Well-designed AP Board Solutions Class 7 Maths Chapter 4 Simple Equations Exercise 4.2 offers step-by-step explanations to help students understand problem-solving strategies.

## Simple Equations Class 7 Exercise 4.2 Solutions – 7th Class Maths 4.2 Exercise Solutions

Question 1.

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

a) x – 1 = 0

Solution:

x – 1 = 0

Adding ‘ 1 ‘ on both sides

x – 1 + 1 = 0 + 1 ⇒ x = 1

b) x + 1 = 0

Solution:

x + 1 = 0

Adding ‘ -1 ‘ on both sides

x + 1 – 1 = 0 – 1 ⇒ x = – 1

c) x – 1 = 5

Solution:

x – 1 = 5

Adding ‘ 1 ‘ on both sides

x – 1 + 1 = 5 + 1 ⇒ x = 6

d) x + 6 = 2

Solution:

x + 6 = 2

Adding ‘ -6 ‘ on both sides

x + 6 – 6 = 2 – 6 ⇒ x = – 4

e) y – 4 = – 7

Solution:

y – 4 = – 7

Adding 4 on both sides

y – 4 + 4 = – 7 + 4 ⇒ y = – 3

f) y – 4 = 4

Solution:

y – 4 = 4

Adding 4 on both sides

y – 4 + 4 = 4 + 4 ⇒ y = 8

g) y + 4 = 4

Solution:

y + 4 = 4

Adding -4 on both șides

y + 4 – 4 = 4 – 4 ⇒ y = 0

h) y + 4 = – 4

Solution:

y + 4 = – 4

Adding -4 on both sides

y + 4 – 4 = – 4 – 4 ⇒ y = – 8.

Question 2.

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

a) 3l = 42

Solution:

3l = 42

Divide by 3 on both sides

\(\frac { 3l }{ 3 }\) = \(\frac { 42 }{ 3 }\) ⇒ l = 14

b) \(\frac { b }{ 2 }\) = 6

Solution:

\(\frac { b }{ 2 }\) = 6

Multiplying by 2 on both sides

\(\frac { b }{ 2 }\) × 2 = 6 × 2 ⇒ b = 12

c) \(\frac { p }{ 7 }\) = 4

Solution:

\(\frac { p }{ 7 }\) = 4

Multiply by 7 on both sides

\(\frac { p }{ 7 }\) × 7 = 4 × 7 ⇒ p = 28

d) 4x = 25

Solution:

4x = 25

Divide by 4 on both sides

\(\frac { 4x }{ 4 }\) = \(\frac { 25 }{ 4 }\) ⇒ x = \(\frac { 25 }{ 4 }\)

e) 8y = 36

Solution:

8y = 36

Divide by 8 on both sides

\(\frac { 8y }{ 8 }\) = \(\frac { 36 }{ 8 }\) ⇒ y = \(\frac { 36 }{ 8 }\) = \(\frac { 18 }{ 4 }\) = \(\frac { 9 }{ 2 }\)

f) \(\frac { z }{ 3 }\) = \(\frac { 5 }{ 4 }\)

Solution:

\(\frac { z }{ 3 }\) = \(\frac { 5 }{ 4 }\)

Multiplying both sides by 3

\(\frac { z }{ 3 }\) × 3 = \(\frac { 5 }{ 4 }\) × 3 ⇒ z = \(\frac { 15 }{ 4 }\)

g) \(\frac { a }{ 5 }\) = \(\frac { 7 }{ 15 }\)

Solution:

\(\frac { a }{ 5 }\) = \(\frac { 7 }{ 15 }\)

Multiplying both sides by 5

\(\frac { a }{ 5 }\) × 5 = \(\frac { 7 }{ 15 }\) × 5 ⇒ a = \(\frac { 7 }{ 3 }\)

h) 20t = – 10

Solution:

20t = – 10

Divide both sides by 20

\(\frac { 20t }{ 20 }\) = \(\frac { -10 }{ 20 }\) ⇒ t = \(\frac { -1 }{ 2 }\)

Question 3.

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

a) 3n – 2 = 46

Solution:

3n – 2 = 46

Adding 2 on both sides

3n – 2 + 2 = 46 + 2 ⇒ 3n = 48

Divide both sides by 3

\(\frac { 3n }{ 3 }\) = \(\frac { 48 }{ 3}\) ⇒ n = 16

b) 5m + 7 = 17

Solution:

5m + 7 = 17

Adding -7 on both sides

5 m + 7 – 7 = 17 – 7 ⇒ 5m = 10

Divide both sides by 5

\(\frac { 5m }{ 5 }\) = \(\frac { 10 }{ 5 }\) ⇒ m = 2

c) \(\frac { 20p }{ 3 }\) = 40

Solution:

\(\frac { 20p }{ 3 }\) = 40

Multiply both sides by 3

\(\frac { 20p }{ 3 }\) × 3 = 40 × 3 ⇒ 20p = 120

Divide both sides by 20

\(\frac { 20p }{ 20 }\) = \(\frac { 120 }{ 20 }\) ⇒ p = 6

d) \(\frac { 3p }{ 10 }\) = 6

Solution:

\(\frac { 3p }{ 10 }\) = 6

Multiply both sides by 10

\(\frac { 3p }{ 10 }\) × 10 = 6 × 10 ⇒ 3p = 60

Divide both sides by 3

\(\frac { 3p }{ 3 }\) = \(\frac { 60 }{ 3 }\) ⇒ p = 20.

Question 4.

Solve the following equations :

a) 10p = 100

Solution:

10p = 100

Divide both sides by 10

\(\frac { 10p }{ 10 }\) = \(\frac { 100 }{ 10 }\) ⇒ p = 10

b) 10p + 10 = 100

Solution:

10p + 10 = 100

Subtract 10 on both sides

10p + 10 – 10 = 100 – 10 ⇒ 10p = 90

Divide both sides by 10

\(\frac { 10p }{ 10 }\) = \(\frac { 90 }{ 10 }\) ⇒ p = 9

c) \(\frac { p }{ 4 }\) = 5

Solution:

\(\frac { p }{ 4 }\) = 5

Multiply both sides by 4

\(\frac { p }{ 4 }\) × 4 = 5 × 4 ⇒ p = 20

d) \(\frac { -p }{ 3 }\) = 5

Solution:

\(\frac { -p }{ 3 }\) = 5

Multiply both sides by 3

\(\frac { -p }{ 3 }\) × 3 = 5 × 3 ⇒ – p = 15

Multiply both sides by -1

-1 × – p = 15 × – 1 ⇒ p = – 15

e) \(\frac { 3p }{ 4 }\) = 6

Solution:

\(\frac { 3p }{ 4 }\) = 6

Multiply both sides by 4

\(\frac { 3p }{ 4 }\) × 4 = 6 × 4 ⇒ 3p = 24

Divide both sides by 3

\(\frac { 3p }{ 3 }\) = \(\frac { 24 }{ 3 }\) ⇒ p = 8.

f) 3s = – 9

Solution:

3s = – 9

Divide both sides by 3

\(\frac { 3s }{ 3 }\) = \(\frac { -9 }{ 3 }\) ⇒ s = -3

g) 3s + 12 = 0

Solution:

3s + 12 = 0

Adding – 12 on both sides

3s + 12 – 12 = 0 – 12

Divide both sides by 3

\(\frac { 3s }{ 3 }\) = \(\frac { -12 }{ 3 }\) ⇒ s = – 4

h) 3s = 0

Solution:

3s = 0

Divide both sides by 3

\(\frac { 3s }{ 3 }\) = \(\frac { 0 }{ 3 }\) ⇒ s = 0

i) 2q = 6

Solution:

2q = 6

Divide both sides by 2

\(\frac { 2q }{ 2 }\) = \(\frac { 6 }{ 2 }\) ⇒ q = 3

j) 2q – 6 = 0

Solution:

2q – 6 = 0

Adding 6 on both sides

2q – 6 + 6 = 0 + 6 ⇒ 2q = 6

Divide both sides by 2

\(\frac { 2q }{ 2 }\) = \(\frac { 6 }{ 2 }\) ⇒ q = 3

k) 2q + 6 = 0

Solution:

2q + 6 = 0

Subtract 6 from both sides

2q + 6 – 6 = 0 – 6 ⇒ 2q = – 6

Divide both sides by 2

\(\frac { 2q }{ 2 }\) = \(\frac { -6 }{ 2 }\) ⇒ q = -3

l) 2q + 6 = 12

Solution:

2q + 6 = 12

Subtract 6 from both sides.

2q + 6 – 6 = 12 – 6 ⇒ 2q = 6

Divide both sides by 2

\(\frac { 2q }{ 2 }\) = \(\frac { 6 }{ 2 }\) ⇒ q = 3