Gujarat Board GSEB Textbook Solutions Class 7 Maths Chapter 4 Simple Equations Ex 4.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 7 Maths Chapter 4 Simple Equations Ex 4.3

Question 1.

Solve the following equations.

(a) 2y + \(\frac { 5 }{ 2 }\) = \(\frac { 37 }{ 2 }\)

(b) 5t + 28 = 10

(c) \(\frac { a }{ 5 }\) + 3 = 2

(d) \(\frac { q }{ 4 }\) + 7 = 5

(e) \(\frac { 5 }{ 2 }\)x = – 10

(f) \(\frac { 5 }{ 2 }\)x = \(\frac { 25 }{ 4 }\)

(g) 7m + \(\frac { 19 }{ 2 }\) = 13

(h) 6z + 10 = – 2

(i) \(\frac { 3l }{ 2 }\) = \(\frac { 2 }{ 3 }\)

Solution:

(a) We have

2y + \(\frac { 5 }{ 2 }\) = \(\frac { 37 }{ 2 }\)

2y = \(\frac { 37 }{ 2 }\) – \(\frac { 5 }{ 2 }\) = \(\frac { 37-5 }{ 2 }\)

Transposing \(\frac { 5 }{ 2 }\) to R.H.S.

= \(\frac { 32 }{ 2 }\) = 16

or y = \(\frac { 16 }{ 2 }\) = 8

Dividing both sides by 2,

Thus, y = 8 is the required solution.

(b) We have 5t + 28 = 10

or 5t = 10 – 28

Transposing 28 to R.H.S.

or 5t = – 18 or t = – \(\frac { 18 }{ 5 }\)

Dividing both sides by 5,

Thus, t = – \(\frac { 18 }{ 5 }\) is the required solution.

(c) We have \(\frac { a }{ 5 }\) + 3 = 2 a

or \(\frac { a }{ 5 }\) = 2 – 3

Transposing 3 to R.H.S.

or \(\frac { a }{ 5 }\) = (- 1) x 5

Multiplying both sides by 5, we have:

or \(\frac { a }{ 5 }\) x 5 = (- 1) x 5

or a = – 5

Thus, a = – 5 is the required solution.

(d) \(\frac { q }{ 4 }\) + 7 = 5

or \(\frac { q }{ 4 }\) = 5 – 7

Transposing 7 to R.H.S.

or \(\frac { q }{ 4 }\) = – 2

Multiplying both sides by 4, we have:

or \(\frac { q }{ 4 }\) x 4 = – 2 x 4

or q = – 8

Thus, q = – 8 is the required solution.

(e) \(\frac { 5 }{ 2 }\)x = – 10

Multiplying both sides by 2, we have

or \(\frac { 5x }{ 2 }\) x 2 = – 10 x 2 = – 20

or 5x = – 20

or \(\frac { 5x }{ 5 }\) = \(\frac { – 20 }{ 5 }\)

Dividing both sides by 5,

i.e. x = – 4

∴ x = – 4 is the required solution.

(f) We have \(\frac { 5 }{ 2 }\)x = \(\frac { 25 }{ 4 }\)

Multiplying both sides by 2, we have

\(\frac { 5 }{ 2 }\)x × 2 = \(\frac { 25 }{ 4 }\) x 2

\(\frac { 5 }{ 2 }\)x × \(\frac { 25 }{ 2 }\)

Dividing both sides by 5, we have

\(\frac { 5x }{ 5 }\) = \(\frac { 25 }{ 2 }\) x \(\frac { 1 }{ 5 }\)

or x = \(\frac { 5 }{ 2 }\)

Thus, x= \(\frac { 5 }{ 2 }\) is the required solution.

(g) We have 7m + \(\frac { 19 }{ 2 }\) = 13

or 7m = 13 – \(\frac { 19 }{ 2 }\)

Transposing \(\frac { 19 }{ 2 }\) from L.H.S. to R.H.S.

or 7m = \(\frac { 26-19 }{ 2 }\)

or 7m = \(\frac { 7 }{ 2 }\)

Dividing both sides by 7, we have

\(\frac { 7m}{ 7 }\) = \(\frac { 7 }{ 2 }\) x \(\frac { 1 }{ 7 }\)

or m = \(\frac { 1 }{ 2 }\)

Thus, m = \(\frac { 1 }{ 2 }\) is the required solution.

(h) We have 6z + 10 = – 2

or 6z = – 2 – 10

Transposing 10 to R.H.S.

or 6z = -12

Dividing both sides by 6, we have

\(\frac { 6z }{ 6 }\) = \(\frac { -12 }{ 6 }\)

or z = – 2

Thus, z = – 2 is the required solution.

(i) We have \(\frac { 3l }{ 2 }\) = \(\frac { 2 }{ 3 }\)

Multiplying both sides by 2, we have

\(\frac { 3l }{ 2 }\) x 2 = \(\frac { 2 }{ 3 }\) x 2

Dividing both sides by 3, we have

\(\frac { 3l }{ 3 }\) = \(\frac { 4 }{ 3 }\) x \(\frac { 1 }{ 3 }\)

or l = \(\frac { 4 }{ 9 }\)

Thus, l = \(\frac { 4 }{ 9 }\) is the required solution.

(j) We have \(\frac { 2b }{ 3 }\) – 5 = 3

or \(\frac { 2b }{ 3 }\)

Transposing – 5 to R.H.S.

Multiplying both sides by \(\frac { 3 }{ 2 }\), we have

∴ \(\frac { 2b }{ 3 }\) x \(\frac { 3 }{ 2 }\) = \(\frac { 8 }{ 2 }\) x 3 = 12

or b = 12

Thus, b = 12 is the required solution.

Question 2.

Solve the following equations.

(a) 2(x + 4) = 12

(b) 3(n – 5) = 21

(c) 3(n – 5) = – 21

(d) – 4(2 + x) = 8

(e) 4(2 – x) = 8

Solution:

(a) We have 2(x + 4) = 12

or x + 4 = \(\frac { 12 }{ 2 }\) = 6

[Dividing both sides by 2]

or x = 6 – 4 = 2

[Transposing 4 to R.H.S.]

∴ x = 2 is the required solution.

(b) We have 3(n – 5) = 21

or n – 5 = \(\frac { 21 }{ 3 }\) = 7

[Dividing both sides by 3]

or n = 7 + 5 = 12

[Transposing – 5 to R.H.S.]

Thus, n = 12 is the required solution.

(c) We have 3(n – 5) = – 21

or n – 5 = \(\frac { – 21 }{ 3 }\) = – 7

[Dividing both sides by 3]

or n = – 7 + 5 = – 2

[Transposing – 5 to R.H.S.]

Thus, n = – 2 is the required solution.

(d) We have – 4(2 + x) = 8

or \(\frac { -4(2+x) }{ -4 }\) = \(\frac { 8 }{ -4 }\)

[Dividing both sides by – 4]

or 2 + x = – 2

x = – 2 – 2

[Transposing 2 to R.H.S.]

x = – 4

Thus, x = – 4 is the required solution.

(e) We have 4(2 – x) = 8

or 2 – x = \(\frac { 8 }{ 4 }\) =2

[Dividing both sides by 4]

or – x = 2 – 2

[Transposing 2 to R.H.S.]

or – x = 0

or x – 0

Thus, x = 0 is the required solution.

Question 3.

Solve the following equations:

(a) 4 = 5(p – 2)

(b) – 4 = 5(p – 2)

(c) 16 = 4 + 3(t + 2)

(d) 4 + 5(p – 1) = 34

(e) 0 = 16 + 4(m – 6)

Solution:

(a) 4 = 5(p – 2)

Interchanging the sides, we have

5(p – 2) = 4

Dividing both sides by 5, we have

\(\frac { 5(p-2) }{ 5 }\) = \(\frac { 4 }{ 5 }\)

or p – 2 = \(\frac { 4 }{ 5 }\) or p = \(\frac { 4 }{ 5 }\) + 2

[Transposing (- 2) from L.H.S. to R.H.S.]

or p = \(\frac { 4+10 }{ 5 }\) = \(\frac { 14 }{ 5 }\)

∴ p = \(\frac { 14 }{ 5 }\) is the required solution.

(b) – 4 = 5(p – 2)

Interchanging the sides, we have

5(p – 2) = – 4

Dividing both sides by 5, we have

\(\frac { 5(p-2) }{ 5 }\) = – \(\frac { 4 }{ 5 }\)

or p – 2 = – \(\frac { 4 }{ 5 }\) or p = – \(\frac { 4 }{ 5 }\) + 2

[Transposing (-2) from L.H.S. to R.H.S.]

= \(\frac { -4+10 }{ 5 }\) = \(\frac { 6 }{ 5 }\)

Thus, p = \(\frac { 6 }{ 5 }\) is the required solution.

(c) 16 = 4 + 3(t + 2)

Interchanging the sides, we have

4 + 3(t + 2) = 16

or 3(t + 2)= 16 – 4 = 12

[Transposing 4 to R.H.S.]

or \(\frac { 3(t+2) }{ 3 }\) = \(\frac { 12 }{ 3 }\)

[Dividing both sides by 3]

or t + 2 =4

or t = 4 – 2

[Transposing 2 to R.H.S.]

or t = 2

Thus, t = 2 is the required solution.

(d) We have 4 + 5(p – 1) = 34

or 5(p – 1) = 34 – 4 = 30

[Transposing 4 to R.H.S.]

or \(\frac { 5(p-1) }{ 5 }\) = \(\frac { 30 }{ 5 }\)

[Dividing both sides by 5]

or p – 1 = 6

or p = 6 + 1 = 7

[Transposing 1 to R.H.S.]

Thus, p = 7 is the required solution.

(e) We have 0 = 16 + 4(m – 6)

Interchanging the sides, we have

16 + 4(m – 6) = 0

or 4(m – 6) = – 16

[Transposing 16 from L.H.S. to R.H.S.]

Dividing both sides by 4, we have

\(\frac { 4(m-6) }{ 4 }\) = \(\frac { – 16 }{ 4 }\)

or m – 6 = – 4

or m = – 4 + 6

[Transposing – 6 from L.H.S. to R.H.S.] or m = 2

Thus, m = 2 is the required solution.

Question 4.

(a) Construct 3 equations starting with x = 2.

(b) Construct 3 equations starting with x = – 2.

Solution:

(a) Starting with x = 2

I. x = 2

Multiplying both sides by 5, we have

5 × x = 5 × 2

or 5x = 10

Subtracting 3 from both sides, we have

5x – 3 = 10 – 3

or 5x – 3 = 7

II. x = 2

Multiplying both sides by 7, we have

7 × x = 7 × 2

or 7x = 14

Adding 5 to both sides, we have

7x + 5 = 14 + 5

or 7x + 5 = 19

III. x = 2

Dividing both sides by 3, we have

\(\frac { x }{ 3 }\) = \(\frac { 2 }{ 3 }\)

Subtracting 4 from both sides, we have

\(\frac { x }{ 3 }\) – 4 = \(\frac { 2 }{ 3 }\) – 4

or \(\frac { x }{ 3 }\) – 4 = \(\frac { 2-12 }{ 3 }\) = \(\frac { -10 }{ 3 }\)

or \(\frac { x }{ 3 }\) – 4 = \(\frac { -10 }{ 3 }\)

(b) Starting with x = – 2

I.x = – 2

Adding 8 to both sides, we have

x + 8 = – 2 + 8 or x + 8 = 6

II. x = – 2

Subtracting 10 from both sides, we have x – 10 = – 2 – 10 or x – 10 = – 12

III. x = – 2

Multiplying both sides by 8, we have

8 × x = (- 2) x 8

or 8x = – 16

Subtracting 2 from both sides, we have

8x – 2 = – 16 – 2

or 8x – 2 = – 18