LHS we get,

(8Â Ã— (2 Ã— (1/2)â€“ 5) â€“ (6Â Ã— (3 Ã— (1/2) -7) = 1

8(1 â€“ 5) â€“ 6(3/2 â€“ 7) = 1

8Ã—Â (-4) â€“ (6Â Ã—Â 3/2) + (6Â Ã—Â 7) = 1

– 32 â€“ 9 + 42 = 1

â€“ 41 + 42 = 1

1 = 1

Therefore LHS = RHS

Hence, verified.

19. 6 (1 â€“ 4x) + 7 (2 + 5x) = 53

Solution:

Given 6 (1 â€“ 4x) + 7 (2 + 5x) = 53

On simplifying the brackets, we get

(6 Ã—1) â€“ (6 Ã— 4x) + (7 Ã— 2) + (7 Ã— 5x) = 53

6 â€“ 24x + 14 + 35x = 53

6 + 14 + 35x â€“ 24x = 53

20 + 11x = 53

Subtracting 20 from both sides, we get 20 + 11x â€“ 20 = 53 â€“ 20

11x = 33

Dividing both sides by 11, we get

11x/11 = 33/11

x = 3

Verification:

Substituting x = 3 in LHS, we get

6(1 â€“ 4 Ã— 3) + 7(2 + 5 Ã— 3) = 53

6(1 â€“ 12) + 7(2 + 15) = 53

6(-11) + 7(17) = 53

– 66 + 119 = 53

53 = 53

Therefore LHS = RHS

Hence, verified.

20. 5 (2 â€“ 3x) -17 (2x -5) = 16

Solution:

Given 5 (2 â€“ 3x) -17 (2x – 5) = 16

On expanding the brackets, we get

(5Â Ã—Â 2) â€“ (5Â Ã—Â 3x) â€“ (17Â Ã—Â 2x) + (17Â Ã—Â 5) = 16

10 â€“ 15x â€“ 34x + 85 = 16

10 + 85 â€“ 34x â€“ 15x = 16

95 â€“ 49x = 16

Subtracting 95 from both sides, we get

â€“ 49x + 95 â€“ 95 = 16 â€“ 95

– 49x = -79

Dividing both sides by â€“ 49, we get

– 49x/ -49 = -79/-49

x = 79/49

Verification:

Substituting x = (79/49) in LHS we get,

5 (2 â€“ 3 Ã— (79/49) – 17 (2 Ã— (79/49) – 5) = 16

(5 Ã— 2) â€“ (5 Ã— 3 Ã— (79/49)) â€“ (17 Ã— 2 Ã— (79/49)) + (17 Ã— 5) = 16

10 â€“ (1185/49) â€“ (2686/49) + 85 = 16

(490 â€“ 1185 â€“ 2686 + 4165)/49 = 16

784/49 = 16

16 = 16

Therefore LHS = RHS

Hence, verified.

21. (x â€“ 3)/5 -2 = -1

Solution:

Given ((x â€“ 3)/5) -2 = -1

Adding 2 to both sides we get,

((x -3)/5) â€“ 2 + 2 = -1 + 2

(x -3)/5 = 1

Multiply both sides by 5 we get

(x â€“ 3)/ 5 Ã— 5 = 1 Ã— 5

x â€“ 3 = 5

Now add 3 to both sides we get,

x â€“ 3 + 3 = 5 + 3

x = 8

Verification:

Substituting x = 8 in LHS we get,

((8 â€“ 3)/5) -2 = -1

(5/5) â€“ 2 = -1

1 -2 = -1

-1 = -1

Therefore LHS = RHS

Hence, verified.

22. 5 (x â€“ 2) + 3 (x +1) = 25

Solution:

Given 5 (x â€“ 2) + 3 (x +1) = 25

On simplifying the brackets, we get

(5Â Ã—Â x) â€“ (5Â Ã—Â 2) +3 Ã—Â x + 3Ã—Â 1 = 25

5x â€“ 10 + 3x + 3 = 25

5x + 3x â€“ 10 + 3 = 25

8x â€“ 7 = 25

Adding 7 to both sides, we get

8x â€“ 7 + 7 = 25 + 7

8x = 32

Dividing both sides by 8, we get

8x/8 = 32/8

x = 4

Verification:

Substituting x = 4 in LHS, we get

5(4 â€“ 2) + 3(4 + 1) = 25

5(2) + 3(5) = 25

10 + 15 = 25

25 = 25

Therefore LHS = RHS

Hence, verified.