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NCERT Solutions for Class 7 Mathematics Chapter 4 -Simple Equations Exercise 4.2

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Please find the detailed Solutions for class 7th Maths Chapter 4 Simple Equations Exercise 4.2 Solutions, you can check all the solutions from here chapter wise

NCERT Class 7th Mathematics Chapter 4 Simple Equations Exercise 4.2 Solutions

1. Give first the step you will use to separate the variable and then solve the equation:
(a) x – 1 = 0

Sol: Adding 1 to both sides of the equation, we get

x-1+1 = 0+1

x = 1

(b) x + 1 = 0

Sol: Subtracting 1 from both sides of the equation, we get

x+1-1 = 0-1

x = -1 

(c) x – 1 = 5

Adding 1 to both sides of the equation, we get

x-1+1 = 5+1

x = 6

(d) x + 6 = 2

Sol: Subtracting 6 from both sides of the equation, we get

x+6-6 = 2-6

x = -4

(e) y – 4 = – 7

Adding 4 to both sides of the equation, we get

y-4+4 = -7+4

y = -3

(f) y – 4 = 4

Adding 4 to both sides of the equation, we get

y-4+4 = 4+4

y = 8

(g) y + 4 = 4

Sol: Subtracting 4 from both sides of the equation, we get

y+4-4 = 4-4

y = 0

(h) y + 4 = – 4

Sol: Subtracting 4 from both sides of the equation, we get

y+4-4 = -4-4

y = -8

2. Give first the step you will use to separate the variable and then solve the equation:
(a) 3l = 42

Sol: Dividing both sides of the equation by 3, we get

3l/3 = 42/3

l = 14

(b) b/2 = 6

Sol: Multiplying both sides of the equation by 2, we get

(b*2)/2 = 6*2

b = 12

(c) p/7 = 4

Sol: Multiplying both sides of the equation by 7, we get

(p/7)*7 = 4*7

p = 28

(d) 4x = 25

Sol: Dividing both sides of the equation by 4, we get

(4x)/4 = 25/4

x = 25/4

(e) 8y = 36

Sol : Dividing both sides of the equation by 8, we get

8y/8 = 36/8

y = 36/8

(f) z/3 = 5/4

Sol: Multiplying both sides of the equation by 3, we get

(z/3) *3 = (5/4) *3

z = 15/4

(g) a/5 = 7/15 

Sol: Multiplying both sides of the equation by 5, we get

a*5 / 5 = (7/15) *5

a = 7/3

(h) 20t = – 10

Sol: Dividing both sides of the equation by 20, we get

20t/20 = -10/20

t = -1/2

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

Sol: Adding 2 to both sides of the equation, we get

3n-2+2 = 46+2

3n = 48

Now, dividing both sides of the equation by 3, we get

3n/3 = 48 /3

n = 16

(b) 5m + 7 = 17

Sol: Subtracting 7 from both the sides of the equation, we get

5m+7-7 = 17-7

5m = 10

Now, dividing both sides of the equation by 5,we get 

5m / 5 =10/5

m = 2

(c) 20p/3 = 40

Sol: Multiplying both sides of the equation by 3, we get

(20p/3) * 3 = 40*3

20p = 120

Now, dividing both sides of the equation by 20, we get

20p/20 = 120/20

p = 6

(d) 3p/10 = 6

Sol: Multiplying both sides of the equation by 10, we get

(3p/10) * 10 = 6*10

3p = 60

Now, dividing both sides of the equation by 3, we get

3p/3 = 60/3

p = 20

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

Sol: Dividing both sides of the equation by 10, we get

10p/10 = 100/10

p= 10

(b) 10p + 10 = 100

Sol:  Subtracting 10 from of the equation both sides, we get

10p+10-10 = 100-10

10p = 90

Now, dividing both sides of the equation by 10, we get

10p/10 = 90/10

p = 9

(c) p/4 = 5

Sol: Multiplying both sides of the equation by 4, we get

(p/4) * 4 = 5*4

p = 20

(d) -p/3 = 5

Sol: Multiplying both of the equation sides by -3, we get

(-p/3) * -3 = 5 * -3

p = -15

(e) 3p/4 = 6

Sol: Multiplying both sides of the equation by 4, we get

(3p/4) * 4 = 6*4

3p = 24

Now dividing both sides of the equation by 3, we get

(3p)/3 = 24/3

p = 8

(f) 3s = –9

Sol: Dividing both sides of the equation by 3, we get

3s/3 = -9/3

s = -3

(g) 3s + 12 = 0

Sol: Subracting 12 from both sides of the equation, we get

3s+12-12 = 0-12

3s = -12

Now, dividing both sides of the equation by 3, we get

3s/3 = -12/3

s = -4

(h) 3s = 0

Sol: Dividing both sides of the equation by 3, we get

3s/3 = 0/3

s = 0

(i) 2q = 6

Sol: Dividing both sides of the equation by 2, we get

2q/2 = 6/2

q = 3

(j) 2q – 6 = 0

Sol: Adding 6 to both sides of the equation, we get

2q-6+6 = 0+6

2q = 6

Now, dividing both sides of the equation by 2, we get

2q/2 = 6/2

q = 3

(k) 2q + 6 = 0

Sol: Subtracting 6 from both sides of the equation, we get
2q+6-6 = 0-6

2q = -6

Now, dividing both sides of the equation by 2, we get

2q/2 = -6/2

q = -3

(l) 2q + 6 = 12

Sol: Subtracting 6 from both sides of the equation, we get
2q+6-6 = 12-6

2q = 6

Now, dividing both sides of the equation by 2, we get

2q/2 = 6/2

q = 3

You can find the solutions for previous exercise from here

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