**NCERT
Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2**

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex
4.2 are the part of NCERT Solutions for Class 7 Maths. Here you can find the
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2.

**NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1****NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2****NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3****NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4**

**Ex 4.2 Class 7 Maths 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

**Solution:
**(a) x – 1 = 0

Adding 1 to both sides, we get

x – 1 + 1 = 0 + 1 ⇒ x = 1

Thus, x = 1 is the required solution.

**Check:**Putting x = 1 in the given equation, we get

x – 1 = 0

1 – 1 = 0

0 = 0

LHS = RHS

Thus, x = 1 is the correct solution.

(b) x + 1 = 0

Subtracting 1 from both sides, we get

x + 1 – 1 = 0 – 1 ⇒ x = -1

Thus, x = -1 is the required solution.

**Check:** Putting x = -1 in the given equation,
we get

-1 + 1 = 0

0 = 0

LHS = RHS

Thus, x = -1 is the correct solution.

(c)
x – 1 = 5

Adding 1 to both sides, we get

x – 1 + 1 = 5 + 1 ⇒ x = 6

Thus, x = 6 is the required solution.

**Check:** x – 1 = 5

Putting x = 6 in the given equation, we get

6 – 1 = 5 ⇒ 5 = 5

LHS = RHS

Thus, x = 6 is the correct solution.

(d) x + 6 = 2

Subtracting 6 from both sides, we get

x + 6 – 6 = 2 – 6 ⇒ x = -4

Thus, x = -4 is the required solution.

**Check:** x + 6 = 2

Putting x = -4 in the given equation, we get

-4 + 6 = 2 ⇒ 2 = 2 LHS = RHS

Thus, x = -4 is the correct solution.

(e) y – 4 = -7

Adding 4 to both sides, we get

y – 4 + 4 = -7 + 4 ⇒ y = -3

Thus, y = -3 is the required solution.

**Check:** y – 4 = -7

Putting y = -3 in the given equation, we get

-3 – 4 = -7 ⇒ -7 = -7

LHS = RHS

Thus, y = -3 is the correct solution.

(f) y – 4 = 4

Adding 4 to both sides, we get

y – 4 + 4 = 4 + 4 ⇒ y = 8

Thus, y = 8 is the required solution.

**Check:** y – 4 = 4

Putting y = 8 in the given equation, we get

8 – 4 = 4 ⇒ 4 = 4

LHS = RHS

Thus, y = 8 is the correct solution.

(g) y + 4 = 4

Subtracting 4 from both sides, we get

y + 4 – 4 = 4 – 4 ⇒ y = 0

Thus, y = 0 is the required solution.

**Check:** y + 4 = 4

Putting y = 0 in the given equation, we get

0 + 4 = 4 ⇒ 4 = 4

LHS = RHS

Thus, y = 0 is the correct solution.

(h) y + 4 = -4

Subtracting 4 from both sides, we get

y + 4 – 4 = -4 – 4 ⇒ y = -8

Thus, y = -8 is the required solution.

**Check:** y + 4 = -4

Putting y = -8 in the given equation, we get

-8 + 4 = -4 ⇒ -4 = -4

LHS = RHS

Thus, y = -8 is the correct solution.

**Ex 4.2 Class 7 Maths Question 2.**

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

**Ex
4.2 Class 7 Maths Question 3.**

Give the steps you will use to separate the
variables and then solve the equation:(a) 3n – 2 = 46

(b) 5m + 7 = 17

(c) 20p/3 = 40

(d) 3p/10 = 6

**Solution:
**(a) 3n – 2 = 46

⇒ 3n – 2 + 2 = 46 + 2 (Adding 2 to both sides)

⇒ 3n = 48

⇒ 3n ÷ 3 = 48 ÷ 3 (Dividing both sides by 3)

⇒ n = 16

(b) 5m + 7 = 17

⇒ 5m + 7 – 7 = 17 – 7 (Subtracting
7 from both sides)

⇒ 5m = 10

⇒ 5m ÷ 5 = 10 ÷ 5 (Dividing
both sides by 5)

⇒ m = 2

(c) 20p/3 = 40

(d) 3p/10 = 6

⇒3p/10 × 10 = 6 × 10 (Multiplying
both sides by 10)

⇒ 3p = 60

⇒ 3p ÷ 3 = 60 ÷ 3 (Dividing both sides by 3)

⇒ p = 20

**Ex
4.2 Class 7 Maths Question 4.**

Solve the following equations:(a) 10p = 100

(b) 10p + 10 = 100

(c) p/4 = 5

(d) −p/3 = 5

(e) 3p/4 = 6

(f) 3s = -9

(g) 3s + 12 = 0

(h) 3s = 0

(i) 2q = 6

(j) 2q – 6 = 0

(k) 2q + 6 = 0

(l) 2q + 6 = 12

**Solution:
**(a) 10p = 100

⇒ 10p ÷ 10 = 100 ÷ 10 (Dividing both sides by 10)

p = 100/10 = 10

Thus, p = 10

(b)
10p + 10 = 100

⇒ 10p + 10 – 10 = 100 – 10 (Subtracting 10 from both sides)

⇒ 10p = 90

⇒ 10p ÷ 10 = 90 ÷ 10 (Dividing both sides by 10)

p = 90/10 = 9

Thus, p = 9

(l) 2q + 6 = 12

⇒ 2q + 6 – 6 = 12 – 6 ( Subtracting 6 from both sides)

⇒ 2q = 6

⇒ 2q ÷ 2 = 6 ÷ 2

⇒ q = 3

Thus, q = 3.

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