**Complete the last column of the table.**

S. No. | Equation | Value | Say, whether the equation is satisfied. (Yes/No) |

(i) | x + 3 = 0 | x = 3 | |

(ii) | x + 3 = 0 | x = 0 | |

(iii) | x + 3 = 0 | x = -3 | |

(iv) | x – 7 = 1 | x = 7 | |

(v) | x – 7 = 1 | x = 8 | |

(vi) | 5x = 25 | x = 0 | |

(vii) | 5x = 25 | x = 5 | |

(viii) | 5x = 25 | x = -5 | |

(ix) | (m/3) = 2 | m = – 6 | |

(x) | (m/3) = 2 | m = 0 | |

(xi) | (m/3) = 2 | m = 6 | |

**Solution:**

**(i) x + 3 = 0**

**Given** **equation:**

**X + 3 = 0**

**Value of X = 3**

**LHS = x + 3**

**By substituting the value of x**

**LHS = 3 + 3 = 6**

**By comparing LHS and RHS**

**LHS ≠ RHS**

**No, the equation is not satisfied.**

**(ii) x + 3 = 0**

**Given equation:**

**X +3 = 0**

**Value of x = 0**

**LHS = x + 3**

**By substituting the value of x**

**LHS = 0 + 3 = 3**

**By comparing LHS and RHS**

**LHS ≠ RHS**

**No, the equation is not satisfied.**

**(iii) x + 3 = 0**

**Given equation:**

**X + 3 = 0**

**Value of x = – 3**

**LHS = x + 3**

**By substituting the value of x**

**LHS = – 3 + 3 = 0**

**By comparing LHS and RHS**

**LHS = RHS**

**Yes, the equation is satisfied**

**(iv) x – 7 = 1**

**LHS = x – 7**

**By substituting the value of x = 7**

**Then,**

**LHS = 7 – 7 = 0**

**By comparing LHS and RHS**

**LHS ≠ RHS**

**No, the equation is not satisfied**

**(v) x – 7 = 1**

**Given equation:**

**X – 7 = 1**

**Value of x = 8**

**LHS = x – 7**

**By substituting the value of x**

**LHS = 8 – 7 = 1**

**By comparing LHS and RHS**

**LHS = RHS**

**Yes, the equation is satisfied.**

**(vi) 5x = 25**

**Given equation:**

**5x = 25**

**Value of x = 0**

**LHS = 5x**

**By substituting the value of x**

**LHS = 5 × 0 = 0**

**By comparing LHS and RHS**

**LHS ≠ RHS**

**No, the equation is not satisfied.**

**(vii) 5x = 25**

**Given equation:**

**5x = 25**

**Value of x = 5**

**LHS = 5x**

**By substituting the value of x**

**LHS = 5 × 5 = 25**

**By comparing LHS and RHS**

**LHS = RHS**

**Yes, the equation is satisfied.**

**(viii) 5x = 25**

**Given equation:**

**5x = 25**

**Value of x = -5**

**LHS = 5x**

**By substituting the value of x**

**LHS = 5 × (-5) = – 25**

**By comparing LHS and RHS**

**LHS ≠ RHS**

**No, the equation is not satisfied.**

**(ix) m/3 = 2**

**Given equation:**

**m/3 = 2**

**value of x = -6**

**LHS = m/3**

**By substituting the value of m**

**LHS = -6/3 = – 2**

**By comparing LHS and RHS**

**LHS ≠ RHS**

**No, the equation is not satisfied.**

**(x) m/3 = 2**

**Given equation:**

**m/3 = 2**

**value of x = 0**

**LHS = m/3**

**By substituting the value of m**

**LHS = 0/3 = 0**

**By comparing LHS and RHS**

**LHS ≠ RHS**

**No, the equation is not satisfied.**

**(xi) m/3 = 2**

**Given equation:**

**m/3 = 2**

**value of x = 6**

**LHS = m/3**

**By substituting the value of m**

**LHS = 6/3 = 2**

**By comparing LHS and RHS**

**LHS = RHS**

**Yes, the equation is satisfied.**

S. No. | Equation | Value | Say, whether the equation is satisfied. (Yes/No) |

(i) | x + 3 = 0 | x = 3 | No |

(ii) | x + 3 = 0 | x = 0 | No |

(iii) | x + 3 = 0 | x = -3 | Yes |

(iv) | x – 7 = 1 | x = 7 | No |

(v) | x – 7 = 1 | x = 8 | Yes |

(vi) | 5x = 25 | x = 0 | No |

(vii) | 5x = 25 | x = 5 | Yes |

(viii) | 5x = 25 | x = -5 | No |

(ix) | (m/3) = 2 | m = – 6 | No |

(x) | (m/3) = 2 | m = 0 | No |

(xi) | (m/3) = 2 | m = 6 | Yes |

**2. Check whether the value given in the brackets is a solution to the given equation or not:**

**(a) n + 5 = 19 (n = 1)**

**Solution:**

**Given equation:**

**n+ 5 = 19**

**Where n = 1**

**LHS = n + 5**

**By substituting the value of n = 1**

**= n + 5**

**= 1 + 5**

**= 6**

**By comparing LHS and RHS**

**6 ≠ 19**

**LHS ≠ RHS**

**Hence, the value of n = 1 is not a solution to the given equation n + 5 = 19.**

**(b) 7n + 5 = 19 (n = – 2)**

**Solution:**

**Given equation:**

**7n + 5 = 19**

**Where n = -2**

**LHS = 7n + 5**

**By substituting the value of n = -2**

**LHS = 7n + 5**

**= (7 × (-2)) + 5**

**= – 14 + 5**

**= – 9**

**By comparing LHS and RHS**

**-9 ≠ 19**

**LHS ≠ RHS**

**Hence, the value of n = -2 is not a solution to the given equation 7n + 5 = 19.**

**(c) 7n + 5 = 19 (n = 2)**

**Solution:**

**Given equation:**

** 7n + 5 = 19**

**Where n = 2**

**LHS = 7n + 5**

**By substituting the value of n = 2**

**LHS = 7n + 5**

**= (7 × (2)) + 5**

**= 14 + 5**

**= 19**

**By comparing LHS and RHS**

**19 = 19**

**LHS = RHS**

**Hence, the value of n = 2 is a solution to the given equation 7n + 5 = 19.**

**(d) 4p – 3 = 13 (p = 1)**

**Solution:**

**Given equation:**

**4p – 3 = 13**

**Where p = 1**

**LHS = 4p – 3**

**By substituting the value of p = 1**

**LHS = 4p – 3**

**= (4 × 1) – 3**

**= 4 – 3**

**= 1**

**By comparing LHS and RHS**

**1 ≠ 13**

**LHS ≠ RHS**

**Hence, the value of p = 1 is not a solution to the given equation 4p – 3 = 13.**

**(e) 4p – 3 = 13 (p = – 4)**

**Solution:**

**Given equation:**

**4p – 3 = 13**

**Where p = -4**

**LHS = 4p – 3**

**By substituting the value of p = – 4**

**LHS = 4p – 3**

**= (4 × (-4)) – 3**

**= -16 – 3**

**= -19**

**By comparing LHS and RHS**

**-19 ≠ 13**

**LHS ≠ RHS**

**Hence, the value of p = -4 is not a solution to the given equation 4p – 3 = 13.**

**(f) 4p – 3 = 13 (p = 0)**

**Solution:**

**Given equation:**

**4p – 3 = 13**

**Where p = 0**

**LHS = 4p – 3**

**By substituting the value of p = 0**

**LHS = 4p – 3**

**= (4 × 0) – 3**

**= 0 – 3**

**= -3**

**By comparing LHS and RHS**

**– 3 ≠ 13**

**LHS ≠ RHS**

**Hence, the value of p = 0 is not a solution to the given equation 4p – 3 = 13.**

**3. Solve the following equations by trial and error method:**

**(i) 5p + 2 = 17**

**Solution:**

Given,

LHS = 5p + 2

Now,

Let p = 0

By substituting the value of p = 0

We get

LHS = 5p + 2

= (5 × 0) + 2

= 0 + 2

**= 2**

By comparing LH**S and RHS**

2 ≠ 17

LHS ≠ RHS

Hence, the value of p = 0 is not a solution to the given equation.

Now,

Let, p = 1

By substituting the value of p = 1

LHS = 5p + 2

= (5 × 1) + 2

= 5 + 2

= 7

By comparing LHS and RHS

7 ≠ 17

LHS ≠ RHS

Hence, the value of p = 1 is not a solution to the given equation.

Now,

Let, p = 2

LHS = 5p + 2

= (5 × 2) + 2

= 10 + 2

= 12

By comparing LHS and RHS

12 ≠ 17

LHS ≠ RHS

Hence, the value of p = 2 is not a solution to the given equation.

Now,

Let p = 3

LHS = 5p + 2

= (5 × 3) + 2

= 15 + 2

= 17

By comparing LHS and RHS

17 = 17

LHS = RHS

Hence, the value of p = 3 is a solution to the given equation.

**(ii) 3m – 14 = 4**

**Solution:**

Given**,**

LHS = 3m – 14

By substituting the value of m = 2

LHS = 3m – 14

= (3 × 2) – 14

= 6 – 14

**= –** **8**

By comparing LHS and RHS

-8 ≠ 4

LHS ≠ RHS

**Hence, the value of m = ****2**** is not a solution to the given equation.**

**Now,**

**Let, m = ****3**

**LHS = 3m – 14**

**= (3 × ****3****) – 14**

**= ****9**** – 14**

**= – ****5**

**By comparing LHS and RHS**

**–****5**** ≠ 4**

**LHS ≠ RHS**

**Hence, the value of m = 4 is not a solution to the given equation.**

**Let, m = 5**

**LHS = 3m – 14**

**= (3 × 5) – 14**

**= 15 – 14**

**= 1**

**By comparing LHS and RHS**

**1 ≠ 4**

**LHS ≠ RHS**

**Hence, the value of m = 5 is not a solution to the given equation.**

**Let, m = 6**

**LHS = 3m – 14**

**= (3 × 6) – 14**

**= 18 – 14**

**= 4**

**By comparing LHS and RHS**

**4 = 4**

**LHS = RHS**

**Hence, the value of m = 6 is a solution to the given equation.**

**4. Write equations for the following statements:**

**(i) The sum of numbers x and 4 is 9.**

**Solution:**

**The above statement can be written as,**

**= x + 4 = 9**

**(ii) 2 subtracted from y is 8.**

**Solution:**

**The above statement can be written as,**

**= y – 2 = 8**

**(iii) Ten times a is 70.**

**Solution:**

**The above statement can be written as,**

**= 10a = 70**

**(iv) The number b divided by 5 gives 6.**

**Solution:**

**The above statement can be written as,**

**= (b/5) = 6**

**(v) Three-fourth of t is 15.**

**Solution:**

**The above statement can be written as,**

**= ¾t = 15**

**(vi) Seven times m plus 7 gets you 77.**

**Solution:**

**The above statement can be written as,**

**= 7m + 7 = 77**

**(vii) One-fourth of a number x minus 4 gives 4.**

**Solution:**

**The above statement can be written as,**

**One-fourth of a number x is x/4**

**Then,**

**= x/4 – 4 = 4**

**(viii) If you take away 6 from 6 times y, you get 60.**

**Solution:**

**The above statement can be written as,**

**6 times of y is 6y**

**Now,**

**= 6y – 6 = 60**

**(ix) If you add 3 to one-third of z, you get 30.**

**Solution:**

**The above statement can be written as,**

**One-third of z is z/3**

**= 3 + z/3 = 30**

**5. Write the following equations in statement forms:**

**(i) p + 4 = 15**

**Solution:**

** The sum of numbers p and 4 gives 15.**

**(ii) m – 7 = 3**

**Solution:**

**7 subtracted from m gives 3.**

**(iii) 2m = 7**

**Solution:**

**Twice of number m gives 7.**

**(iv) m/5 = 3**

**Solution:**

**The number m divided by 5 gives 3.**

**(v) (3m)/5 = 6**

**Solution:**

**Three times of m divided by 5 gives 6.**

**(vi) 3p + 4 = 25**

**Solution:**

**Three times p plus 4 gives us 25.**

**(vii) 4p – 2 = 18**

**Solution:**

**Four times p minus 2 gives us 18.**

**(viii) p/2 + 2 = 8**

**Solution:**

**When we add half of a number p to 2, we get 8.**

**6. Set up an equation in the following cases:**

**(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmit’s marbles.)**

Solution:

Let the Parmit’s marble be m

Then, Irfan’s marble = 5m + 7

The total number of marble = 37

Hence, the required equation is:

5m + 7 = 37

**(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)**

Solution:

Let the age of Laxmi is y

Age of Laxmi’s Father = 3y + 4

but the age of Laxmi’s father is given by 49

Hence the required equation:

3y + 4 = 49

**(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be l.)**

**Solution:**

Let the lowest score be l

then, the highest score = 2l + 1

but the given highest score = 87

Hence, the required equation:

2l + 1 = 87

**(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees).**

**Solution:**

Let the base of each angle be ‘b’ degree

then, the vertex angle of the triangle = 2b

We know that the sum of the angles of a triangle is 180^{o}

Hence the required equation is:

b + b + 2b = 180^{o} or 4b

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