Question

The length of a rectangle is 3 more than twice the breadth of the rectangle. If
the length is reduced by 5 units and the breadth is increased by 3 units,
the area remains same . find the dimension s of the original rectangle
and also the area.

Answer 1

Final Answer : Length = 15units , Breadth = 6 units
Area = 90 sq. units

Steps: 1)
Let the length and breadth of the rectangle be
x, y respectively.

According to the question,
$x = 2y + 3 - - - - (1)$
and
If length is reduced by 5 units, and breadth is increased by 3 units, then Area remains same.
New Length = (x-5)
New Breadth = (y+3)
And x, y are Original Length.

2)
Area Original = New Area
$= > (x - 5)(y + 3) = xy \\ = > xy + 3x - 5y - 15 = xy \\ = > 3x - 5y - 15 = 0 - - - - (2)$
3) Using equation 1 and 2,
x = 15 , y = 6

Area = xy = 12 * 6 = 90 sq. units

Therefore, Dimension of Rectangle :
Length = 15 units
Breadth = 6 units.

Answer 2

Let the original length and the breadth of the Rectangle be l and b respectively. 

   ∴ l = 3 + 2b   ---------eq(i). 

Area of the Rectangle = l × b

Now, If length is reduced by the 5 units and the breadth is increased by the 3 units, then area will be, 
  Area = (l - 5)(b + 3)

Now, Area in both the cases is same. 
∴ lb = (l - 5)(b + 3)
⇒ lb = lb + 3l - 5b - 15
⇒ 3l - 5b = 15
3(3 + 2b) - 5b = 15  [From eq(i)] 
9 + 6b - 5b = 15
b = 15 - 9
∴ b = 6 units. 

Thus, l = 3 + 2(6)
 l = 3 + 12
 l = 15 units. 


Hope it helps.