RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions for Class 7 Mathematics

Created by: Abhishek Kapoor

Class 7 : RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

The document RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
All you need of Class 7 at this link: Class 7

Question 1:

Solve each of the following equations and check your answers:
x − 3 = 5

Answer 1:

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.

Question 2:

Solve each of the following equations and check your answers:
x +9  = 13

Answer 2:

x + 9 = 13
Subtracting 9 from both sides, 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.

Question 3:

Solve each of the following equations and check your answers:

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Answer 3:

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Verification:
Substituting x = 2 in LHS, we get

LHS = RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

LHS = RHS
Hence, verified.

Question 4:

Solve each of the following equations and check your answers:
 3x = 0

Answer 4:

3x = 0
Dividing both sides by 3, we get

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Verification:

Substituting x = 0 in LHS = 3x, we get
LHS = 3 ×× 0 = 0 and RHS = 0
LHS = RHS
Hence, verified.

 

Question 5:

Solve each of the following equations and check your answes: x/2=0

Answer 5:

x/2=0

Multiplying both sides by 2, we get

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Verification:

Substituting x= 0 in LHS, we get
LHS =0/2= 0 and RHS = 0
LHS = 0 and RHS = 0
LHS = RHS
Hence, verified.

Question 6:

Solve each of the following equations and check your answers:

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Answer 6:

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Verification:

Substituting x= 1 in LHS, we get

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

LHS = RHS
Hence, verified.

Question 7:

Solve each of the following equations and check your answers:

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Answer 7:

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Verification:

Substituting x = 3 in LHS, we get

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Question 8:

Solve each of the following equations and check your answers:
 10 − y = 6

Answer 8:

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
LHS = 10 − y = 10−4 = 6 and RHS = 6
LHS = RHS
Hence, verified.

Question 9:

Solve each of the following equations and check your answers:
 7 + 4y = −5

Answer 9:

7 + 4y = −5
Subtracting 7 from both sides, we get
⇒ 7 + 4y − 7 = −5 − 7
⇒  4y  = −12
Dividing both sides by 4, we get 

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Verification :
Substituting y = −3 in LHS, we get
LHS = 7 + 4y = 7  + 4(−3) = 7 − 12 = −5, and RHS = −5
LHS = RHS
Hence, verified.

Question 10:

Solve each of the following equations and check your answers:

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Answer 10:

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Subtracting 4/5 from both sides, we get

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Multiplying both sides by -1, we ge

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Verification :

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

LHS = RHS
Hence, verified.

Question 11:

Solve each of the following equations and check your answers:

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Answer 11:

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Dividing both sides by 2, we get

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Verification :

Substituting y = 1/12 in LHS, we get

LHS = RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

LHS = RHS
Hence, verified. 

Question 12:

Solve each of the following equations and check your answers:

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

ANSWER 12:

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Adding 8 to both sides, we get

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Multiplying both sides by 10, we get

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Dividing both sides by 7, we get

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

Verification :

Substituting x = 220/7 in RHS, we get

LHS = 14, and RHS =  RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev

LHS = RHS
Hence, verified.

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