Table of Contents

**Chapter 8: Linear Equations in One Variable Exercise – 8.1**

In this chapter, we provide RD Sharma Class 7 ex 8.1 Solutions Chapter 8 Linear Equations in One Variable for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 7 ex 8.1 Solutions Chapter 8 Linear Equations in One Variable Maths pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 7 |

Subject | Maths |

Chapter | Chapter 8 |

Chapter Name | Linear Equations in One Variable |

Exercise | Ex 8.1 |

**RD Sharma Solutions for Class 7 Chapter 8 Linear Equations in One Variable** **Ex 8.1 Download PDF**

**Question: 1**

Verify by substitution that:

(i) x = 4 is the root of 3x – 5 = 7

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

(iii) x = 2 is the root of 3x – 2 = 8x – 12

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

(v) y = 2 is the root of y – 3 = 2y – 5

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

**Solution:**

(i) x = 4 is the root of 3x – 5 = 7.

Now, substituting x = 4 in place of ‘x’ in the given equation 3x – 5 = 7,

3(4) – 5 = 7

12 – 5 = 7

7 = 7

Since, LHS = RHS

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

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

Now, substituting x = 3 in place of ‘x’ in the given equation 5 + 3x = 14,

5 + 3(3) = 14

5 + 9 = 14

14 = 14

Since, LHS = RHS

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

(iii). x = 2 is the root of 3x – 2 = 8x – 12.

Now, substituting x = 2 in place of ‘x’ in the given equation 3x –2 = 8x – 12,

3(2) – 2 = 8(2) – 12

6 – 2 = 16 – 12

4 = 4

Since, LHS = RHS

Hence, x = 2 is the root of 3x – 2 = 8x – 12.

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

Now, substituting x = 4 in place of ‘x’ in the given equation 3x/2 = 6,

(3× 4)/2 = 6

12/2 = 6

6 = 6

Since, LHS = RHS

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

(v). y = 2 is the root of y – 3 = 2y – 5.

Now, substituting y = 2 in place of ‘y’ in the given equation y – 3 = 2y – 5,

2 – 3 = 2(2) – 5

-1 = 4 – 5

-1 = -1

Since, LHS = RHS

Hence, y = 2 is the root of y – 3 = 2y – 5.

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

Now, substituting x = 8 in place of ‘x’ in the given equation 12x + 7 = 11,

12(8) + 7 =11

4 + 7 = 11

11 = 11

Since, LHS = RHS

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

**Question: 2**

Solve each of the following equations by trial and 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

(viii) x/18 = 20

**Solution:**

(i) 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) x – 7 = 10

Here, LHS = x – 7 and RHS = 10.

x | LHS | RHS | Is LHS = RHS |

9 | 9 – 7 = 2 | 10 | No |

10 | 10 – 7 = 3 | 10 | No |

11 | 11 – 7 = 4 | 10 | No |

12 | 12 – 7 = 5 | 10 | No |

13 | 13 – 7 = 6 | 10 | No |

14 | 14 – 7 = 7 | 10 | No |

15 | 15 – 7 = 8 | 10 | No |

16 | 16 – 7 = 9 | 10 | No |

17 | 17 – 7 = 10 | 10 | Yes |

Therefore, if x = 17, LHS = RHS.

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

(iii) 4x = 28

Here, LHS = 4x and RHS = 28.

x | LHS | RHS | Is LHS = RHS |

1 | 4 x 1 = 4 | 28 | No |

2 | 4 x 2 = 8 | 28 | No |

3 | 4 x 3 = 12 | 28 | No |

4 | 4 x 4 = 16 | 28 | No |

5 | 4 x 5 = 20 | 28 | No |

6 | 4 x 6 = 24 | 28 | No |

7 | 4 x 7 = 28 | 28 | Yes |

Therefore, if x = 7, LHS = RHS

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

(iv) 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 = 8 | 11 | No |

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

6 | 6/2 + 7 = 10 | 11 | No |

8 | 8/2 + 7 = 11 | 11 | Yes |

Therefore, if x = 8, LHS = RHS.

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

(v) 2x + 4 = 3x

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

x | LHS | RHS | Is LHS = RHS |

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

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

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

4 | 2(4) + 5 = 12 | 3(4) = 12 | Yes |

Therefore, if x = 4, LHS = RHS.

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

(vi) 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 solution to this equation.

(vii) 15x = 3

**All Chapter RD Sharma Solutions For Class 7 Maths**

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