# Mathematics Class 7 – Chapter 4 – Simple Equation – Exercise – 4.3 – NCERT Exercise Solution

1. Solve the following equations:

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

Solution:

Transposing (5/2) from LHS to RHS, then it becomes (-5/2)

2y = (37/2) – (5/2)

2y = (37-5)/2

2y = 32/2

2y = 16

Dividing both side by 2,

2y/2 = (16/2)

y = (16/2)

y = 8

(b) 5t + 28 = 10

Solution:

Transposing 28 from LHS to RHS, then it becomes (-28)

5t = 10 – 28

5t = – 18

Dividing both side by 5,

5t/5= -18/5

t = -18/5

(c) (a/5) + 3 = 2

Solution:

Transposing 3 from LHS to RHS, then it becomes (-3)

a/5 = 2 – 3

a/5 = – 1

Multiplying both side by 5,

(a/5) × 5= -1 × 5

a = -5

(d) (q/4) + 7 = 5

Solution:

Transposing 7 from LHS to RHS, then it becomes -7

q/4 = 5 – 7

q/4 = – 2

Multiplying both side by 4,

(q/4) × 4= -2 × 4

a = -8

(e) (5/2) x = -5

Solution:

Multiplying both the side by 2,

(5x/2) × 2 = – 5 × 2

5x = – 10

Dividing both the side by 5,

we get,

5x/5 = -10/5

x = -2

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

Solution:

Multiplying both the side by 2,

(5x/2) × 2 = (25/4) × 2

5x = (25/2)

Dividing both the side by 5,

we get,

5x/5 = (25/2)/5

x = (25/2) × (1/5)

x = (5/2)

(g) 7m + (19/2) = 13

Solution:

Transposing (19/2) from LHS to RHS it becomes -19/2

Now,

7m = 13 – (19/2)

7m = (26 – 19)/2

7m = 7/2

Dividing both side by 7,

7m/7 = (7/2)/7

m = (7/2) × (1/7)

m = ½

(h) 6z + 10 = – 2

Solution:

Transposing 10 from LHS to RHS it becomes – 10

Now,

6z = -2 – 10

6z = – 12

Dividing both side by 6,

6z/6 = -12/6

m = – 2

(i) (3/2) l = 2/3

Solution:

Multiplying both the side by 2,

(3l/2) × 2 = (2/3) × 2

3l = (4/3)

Dividing both the side by 3,

we get,

3l/3 = (4/3)/3

l = (4/3) × (1/3)

x = (4/9)

(j) (2b/3) – 5 = 3

Solution:

Transposing -5 from LHS to RHS it becomes 5

Now,

2b/3 = 3 + 5

2b/3 = 8

Multiplying both side by 3,

(2b/3) × 3= 8 × 3

2b = 24

Divide both side by 2,

2b/2 = 24/2

b = 12

2. Solve the following equations:

(a) 2(x + 4) = 12

Solution:

Let us divide both the side by 2,

(2(x + 4))/2 = 12/2

x + 4 = 6

By transposing 4 from LHS to RHS it becomes -4

x = 6 – 4

x = 2

(b) 3(n – 5) = 21

Solution:

Let us divide both the side by 3,

(3(n – 5))/3 = 21/3

n – 5 = 7

By transposing -5 from LHS to RHS it becomes 5

n = 7 + 5

n = 12

(c) 3(n – 5) = – 21

Solution:

Let us divide both the side by 3,

(3(n – 5))/3 = – 21/3

n – 5 = -7

By transposing -5 from LHS to RHS it becomes 5

n = – 7 + 5

n = – 2

(d) – 4(2 + x) = 8

Solution:

Let us divide both the side by -4,

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

2 + x = -2

By transposing 2 from LHS to RHS it becomes – 2

x = -2 – 2

x = – 4

(e) 4(2 – x) = 8

Solution:

Let us divide both the side by 4,

(4(2 – x))/ 4 = 8/ 4

2 – x = 2

By transposing 2 from LHS to RHS it becomes – 2

– x = 2 – 2

– x = 0

x = 0

3. Solve the following equations:

(a) 4 = 5(p – 2)

Solution:

Dividing both the side by 5,

4/5 = (5(p – 2))/5

4/5 = p -2

Transposing (– 2) from RHS to LHS it becomes (2)

(4/5) + 2 = p

(4 + 10)/ 5 = p

p = 14/5

(b) – 4 = 5(p – 2)

Solution:

Let us divide both the side by 5,

– 4/5 = (5(p – 2))/5

– 4/5 = p -2

Transposing – 2 from RHS to LHS it becomes 2

– (4/5) + 2 = p

(- 4 + 10)/ 5 = p

p = 6/5

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

Solution:

Transposing (4) from RHS to LHS it becomes (– 4)

16 – 4 = 3(t + 2)

12 = 3(t + 2)

Divide both the side by 3,

12/3 = (3(t + 2))/ 3

4 = t + 2

Transposing 2 from RHS to LHS it becomes – 2

4 – 2 = t

t = 2

(d) 4 + 5(p – 1) =34

Solution:

Transposing 4 from LHS to RHS it becomes – 4

5(p – 1) = 34 – 4

5(p – 1) = 30

Divide both the side by 5,

(5(p – 1))/ 5 = 30/5

p – 1 = 6

Transposing – 1 from RHS to LHS it becomes 1

p = 6 + 1

p = 7

(e) 0 = 16 + 4(m – 6)

Solution:

Transposing 16 from RHS to LHS it becomes – 16

0 – 16 = 4(m – 6)

– 16 = 4(m – 6)

Divide both the side by 4,

– 16/4 = (4(m – 6))/ 4

– 4 = m – 6

Transposing – 6 from RHS to LHS it becomes 6

– 4 + 6 = m

m = 2

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

Solution:

First equation is,

X = 2

Multiplying both side by 6

6x = 12 … (equation 1)

Second equation is,

6x = 12

Subtracting 4 from both side,

6x – 4 = 12 -4

6x – 4 = 8 … (equation 2)

Third equation is,

6x – 4 = 8

Dividing both side by 6

(6x/6) – (4/6) = (8/6)

x – (4/6) = (8/6) … (equation 3)

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

Solution:

First equation is,

X = -10

Multiply both side by 5

5x = -10 … (equation 1)

Second equation is,

5x = -10

Subtracting 3 from both side,

5x – 3 = – 10 – 3

5x – 3 = – 13 … (equation 2)

Third equation is,

5x – 3 = -13

Dividing both sides by 2

(5x/2) – (3/2) = (-13/2) … (equation 3)

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