## MP Board Class 7th Maths Solutions Chapter 4 Simple Equations Ex 4.2

**Class 7 Maths Chapter 4 Exercise 4.2 Question 4 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:

On adding 1 to both sides of the given equation, we obtain

x – 1 + 1 = 0 + 1 ⇒ x = 1

(b) x + 1 = 0

On subtracting 1 from both sides of the given equation, we obtain

x + 1 – 1 = 0 – 1 ⇒ x = -1

(c) x – 1 = 5

On adding 1 to both sides of the given equation, we obtain

x – 1 + 1 = 5 + 1 ⇒ x = 6

(d) x + 6 = 2

On subtracting 6 from both sides of the given equation, we obtain

x + 6 – 6 = 2 – 6 ⇒ x = -4

(e) y – 4 = – 7

On adding 4 to both sides of the given equation, we obtain

y – 4 + 4 = -7 + 4 ⇒ y = -3

(f) y – 4 = 4

On adding 4 to both sides of the given equation, we obtain

y – 4 + 4 = 4 + 4 ⇒ y = 8

(g) y + 4 = 4

On subtracting 4 from both sides of the given equation, we obtain

y + 4 – 4 = 4 – 4 ⇒ y = 0

(h) y + 4 = -4

On subtracting 4 from both sides of the given equation, we obtain

y + 4 – 4 = – 4 – 4 ⇒ y = -8

**Class 7th Maths Chapter 4 Exercise 4.2 Question 2.**

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

(a) 3l = 42

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

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

(d) 4x = 25

(e) 8y = 36

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

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

(h) 20t = -10

Solution:

(a) 3l = 42

On dividing both sides of the given equation by 3, we obtain

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

On multiplying both sides of the given equation by 2, we obtain

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

On multiplying both sides of the given equation by 7, we obtain

(d) 4x = 25

On dividing both sides of the given equation by 4, we obtain

(e) 8y = 36

On dividing both sides of the given equation by 8, we obtain

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

On multiplying both sides of the given equation by 3, we obtain

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

On multiplying both sides of the given equation by 5, we obtain

(h) 20t = -10

On dividing both sides of the given equation by 20, we obtain

**Exercise 4.2 Class 7 Solutions Question 3.**

Give the steps you will use to separate the variable 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

On adding 2 to both sides, we obtain

3n – 2 + 2 = 46 + 2 ⇒ 3n = 48

On dividing both sides of the given equation by 3, we obtain

(b) 5m + 7 = 17

On subtracting 7 from both sides, we obtain

5m + 7 – 7 = 17 – 7 ⇒ 5m = 10

On dividing both sides of the given equation by 5, we obtain

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

On multiplying both sides of the given equation by 3, we obtain

On dividing both sides of the given equation by 20, we obtain

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

On multiplying both sides of the given equation by 10, we obtain 3pxl0

On dividing both sides of the given equation by 3, we obtain

**Class 7 Maths Chapter 4 Exercise 4.2 Question 3 Question 4.**

Solve the following equations:

(a) 10p = 100

(b) 10p + 10 = 100

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

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

(e) \(\frac{39}{4}\) = 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: