CBSE students can refer to NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 Textbook Questions and Answers are provided by experts in order to help students secure good marks in exams.

## Class 7 Maths NCERT Solutions 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)**We have, x – 1 = 0

In order to solve this equation, we have to get x by itself on the L.H.S.

To get x by itself on the L.H.S., we need to shift -1.

This can be done by adding 1 to both sides of the given equation.

x – 1 + 1 = 0 + 1 [ Adding 1 to both sides]

x = 1 [∵ -1 + 1 = 0 and 0 + 1 = 1]

**(b)** We have, x + 1 = 0

In order to get x by itself on the L.H.S., we need to shift 1. This can be done by subtracting 1 from both sides of the given equation.

x + 1 – 1 = 0 – 1 [Subtracting 1 from both sides]

⇒ x = -1 [∵ 1 – 1 = 0, 0 – 1 = -1 ]

So, x = -1 is the solution of the given equation.

**(c)** We have, x – 1 = 5

In order to solve this equation, we have to get x by itself on the L.H.S. To get x by itself on the L.H.S., we need to shift -1.

This can be done by adding 1 to both sides of the given equation.

**(d)** We have, x + 6 = 2

In order to solve this equation, we have to get x by itself on the L.H.S. To get x by itself on the L.H.S., we need to shift 6.

This can be done by subtracting 6 from both sides of the given equation.

**(e)** We have, y – 4 = -7

In order to solve this equation, we have to get y by itself on the L.H.S. To get y by itself on the L.H.S., we need to shift -4.

This can be done by adding 4 to both sides of the given equation.

**(f)** We have, y – 4 = 4

In order to solve this equation, we have to get y by itself on the L.H.S. To get y by itself on the L.H.S., we need to shift -4.

This can be done by adding 4 to both sides of the given equation.

**(g)** We have, y + 4 = 4

In order to solve this equation, we have to get y by itself on the L.H.S. To get y by itself on the L.H.S., we need to shift 4.

This can be done by subtracting 4 from both sides of the given equation.

**(h)** We have, y + 4 = -4

In order to solve this equation, we have to get y by itself on the L.H.S. To get y by itself on the L.H.S., we need to shift 4.

This can be done by subtracting 4 from both sides of the given equation.

**Ex 4.2 Class 7 Maths 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)** We have, 3l = 42

In order to solve this equation, we have to get l by itself on the L.H.S. For this, 3 has to be removed from the L.H.S.

This can be done by dividing both sides of the equation by 3.

**(b)** We have, \(\frac { b }{ 2 } \) = 6

In order to solve this equation, we have to get b by itself on the L.H.S. To get b by itself on L.H.S., we have to remove 2 from L.H.S. This can be done by multiplying both sides of the equation by 2. Thus, we have

**(c)** We have, \(\frac { p }{ 7 } \) = 4

In order to solve this equation, we have to get p by itself on the L.H.S. To get p by itself on L.H.S., we have to remove 7 from L.H.S. This can be done by multiplying both sides of the equation by 7. Thus, we have

**(d)** We have, 4x = 25

In order to solve this equation, we have to get x by itself on the L.H.S. For this, 4 has to be removed from the L.H.S. This can be done by dividing both sides of the equation by 4.

**(e)** We have, 8y = 36

In order to solve this equation, we have to get y by itself on the L.H.S. For this, 8 has to be removed from the L.H.S. This can be done by dividing both sides of the equation by 8.

**(f)** We have, \(\frac { z }{ 3 } \) = \(\frac { 5 }{ 4 } \)

In order to solve this equation,we have to get z by itself on the L.H.S. To get z by itself on L.H.S., we have to remove 3 from L.H.S. This can be done by multiplying both sides of the equation by 3.

Thus, we have

**(g)** We have, \(\frac { a }{ 5 } \) = \(\frac { 7 }{ 15 } \)

In order to solve this equation, we have to get a by itself on the L.H.S. To get a by itself on L.H.S., we have to remove 5 from L.H.S. This can be done by multiplying both sides of the equation by 5.

**(h)** We have, 20t = -10

In order to solve this equation, we have to get t by itself on the L.H.S. For this, 20 has to be removed from the L.H.S. This can be done by dividing both sides of the equation by 20.

**Ex 4.2 Class 7 Maths 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)** \(\frac { 20p }{ 3 } \) = 40

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

**Solution:
**

**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)** \(\frac { 3p }{ 4 } \) = 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:**