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

- Class 7 Maths Simple Equations Exercise 4.1
- Class 7 Maths Simple Equations Exercise 4.2
- Class 7 Maths Simple Equations Exercise 4.3
- Class 7 Maths Simple Equations Exercise 4.4

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

Ex 4.2 Class 7 Maths Question 1.

Given 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 + l = 0 + 1 ⇒ x = 1

Thus, x = 1 is the required solutions.

Check: Put x = 1 in the given equations

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: Put x = -1 in the given equation

-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

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

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

Solution:

(a) 3n – 2 = 46

⇒ 3n- 2 + 2 = 46+ 2 (adding 2 to both sides)

⇒ 3n = 48

⇒ 3n + 3 = 48 ÷ 3

(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)

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

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

⇒\(\frac{3 p}{10} \times 10=6 \times 10\) (Multiplying both sides by 10)

⇒ 3p = 60

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

Ex 4.2 Class 7 Maths Question 4.

Solve the following equations:

(a) 10p = 100

(b) 10p + 10 = 100

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

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

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

(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)

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

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 side by 10)

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

Thus p = 9

(l) 2q + 6 = 12

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

⇒ 2q = 6

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