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

## 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.

### 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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