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 available here. The BYJU’S subject experts in Maths have collated solutions to the questions present in this exercise. This exercise explains solving linear equations. One of the methods is the systematic method, which includes
- Equation involving addition
- 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 be able to solve problems effortlessly.
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
Solve each of the following equations and check your answers:
1. x – 3 = 5
Solution:
Given x – 3 = 5
Adding 3 to both sides we get,
x – 3 + 3 = 5 + 3
x = 8
Verification:
Substituting x = 8 in LHS, we get
LHS = x – 3 and RHS = 5
LHS = 8 – 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 – 9 = 13 – 9
x = 4
Verification:
Substituting x = 4 on LHS, we get
LHS = 4 + 9 = 13 = RHS
LHS = RHS
Hence, verified.
3. x – (3/5) = (7/5)
Solution:
Given x – (3/5) = (7/5)
Add (3/5) to both sides, we get
x – (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 – (3/5) = (7/5)
(10 – 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/2 = 0
Multiplying both sides by 2, we get
(x/2) × 2 = 0 × 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 – (1/3) = (2/3)
Solution:
Given x – (1/3) = (2/3)
Adding (1/3) to both sides, we get
x – (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 – (1/3) = (2/3)
(3 – 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) – (1/2) = (7/2) – (1/2)
x = (7 – 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 – y = 6
Solution:
Given 10 – y = 6
Subtracting 10 from both sides, we get
10 – y – 10 = 6 – 10
-y = -4
Multiplying both sides by -1, we get
-y × -1 = – 4 × – 1
y = 4
Verification:
Substituting y = 4 in LHS, we get
10 – y = 10 – 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 – 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 – 12 = -5, and RHS = -5
Therefore LHS = RHS
Hence, verified.
10. (4/5) – x = (3/5)
Solution:
Given (4/5) – x = (3/5)
Subtracting (4/5) from both sides, we get
(4/5) – x – (4/5) = (3/5) – (4/5)
– x = (3 -4)/5
– x = (-1/5)
x = (1/5)
Verification:
Substituting x = (1/5) in LHS we get
(4/5) – (1/5) = (3/5)
(4 -1)/5 = (3/5)
(3/5) = (3/5)
Therefore LHS =RHS
Hence, verified.
11. 2y – (1/2) = (-1/3)
Solution:
Given 2y – (1/2) = (-1/3)
Adding (1/2) from both the sides, we get
2y – (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) – (1/2) = (-1/3)
(1/6) – (1/2) = (-1/3)
(2 – 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) – 8
Solution:
Given 14 = (7x/10) – 8
Adding 8 to both sides we get,
14 + 8 = (7x/10) – 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) × (220/7) – 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) × 4 = (7/8) × 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) – 2x = 0
Solution:
Given (1/3) – 2x = 0
Subtract (1/3) from both sides we get,
(1/3) – 2x – (1/3) = 0 – (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) – 2 (1/6) = 0
(1/3) – (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 – 6 = 8 – 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) – 2 (x – 1) = 7
Solution:
Given 3 (x + 2) – 2 (x – 1) = 7
On simplifying the brackets, we get
3 × x + 3 × 2 – 2 × x + 2 × 1 = 7
3x + 6 – 2x + 2 = 7
3x – 2x + 6 + 2 = 7
x + 8 = 7
Subtracting 8 from both sides, we get
x + 8 – 8 = 7 – 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
(3×1) – (2×-2) = 7
3 + 4 = 7
Therefore LHS = RHS
Hence, verified.
18. 8 (2x – 5) – 6(3x – 7) = 1
Solution:
Given 8 (2x – 5) – 6(3x – 7) = 1
On simplifying the brackets, we get
(8 × 2x) – (8 × 5) – (6 × 3x) + (-6) × (-7) = 1
16x – 40 – 18x + 42 = 1
16x – 18x + 42 – 40 = 1
-2x + 2 = 1
Subtracting 2 from both sides, we get
-2x+ 2 – 2 = 1 -2
-2x = -1
Multiplying both sides by -1, we get
-2x × (-1) = -1× (-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,
(8 × (2 × (1/2)– 5) – (6 × (3 × (1/2) -7) = 1
8(1 – 5) – 6(3/2 – 7) = 1
8× (-4) – (6 × 3/2) + (6 × 7) = 1
– 32 – 9 + 42 = 1
– 41 + 42 = 1
1 = 1
Therefore LHS = RHS
Hence, verified.
19. 6 (1 – 4x) + 7 (2 + 5x) = 53
Solution:
Given 6 (1 – 4x) + 7 (2 + 5x) = 53
On simplifying the brackets, we get
(6 ×1) – (6 × 4x) + (7 × 2) + (7 × 5x) = 53
6 – 24x + 14 + 35x = 53
6 + 14 + 35x – 24x = 53
20 + 11x = 53
Subtracting 20 from both sides, we get 20 + 11x – 20 = 53 – 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 – 4 × 3) + 7(2 + 5 × 3) = 53
6(1 – 12) + 7(2 + 15) = 53
6(-11) + 7(17) = 53
– 66 + 119 = 53
53 = 53
Therefore LHS = RHS
Hence, verified.
20. 5 (2 – 3x) -17 (2x -5) = 16
Solution:
Given 5 (2 – 3x) -17 (2x – 5) = 16
On expanding the brackets, we get
(5 × 2) – (5 × 3x) – (17 × 2x) + (17 × 5) = 16
10 – 15x – 34x + 85 = 16
10 + 85 – 34x – 15x = 16
95 – 49x = 16
Subtracting 95 from both sides, we get
– 49x + 95 – 95 = 16 – 95
– 49x = -79
Dividing both sides by – 49, we get
– 49x/ -49 = -79/-49
x = 79/49
Verification:
Substituting x = (79/49) in LHS we get,
5 (2 – 3 × (79/49) – 17 (2 × (79/49) – 5) = 16
(5 × 2) – (5 × 3 × (79/49)) – (17 × 2 × (79/49)) + (17 × 5) = 16
10 – (1185/49) – (2686/49) + 85 = 16
(490 – 1185 – 2686 + 4165)/49 = 16
784/49 = 16
16 = 16
Therefore LHS = RHS
Hence, verified.
21. (x – 3)/5 -2 = -1
Solution:
Given ((x – 3)/5) -2 = -1
Adding 2 to both sides we get,
((x -3)/5) – 2 + 2 = -1 + 2
(x -3)/5 = 1
Multiply both sides by 5 we get
(x – 3)/ 5 × 5 = 1 × 5
x – 3 = 5
Now add 3 to both sides we get,
x – 3 + 3 = 5 + 3
x = 8
Verification:
Substituting x = 8 in LHS we get,
((8 – 3)/5) -2 = -1
(5/5) – 2 = -1
1 -2 = -1
-1 = -1
Therefore LHS = RHS
Hence, verified.
22. 5 (x – 2) + 3 (x +1) = 25
Solution:
Given 5 (x – 2) + 3 (x +1) = 25
On simplifying the brackets, we get
(5 × x) – (5 × 2) +3 × x + 3× 1 = 25
5x – 10 + 3x + 3 = 25
5x + 3x – 10 + 3 = 25
8x – 7 = 25
Adding 7 to both sides, we get
8x – 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 – 2) + 3(4 + 1) = 25
5(2) + 3(5) = 25
10 + 15 = 25
25 = 25
Therefore LHS = RHS
Hence, verified.
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