### Exercise-4.1

#### Question 1:

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 |

**Answer:**

(i) The equation we have is: x + 3 = 0

Note: To check whether x =3 satisfies the given equation, we put x = 3 in the equation.

L.H.S. = 3 + 3 = 6 ≠ RHS

Thus, the answer is No i.e. the equation is not satisfied.

(ii) The equation we have is: x + 3 = 0

Note: To check whether the x = 0 satisfies the given equation, we put x = 0 in the given equation.

L.H.S. = 0 + 3 = 3 ≠ RHS

Thus, the answer is No i.e. the equation is not satisfied.

(iii) The equation we have is: x + 3 = 0**Note: **To check whether the x = -3 satisfies the given equation, we put x = -3 in the given equation.

L.H.S. = -3 + 3 = 0 = RHS

Thus, the answer is Yes i.e. the equation is satisfied.

(iv) The equation we have is: x – 7 = 1**Note: **To check whether the x = 7 satisfies the given equation, we put x = 7 in the given equation.

L.H.S. = 7 – 7 = 0 ≠ RHS

Thus, the answer is No i.e. the equation is not satisfied.

(v) The equation we have is: x – 7 = 1

Note: To check whether the x = 8 satisfies the given equation, we put x = 8 in the given equation.

L.H.S. = 8 – 7 = 1 = RHS

Thus, the answer is Yes i.e. the equation is satisfied.

(vi) The equation we have is: 5x = 25**Note: **To check whether the x = 0 satisfies the given equation, we put x = 0 in the given equation.

L.H.S. = 5(0) = 0 ≠ RHS

Thus, the answer is No i.e. the equation is not satisfied.

(vii) The equation we have is: 5x = 25**Note: **To check whether the x = 5 satisfies the given equation, we put x = 5 in the given equation.

L.H.S. = 5(5) = 25 = RHS

Thus, the answer is Yes i.e. the equation is satisfied.

(viii) The equation we have is: 5x = 25**Note: **To check whether the x = -5 satisfies the given equation, we put x = -25 in the given equation.

L.H.S. = 5(-5) = -25 = RHS

Thus, the answer is No i.e. the equation is not satisfied.

### Exercise-4.2

**Question 1:**

Give first the step you will use to separate the variable and then solve the equation:

(a) x – 1 = 0

(b) x + 1 = 0

(c) x – 1 = 5

(d) x + 6 = 2

(e) y – 4 = – 7

(f) y – 4 = 4

(g) y + 4 = 4

(h) y + 4 = – 4

**Answer:** (a) x – 1 = 0

**Solution: **We have to separate the variables

Thus, x – 1 = 0

Now,

Adding 1 on both side of the equation, We get,

x – 1 + 1 = 0 + 1

x = 1

(b) x + 1 = 0

**Solution:** Here, we have to separate the variables

We have,

x + 1 = 0

Now,

Subtracting 1 on both side of the equation,We get,

x – 1 + 1 = 0 – 1

x = -1

(c) x – 1 = 5

**Solution: **Here, we have to separate the variables

We have,

x – 1 = 5

Now,Adding 1 on both side of the equation,We get,

x – 1 + 1 = 5 + 1

x = 6

(d) x + 6 = 2

**Solution:** Here, we have to separate the variables

We have,

x + 6 = 2

Now, subtracting 6 on both side of the equation,

We get,

x + 6 – 6 = 2 – 6

x = -4

(e)y – 4 = – 7

**Solution: **Here, we have to separate the variables

We have,

y – 4 = -7

Now,

Adding 4 on both side of the equation,

We get,

x – 4 + 4 = -7 + 4

x = -3

(f) y – 4 = 4

**Solution:** Here, we have to separate the variables

We have,

y – 4 = 4

Now,Adding 4 on both side of the equation,We get,

x – 4 + 4 = 4 + 4

x = 8

(g) y + 4 = 4

**Solution:** Here, we have to separate the variables

Thus,

We have, y + 4 = 4

Now,

Subtracting 4 on both side of the equation, We get,

x + 4 – 4 = 4 – 4

x = 0

(h )y + 4 = – 4

**Solution:** Here,We have to separate the variables

Thus,We have,

y + 4 = -4

Now,

Subtracting 4 on both side of the equation,We get,

x + 4 – 4 = -4 – 4

x = -8

**Question 2:****Check whether the value given in the brackets is a solution to the given equation or not:(a)3l = 42(b )b/2 = 6(c)p/7 = 4(f)(h)20t = -10**

The given parts of the question are solved below:

Here,

Thus,

Dividing both sides by 3, we get, l=423

We have to separate the variables

Multiplying both sides by 2, we get,

b = 12

(c) p/7 = 4

We have, p7=4

p/7×7 = 4 × 7

### Exercise-4.3

### Exercise-4.4

Question 1:

Set up equations and solve them to find the unknown numbers in the following cases:

(a) Add 4 to eight times a number you get 60.

(b) One-fifth of a number minus 4 gives 3.

(c) If I take three-fourths of a number and add 3 to it, I get 21.

(d) When I subtracted 11 from twice a number, the result was 15.

(e) Munna subtracts thrice the number of

(f) Notebooks he has from 50, he finds the result to be 8.

(g) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.

(h) Anwar thinks of a number. If he takes away 7 from **5/2** of the number, the result is 23.

Answer:

The parts of this given questions are solved below:

(a) Add 4 to eight times a number you get 60.We have to frame the equation on the basis of given statement

And then solve the so framed equation

Thus,

Let the number be a

We know that,

Eight times of a is 8a

Therefore,

As per the statement,

We can write the equation as,

8a + 4 = 60

8a = 60 – 48a = 56

Now,

Dividing 8 on both the sides, we get

8a/8=56/8

Therefore,

a = 7

(b) One-fifth of a number minus 4 gives 3.

We have to frame the equation on the basis of given statement

And then solve the so framed equation

Thus,

Let the number be a

We know that,

One fifth of a is x/5

Therefore,

As per the statement,

We can write the equation as,

x/5-4=3

x/5 = 3 + 4

x/5 = 7

Now,

Multiplying 5 to both the sides, we get

a/5×5=7×5