NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 are the part of NCERT Solutions for Class 7 Maths. Here you can find the NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3.

Ex 4.3 Class 7 Maths Question 1.

Solve the following equations:

Solution:

Ex 4.3 Class 7 Maths Question 2.

Solve the following equations:
(а) 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) 2(x + 4) = 12
⇒ 2(x + 4)/2 = 12/2 (Dividing both sides by 2)
⇒ x + 4 = 6
⇒ x = 6 – 4 (Transposing 4 to RHS)
⇒ x = 2
Check: Putting x = 2 in LHS, we get
2(2 + 4) = 2 × 6 = 12 RHS as required.

(b) 3(n – 5) = 21
⇒ 3(n − 5)/3 = 21/3 (Dividing both sides by 3)
⇒ n – 5 = 7
⇒ n = 7 + 5 (Transposing 5 to RHS)
n = 12
Check: Putting n = 12 in LHS, we get
3(12 – 5) = 3 × 7 = 21 RHS as required.

(c) 3(n – 5) = -21
⇒ 3(n − 5)/3= −21/3 (Dividing both sides by 3)
⇒ n – 5 = -7
⇒ n = -7 + 5 (Transposing 5 to RHS)
⇒ n = -2
Check: Putting n = -2 in LHS, we get
3(-2 – 5) = 3 × -7 = -21 RHS as required.

(d) -4(2 + x) = 8
⇒ −4(2 + x)/−4 = 8/−4 (Dividing both sides by -4)
⇒ 2 + x = -2
⇒ x = -2 – 2 (Transposing 2 to RHS)
⇒ x = -4
Check: Putting x = -4 in LHS, we get
-4(2 – 4) = -4 × -2 = 8 RHS as required.

(e) 4(2 – x) = 8
⇒ 4(2 – x)/4 = 8/4 (Dividing both sides by 4)
⇒ 2 – x = 2

⇒ –x = 2 – 2 (Transposing 2 to RHS)
⇒ -x = 0
∴ x = 0 (Multiplying both sides by -1)
Check: Putting x = 0 in LHS, we get
4(2 – 0) = 4 × 2 = 8 RHS as required.

Ex 4.3 Class 7 Maths 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:

(c) 16 = 4 + 3(t + 2)
⇒ 16 – 4 = 3(t + 2)      (Transposing 4 to LHS)
⇒ 12 = 3(t + 2)
⇒ 12/3 = 3(t + 2)/3     (Dividing both sides by 3)
⇒ 4 = t + 2
⇒ 4 – 2 = t                   (Transposing 2 to LHS)
⇒ 2 = t or t = 2
Check: Putting t = 2 in RHS, we get
4 + 3(2 + 2) = 4 + 3 × 4 = 4 + 12 = 16 LHS as required.

(d) 4 + 5(p – 1) = 34
⇒ 5(p – 1) = 34 – 4 (Transposing 4 to RHS)
⇒ 5(p – 1) = 30
⇒ 5(p − 1)/5 = 30/5 (Dividing both sides by 5)
⇒ p – 1 = 6
⇒ P = 7
Check: Putting p = 7 in LHS, we get
4 + 5(7 – 1) = 4 + 5 × 6 = 4 + 30 = 34 RHS as required.

(e) 0 = 16 + 4(m – 6)
⇒ 0 – 16 = 4(m – 6)        (Transposing 16 to LHS)
⇒ -16 = 4(m – 6)
⇒ −16/4 = 4(m − 6)/4      (Dividing both sides by 4)
⇒ -4 = m – 6
⇒ -4 + 6 = m                     (Transposing 6 to LHS)
⇒ 2 = m
or m = 2
Check: Putting m = 2 in RHS, we get
16 + 4(2 – 6) = 16 + 4 × (-4) = 16 – 16 = 0 LHS as required.

Ex 4.3 Class 7 Maths Question 4.

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

Solution:
(a) Three possible equations whose solution is x = 2 are:
10x + 5 = 25; 2x = 4; 3x – 1 = 5
(b) Three possible equations whose solution is x = -2 are:
4x = -8; 5x + 11 = 1; 2x + 8 = 4

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