Chapter 4 Simple Equations Exercise 4.2


Question 1.
Given 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) x – 1 = 0
Adding 1 to both sides, we get
x – 1 + l = 0 + 1 ⇒ x = 1
Thus, x = 1 is the required solutions.
Check: Put x = 1 in the given equations
x – 1 = 0
1 – 1 = 0
0 = 0
LHS = RHS
Thus x = 1 is the correct solution.

(b) x + 1 = 0
Subtracting 1 from both sides, we get
x + 1 – 1 = 0 – 1 ⇒ x = -1
Thus x = -1 is the required solution.
Check: Put x = -1 in the given equation
-1 + 1 = 0
0 = 0
LHS = RHS
Thus x = -1 is the correct solution.

(c) x – 1 = 5
Adding 1 to both sides, we get
x – 1 + 1 = 5 + 1 ⇒ x = 6
Thus x = 6 is the required solution.
Check: x – 1 = 5
Putting x = 6 in the given equation
6 – 1 = 5 ⇒ 5 = 5
LHS = RHS
Thus, x = 6 is the correct solution.

(d) x + 6 = 2
Subtracting 6 from both sides, we get
x + 6 – 6 = 2 – 6 ⇒ x = -4
Thus, x = -4 is the required solution.
Check: x + 6 = 2
Putting x = -4, we get
-4 + 6 = 2 ⇒ 2 = 2 LHS = RHS
Thus x = -4 is the correct solution.

(e) y – 4 = -7
Adding 4 to both sides, we get
y – 4 + 4 = -7 + 4 ⇒ y = -3
Thus, y = -3 is the required solution.
Check: y – 4 = -7
Putting y = -3, we get
-3 – 4 = -7 ⇒ -7 = -7
LHS = RHS
Thus, y = -3 is the correct solution.

(f) y – 4 = 4
Adding 4 to both sides, we get
y – 4 + 4 = 4 + 4 ⇒ y = 8
Thus, y = 8 is the required solution.
Check: y – 4 = 4
Putting y = 8, we get
8 – 4 = 4 ⇒ 4 = 4
LHS = RHS
Thus y = 8 is the correct solution.

(g) y + 4 = 4
Subtracting 4 from both sides, we get
y + 4 – 4 = 4 – 4 ⇒ y = 0
Thus y = 0 is the required solution.
Check: y + 4 = 4
Putting y = 0, we get
0 + 4 = 4 ⇒ 4 = 4
LHS = RHS
Thus y = 0 is the correct solution.

(h) y + 4 =-4
Subtracting 4 from both sides, we get
y + 4 – 4 = -4 – 4 ⇒ y = -8
Thus, y = -8 is the required solution.
Check: y + 4 = -4
Putting y = -8, we get
-8 + 4 = -4 ⇒ -4 = -4
LHS = RHS
Thus, y = -8 is the correct solution.

Question 2.
Give first the step you will use to separate the variable and then solve the following equation:


Solution:
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 2
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 3

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

Question 3.
Give the steps you will use to separate the variables and then solve the equation:
(a) 3n – 2 = 46
(b) 5m + 7 = 17


Solution:
(a) 3n – 2 = 46
⇒ 3n- 2 + 2 = 46+ 2 (adding 2 to both sides)
⇒ 3n = 48
⇒ 3n + 3 = 48 ÷ 3

(b) 5m + 7 = 17
⇒ 5m+ 7 – 7 = 17 – 7 (Subtracting 7 from both sides)
⇒ 5m = 10
⇒ 5m + 5 = 10 ÷ 5 (Dividing both sides by 5)

(c) 20p3=40
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 7


⇒ 3p = 60
⇒ 3p ÷ 3 = 60 ÷ 3 (Dividing both sides by 3)

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 9
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 10
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 11
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 14

(l) 2q + 6 = 12
⇒ 2q + 6 – 6 = 12 – 6 ( Subtracting 6 from both sides)
⇒ 2q = 6
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 13

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 Q1
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 Q2
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 Q2.1
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 Q3
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 Q3.1
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 Q3.2
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 Q3.3
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