**ML Aggarwal CBSE Solutions Class 6 Math 12th Chapter Algebra Exercise 12.4**

**Ex 12.4**

**(1) State which of the following are equations with a variable. In case of an equation with a variable, identify the variable.**

**(i) 17 + x = 5**

**Solution: **It is an equation with variable x.

**(ii) 2b – 3 = 7**

**Solution: **It is an equation with variable b.

**(iii) (y -7) > 5**

**Solution: **it is not an equation.

**(iv) 9/3 = 3**

**Solution: **It is a numerical equation. It has no variable.

**(v) 7 x 3 – 19 = 2**

**Solution: **It is a numerical equation. It has no variable.

**(vi) 5 x 4 – 8 = 3t**

**Solution: **It is an equation with variable t.

**(vii) 2p < 15**

**Solution: **It is not an equation.

**(viii) 7 = 11 x 5 – 12 x 4**

**Solution: **It is a numerical equation. It has no variable.

**(ix) 3/2 q = 5**

**Solution: **It is an equation with variable q.

**(2) (i) x + 12 = 20 (12, 8, 20, 0)**

**Solution: **The equation is x + 12 = 20

For x = 12, LHS = 12 + 12 = 24 ≠ RHS Therefore, x = 12 is not a solution

For x = 8, LHS 8 + 12 = 20 = RHS Therefore, x = 8 is a solution

For x = 20, LHS 20 + 12 = 32 ≠ RHS Therefore x = 20 is not a solution

For x = 0, LHS 0 + 12 = 12 ≠ RHS Therefore x = 0 is not a solution.

Hence, x = 8 is the solution of the given equation.

(**ii) y/2 = 7 (7, 2, 10, 14)**

**Solution: **The given equation y/2 = 7

For y = 7, LHS = 7/2 = 3.5 ≠ RHS Therefore y = 7 is not a solution.

For y = 2, LHS = 2/2 = 1 ≠ RHS Therefore, y = 2 is not a solution

For y = 10, LHS = 10/2 = 5 ≠ RHS Therefore y = 10 is not a solution

For y = 14, LHS = 14 / 2 = 7 = RHS Therefore y = 14 is a solution

Hence, y = 14 is the solution of the given equation.

**(iii) p – 4 = 0 (4. -4, 8, 0)**

**Solution: **The given equation p – 4 = 0

For p = 4, LHS = 4 – 4 = 0 = RHS Therefore p = 4 is the solution

For p = -4, LHS = -4 – 4 = -8 ≠ RHS Therefore, p = -4 is not a solution

For p = 8, LHS = 8 – 4 = 4 ≠ RHS Therefore p = 8 is not a solution

For p = 0, LHS = 0 – 4 = -4 ≠ RHS Therefore p = o is not a solution

Hence, p = 4 is the solution of the given equation

**(iv) n + 4 = 2 (-2, 0, 2, 4)**

**Solution: **The given equation n + 4 = 2

For n = -2, LHS = -2 + 4 = 2 = RHS Therefore n = -2 is the solution

For n = 0, LHS = 0 + 4 = 4 ≠ RHS Therefore n = 0 is not a solution

For n = 2, LHS = 2 + 4 = 6 ≠ RHS Therefore n = 2 is not a solution

For n = 4, LHS = 4 + 4 = 8 ≠ RHS Therefore n = 4 is not a solution

Hence, n = -2 is the solution of the given equation.

**(v) 2y – 3 = 7 (0,3,5,7)**

**Solution: **For y = 0, LHS = 2×0 – 3 = 0 – 3 = -3 ≠ RHS Therefore y = 0 is not a solution

For y = 3, LHS = 2×3 – 3 = 6 – 3 = 3 ≠ RHS Therefore, y = 3 is not a solution.

For y = 5, LHS = 2 x 5 – 3 = 10 – 3 = 7 = RHS Therefore, y = 5 is a solution.

For y = 7, LHS = 7 x 5 – 3 = 35 – 3 = 32 ≠ RHS Therefore, y = 7 is not a solution.

**(3) Complete the entries in the fourth of the following table:**

**Solution:**

S. No. |
Equation |
Value of Variable |
Equation Satisfied Yes / No |

(i) | 5n = 40 | 10 | No |

(ii) | 5n = 40 | 8 | Yes |

(iii) | 5n = 40 | 5 | No |

(iv) | X + 10 = 30 | 10 | No |

(v) | X + 10 | 20 | Yes |

(vi) | X + 20 = 30 | 30 | No |

(vii) | 2p – 3 = 7 | 5 | Yes |

(viii) | 2p – 3 = 7 | 10 | No |

(ix) | 2p – 3 = 7 | 15 | No |

(x) | Y + 3 = 1 | 1 | No |

(xi) | Y + 3 = 1 | 0 | No |

(xii) | Y + 3 = 1 | -1 | No |

(xiii) | Y + 3 = 1 | -2 | yes |

**(5) Complete the following table and by inspection of the table, find the solution of the equation 2x + 3 = 17**

**(i) Solution: **The given equation is 2x + 3 = 17. We complete the table as under:

x | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | …. |

2x + 3 | 5 | 7 | 9 | 11 | 13 | 15 | 17 | 19 | 21 | …. |

**(ii) Solution: **

x | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | …. |

25 – 3x | 16 | 13 | 10 | 7 | 4 | 1 | -2 | -5 | -8 | …. |

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It’s is right solution but teachers are get wrong it is wrong solution

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