#### Exercise 8.1

*Q 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 \(\frac{1}{2}x+7\) = 11*

**SOLUTION :**

(i). x = 4 is the root of 3x â€” 5 = 7.

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

3(4) – Â 5 = 7

12 – 5 = 7

7 = 7

Since, LHS = RHS

Hence, x = 4 is the root of 3x â€” 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 â€“ 12.

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

3(2) – 2 = 8(2) â€“ 12

6 – 2 = 16 – 12

4 =Â 4

Since , LHS = RHS

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

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

Now, substituting x = 4 in place of ‘xâ€™ 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 â€“ 5.

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

2 – 3 = 2(2) â€“ 5

-1 = 4 â€“ 5

-1 = -1

Since, Â LHS = RHS

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

(vi). x = 8 is the root ofÂ \(\frac{1}{2}x+7\)

Now, substituting x = 8 in place of ‘xâ€™ in the given equation \(\frac{1}{2}x+7\)

\(\frac{1}{2}(8)+7\)

4 + 7 = 11

11 = 11

Since, Â LHS = RHS

Hence, x = 8 is the root of \(\frac{1}{2}x+7\)

**Q 2. Solve each of the following equations by trial and error method :**

(i). x + 3 = 12

(ii). x â€“ 7 = 10

(iii). 4x = 28

(iv). \(\frac{x}{2}+7\)

(v). 2x + 4 = 3x

(vi). \(\frac{x}{4}\)

(vii). \(\frac{15}{x}\)

(viii). \(\frac{x}{18}\)

**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 â€“ 7 = 3 | 10 | No |

11 | 11 â€“ 7 =4 | 10 | No |

12 | 12 â€“ 7 =5 | 10 | No |

13 | 13 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). 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, LHSÂ = RHS

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

(iv). \(\frac{x}{2}+7\)

Here, LHS = \(\frac{x}{2}+7\)

Since RHS is a natural number, \(\frac{x}{2}\)

x | LHS | RHS | Is LHS = RHS |

2 | \(\frac{2}{2}+7\) |
11 | No |

4 | \(\frac{4}{2}+7\) |
11 | No |

6 | \(\frac{6}{2}+7\) |
11 | No |

8 | \(\frac{8}{2}+7\) |
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 | Â 3(4)= 12 | Yes |

Therefore, if x = 4, LHS = RHS.

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

(vi). \(\frac{x}{4}\)

Here, LHS = \(\frac{x}{4}\)

Since RHS is a natural number, \(\frac{x}{4}\)

X | LHS | RHS | Is LHS = RHS |

16 | \(\frac{16}{4}\) |
12 | NO |

20 | \(\frac{20}{4}\) |
12 | NO |

24 | \(\frac{24}{4}\) |
12 | NO |

28 | \(\frac{28}{4}\) |
12 | NO |

32 | \(\frac{32}{4}\) |
12 | NO |

36 | \(\frac{36}{4}\) |
12 | NO |

40 | \(\frac{40}{4}\) |
12 | NO |

44 | \(\frac{44}{4}\) |
12 | NO |

48 | \(\frac{48}{4}\) |
12 | Yes |

Therefore, if x = 48, LHS = RHS.

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

(vii). \(\frac{15}{x}\)

Here, LHS = \(\frac{15}{x}\)

Since RHS is a natural number, \(\frac{15}{x}\)

x | LHS | RHS | Is LHS = RHS |

1 | \(\frac{15}{1}\) |
3 | No |

3 | \(\frac{15}{3}\) |
3 | No |

5 | \(\frac{15}{5}\) |
3 | Yes |

Therefore, if x = 5, LHS = RHS.

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

(viii). \(\frac{x}{18}\)

Here, LHS = \(\frac{x}{18}\)

Since RHS is a natural number, \(\frac{x}{18}\)

X | LHS | RHS | Is LHS = RHS |

324 | \(\frac{324}{18}\) |
20 | No |

342 | \(\frac{342}{18}\) |
20 | No |

360 | \(\frac{360}{18}\) |
20 | Yes |

Therefore, if x = 360, LHS = RHS.

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