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