NCERT Exercise 4.3 solution-2
Simple Equations 7th Math
NCERT Exercise 4.3 solution-2
Question (2) Solve the following equations.
Question(2)(a) 2 (x + 4) = 12
Solution
Given, 2 (x + 4) = 12
After dividing both sides by 2, we get
⇒ 2(x + 4)/2 = 12/2
⇒ x + 4 = 6
After subtracting 4 from both sides
⇒ x + 4 –4 = 6 – 4
⇒ x = 2 Answer
Alternate Method to solve an equation
Given, 2 (x + 4) = 12
After transposing 2 to RHS
[When a digit or variable in multiple goes to division after transposing to other side]
Here, 2 will become in division, i.e. goes to denominator after transposing to RHS
x + 4 = 12/2
⇒ x + 4 = 6
After transposing 4 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, + 4 will become –4 when transposed to RHS
⇒ x = 6 – 4
⇒ x = 2 Answer
Question (2)(b) 3 (n – 5) = 21
Solution
Given, 3 (n – 5) = 21
After dividing both sides by 3, we get
3(n – 5)/3 = 21/3
⇒ n –5 = 7
After adding 5 to both sides, we get
⇒ n – 5 + 5 = 7 + 5
Alternate Method to solve an equation
Given, 3 (n – 5) = 21
After transposing 3 to RHS
[When a digit or variable in multiple goes to division after transposing to other side]
Here, 3 will become in division, i.e. goes to denominator after transposing to RHS
⇒ n – 5 = 21/3
⇒ n – 5 = 7
After transposing –5 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, – 5 will become +5 when transposed to RHS
⇒ n = 7 + 5
⇒ n = 12 Answer
Question (2)(c) 3 (n – 5) = – 21
Solution
Given, 3 (n – 5) = – 21
After dividing both sides by 3, we get
⇒ 3(n – 5)/3 = – 21/3
⇒ n – 5 = –7
After adding 5 to both sides, we get
⇒ n – 5 + 5 = – 7 + 5
⇒ n = –2 Answer
Alternate Method to solve an equation
Given, 3 (n – 5) = – 21
After transposing 3 to RHS
[When a digit or variable in multiple goes to division after transposing to other side]
Here, 3 will become in division, i.e. goes to denominator after transposing to RHS
⇒ n – 5 = – 21/3
⇒ n – 5 = –7
After transposing –5 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, –5 will become +5 when transposed to RHS
⇒ n = – 7 + 5
⇒ n = – 2 Answer
Question (2)(d) 3 – 2(2 – y) = 7
Solution
Given, 3 – 2(2 – y) = 7
After subtracting 3 to both sides, we get
⇒ 3 – 2(2 – y) – 3 = 7 –3
⇒ – 2(2 – y) = 4
After dividing both sides by 2, we get
⇒ – 2(2 – y)/2 = 4/2
⇒ – (2 – y) = 2
⇒ – 2 + y = 2
After adding 2 to both sides, we get
⇒ – 2 + y + 2 = 2 + 2
⇒ y = 4 Answer
Alternate Method to solve an equation
Given, 3 – 2(2 – y) = 7
After transposing 3 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, +3 will become – 3 when transposed to RHS
⇒ – 2(2 – y) = 7 – 3
⇒ – 2(2 – y) = 4
After transposing 2 to RHS
[When a digit or variable in multiple goes to division after transposing to other side]
Here, 2 will become in division, i.e. goes to denominator after transposing to RHS
⇒ – (2 – y) = 4/2
⇒ –2 + y = 2
After transposing –2 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, – 2 will become +2 when transposed to RHS
⇒ y = 2 + 2
⇒ y = 4 Answer
Question (2)(e) –4(2 – x) = 9
Solution
Given, –4 (2 – x) = 9
After dividing both sides by 4, we get
⇒ – 4(2 – x)/4 = 9/4
⇒ – (2 – x) = 9/4
⇒ – 2 + x = 9/4
After adding 2 to both sides, we get
⇒ – 2 + x + 2 = 9/4 + 2
⇒ x = 9 + 8/4
⇒ x = 17/4 Answer
Alternate Method to solve an equation
Given, – 4 (2 – x) = 9
After transposing 4 to RHS
[When a digit or variable in multiple goes to division after transposing to other side]
Here, 4 will become in division, i.e. goes to denominator after transposing to RHS
⇒ – (2 – x) = 9/4
⇒ – 2 + x = 9/4
After transposing – 2 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, – 2 will become +2 when transposed to RHS
⇒ x = 9/4 + 2
⇒ x = 9 + 8/4
⇒ x = 17/4 Answer
Question (2)(f) 4 (2 – x) = 9
Solution
Given, 4 (2 – x) = 9
After dividing both sides by 4, we get
⇒ 4(2 – x)/4 = 9/4
⇒ 2 – x = 9/4
After subtracting 2 to both sides, we get
⇒ – 2 – x – 2 = 9/4 – 2
⇒ – x =9 – 8/4
⇒ – x = 1/4
⇒ x = – 1/4 Answer
Alternate Method to solve an equation
Given, 4 (2 – x) = 9
After transposing 4 to RHS
[When a digit or variable in multiple goes to division after transposing to other side]
Here, 4 will become in division, i.e. goes to denominator after transposing to RHS
⇒ 2 – x = 9/4
After transposing 2 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, 2 will become – 2 when transposed to RHS
⇒ – x = 9/4 – 2
⇒ x = 9 – 8/4
⇒ x = 1/4
⇒ x = – 1/4 Answer
Question (2)(g) 4 + 5 (p – 1) = 34
Solution
Given, 4 + 5 (p – 1) = 34
After subtracting 4 from both sides, we get
⇒ 4 + 5(p – 1) –4 = 34 – 4
⇒ 5 (p – 1) = 30
After dividing both sides by 5, we get
⇒ 5(p – 1)/5 = 30/5
⇒ p – 1 = 6
After adding 1 to both sides, we get
⇒ p – 1 + 1 = 6 + 1
⇒ p = 7 Answer
Alternate Method to solve an equation
Given, 4 + 5 (p – 1) = 34
After transposing 4 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, +4 will become –4 when transposed to RHS
⇒ 5 (p – 1) = 34 – 4
⇒ 5 (p – 1) = 30
After transposing 5 to RHS
[When a digit or variable in multiple goes to division after transposing to other side]
Here, 5 will become in division, i.e. goes to denominator after transposing to RHS
⇒ p – 1 = 30/5
⇒ p – 1 = 6
After transposing –1 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, – 1 will become +1 when transposed to RHS
⇒ p = 6 + 1
⇒ p = 7 Answer
Question (2)(h) 34 – 5(p – 1) = 4
Solution
Given, 34 – 5(p – 1) = 4
After subtracting 34 to both sides, we get
⇒ 34 – 5(p – 1) – 34 = 4 – 34
⇒ – 5(p – 1) = – 30
After dividing both sides by 5, we get
⇒ – 5(p – 1)/5 = – 30/5
⇒ –(p – 1) = –6
⇒ p –1 = 6
After adding 1 to both sides, we get
⇒ p – 1 + 1 = 6 + 1
⇒ p = 7 Answer
Alternate Method to solve an equation
Given, 34 – 5(p – 1) = 4
After transposing 34 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, +34 will become –34 when transposed to RHS
⇒ –5(p – 1) = 4 – 34
⇒ –5 (p – 1) = –30
⇒ 5(p – 1) = 30
After transposing 5 to RHS
[When a digit or variable in multiple goes to division after transposing to other side]
Here, 5 will become in division, i.e. goes to denominator after transposing to RHS
⇒ p – 1 = 30/5
⇒ p – 1 = 6
After transposing –1 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, – 1 will become +1 when transposed to RHS
⇒ p = 6 + 1
⇒ p = 7 Answer
Question(3) Solve the following equations
Question(3)(a) 4 = 5(p – 2)
Solution
Given, 4 = 5(p – 2)
⇒ 5(p – 2) = 4
After dividing both sides by 5, we get
⇒ 5(p – 2)/5 = 4/5
⇒ p – 2 = 4/5
After adding 2 to both sides, we get
⇒ p – 2 + 2 = 4/5 + 2
p = 4 + 10/2
⇒ p = 14/2
⇒ p = 7 Answer
Alternate Method to solve an equation
Given, 4 = 5(p – 2)
⇒ 5(p – 2) = 4
After transposing 5 to RHS
[When a digit or variable in multiple goes to division after transposing to other side]
Here, 5 will become in division, i.e. goes to denominator after transposing to RHS
⇒ p – 2 = 4/5
After transposing – 2 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, – 2 will become +2 when transposed to RHS
⇒ p = 4/5 + 2
⇒ p = 4 + 10/2
⇒ p = 14/2
⇒ p = 7 Answer
Question(3)(b) –4 = 5(p – 2)
Solution
Given, – 4 = 5(p – 2)
⇒ 5(p – 2) = – 4
After dividing both sides by 5, we get
⇒ 5(p – 2)/5 = – 4/5
⇒ p – 2 = – 4/5
After adding 2 to both sides, we get
⇒ p – 2 + 2 = – 4/5 + 2
⇒ p = – 4 + 10/2
⇒ p = 6/2
⇒ p = 3 Answer
Alternate Method to solve an equation
Given, – 4 = 5(p – 2)
⇒ 5(p – 2) = – 4
After transposing 5 to RHS
[When a digit or variable in multiple goes to division after transposing to other side]
Here, 5 will become in division, i.e. goes to denominator after transposing to RHS
⇒ p – 2 = – 4/5
After transposing –2 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, – 2 will become + 2 when transposed to RHS
⇒ p = – 4/5 + 2
⇒ p = – 4 + 10/2
⇒ p = 6/2
⇒ p = 3 Answer
Question(3)(c) – 16 = – 5(2 – p)
Solution
Given, – 16 = – 5(2 – p)
⇒ 16 = 5(2 – p)
⇒ 5 (2 – p) = 16
After dividing both sides by 5, we get
⇒ 5(2 – p)/5 = 16/5
⇒ 2 – p = 16/5
After subtracting 2 from both sides, we get
⇒ 2 – p – 2 = 16/5 – 2
⇒ – p = 16 –10/5
⇒ p = 6/5
⇒ p = – 6/5 Answer
Alternate Method to solve an equation
Question(3)(d) 10 = 4 + 3(t + 2)
Solution
Given, 10 = 4 + 3(t + 2)
⇒ 4 + 3(t + 2) = 10
After subtracting 4 from both sides, we get
⇒ 4 + 3(t + 2) – 4 = 10 – 4
⇒ 3(t + 2) = 6
After dividing both sides by 3, we get
⇒ 3(t + 2)/3 = 6/3
⇒ t + 2 = 2
After subtracting 2 from both sides
⇒ t + 2 – 2 = 2 – 2
⇒ t = 0 Answer
Alternate Method to solve an equation
Given, 10 = 4 + 3(t + 2)
⇒ 4 + 3(t + 2) = 10
After transposing 4 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, +4 will become – 4 when transposed to RHS
⇒ 3(t + 2) = 10 – 4
⇒ 3(t + 2) = 6
After transposing 3 to RHS
[When a digit or variable in multiple goes to division after transposing to other side]
Here, 3 will become in division, i.e. goes to denominator after transposing to RHS
⇒ t + 2 = 6/3
⇒ t + 2 = 2
After transposing 2 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, +2 will become – 2 when transposed to RHS
⇒ t = 2 – 2
⇒ t = 0 Answer
Question(3)(e) 28 = 4 + 3(t + 5)
Solution
Given, 28 = 4 + 3(t + 5)
⇒ 4 + 3(t + 5) = 28
After subtracting 4 from both sides, we get
⇒ 4 + 3(t + 5) = 28 – 4
⇒ 3(t + 5) = 24
After dividing both sides by 3, we get
⇒ 3(t +5 )/3 = 24/3
⇒ t + 5 = 8
After subtracting 5 from both sides, we get
⇒ t + 5 – 5 = 8 – 5
⇒ t = 3 Answer
Alternate Method to solve an equation
Given, 28 = 4 + 3(t + 5)
⇒ 4 + 3(t + 5) = 28
⇒ After transposing 4 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, +4 will become – 4 when transposed to RHS
⇒ 3( t + 5) = 28 – 4
⇒ 3(t + 5) = 24
After transposing 3 to RHS
[When a digit or variable in multiple goes to division after transposing to other side]
Here, 3 will become in division, i.e. goes to denominator after transposing to RHS
⇒ t+ 5 = 24/3
⇒ t + 5 = 8
After transposing 5 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, +5 will become –5 when transposed to RHS
⇒ t = 8 – 5
⇒ t = 3 Answer
(f) 0 = 16 + 4(m – 6)
Solution
Given, 0 = 16 + 4(m – 6)
⇒ 16 + 4(m – 6) = 0
After subtracting 16 from both sides, we get
⇒ 16 + 4( m – 6) – 16 = 0 – 16
⇒ 4(m – 6) = –16
After dividing both sides by 4, we get
⇒ 4(m – 6)/4 = – 16/4
⇒ m – 6 = – 4
After adding 6 to both sides, we get
⇒ m – 6 + 6 = –4 + 6
⇒ m = 2 Answer
Alternate Method to solve an equation
Given, 0 = 16 + 4(m – 6)
⇒ 16 + 4(m – 6) = 0
After transposing 16 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, +16 will become – 16 when transposed to RHS
⇒ 16 + 4(m – 6) = 0 – 16
⇒ 4(m – 6) = –16
After transposing 4 to RHS
[When a digit or variable in multiple goes to division after transposing to other side]
Here, 4 will become in division, i.e. goes to denominator after transposing to RHS
⇒m – 6 = – 16/4
⇒ m – 6 = – 4
After transposing –6 to RHS
[When a digit or variable is transposed to other side, its sign changes]
Thus, – 6 will become +6 when transposed to RHS
⇒ m = – 4 + 6
⇒ m = 2 Answer
Question (4)(a) Construct 3 equations starting with x = 2
Solution
Given, x = 2
After multiplying 2 to both sides, we get
⇒ 2 × x = 2 × 2
⇒ 2 x = 4
After adding 10 to both sides, we get
⇒ 2x + 10 = 4 + 10
⇒ 2x + 10 = 14 equation (1)
After dividing both sides with 7, we get
⇒2x/7 + 10/7 = 14/7
⇒ 2/7 x + 10/7 = 2 Equations (2)
Given, x = 2
After multiplying both sides by 5, we get
x × 5 = 2 × 5
⇒ 5 x = 10
After adding 2 to both sides, we get
⇒ 5 x + 2 = 10 + 2
⇒ 5 x + 2 = 12 Equation (3)
Thus, three equations constructed from x = 2 are
(i) 2 x + 10 = 14
(ii) 2/7 x + 10/7 = 2
(iii) 5 x + 2 = 12
Question (4)(b) Construct 3 equations starting with x = –2
Solution
Given, x = –2
After multiplying both sides with 4, we get
⇒ × 4 = × 2 × 4
⇒ 4 x = 8
After adding 15 to both sides, we get
⇒ 4x + 15 = 8 + 15
⇒ 4x + 15 = 23 Equation (1)
Given, x = – 2
After dividing both sides by 3, we get
⇒ x/3 = – 2/3
After adding 2 to both sides, we get
⇒ x/3 + 2 = – 2/3 + 2
⇒ x/3 + 2 = – 2 + 6/3
⇒ x/3 + 2 = 4/3 Equation (2)
Given, x = – 2
After adding 1 to both sides, we get
⇒ x + 1 = – 2 + 1
⇒ x + 1 = – 1 Equation (3)
Thus three equation constructed starting with x = – 2 are
(i) 4 x + 15 = 23
(ii) x/3 + 2 = 4/3
(iii) x + 1 = – 1
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