Gujarat BoardĀ 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.

ā“ 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.

ā“ 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.

ā“ 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.

ā“ 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.

ā“ 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.

ā“ 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

ā“ 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

ā“ b = 12 is the required solution.

Check : L.H.S. = \(\frac { b }{ 2 }\) = \(\frac { 12 }{ 2 }\) = 6 = R.H.S.

ā“ 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

ā“ p = 28 is the required solution.

Check : L.H.S . = \(\frac { p}{ 7 }\) = \(\frac { 28 }{ 7 }\)

= 4 = R.H.S.

ā“ p = 28 is the correct solution.

(d) 4x = 25

Dividing both sides by 4, we have

\(\frac { 4x }{ 4 }\) = \(\frac { 25}{ 4 }\) or x = \(\frac { 25}{ 4 }\)

ā“ x = \(\frac { 25 }{ 4 }\) is the required solution.

Check : L.H.S = 4x = 4 x \(\frac { 25 }{ 4 }\)

= 25 = R.H.S.

ā“ 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 = \(\frac { 9 }{ 2 }\)

ā“ y = \(\frac { 9 }{ 2 }\) is the required solution.

Check : L.H.S. = 8y = 8 x \(\frac { 9 }{ 2 }\)

= 36 = R.H.S.

ā“ 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 = \(\frac { 15 }{ 4 }\)

ā“ 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.

ā“ 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 = \(\frac { 7 }{ 3 }\)

ā“ 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.

ā“ 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 = – \(\frac { 1 }{ 2 }\)

ā“ t = – \(\frac { 1 }{ 2 }\)

Check : L.H.S. = 20t = 20 x ( – \(\frac { 1 }{ 2 }\) )

= – 10 = R.H.S.

ā“ 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

ā“ 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

ā“ 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

ā“ 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

ā“ 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

\(\frac { 3s }{ 3 }\) = \(\frac { 0 }{ 3 }\)

ā“ 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

ā“ 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

ā“ 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.