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

**NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 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

⇒ \(\frac{2(x+4)}{2}=\frac{12}{2}\) (Dividing both sides by 2)

⇒ x + 4 = 6

⇒ x = 6 – 4 (Transposing 4 to RHS)

⇒ x – 2

Check: Put x = 2 in LHS

2(2 + 4) = 2 × 6 = 12 RHS as required

(b) 3(n – 5) = 21

⇒ \(\frac{3(n-5)}{3}=\frac{21}{3}\) (Dividing both sides by 3)

⇒ n – 5 = 7

⇒ n = 7 + 5 (Transposing 5 to RHS)

n = 12

Check: Put n = 12 in LHS

3(12 – 5) = 3 × 7 = 21 RHS as required

(c) 3(n – 5) = -21

⇒ \(\frac{3(n-5)}{3}=\frac{-21}{3}\) (Dividing both sides by 3)

⇒ n – 5 = -7

⇒ n = -7 + 5 (Transposing 5 to RHS)

⇒ n = -2

Check: Put n = -2 in LHS

3(-2 – 5) = 3 × -7

= -21 RHS as required

(d) -4(2 + x) = 8

⇒ \(\frac{-4(2+x)}{-4}=\frac{8}{-4}\) (Dividing both sides by -4)

⇒ 2 + x = -2

⇒ x = -2 – 2 (Transposing 2 to RHS)

⇒ x = —4

Check: Put x = -4 in LHS

-4(2 – 4) = -4 × -2 = 8 RHS as required

(e) 4(2-x) = 8

⇒ \(\frac{4(2-x)}{4}=\frac{8}{4}\) (Dividing both sides by 4)

⇒ 2 – x = 2 – 2 (Transposing 2 to RHS)

⇒ -x = 0

∴ x = 0 (Multiplying both sides by -1)

Check: Put x = 0 in LHS

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)

⇒ \(\frac{12}{3}=\frac{3(t+2)}{3}\) (Dividing both sides by 3)

⇒ 4 = t + 2

⇒ 4 – 2 = t (Transposing 2 to LHS)

⇒ 2 = t or t = 2

Check: Put t = 2 in RHS

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

⇒ \(\frac{5(p-1)}{5}=\frac{30}{5}\) (Dividing both sides by 5)

⇒ p – 1 = 6

⇒ P = 7

Check: Put p = 7 in LHS

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)

⇒ \(-\frac{16}{4}=\frac{4(m-6)}{4}\) (Dividing both sides by 4)

⇒ -4 = m – 6

⇒ -4 + 6 = m (Transposing 6 to LHS)

⇒ 2 = m

or m = 2

Check: Put m = 2 in RHS

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) Possible equations are:

10x + 2 = 22; \(\frac{x}{5}=\frac{2}{5}\) ; 5x – 3 = 7

(b) Possible equations are:

3x= -6; 3x + 7 = 1; 3x + 10 = 4