Get free PDF of RD Sharma Solutions for Class 7 Maths Exercise 8.1 of Chapter 8 Linear Equation in One Variable from the provided links. These PDFs can be easily downloaded from the students. Our experts have uniquely formulated these questions to approach a studentâ€™s mentality. RD Sharma Solutions for Class 7 is one of the best study material designed for CBSE students. This exercise includes linear equations. Some of the topics covered in this exercise are listed below:

- Equation – A statement of equality which involves one or more literals
- Linear equation – An equation in which the highest power of the variables involved is 1
- Solution of an linear equation
- Solving a linear equation by trial-and-error method

## Download the PDF of RD Sharma Solutions For Class 7 Chapter 8 – Linear Equations in One variable Exercise 8.1

### Access answers to Maths RD Sharma Solutions For Class 7 Chapter 8 – Linear Equation in One Variable Exercise 8.1

**1. Verify by substitution that:**

**(i) x = 4 is the root of 3x â€“ 5 = 7**

**(ii) x = 3 is the root of 5 + 3x = 14**

**(iii) x = 2 is the root of 3x â€“ 2 = 8x â€“ 12**

**(iv) x = 4 is the root of (3x/2) = 6**

**(v) y = 2 is the root of y â€“ 3 = 2y â€“ 5**

**(vi) x = 8 is the root of (1/2)x + 7 = 11**

**Solution:**

(i) Given x = 4 is the root of 3x – 5 = 7

Now, substituting x = 4 in place of â€˜xâ€™ in the given equation, we get

= 3(4) â€“ 5 = 7

= 12 â€“ 5 = 7

7 = 7

Since, LHS = RHS

Hence, x = 4 is the root of 3x – 5 = 7.

(ii) Given x = 3 is the root of 5 + 3x = 14.

Now, substituting x = 3 in place of â€˜xâ€™ in the given equation, we get

= 5 + 3(3) = 14

= 5 + 9 = 14

14 = 14

Since, LHS = RHS

Hence, x = 3 is the root of 5 + 3x = 14.

(iii) Given x = 2 is the root of 3x â€“ 2 = 8x â€“ 12.

Now, substituting x = 2 in place of â€˜xâ€™ in the given equation, we get

= 3(2) â€“ 2 = 8(2) â€“ 12

= 6 â€“ 2 = 16 â€“ 12

4 =Â 4

Since, LHS = RHS

Hence, x = 2 is the root of 3x â€“ 2 = 8x â€“ 12.

(iv) Given x = 4 is the root of 3x/2 = 6.

Now, substituting x = 4 in place of â€˜xâ€™ in the given equation, we get

= (3 Ã— 4)/2 = 6

= (12/2) = 6

6 = 6

Since, LHS = RHS

Hence, x = 4 is the root of (3x/2) = 6.

(v) Given y = 2 is the root of y â€“ 3 = 2y â€“ 5.

Now, substituting y = 2 in place of â€˜yâ€™ in the given equation, we get

= 2 â€“ 3 = 2(2) â€“ 5

= -1 = 4 â€“ 5

-1 = -1

Since, LHS = RHS

Hence, y = 2 is the root of y â€“ 3 = 2y â€“Â 5.

(vi) Given x = 8 is the root ofÂ 12x + 7 = 11.

Now, substituting x = 8 in place of â€˜xâ€™ in the given equation, we get

= 12(8) + 7 =11

= 4 + 7 = 11

= 11 = 11

Since, LHS = RHS

Hence, x = 8 is the root of 12x + 7 = 11.

**2. Solve each of the following equations by trial â€“ and â€“ error method:**

**(i) x + 3 =12**

**(ii) x -7 = 10**

**(iii) 4x = 28**

**(iv) (x/2) + 7 = 11**

**(v) 2x + 4 = 3x**

**(vi) (x/4) = 12**

**(vii) (15/x) = 3**

**(vii) (x/18) = 20**

**Solution:**

(i) Given x + 3 =12

Here LHS = x +3 and RHS = 12

x |
LHS |
RHS |
Is LHS = RHS |

1 |
1 + 3 = 4 |
12 |
No |

2 |
2 + 3 = 5 |
12 |
No |

3 |
3 + 3 = 6 |
12 |
No |

4 |
4 + 3 = 7 |
12 |
No |

5 |
5 + 3 = 8 |
12 |
No |

6 |
6 + 3 = 9 |
12 |
No |

7 |
7 + 3 = 10 |
12 |
No |

8 |
8 + 3 = 11 |
12 |
No |

9 |
9 + 3 = 12 |
12 |
Yes |

Therefore, if x = 9, LHS = RHS.

Hence, x = 9 is the solution to this equation.

(ii) Given x -7 = 10

Here LHS = x -7 and RHS = 10

x |
LHS |
RHS |
Is LHS = RHS |

9 |
9 â€“ 7 = 2 |
10 |
No |

10 |
10 -7 = 3 |
10 |
No |

11 |
11 â€“ 7 = 4 |
10 |
No |

12 |
12 â€“ 7 = 5 |
10 |
No |

13 |
19 â€“ 7 = 6 |
10 |
No |

14 |
14 â€“ 7 = 7 |
10 |
No |

15 |
15 â€“ 7 = 8 |
10 |
No |

16 |
16 â€“ 7 = 9 |
10 |
No |

17 |
17 â€“ 7 = 10 |
10 |
Yes |

Therefore if x = 17, LHS = RHS

Hence, x = 17 is the solution to this equation.

(iii) Given 4x = 28

Here LHS = 4x and RHS = 28

x |
LHS |
RHS |
Is LHS = RHS |

1 |
4 Ã— 1 = 4 |
28 |
No |

2 |
4 Ã— 2 = 8 |
28 |
No |

3 |
4 Ã— 3 = 12 |
28 |
No |

4 |
4 Ã— 4 = 16 |
28 |
No |

5 |
4 Ã— 5 = 20 |
28 |
No |

6 |
4 Ã— 6 = 24 |
28 |
No |

7 |
4 Ã— 7 = 28 |
28 |
Yes |

Therefore if x = 7, LHS = RHS

Hence, x = 7 is the solution to this equation.

(iv) Given (x/2) + 7 = 11

Here LHS = (x/2) + 7 and RHS = 11

Since RHS is a natural number, (x/2) must also be a natural number, so we must substitute values of x that are multiples of 2.

x |
LHS |
RHS |
Is LHS = RHS |

2 |
(2/2) + 7 = 1 + 7 =8 |
11 |
No |

4 |
(4/2) + 7 = 2 + 7 = 9 |
11 |
No |

6 |
(6/2) + 7 = 3 + 7 = 10 |
11 |
No |

8 |
(8/2) + 7 = 4 + 7 = 11 |
11 |
Yes |

Therefore if x = 8, LHS = RHS

Hence, x = 8 is the solutions to this equation.

(v) Given 2x + 4 = 3x

Here LHS = 2x + 4 and RHS = 3x

x |
LHS |
RHS |
Is LHS = RHS |

1 |
2 (1) + 4 = 2 + 4 = 6 |
3 (1) = 3 |
No |

2 |
2 (2) + 4 = 4 + 4 = 8 |
3 (2) = 6 |
No |

3 |
2 (3) + 4 = 6 + 4 = 10 |
3 (3) = 9 |
No |

4 |
2 (4) + 4 = 8 + 4 = 12 |
3 (4) = 12 |
Yes |

Therefore if x = 4, LHS = RHS

Hence, x = 4 is the solutions to this equation.

(vi) Given (x/4) = 12

Here LHS = (x/4) and RHS = 12

Since RHS is a natural number, x/4 must also be a natural number, so we must substitute values of x that are multiples of 4.

x |
LHS |
RHS |
Is LHS = RHS |

16 |
(16/4) = 4 |
12 |
No |

20 |
(20/4) = 5 |
12 |
No |

24 |
(24/4) = 6 |
12 |
No |

28 |
(28/4) = 7 |
12 |
No |

32 |
(32/4) = 8 |
12 |
No |

36 |
(36/4) = 9 |
12 |
No |

40 |
(40/4) = 10 |
12 |
No |

44 |
(44/4) = 11 |
12 |
No |

48 |
(48/4) = 12 |
12 |
Yes |

Therefore if x = 48, LHS = RHS

Hence, x = 48 is the solutions to this equation.

(vii) Given (15/x) = 3

Here LHS = (15/x) and RHS = 3

Since RHS is a natural number, 15x must also be a natural number, so we must substitute values of x that are factors of 15.

x |
LHS |
RHS |
Is LHS = RHS |

1 |
(15/1) = 15 |
3 |
No |

3 |
(15/3) = 5 |
3 |
No |

5 |
(15/5) = 3 |
3 |
Yes |

Therefore if x = 5, LHS = RHS

Hence, x = 5 is the solutions to this equation.

(viii) Given (x/18) = 20

Here LHS = (x/18) and RHS = 20

Since RHS is a natural number, (x/18) must also be a natural number, so we must substitute values of x that are multiples of 18.

x |
LHS |
RHS |
Is LHS = RHS |

324 |
(324/18) = 18 |
20 |
No |

342 |
(342/18) = 19 |
20 |
No |

360 |
(360/18) = 20 |
20 |
Yes |

Therefore if x = 360, LHS = RHS

Hence, x = 360 is the solutions to this equation.

8Ã—8Ã—8Ã—8Ã—8=2x what is x?

8Ã—8Ã—8Ã—8Ã—8=32768

So, 2x=32768

x=16384

Answer is:-

x=16384