Class 7 Maths NCERT Solutions Chapter – 4 Simple Equations EX – 4.2

Simple Equations

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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 7

(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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 8

(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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 9

(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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 10

(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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 11

(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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 12

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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 13

(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
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 14

(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
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 15

(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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 16

(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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 17

(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
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 18

(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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 19

(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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 20

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:

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 21
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 22
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 23

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:
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 24
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 25
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 26
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 27