**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.

**NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1****NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2****NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3****NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4**

**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:(Ð°) 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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