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:
