NCERT Solutions Class 7 Maths Chapter 4 Simple Equations Exercise 4.3

Simple Equations

Exercise 4.3

1). Solve the following equations:

(a) 2 y + 5/2 = 37/2                             (b) 5t + 28 = 10

(c) a/5 + 3 = 2                                   (d) q/4 + 7 = 5

(e) 5/2 x = –5                                     (f) 5/2 x = 25/4

(g) 7m+ 19/2 =13                               (h) 6z + 10 = –2

(i) 3 l /2 =2/3                                      (j) 2b/3 – 5 = 3

Solution

(a) 2 y + 5/2 = 37/2

(b) 5t + 28 = 10

 

(c) a/5 + 3 = 2

                          

(d) q/4 + 7 = 5   

(e) 5/2 x = –5

(f) 5/2 x = 25/4

 

(g) 7m+ 19/2 =13

                             

(h) 6z + 10 = –2

(i) 3 l /2 =2/3  

                                    

(j) 2b/3 – 5 = 3

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

Solution

(a) 2(x + 4) = 12

By removing the bracket we get

2x + 8 = 12

Subtracting 8 from both the sides

2x + 8 – 8 = 12 – 8

2x = 4

Dividing both the sides by 2

 2x/2  = 4/2

x  = 2

(b) 3(n – 5) = 21

By removing the bracket we get

3n – 15 = 21

Adding 15 to both the sides

3n – 15 + 15 = 21 + 15

3n = 36

Dividing both the sides by 3

 3n/3  = 36/3

n  = 12

(c) 3(n – 5) = – 21

By removing the bracket we get

3n – 15 = -21

Adding 15 to both the sides

3n – 15 + 15 = -21 + 15

3n = -6

Dividing both the sides by 3

 3n/3  = -6/3

n  = -2                                 

(d) – 4(2 + x) = 8

By removing the bracket we get

-8 – 4x = 8

Adding 8 to both the sides

-8 – 4x + 8 = 8 + 8

– 4x = 16

Dividing both the sides by 4

 -4x/4  = 16/4

x  = 4

x  = -4

(e) 4(2 – x) = 8

By removing the bracket we get

8 – 4x = 8

Subtracting 8 from both the sides

8 – 4x – 8 = 8 – 8

– 4x = 0

Dividing both the sides by 4

 -4x/4  = 0/4

x  = 0

x  = 0

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:

(a) 4 = 5(p – 2)

Removing brackets we have

4 = 5p – 10

Transposing -10 to LHS

4 +10 = 5p

14 = 5p

Transposing 5 to LHS

14/5 = p

 p =14/5                          

(b) – 4 = 5(p – 2)

Removing brackets we have

-4 = 5p – 10

Transposing -10 to LHS

-4 +10 = 5p

6 = 5p

Transposing 5 to LHS

6/5 = p

 p =6/5

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

Removing brackets we have

16 = 4 + 3t + 6

16 = 10 + 3t

Transposing 10 to LHS

16 -10 = 3t

6 = 3t

Transposing 3 to LHS

6/3 = t

2 = t

 t =2                          

(d) 4 + 5(p – 1) =34

Removing brackets we have

4 + 5p – 5 = 34

5p – 1 = 34

Transposing -1 to RHS

5p  = 34 + 1

5p  = 35

Transposing 5 to RHS

p =35/5

p = 7

(e) 0 = 16 + 4(m – 6)

Removing brackets we have

0 = 16 + 4m – 24

0 = 4m – 8

Transposing -8 to LHS

0 + 8 = 4m

8 = 4m

Transposing 4 to LHS

8/4 = m

2 = m

m = 2

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

x = 2

 

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

Click here for the solutions of 

Exercise 4.1

Exercise 4.2

Exercise 4.3

Exercise 4.4

Exercise 3.1

Exercise 3.2

Exercise 3.3

Exercise 3.4

Loading

Leave a Comment

Your email address will not be published. Required fields are marked *

error: Content is protected !!