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CBSE Class 7  >  Class 7 Notes  >  Mathematics (Maths) (Old NCERT)  >  RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2)

RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2)

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 Page 1


 
 
 
 
 
 
 
Exercise 8.2        Page No: 8.12 
 
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) 
Page 2


 
 
 
 
 
 
 
Exercise 8.2        Page No: 8.12 
 
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. 
Page 3


 
 
 
 
 
 
 
Exercise 8.2        Page No: 8.12 
 
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: 
Page 4


 
 
 
 
 
 
 
Exercise 8.2        Page No: 8.12 
 
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) 
Page 5


 
 
 
 
 
 
 
Exercise 8.2        Page No: 8.12 
 
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)
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Document Description: RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) for Class 7 2026 is part of Mathematics (Maths) Class 7 (Old NCERT) preparation. The notes and questions for RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) have been prepared according to the Class 7 exam syllabus. Information about RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) covers topics like and RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2) Example, for Class 7 2026 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.2).
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