Question 1. Each equation is followed by the values of the variable. Decide whether these values are the solutions of that equation.

i. x – 4 = 3, x = – 1, 7, – 7

ii. 9m = 81, m = 3, 9, -3

iii. 2a + 4 = 0, a = 2, – 2, 1

iv. 3 – y = 4, y = – 1, 1, 2

Solution:

i. x – 4 = 3 ….(i)

Substituting x = – 1 in L.H.S. of equation (i),

L.H.S. = (-1) – 4

= – 5

R.H.S. = 3

∴ L.H.S. ≠ R.H.S.

∴ x = – 1 is not the solution of the given equation.

Substituting x = 7 in L.H.S. of equation (i),

L.H.S. = (7) – 4

= 3

R.H.S. = 3

∴ L.H.S. = R.H.S.

∴ x = 7 is the solution of the given equation.

Substituting x = – 7 in L.H.S. of equation (i),

L.H.S. = (- 7) – 4

= -11

R.H.S. = 3

∴ L.H.S. ≠ R.H.S.

∴ x = – 7 is not the solution of the given equation.

ii. 9m = 81 …(i)

Substituting m = 3 in L.H.S. of equation (i),

L.H.S. = 9 × (3)

= 27

R.H.S. = 81

∴L.H.S. ≠ R.H.S.

∴m = 3 is not the solution of the given equation.

Substituting m = 9 in L.H.S. of equation (i),

L.H.S. = 9 × (9)

= 81

R.H.S. = 81

∴L.H.S. = R.H.S.

∴m = 9 is the solution of the given equation.

Substituting m = – 3 in L.H.S. of equation (i),

L.H.S. = 9 × (- 3)

= -27

R.H.S. = 81

∴L.H.S. ≠ R.H.S.

∴m = – 3 is not the solution of the given equation.

iii. 2a + 4 = 0 …..(i)

Substituting a = 2 in L.H.S. of equation (i),

L.H.S. = 2 (2) + 4

= 4 + 4

= 8

R.H.S. = 0

∴L.H.S. ≠ R.H.S.

∴a = 2 is not the solution of the given equation.

Substituting a = – 2 in L.H.S. of equation (i),

L.H.S. = 2 (-2)+ 4

= -4 + 4

= 0

R.H.S. = 0

∴L.H.S. = R.H.S.

∴a = – 2 is the solution of the given equation.

Substituting a = 1 in L.H.S. of equation (i),

L.H.S. = 2(1)+ 4

= 2 + 4

= 6

R.H.S. = 0

∴ L.H.S. ≠ R.H.S.

∴a = 1 is not the solution of the given equation.

iv. 3 – y = 4 …(i)

Substituting y = -1 in L.H.S. of equation (i),

L.H.S. = 3 – (- 1)

= 3 + 1

= 4

R.H.S. = 4

∴L.H.S. = R.H.S.

∴y = – 1 is the solution of the given equation.

Substituting y = 1 in L.H.S. of equation (i),

L.H.S. = 3-(1)

= 2

R.H.S. = 4

∴L.H.S. ≠ R.H.S.

∴y = 1 is not the solution of the given equation.

Substituting y = 2 in L.H.S. of equation (i),

L.H.S. = 3-(2)

= 1

R.H.S. = 4

∴L.H.S. ≠ R.H.S.

∴y = 2 is not the solution of the given equation.

Question 2.

Solve the following equations:

Solution:

i. 17p – 2 = 49

∴ 17p – 2 + 2 = 49 + 2

…[Adding 2 on both the sides]

∴ 17p = 51

p = 3

ii. 2m + 7 = 9

∴ 2m + 7 – 7 = 9 – 7

…[Subtracting 7 from both the sides]

∴ 2m = 2

∴ m = 1

iii. 3x + 12 = 2x – 4

∴ 3x + 12 – 12 = 2x – 4 – 12

…[Subtracting 12 from both the sides]

∴ 3x = 2x – 16

∴ 3x – 2x = 2x – 16 – 2x

…[Subtracting 2x from both the sides]

∴ x = – 16

iv. 5 (x – 3) = 3 (x + 2)

∴ 5x – 15 = 3x + 6

∴ 5x – 15 + 15 = 3x + 6 + 15

…[Adding 15 on both the sides]

∴ 5x = 3x + 21

∴ 5x – 3x = 3x + 21 – 3x

…[Subtracting 3x from both the sides]

∴ 2x = 21

viii. 3 (y + 8) = 10 (y – 4) + 8

∴ 3y + 24 = 10y – 40 + 8

∴ 3y + 24 = 10y – 32

∴ 3y + 24 – 24 = 10y – 32 – 24

…[Subtracting 24 from both the sides]

∴ 3y = 10y – 56

∴ 3y – 10y = 10y – 56

…[Subtracting 10y from both the sides]

∴ – 7y = – 56

…[Multiplying both the sides by 7 (x – 5)]

∴7 (x – 9) = 5 (x – 5)

∴7x – 63 = 5x – 25

∴7x – 63 + 63 = 5x – 25 + 63

…[Adding 63 on both the sides]

∴7x = 5x + 38

∴7x – 5x = 5x + 38 – 5x

…[Subtracting 5x from both the sides]

∴ 2x = 38

…[Multiplying both the sides by 4]

∴b + b + 1 + b + 2 = 84

∴3b + 3 = 84

∴3b + 3 – 3 = 84 – 3

…[ Subtracting 3 from both the sides]

∴3b = 81

∴\(\frac{3 b}{3}=\frac{81}{3}[/latex …[Dividing both the sides by 3]

∴b = 27

**Intext Questions and Activities**

Question 1.

Fill in the boxes to solve the following equations. (Textbook pg. no. 75)

i. x + 4 = 9

∴x + 4 – __ = 9 – __

… [Subtracting 4 from both the sides]

∴ x = __

ii. x – 2 = 7

∴x – 2 + __ = 7 + __

… [Adding 2 on both the sides]

∴x = __

iii. [latex]\frac { x }{ 3 }=4\)

∴ × __ = 4 ×__

∴x = __

iv. 4x = 24

∴ __ = __

∴x = __

Solution:

i. x + 4 = 9

∴x + 4 – 4 = 9 – 4

… [Subtracting 4 from both the sides]

∴ x = 5

ii. x – 2 = 7

∴x – 2 + 2 = 7 + 2

… [Adding 2 on both the sides]

∴x = 9

iii. x/3=4

∴x/3× 3 = 4 × 3

… [Multiplying both the sides by 3]

∴x = 12

iv. 4x = 24

… [Dividing both the sides by 4]

∴x = 6