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