## NCERT MATHS SOLUTION CLASS 7 EXERCISE 4.3 | SIMPLE EQUATIONS

Find here step-by-step NCERT Maths Solution Class 7 Exercise 4.3 | Solving Equations | Chapter 4 Simple Equations from CBSE NCERT textbook.

Before moving to maths solution class 7 exercise 4.3 , it is advised to go through the following topics:
• Simple Equations

## NCERT Maths Solution Class 7 Exercise 4.3

Problems in Exercise 4.3 require the knowledge of finding the Solution of a Simple Equation, Transposing etc.

The properties of equality gives a systematic method of solving an equation. We carry out a series of identical mathematical operations on the two sides of the equation in such a way that on one of the sides we get just the variable. The last step is the solution of the equation.

Transposing means moving to the other side. Transposition of a number has the same effect as adding same number to (or subtracting the same number from) both sides of the equation. When we transpose a number from one side of the equation to the other side, we change its sign.

### Exercise 4.3 Solving Simple Equations | Transposing

QUESTION 1

Solve the following equations:

(a)

(b) 5t + 28 = 10

(c)

(d)

(e)

(f)

(g)

(h) 6z + 10 = –2

(i)

(j)

(a)

(b) 5t + 28 = 10

=> 5t = 10 – 28

=> t = – 18/5

=> t = – 3.6

(c)

(d)

(e)

(f)

(g)

(h) 6z + 10 = –2

(i)

(j)

[ NCERT Maths Solution Class 7 Exercise 4.3 ]

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

(a) 2(x + 4) = 12

=> x + 4 =

=> x = 6 – 4

=> x = 2

(b) 3(n – 5) = 21

=> n – 5 =

=> n = 7 + 5

=> n = 12

(c) 3(n – 5) = – 21

=> n – 5 =

=> n = – 7 + 5

=> n = – 2

(d) – 4(2 + x) = 8

=> 2 + x =

=> x = – 2 – 2

=> x = – 4

(e) 4(2 – x) = 8

=> 2 – x =

=> – x = 2 – 2

=> x = 0

[ NCERT Maths Solution Class 7 Exercise 4.3 ]

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)

(a) 4 = 5(p – 2)

=> 5(p – 2) = 4

=> p – 2 =

=> p = + 2

=> p =

=> p =

=> p = 2.8

(b) – 4 = 5(p – 2)

=> 5(p – 2) = – 4

=> p – 2 =

=> p = + 2

=> p =

=> p =

=> p = 1.2

(c) 16 = 4 + 3(t + 2)

=> 4 + 3(t + 2) = 16

=> 3(t + 2) = 16 – 4

=> (t + 2) =

=> t = 4 – 2

=> t = 2

(d) 4 + 5(p – 1) = 34

=> 5(p – 1) = 34 – 4

=> (p – 1) =

=> p = 6 + 1

=> p = 7

(e) 0 = 16 + 4(m – 6)

=> 16 + 4(m – 6) = 0

=> 4(m – 6) = – 16

=> (m – 6) =

=> m = – 4 + 6

=> m = 2

[ NCERT Maths Solution Class 7 Exercise 4.3 ]

QUESTION 4

(a) Construct 3 equations starting with x = 2

(b) Construct 3 equations starting with x = – 2

(a) x = 2

Multiplying both sides by 3

3x = 6 … (i)

Subtracting one from both sides

3x – 1 = 5 … (ii)

Dividing both sides by 3

… (iii)

(b) x = – 2