Linear Equation in One Variable Exercise 9.1 |

Linear Equation in One Variable Exercise 9.2 |

**Question 1:**

**Write each of the following statements as an equation:**

**(i) 5 times a number equals 40.**

**(ii) A number increased by 8 equals 15.**

**(iii) 25 exceeds a number by 7.**

**(iv) A number exceeds 5 by 3.**

**(v) 5 subtracted from thrice a number is 16.**

**(vi) If 12 is subtracted from a number, the result is 24.**

**(vii) Twice a number subtracted from 19 is 11.**

**(ix) 3 less than 4 times a number is 17.**

**(x) 6 times a number is 5 more than the number.**

**Solution:**

(i) Let the required number be x.

So, five times the number will be 5x.

Therefore, 5x = 40

(ii) Let the required number be x.

So, when it is increased by 8, we get x + 8.

Therefore, x + 8 = 15

(iii) Let the required number be x.

So, when 25 exceeds the number, we get 25 – x.

Therefore, 25 – x = 7

(iv) Let the required number be x.

So, when the number exceeds 5, we get x – 5.

Therefore, x – 5 = 3.

(v) Let the required number be x.

So, thrice the number will be 3x.

Therefore, 3x – 5 = 16

(vi) Let the required number be x.

So, 12 subtracted from the number will be x – 12.

Therefore, x – 12 = 24

(vii) Let the required number be x.

So, twice the number will be 2x.

Therefore, 19 – 2x = 11

(viii) Let the required number be x.

So, the number when divided by 8 will be \(\frac{x}{8}\).

Therefore,\(\frac{x}{8}=7\)

(ix) Let the required number be x.

So, four times the number will be 4x.

Therefore, 4x – 3 = 17

(x) Let the required number be x.

So, 6 times the number will be 6x.

Therefore, 6x = x + 5

**Question 2:**

**Write a statement for each of the equations, given below:**

**(i) x – 7 = 14**

**(ii) 2y = 18**

**(iii) 11 + 3x = 17**

**(iv) 2x – 3 = 13**

**(v) 12y – 30 = 6 **

**(vi) \(\frac{2z}{3}=8\)**

**Solution:**

(i) 7 less than the number x equals 14.

(ii) Twice the number y equals 18.

(iii) 11 more than thrice the number x equals 17.

(iv) 3 less than twice the number x equals 13.

(v) 30 less than 12 times the number y equals 6.

(vi) When twice the number z is divided by 3, it equals 8.

**Question 3:**

**Verify by substitution that**

**(i) the root of 3x – 5 = 7 is x = 4**

**(ii) the root of 3 + 2x = 9 is x = 3**

**(iii) the root of 5x – 8 = 2x – 2 is x = 2**

**(iv) the root of 8 – 7y = 1 is y = 1**

**(v) the root of \(\frac{z}{7}=8\) is z = 56**

**Solution:**

(i) 3x – 5 = 7

Substituting x = 4 in the given equation:

L.H.S. : 3 x 4 – 5

or, 12 – 5 = 7 = R.H.S.

Hence, x = 4 is the root of the given equation.

(ii) 3 + 2x = 9

Substituting x = 3 in the given equation:

L.H.S. :

3 + 2 x 3

or, 3 + 6 = 9 = R.H.S.

L.H.S. = R.H.S.

Hence, x = 3 is the root of the given equation.

(iii) 5x – 8 = 2x – 2

Substituting x = 2 in the given equation:

L.H.S. :

5 x 2 – 8

or, 10 – 8 = 2

R.H.S. :

2 x 2 – 2

= 4 – 2 = 2

L.H.S. = R.H.S.

Hence, x = 2 is the root of the given equation.

(iv) 8 – 7y = 1

Substituting y = 1 in the given equation:

L.H.S. :

8 – 7 x 1

or, 8 – 7 = 1 = R.H.S.

L.H.S. = R.H.S.

Hence, y = 1 is the root of the given equation.

(v)\(\frac{z}{7}=8\)

Substituting z = 56 in the given equation:

L.H.S. :

\(\frac{56}{7}=8\) = R.H.S.

L.H.S. = R.H.S.

Hence, z = 56 is the root of the given equation.

**Question 4:**

**Solve each of the following equations by the trial-and-error method:**

**(i) y + 9 = 13**

**(ii) x – 7 = 10**

**(iii) 4x = 28**

**(iv) 3y = 36**

**(v) 11 + x = 19**

**(vi) \(\frac{x}{3}=4\)**

**(vii) 2x – 3 = 9**

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

**(ix) 2y + 4 = 3y**

**(x) z – 3 = 2z – 5 **

**Solution:**

(i) y + 9 = 13

We try several values of y until we get the L.H.S. equal to the R.H.S.

y | L.H.S. | R.H.S. | Is L.H.S. = R.H.S.? |

1 | 1 + 9 = 10 | 13 | No |

2 | 2 + 9 = 11 | 13 | No |

3 | 3 + 9 = 12 | 13 | No |

4 | 4 + 9 = 13 | 13 | Yes |

Therefore, y = 4

(ii) x – 7 = 10

We try several values of x until we get the L.H.S. equal to the R.H.S.

x | L.H.S. | R.H.S. | Is L.H.S. = R.H.S.? |

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, x = 17

(iii) 4x = 28

We try several values of x until we get the L.H.S. equal to the R.H.S.

x | L.H.S. | R.H.S. | Is L.H.S. = R.H.S.? |

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, x = 7

(iv) 3y = 36

We try several values of y until we get the L.H.S. equal to the R.H.S.

y | L.H.S. | R.H.S. | Is L.H.S. = R.H.S.? |

6 | 3 x 6 = 18 | 36 | No |

7 | 3 x 7 = 21 | 36 | No |

8 | 3 x 8 = 24 | 36 | No |

9 | 3 x 9 = 27 | 36 | No |

10 | 3 x 10 = 30 | 36 | No |

11 | 3 x 11 = 33 | 36 | No |

12 | 3 x 12 = 36 | 36 | Yes |

Therefore, y = 12

(v) 11 + x = 19

We try several values of x until we get the L.H.S. equal to the R.H.S.

x | L.H.S. | R.H.S. | Is L.H.S. = R.H.S.? |

1 | 11 + 1 = 12 | 19 | No |

2 | 11 + 2 = 13 | 19 | No |

3 | 11 + 3 = 14 | 19 | No |

4 | 11 + 4 = 15 | 19 | No |

5 | 11 + 5 = 16 | 19 | No |

6 | 11 + 6 = 17 | 19 | No |

7 | 11 + 7 = 18 | 19 | No |

8 | 11 + 8 = 19 | 19 | Yes |

Therefore, x = 8

(vi)\(\frac{x}{3}=4\)

Since R.H.S. is an natural number so L.H.S. must also be a natural number. Thus, x has to be a multiple of 3.

x | L.H.S. | R.H.S. | Is L.H.S. = R.H.S.? |

3 | \(\frac{3}{3}=1\) | 4 | No |

6 | \(\frac{6}{3}=2\) | 4 | No |

9 | \(\frac{9}{3}=3\) | 4 | No |

12 | \(\frac{12}{3}=4\) | 4 | Yes |

Therefore, x = 12

(vii) 2x – 3 = 9

We try several values of x until we get the L.H.S. equal to the R.H.S.

x | L.H.S. | R.H.S. | Is L.H.S. = R.H.S.? |

1 | 2 x 1 – 3 = -1 | 9 | No |

2 | 2 x 2 – 3 = 1 | 9 | No |

3 | 2 x 3 – 3 = 3 | 9 | No |

4 | 2 x 4 – 3 = 5 | 9 | No |

5 | 2 x 5 – 3 = 7 | 9 | No |

6 | 2 x 6 – 3 = 9 | 9 | Yes |

Therefore, x = 6

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

Since, R.H.S. is a natural number so L.H.S. must be a natural number. Thus, we will try values if x which are multiples of 2

x | L.H.S. | R.H.S. | Is L.H.S. = R.H.S.? |

2 | \(\frac{2}{2}+7=8\) | 11 | No |

4 | \(\frac{4}{2}+7=9\) | 11 | No |

6 | \(\frac{6}{2}+7=10\) | 11 | No |

8 | \(\frac{8}{2}+7=11\) | 11 | Yes |

Therefore, x = 8

(ix) 2y + 4 = 3y

We try several values of y until we get the L.H.S. equal to the R.H.S.

y | L.H.S. | R.H.S. | Is L.H.S. = R.H.S.? |

1 | 2 x 1 + 4 = 6 | 3 x 1 = 3 | No |

2 | 2 x 2 + 4 = 8 | 3 x 2 = 6 | No |

3 | 2 x 3 + 4 = 10 | 3 x 3 = 9 | No |

4 | 2 x 4 + 4 = 12 | 3 x 4 = 12 | Yes |

Therefore, y = 4

(x) z – 3 = 2z – 5

We try several values of z till we get the L.H.S. equal to the R.H.S.

z | L.H.S. | R.H.S. | Is L.H.S. = R.H.S.? |

1 | 1 – 3 = -2 | 2 x 1 – 5 = -3 | No |

2 | 2 – 3 = -1 | 2 x 2 – 5 = -1 | Yes |

Therefore, z = 2