# GSEB Solutions Class 7 Maths Chapter 4 Simple Equations Ex 4.2

Gujarat Board GSEB Textbook Solutions Class 7 Maths Chapter 4 Simple Equations Ex 4.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions 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
(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 have
x – 1 + 1 = 0 + 1
or x = 1,
which is the required solution.
Check : L.H.S = x – 1
= 1 – 1
= 0 = R.H.S
Thus, x = 1 is the correct solution.

(b) x + 1 = 0
Subtracting 1 from both sides, we have x + 1 – 1 = 0 – 1
or x = – 1,
which is the required solution.
Check : L.H.S. = x + 1
= (- 1) + 1
= 0 = R.H.S.
∴ The solution x = – 1 is correct.

(c) x – 1 = 5
Adding 1 to both sides, we have
x – 1 + 1 = 5 + 1.
or x = 6,
which is the required solution.
Check : L.H.S. = x – 1
= 6 – 1
= 5
= R.H.S.
∴ The solution x = 6 is correct.

(d) x + 6 = 2
Subtracting 6 from both sides, we have
x + 6 – 6 = 2 – 6
or x = -4,
which is the required solution.
Check :
L.H.S. = x + 6
= – 4 + 6
= 2
= R.H.S.
∴ The solution x = – 4 is correct.

(e) y – 4 = – 7
Adding 4 to both sides, we have y – 4 + 4 = – 7 + 4 or y = – 3,
which is the required solution.
Check :
L.H.S. = y – 4
= – 3 – 4
= – 7
= R.H.S.
∴ The solution y = – 3 is correct.

(f) y – 4 = 4
Adding 4 to both sides, we have y – 4 + 4 = 4 + 4 or y = 8,
which is the required solution.
Check :
L.H.S. =y – 4
= 8 – 4
= 4
= R.H.S.
∴ y = 8 is the correct solution.

(g) y + 4 = 4
Subtracting 4 from both sides, we have
y + 4 – 4 = 4 – 4
or y = 0,
which is the required solution.
Check :
L.H.S. =y + 4
= 0 + 4
= 4
= R.H.S.
∴ y = 0 is the correct solution.

(h) y + 4 = – 4
Subtracting 4 from both sides, we have
y + 4 – 4 = – 4 – 4 or y = – 8,
which is the required solution.
Check:
L.H.S. = y + 4
= – 8 + 4 = – 4
= R.H.S.
y = – 8 is the correct solution. 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) 3l = 42
Dividing both sides by 3, we have
= $$\frac { 3l }{ 3 }$$ = $$\frac { 42 }{ 3 }$$ or l = 14
∴ l = 14 is the required solution.
Check : L.H.S. =3l
= 3 x 14 = 42 = R.H.S.
Thus, l = 14 is the correct solution.

(b) $$\frac { b }{ 2 }$$ = 6
Multiplying both sides by 2, we have b
$$\frac { b }{ 2 }$$ x 2 =6 x 2 or 6 = 12 2
∴ b = 12 is the required solution.
Check : L.H.S. = $$\frac { b }{ 2 }$$ = $$\frac { 12 }{ 2 }$$ = 6 = R.H.S.
∴ p = 12 is the correct solution.

(c) $$\frac { p}{ 7 }$$ = 4
Multiplying both sides by 7, we have
$$\frac { p}{ 7 }$$ x 7 = 4 x 7 or p = 28
∴ p = 28 is the required solution.
Check : L.H.S . = $$\frac { p}{ 7 }$$ = $$\frac { 28 }{ 7 }$$
= 4 = R.H.S.
∴ p = 28 is the correct solution.

(d) 4x = 25
Dividing both sides by 4, we have
$$\frac { 4x }{ 4 }$$ = $$\frac { 25}{ 4 }$$ or x = $$\frac { 25}{ 4 }$$
∴ x = $$\frac { 25 }{ 4 }$$ is the required solution.
Check : L.H.S = 4x = 4 x $$\frac { 25 }{ 4 }$$
= 25 = R.H.S.
∴ x = $$\frac { 25 }{ 4 }$$ is the correct solution.

(e) 8y = 36
Dividing both sides by 8, we have
$$\frac { 8y }{ 8 }$$ = $$\frac { 36 }{ 8 }$$ or y = $$\frac { 9 }{ 2 }$$
∴ y = $$\frac { 9 }{ 2 }$$ is the required solution.
Check : L.H.S. = 8y = 8 x $$\frac { 9 }{ 2 }$$
= 36 = R.H.S.
∴ y = $$\frac { 9 }{ 2 }$$ is the correct solution.

(f) $$\frac { z }{ 3 }$$ = $$\frac { 5 }{ 4 }$$
Multiplying both sides by 3, we have
$$\frac { z }{ 3 }$$ x 3 = $$\frac { 5 }{ 4 }$$ x 3 or z = $$\frac { 15 }{ 4 }$$
∴ z = – y is the required solution.
Check : L.H.S. = $$\frac { z }{ 3 }$$ = $$\frac { 15 }{ 4 }$$ x $$\frac { 1 }{ 3 }$$
= $$\frac { 5 }{ 4 }$$ = R.H.S.
∴ z = $$\frac { 15 }{ 4 }$$ is the correct solution.

(g) $$\frac { a }{ 5 }$$ = $$\frac { 7 }{ 15 }$$
Multiplying both sides by 5, we have
$$\frac { a }{ 5 }$$ x 5 = $$\frac { 7 }{ 15 }$$ x 5 or a = $$\frac { 7 }{ 3 }$$
∴ a = $$\frac { 7 }{ 3 }$$ is the required solution.
Check : L.H.S. = $$\frac { a }{ 5 }$$ = $$\frac { 7 }{ 3 }$$ x $$\frac { 1 }{ 5 }$$
= $$\frac { 7 }{ 15 }$$ = R.H.S.
∴ a = $$\frac { 7 }{ 3 }$$ is the correct solution.

(h) 20t = – 10
Dividing both sides by 20, we have
$$\frac { 20t }{ 20 }$$ = $$\frac { -10 }{ 20 }$$ or t = – $$\frac { 1 }{ 2 }$$
∴ t = – $$\frac { 1 }{ 2 }$$
Check : L.H.S. = 20t = 20 x ( – $$\frac { 1 }{ 2 }$$ )
= – 10 = R.H.S.
∴ t = – $$\frac { 1 }{ 2 }$$ is the correct 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:
(a) 3n – 2 = 46
Step I : Adding 2 to both sides, we have
3n – 2 + 2 = 46 + 2
or
3n = 48

Step II :
Dividing both sides by 3, we have
$$\frac { 3n}{ 3 }$$ = $$\frac { 48 }{ 3 }$$
or n = 16
∴ n = 16 is the required solution.

(b) 5m + 7 = 17
Step I: Subtracting 7 from both sides, we have
5m + 7 – 7 = 17 – 7 or 5m = 10

Step II: Dividing both sides by 5, we have
$$\frac { 5m }{ 5 }$$ = $$\frac { 10 }{ 5 }$$
or m = 2
∴ m = 2 is the required solution

(c) $$\frac { 20p }{ 3 }$$ = 40
Step I : Multiplying both sides by 3, we have
$$\frac { 20p }{ 3 }$$ x 3 = 40 x 3
or 20 p =120

Step II : Dividing both sides by 20, we have
$$\frac { 20p }{ 20 }$$ = $$\frac { 120 }{ 20 }$$
or p = 6
Thus, p = 6 is the required solution.

(d) $$\frac { 3p }{ 10 }$$
Step I : Multiplying both sides by 10, we have
$$\frac { 3p }{ 10 }$$ x 10 = 6 x 10
or 3p = 60

Step II : Dividing both sides by 3, we have
$$\frac { 3p }{ 3 }$$ = $$\frac { 60 }{ 3 }$$
or p = 20
Thus, p = 20 is the required solution. 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) $$\frac { -p }{ 3 }$$ = 5
(g) 3s + 12 = 0
(h) 3s = 0
(i) 2q = 6
(j) 2q – 6 = 0
(k) 2q + 6 = 0
(l) 2q + 6 = 12
Solution:
(a) 1o p = 100
Dividing both sides by 10, We have:
$$\frac { 10p }{ 10 }$$ = $$\frac { 100 }{ 10 }$$
or p = 10
Thus, p = 10 is the required solution.

(b) 10p + 10 = 100
Subtracting 10 from both sides, we have
10p + 10 – 10 = 100 – 10
or 10p = 90
Dividing both sides by 10, we have
$$\frac { 10p }{ 10 }$$ = $$\frac { 90 }{ 10 }$$
or p = 9
∴ p = 9 is the required solution.

(c) $$\frac { p }{ 4 }$$ = 5
Multiplying both sides by 4, we have
$$\frac { p }{ 4 }$$
or p = 20
Thus, p = 20 is the required solution.

(d) $$\frac { -p }{ 3 }$$ = 5
Multiplying both sides by 3, we have
$$\frac { -p }{ 3 }$$ x 3 =5 x 3
or – p = 15
or – p x (- 1) = 15 x (- 1)
[Multiplying both sides by (-1)]
Thus, p = – 15 is the required solution.

(e) $$\frac { 3p }{ 4 }$$ = 6
Multiplying both sides by 4, we have
$$\frac { 3p }{ 4 }$$ x 4 = 6 x 4
3p = 24
both sides by 3, we have
$$\frac { 3p }{ 3 }$$ = $$\frac { 24 }{ 3 }$$
p = 8
Thus p = 8 is the required solution.

(f) 3s = – 9
Dividing both sides by 3, we have
$$\frac { 3s }{ 3 }$$ = $$\frac { – 9 }{ 3 }$$
or s = – 3
∴ s = – 3 is the required solution.

(g) 3s + 12 = 0
Subtracting 12 from both sides, we have
3s + 12 – 12 = 0 – 12
3s = – 12
Dividing both sides by 3, we have
$$\frac { 3s }{ 3 }$$ = $$\frac { -12 }{ 3 }$$
s = – 4
Thus, s = – 4 is the required solution.

(h) 3s = 0
Dividing both sides by 3, we have
$$\frac { 3s }{ 3 }$$ = $$\frac { 0 }{ 3 }$$
∴ s = 0 is the required solution.

(i) 2q = 6
Dividing both sides by 2, we have
$$\frac { 2q }{ 2 }$$ = $$\frac { 6 }{ 2 }$$
or q = 3
∴ q = 3 is the required solution.

(j) 2q – 6 = 0
Adding 6 to both sides, we have
2q – 6 + 6 = 0 + 6
or 2q =6
Dividing both sides by 2, we have
$$\frac { 2q }{ 2 }$$ = $$\frac { 6 }{ 2 }$$ or q = 3
∴ q = 3 is the required solution.

(k) 2q + 6 = 0
Subtracting 6 from both sides, we have
2q + 6 – 6 = 0 – 6
or 2q = – 6
Dividing both sides by 2, we have:
$$\frac { 2q }{ 2 }$$ = $$\frac { -6 }{ 2 }$$ or q = – 3
Thus, q = – 3 is the required solution.

(l) 2q + 6 = 12
Subtracting 6 from both sides, we have
2q + 6 – 6 = 12 – 6 or 2q = 6
Dividing both sides by 2, we have
$$\frac { 2q }{ 2 }$$ = $$\frac { 6 }{ 2 }$$
or q = 3
Thus, q = 3 is the required solution. 