Class 7 Exam  >  Class 7 Notes  >  RD Sharma Solutions for Class 7 Mathematics  >  RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math

RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics PDF Download

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 | RD Sharma Solutions for Class 7 Mathematics

Answer 3:

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

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

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 | RD Sharma Solutions for Class 7 Mathematics

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 | RD Sharma Solutions for Class 7 Mathematics

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 | RD Sharma Solutions for Class 7 Mathematics

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 | RD Sharma Solutions for Class 7 Mathematics

Answer 6:

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

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

Verification:

Substituting x= 1 in LHS, we get

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

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 | RD Sharma Solutions for Class 7 Mathematics

Answer 7:

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

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

Verification:

Substituting x = 3 in LHS, we get

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

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 | RD Sharma Solutions for Class 7 Mathematics

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 | RD Sharma Solutions for Class 7 Mathematics

Answer 10:

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

Subtracting 4/5 from both sides, we get

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

Multiplying both sides by -1, we ge

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

Verification :

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

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 | RD Sharma Solutions for Class 7 Mathematics

Answer 11:

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

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

Dividing both sides by 2, we get

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

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 | RD Sharma Solutions for Class 7 Mathematics

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 | RD Sharma Solutions for Class 7 Mathematics

ANSWER 12:

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

Adding 8 to both sides, we get

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

Multiplying both sides by 10, we get

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

Dividing both sides by 7, we get

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

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 | RD Sharma Solutions for Class 7 Mathematics

LHS = RHS
Hence, verified.

The document RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
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FAQs on RD Sharma Solutions (Part - 1) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math - RD Sharma Solutions for Class 7 Mathematics

1. What are linear equations in one variable?
Ans. Linear equations in one variable are algebraic equations that involve only one variable raised to the first power and have a degree of one. These equations can be represented in the form of ax + b = 0, where 'a' and 'b' are constants and 'x' is the variable.
2. How do you solve linear equations in one variable?
Ans. To solve linear equations in one variable, follow these steps: 1. Simplify both sides of the equation by combining like terms. 2. Move all the variable terms to one side of the equation and the constant terms to the other side. 3. If necessary, multiply or divide both sides of the equation by a constant to isolate the variable. 4. Solve for the variable by performing any remaining operations. 5. Check the solution by substituting it back into the original equation.
3. Can linear equations in one variable have more than one solution?
Ans. No, linear equations in one variable can have only one solution. The solution represents the value of the variable that satisfies the equation. If there were more than one solution, it would imply that the equation has multiple values that satisfy it, which contradicts the definition of a linear equation.
4. What is the importance of solving linear equations in one variable?
Ans. Solving linear equations in one variable is important in various real-life scenarios and mathematical problems. It helps in determining unknown quantities, finding the value of a variable in equations involving proportions, calculating rates of change, solving problems related to distance, time, and speed, and analyzing patterns or relationships between variables.
5. Are there any alternative methods to solve linear equations in one variable?
Ans. Yes, there are alternative methods to solve linear equations in one variable. Some of these methods include graphing the equation and finding the point of intersection, using matrices and determinants, applying the substitution method, using the elimination method, or using the graphical method. These methods provide different approaches to solving linear equations and can be used based on the specific problem or preference.
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