NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2.

 Board CBSE Textbook NCERT Class Class 7 Subject Maths Chapter Chapter 4 Chapter Name Simple Equations Exercise Ex 4.2 Number of Questions Solved 4 Category NCERT Solutions

## NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2

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
(e) y – 4 = – 7
(f) y – 4 = 4
(g) y + 4 = 4
(h) y + 4 = – 4
Solution:
(a) The given equation is x – 1 = 0
x – 1 + 1 = 0 + 1 ⇒ x = 1
It is the required solution.
Check. Put the solution x = 1 back into the equation.
L.H.S. = x – 1 = 1 = 1 – 0 = R.H.S.
The solution is thus checked for its correctness.

(b) The given equation is x + 1 = 0
Subtract 1 from both sides, x + 1 – 1 = 0 – 1 ⇒ x = – 1
It is the required solution.
Check. Put the solution x = – 1 back into the equation.
L.H.S. = x + 1 = (-1)+1
= 0 = R.H.S.
The solution is thus checked for its correctness.

(c) The given equation is
x – 1 = 5
x + 1 – 1 = 5 + 1 ⇒ x = 6
It is the required solution
Check. Put the solution x = 6 back into the equation.
L.H.S. = x – 1 = 6 – 1 = 5 = R.H.S.
The solution is thus checked for its correctness.

(d) The given equation is x + 6 = 2
Subtract 6 from both sides, x + 6 – 6 = 2 – 6 ⇒ x = – 4
It is the required solution.
Check. Put the solution x = – 4 back into the equation.
L.H.S. = x + 6 = – 4 + 6 = 2 = R.H.S.
The solution is thus checked for its correctness.

(e) The given equation is y – 4 = – 7
Add 4 to both sides, y – 4 + 4 = – 7 + 4 ⇒ y = – 3
It is the required solution.
Check. Put the solution
L.H.S. = y – 4 = – 3 – 4 = – 7 = R.H.S.
The solution is thus checked for its correctness.

(f) The given equation is y – 4 = 4
y – 4 + 4 = 4 + 4 ⇒ y = 8
It is the required solution.
Check. Put the solution y = 8 back into the equation.
L.H.S. = y – 4 = 8 – 4 = 4 = R.H.S.
The solution is thus checked for its correctness.

(g) The given equation is y + 4 = 4
Subtract 4 from both sides, y + 4 – 4 = 4 – 4 ⇒ y = 0
It is the required solution.
Check. Put the solution y = 0 back into the equation.
L.H.S. =y + 4 = 0 + 4 = 4 = R.H.S.
The solution is thus checked for its correctness.

(h) The given equation is y + 4 = – 4
Subtract 4 from both sides, y + 4 – 4 = – 4 – 4 ⇒ y = -8
It is the required solution.
Check. Put the solution y = – 8 back into the equation.
L.H.S. = y + 4 = -8 + 4 = – 4 = R.H.S.
The solution is thus checked for its correctness.

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:

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:

Question 4.
Solve the following equations:

Solution:

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