Class 7 Exam  >  Class 7 Notes  >  RD Sharma Solutions for Class 7 Mathematics  >  RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics PDF Download

Question 13:

Solve each of the following equations and check your answers:
 3 (x + 2) = 15

Answer 13:

3 (x + 2) = 15
Dividing both sides by 3, we get

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Subtracting 2 from both sides, we get
⇒ x + 2 −- 2 = 5 −- 2
⇒ x = 3
Verification:
Substituting x = 3 in LHS, we get
LHS = 3 (x + 2)= 3 (3+2) = 3××5 = 15, and RHS = 15
LHS = RHS
Hence, verified.

Question 14:

Solve each of the following equations and check your answers:

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Answer 14:

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Multiplying both sides by 4, we get

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Verification:

Substituting x = 7/2 in LHS, we get
LHS = RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

LHS = RHS
Hence, verified.

Question 15:

Solve each of the following equations and check your answers:

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Answer 15:

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Subtracting 1/3 from both sides, we get

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Multiplying both sides by −-1, we get

RD Sharma Solutions (Part - 2) - 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 - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Verification:

Substituting x = 1/6 in LHS, we get
LHS = RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

LHS = RHS
Hence, verified.

Question 16:

Solve each of the following equations and check your answers:
 3(x + 6) = 24

Answer 16:

3(x + 6) = 24

Dividing both sides by 3, we get

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Subtracting 6 from both sides, we get
⇒ x + 6 − 6 = 8 − 6
⇒ x = 2
Verification:
Substituting x = 2 in LHS, we get
LHS = 3 (x + 6) = 3 (2 + 6) = 3×8 = 24, and RHS = 24
LHS = RHS
Hence, verified.

Question 17:

Solve each of the following equations and check your answers:
 3(x + 2) − 2(x − 1) = 7

Answer 17:

3(x + 2) − 2(x − 1) = 7
On expanding the brackets, we get 
⇒ (3× x) +( 3 × 2) − (2 × x) + (2 × 1) = 73× 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
LHS = 3 (x + 2) −2(x −1), and RHS = 7
LHS = 3 (−1 + 2) −2(−1−1) = (3×1) − (2×−2) = 3 + 4  = 7, and RHS = 7
LHS = RHS
Hence, verified.

Question 18:

Solve each of the following equations and check your answers:
 8(2x − 5) − 6(3x − 7) = 1

Answer 18:

8(2x − 5) − 6(3x − 7) = 1
On expanding 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

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Verification:

Substituting x = 1/2 in LHS, we get

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

LHS = RHS
Hence, verified

Question 19:

Solve each of the following equations and check your answers:
 6(1 − 4x) + 7(2 + 5x) = 53

Answer 19:

6(1 − 4x) + 7(2 + 5x) = 53
On expanding 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)
= 6(1 − 12) + 7(2 + 15)
= 6(−11) + 7(17)
= −66 + 119 = 53 = RHS
LHS = RHS
Hence, verified.

Question 20:

Solve each of the following equations and check your answers:
 5(2 − 3x) − 17(2x − 5) = 16

Answer 20:

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

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Verification:
Substituting x = 79/49  in LHS, we get

 RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

= RHS
So, LHS = RHS
Hence, verified.

Question 21:

Solve each of the following equations and check your answers:

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Answer 21:

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Adding 2 to both sides, we get

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Multiplying both sides by 5, we get

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Adding 3 to both sides, we get
⇒ x −3 + 3  = 5 + 3
⇒ x = 8
Verification:
Substituting x = 8  in LHS, we get

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

= RHS
LHS = RHS
Hence, verified.

Question 22:

Solve each of the following equations and check your answers:
 5(x − 2) + 3(x + 1) = 25

Answer 22:

5(x − 2) + 3(x + 1) = 25
On expanding 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

RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Verification:
Substituting x = 4  in LHS, we get

= 5(4 − 2) + 3(4 + 1)
= 5(2) + 3(5)
= 10 + 15
= 25
= RHS
LHS = RHS
Hence, verified.

The document RD Sharma Solutions (Part - 2) - 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 - 2) - 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 power of one. They can be represented in the form ax + b = 0, where "a" and "b" are constants and "x" is the variable.
2. How can I solve linear equations in one variable?
Ans. To solve linear equations in one variable, you need to isolate the variable on one side of the equation. You can do this by performing various operations such as addition, subtraction, multiplication, and division on both sides of the equation. The goal is to simplify the equation until you are left with the variable on one side and the constant on the other side.
3. Can linear equations in one variable have multiple solutions?
Ans. Yes, linear equations in one variable can have multiple solutions. If the equation simplifies to a true statement, such as 2 = 2 or 3 = 3, it means that every value of the variable will satisfy the equation. In such cases, the equation has infinitely many solutions.
4. Are there any real-life applications of linear equations in one variable?
Ans. Yes, linear equations in one variable have several real-life applications. They can be used to solve problems related to distance, speed, time, money, and proportions. For example, if you want to calculate how long it will take to travel a certain distance at a given speed, you can use a linear equation in one variable to find the answer.
5. Are there any tips or tricks to solve linear equations in one variable more efficiently?
Ans. Yes, there are a few tips that can help you solve linear equations in one variable more efficiently. One tip is to try and simplify the equation by combining like terms before performing any operations. Another tip is to always perform the same operation on both sides of the equation to maintain equality. Additionally, if the equation involves fractions, you can multiply both sides by the least common denominator to eliminate the fractions and make the calculations easier.
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