Step-by-step explanation:

Answer:

Let the length of the rectangle be x m and the breadth of the rectangle be y m.

Then,

Area of rectangle = xy m²

★ If its length is reduced by 5 m and breadth is increased by 3 m,

Then,

Length = (x-5) m

Breadth = (y+3) m

A/Q, [ 1st case]

(x-5)(y+3)=xy-9

→ xy+3x-5y-15=xy-9

→ 3x-5y=15-9

→ 3x-5y = 6 .................... [ equation 1 ] ×2

★ If the length is increased by 3 m and the breadth is increased by 2 m,

Then,

Length = (x+3) m

Breadth = (y+2) m

A/Q, [ 2nd case]

(x+3)(y+2) = xy+67

→ xy +2x + 3y +6 = xy + 67

→ 2x+3y = 61................... [ equation 2] × 3

Now,

2 equations will be,

6x -10y = 12..........[1]

6x + 9y = 183.........[2]

Now subtract eq (1) from eq(2).

6x+9y-6x+10y = 183-12

→ 19y = 171

→ y = 9

Breadth = 9 m

Now out y=9 in eq(1).

3x-5y=6

→ 3x - 5×9 = 6

→ 3x = 6+45

→ x = 17

Length = 17 m