# RS Aggarwal Solutions for Class 6 Chapter 9 Linear Equations in One Variable Exercise 9B

The fundamental concepts which are explained in Class 6 are important as they are continued in higher grades also. RS Aggarwal Solutions which are prepared by faculty at BYJUâ€™S are relevant to the CBSE guidelines of Class 6. Exercise 9B contains problems which help in solving equations using a systematic method. The students can practice the exercise wise problems using the PDF of solutions to ace the exam. Students can download RS Aggarwal Solutions for Class 6 Chapter 9 Linear Equations in One Variable Exercise 9B PDF here.

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### Access answers to Maths RS Aggarwal Solutions for Class 6 Chapter 9 Linear Equations in One Variable Exercise 9B

Solve each of the following equations and verify the answer in each case:

1. x+ 5 = 12

Solution

Given x + 5 = 12

Subtracting -5 from both sides

x + 5 â€“ 5 = 12 â€“ 5

x = 7

Check

Substituting x = 7 in equation x + 5 = 12

We get

7 + 5 = 12

12 = 12

LHS = RHS

âˆ´ LHS = RHS, when x = 7

2. x + 3 = -2

Solution

Given

x + 3 = – 2

Subtracting -3 from both sides

x + 3 â€“ 3 = -2 â€“ 3

x = -5

Check

Substituting x = -5 in equation x + 3 = – 2

We get,

x + 3 = -2

-5 + 3 = -2

-2 = -2

LHS = RHS

âˆ´ LHS = RHS, when x = -5

3. x â€“ 7 = 6

Solution

Given

x â€“ 7 = 6

x â€“ 7 + 7 = 6 + 7

x = 13

Check

Substituting x = 13 in equation x -7 = 6

We get,

x â€“ 7 = 6

13 â€“ 7 = 6

6 = 6

LHS = RHS

âˆ´ LHS = RHS, when x = 13

4. x â€“ 2 = -5

Solution

Given

x â€“ 2 = -5

x â€“ 2 + 2 = -5 + 2

x = -3

Check

Substituting x = -3 in equation x â€“ 2 = -5

We get,

x â€“ 2 = -5

-3 â€“ 2 = -5

-5 = -5

LHS = RHS

âˆ´ LHS = RHS, when x = -3

5. 3x â€“ 5 = 13

Solution

Given

3x â€“ 5 = 13

3x â€“ 5 + 5 = 13 + 5

3x = 18

x = 18 / 3

x = 6

Check

Substituting x = 6 in equation 3x – 5 = 13

We get,

3x â€“ 5 = 13

3 (6) â€“ 5 = 13

3 Ã— 6 â€“ 5 = 13

18 â€“ 5 = 13

13 = 13

LHS = RHS

âˆ´ LHS = RHS, when x = 6

6. 4x + 7 = 15

Solution

Given

4x + 7 = 15

Subtracting 7 from both sides

4x + 7 â€“ 7 = 15 â€“ 7

4x = 8

x = 8 / 4

x = 2

Check

Substituting x = 2 in equation 4x + 7 = 15

We get,

4x + 7 = 15

4 (2) + 7 = 15

4 Ã— 2 + 7 = 15

8 + 7 = 15

15 = 15

LHS = RHS

âˆ´ LHS = RHS, when x = 2

7. x / 5 = 12

Solution

Given

x / 5 = 12

Multiplying both sides by 5

x / 5 Ã— 5 = 12 Ã— 5

x = 60

Check

Substitute x = 60 in equation x / 5 = 12

60 / 5 = 12

12 = 12

LHS = RHS

âˆ´ LHS = RHS, when x = 60

8. 3x / 5 = 15

Solution

Given

3x / 5 = 15

Multiplying both sides by 5

3x / 5 Ã— 5 = 15 Ã— 5

3x = 75

x = 75 / 3

x = 25

Check

Substitute x = 25 in equation 3x / 5 = 15

3x / 5 = 15

3 Ã— 25 / 5 = 15

3 Ã— 5 = 15

15 = 15

LHS = RHS

âˆ´ LHS = RHS, when x = 25

9. 5x â€“ 3 = x + 17

Solution

Given

5x â€“ 3 = x + 17

Transposing x to LHS and -3 to RHS

5x â€“ x = 17 + 3

4x = 20

x = 20 / 4

x = 5

Check

Substituting x = 5 in equation 5x â€“ 3 = x + 17

5x -3 = x + 17

5 (5) â€“ 3 = 5 + 17

5 Ã— 5 â€“ 3 = 22

25 â€“ 3 = 22

22 = 22

LHS = RHS

âˆ´ LHS = RHS, when x = 5

10. 2x â€“ 1 / 2 = 3

Solution

Given

2x â€“ 1 / 2 = 3

Adding 1 / 2 to both sides

2x â€“ 1 / 2 + 1 / 2 = 3 + 1 / 2

2x â€“ 0 = (6 +1) / 2 [By taking LCM]

2x = 7 / 2

Dividing both sides by 2

2x / 2 = 7 / 2 Ã— 2

x = 7 / 4

Check

Substituting x = 7 / 4 in equation 2x â€“ 1 / 2 = 3

2x â€“ 1 / 2 = 3

2 (7 / 4) â€“ 1 / 2 = 3

2 Ã— 7 / 4 â€“ 1 / 2 = 3

7 / 2 â€“ 1 / 2 = 3

(7 â€“ 1) / 2 = 3

6 / 2 = 3

3 = 3

LHS = RHS

âˆ´ LHS = RHS, when x = 7 / 4

11. 3(x + 6) = 24

Solution

Given

3(x + 6) = 24

3x + 18 = 24 [removing parentheses]

Subtracting 18 from both sides

3x + 18 â€“ 18 = 24 â€“ 18

3x = 6

x = 6 / 3

x = 2

Check

Substituting x = 2 in equation 3(x + 6) = 24

3(x + 6) = 24

3(2 + 6) = 24

3 (8) = 24

3 Ã— 8 = 24

24 = 24

LHS = RHS

LHS = RHS

âˆ´ LHS = RHS, when x = 2

12. 6x + 5 = 2x + 17

Solution

Given

6x + 5 = 2x + 17

Transposing 2x to LHS and 5 to RHS

6x â€“ 2x = 17 â€“ 5

4x = 12

x = 12 / 4

x = 3

Check

Substituting x = 3 in equation 6x + 5 = 2x + 17

LHS = 6x + 5

= 6 (3) + 5

= 6 Ã— 3 + 5

= 18 + 5

= 23

RHS = 2x + 17

= 2 (3) + 17

= 2 Ã— 3 + 17

= 6 + 17

= 23

LHS = RHS

âˆ´ LHS = RHS, when x = 3

13. x / 4 â€“ 8 = 1

Solution

Given

x / 4 – 8 = 1

x / 4 â€“ 8 + 8 = 1 + 8

x / 4 = 9

Multiplying both sides by 4

x / 4 Ã— 4 = 9 Ã— 4

x = 36

Check

Substituting x = 36 in equation x / 4 â€“ 8 = 1

x / 4 â€“ 8 = 1

36 / 4 â€“ 8 = 1

9 â€“ 8 = 1

1 = 1

LHS = RHS

âˆ´ LHS = RHS, when x = 36