# Simple Equations NCERT Solution for Class 7 Maths Chapter 4 Exercise 4.2

Simple Equations Class 7 Math EX-4.1

## Simple Equations for Class 7, Chapter 4 Maths NCERT Solution ;

Simple Equations Class 7 Ex. 4.2;

Exercise 4.2

Simple Equations Chapter 4 NCERT Math Ex.-4.2

Question 1:- Gives, the 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 1:-
(a) x – 1 = 0
Adding ”1” to both sides, we get
x – 1 + 1 = 0 + 1
x = 1
Therefore, the necessary solution is x = 1.
Verify: Enter x = 1 in the provided equations.
x – 1 = 0
1 – 1 = 0
0 = 0
L.H.S = R.H.S
Thus x = 1 is required solution.

(b) x + 1 = 0
Subtracting 1 from both sides, we get
x + 1 – 1 = 0 – 1
x = -1
Therefore, the necessary solution is x = -1.
Verify: Enter x = -1 in the provided equations.
x+1=0
-1 + 1 = 0
0 = 0
L.H.S = R.H.S
Thus, the needed solution is x = -1.

(c) x – 1 = 5
Adding 1 to both sides, we get
x – 1 + 1 = 5 + 1
x = 6
Therefore, the necessary solution is x = 6.
Verify: Enter x = 6 in the provided equations.
x – 1 = 5
6 – 1 = 5
5 = 5
L.H.S = R.H.S
Thus, the right answer is x = 6.

(d) x + 6 = 2
Subtracting 6 from both sides, we get
x + 6 – 6 = 2 – 6
x = -4
Therefore, the necessary solution is x = -4.
Verify: Enter x = -4 in the provided equations.
x + 6 = 2
Putting x = -4, we get
-4 + 6 = 2
2 = 2
L.H.S = R.H.S
Therefore, the necessary solution is x = -4.

(e) y – 4 = -7
Adding 4 to both sides, we get
y – 4 + 4 = -7 + 4
y = -3
Therefore, the necessary solution is y = -3.
Verify: Enter y = -3 in the provided equations.
y – 4 = -7
Putting y = -3, we get
-3 – 4 = -7
⇒ -7 = -7
L.H.S = R.H.S
Consequently, the necessary solution is y = -3.

(f) y – 4 = 4
Adding 4 to both sides, we get
y – 4 + 4 = 4 + 4
y = 8
Therefore, the necessary solution is y = 8.
Verify: Enter y = 8 in the provided equations.
y – 4 = 4
8 – 4 = 4
4 = 4
L.H.S = R.H.S
Thus, the right answer is y = 8.

(g) y + 4 = 4
Subtracting 4 from both sides, we get
y + 4 – 4 = 4 – 4
y = 0

Therefore, the necessary solution is y = 0.
Verify: Enter y = 0 in the provided equations.
y + 4 = 4
0 + 4 = 4
4 = 4
L.H.S = R.H.S
Therefore, y = 0 is the required solution.

(h) y + 4 =-4
Subtracting 4 from both sides, we get
y + 4 – 4 = -4 – 4
y = -8
Therefore, the necessary solution is y = -8.
Verify: Enter y = -8 in the provided equations.
y + 4 = -4
Putting y = -8, we get
-8 + 4 = -4
-4 = -4
L.H.S = R.H.S
Therefore, y = -8 is the required solution.

Simple Equations Chapter 4 NCERT Math Ex.-4.2

Question 2 :- Give first the step you will use to separate the variable and then solve the following equation:
(a) 3L = 42
(b)b/2=6
(c)p/7=4
(d)4x=25
(e)8y=36
(f)z/3=5/4
(g)a/5=7/15
(h)20t=-10

Solution 2:-
(a) 3L = 42
Dividing both side by 3, we get
3L/3 = 42/3
L = 14
Thus, L = 14 is the required solution.
Check: 3L = 42
Putting y = 14, we get
3×14 = 42 ⇒ 42 = 42
L.H.S = R.H.S
Thus, 14 = 14 is the correct solution.

(b) b/2 = 6
Multiply both side by 2, we get
(b/2)x2 = 6×2
b = 12
Thus, b = 12 is the required solution.
Check: b = 12
Putting b = 14, we get
3×14 = 42 ⇒ 42 = 42
L.H.S = R.H.S
Thus, 14 = 14 is the correct solution.

(c) p/7 = 4
Multiply both side by 7, we get
(p/7)x7 = 4×7
p = 28
Thus, p = 28 is the required solution.
Check: p/7 = 4
Putting p = 28, we get
28/7 = 4 ⇒ 4 = 4
L.H.S = R.H.S
Thus, 4 = 4 is the correct solution.

(d) 4x = 25
Dividing both side by 4, we get
(4x)/4 = 25
x = 25
Thus, x = 25 is the required solution.
Check: 4x = 25
Putting x = 25, we get
25 = 25 ⇒ 25 = 25
L.H.S = R.H.S
Thus, 25 = 25 is the correct solution.

(e) 8y = 36
Dividing both side by 8, we get
(8y)/8 = 36/8
y = 9/2
Thus, y = 9/2 is the required solution.
Check: 8y = 36
Putting y = 9/2, we get
8x(9/2) = 36 ⇒ 36 = 36
L.H.S = R.H.S
Thus, 36 = 36 is the correct solution.

(f) z/3 = 5/4
Multiply both side by 3, we get
(z/3)x3 = (5/4)x(3)
z = 15/4
Thus, z = 15/4 is the required solution.
Check: z/3 = 5/4
Putting z = 15/4, we get
(15/4)x(1/3) = 5/4
5/4 = 5/4
L.H.S = R.H.S
Thus, 5/4 = 5/4 is the correct solution.

(g) a/5 = 7/15
Multiply both side by 5, we get
(a/5)x5 = (7/15)x5
a = 7/3
Thus, a = 7/15 is the required solution.
Check: a/5 = 7/15
Putting a = 7/3, we get
(7/3)x(1/5) = 7/15 ⇒ 7/15 = 7/15
L.H.S = R.H.S
Thus, 7/15 = 7/15 is the correct solution.

(h) 20t = -10
Dividing both side by 20, we get
(20t)/20 = -(10/20)
t = -(1/2)
Thus, t = -(1/2) is the required solution.
Check: 20t = -10
Putting t = -(1/2), we get
20x-(1/2) = -10 ⇒ -10 = -10
L.H.S = R.H.S
Thus, -10 = -10 is the correct solution. Simple Equations Chapter 4 NCERT Math Ex.-4.2

Question 3 :- Gives, the steps you will use to separate the variables and then solve the equation:
(a) 3n – 2 = 46
(b) 5m + 7 = 17
(c) 20p/3=40
(d) 3p/10=6

Solution 3:-
(a) 3n – 2 = 46
⇒ 3n – 2 + 2 = 46+ 2                     [Since, adding 2 to both sides]
⇒ 3n = 48
⇒ 3n ÷ 3 = 48 ÷ 3                          [Since, Dividing both sides by 3]
n = 16
Check: 3n-2 = 46
Putting n = 16, we get
3x(16) – 2 = 46
48-2 = 46
46=46
L.H.S = R.H.S
Thus, 46 = 46 is the correct solution.

(b) 5m + 7 = 17
⇒ 5m + 7 – 7 = 17- 7                     [Since, Subtracting 7 to both sides]
⇒ 5m = 10
⇒ 5m/5 = 10/5                                 [Since, Dividing both sides by 5]
m = 2
Check: 5m+7 = 17
Putting m = 2, we get
5x(2) + 7 = 17
10+7 = 17
17=17
L.H.S = R.H.S
Thus, 17 = 17 is the correct solution.

(c) 20p/3 = 40
⇒ (20p/3)x3 = 40×3                     [Since, Multiplying 3 to both sides]
⇒ 20p = 120
⇒ 20p/20 = 120/20                           [Since, Dividing both sides by 20]
p = 6
Check: 20p/3 = 40
Putting p = 6, we get
20x(6)/3 = 40
40 = 40
40=40
L.H.S = R.H.S
Thus, 40 = 40 is the correct solution.

(d) 3p/10 = 6
⇒ (3p/10)x10 = 6×10                     [Since, Multiplying 10 to both sides]
⇒ 3p = 60
⇒ 3p/3 = 60/3                           [Since, Dividing both sides by 3]
p = 20
Check: 3p/10 = 6
Putting p = 20, we get
3x(20)/10 = 6
60/10 = 6
6=6
L.H.S = R.H.S
Thus, 6 = 6 is the correct solution.

Simple Equations Chapter 4 NCERT Math Ex.-4.2

Question 4:- Solve the following equations:
(a) 10p = 100
(b) 10p + 10 = 100
(c) p/4=5
(d) −p/3=5
(e) 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 4:-
(a) 10p = 100
⇒ 10p/10 = 100/10                   [Since, Dividing both sides by 10]
p=100/10=10
Thus p= 10

(b) 10p + 10 = 100
⇒ 10p + 10 – 10 = 100 -10           [Since, Subtracting 10 from both sides]
⇒ 10p = 90
⇒ 10p/10 = 90/10                   [Since, Dividing both side by 10]
p=90/10=9
Thus p = 9

(c) p/4 = 5
⇒ (p/4)x(4) = 5×4            [Since, Multiplying 4 from both sides]
⇒ p = 20

(d) -p/3 = 5
⇒ -(p/3)x(3) = 5×3            [Since, Multiplying 3 from both sides]
⇒ p = -15

(e) 3p/4 = 6
⇒ (3p/4)x(4) = 6×4            [Since, Multiplying 4 from both sides]
⇒ 3p = 24                          [Dividing both side by 3]
3p/3 = 24/3
p=8

(f) 3s = -9
⇒ (3s)/(3) = -9/3            [Since, Dividing 3 from both sides]
⇒ s = -3

(g) 3s+12 = 0
⇒ 3s+12-12 = 0-12            [Since, Subtracting 12 from both sides]
⇒ 3s = -12                          [Dividing both side by 3]
3s/3 = -12/3
s=-4

(h) 3s = 0
⇒ 3s/3 = 0/3            [Since, Dividing 3 from both sides]
⇒ s = 0

(i) 2q = 6
⇒ (2q )/(2) = 6/2            [Since, Dividing 2 from both sides]
⇒ q = 3

(j) 2q-6 = 0
⇒ 2q -6+6 = 0+6            [Since, Adding 6 from both sides]
⇒ 2q = 6                          [Dividing both side by 2]
2q/2=6/2
q=3

(k) 2q+6 = 0
⇒ 2q +6-6 = 0-6            [Since, Subtracting 6 from both sides]
⇒ 2q = -6
2q/2=-6/2              [Dividing both side by 2]
q=-3

(l) 2q + 6 = 12
⇒ 2q + 6 – 6 = 12 – 6   [Since, Subtracting 6 from both sides]
⇒ 2q = 6
2q/2=6/2                [Dividing both side by 2]
q=3

Simple Equations Chapter 4 NCERT Math Ex-4.2

Simple Equations

Simple Equations

Simple Equations

Simple Equations

Simple Equations

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