To solve this problem, let's break down the information given.

Problem Statement

1. Initial Information:

   • Let the breadth of the rectangle be $b$.
   • The length $l$ of the rectangle is 2 more than 3 times the breadth. Therefore, $l = 3b + 2$.

2. Original Perimeter:

   • The perimeter $P$ of a rectangle is given by $P = 2(l + b)$.
   • Substituting for $l$: $P = 2((3b + 2) + b) = 2(4b + 2) = 8b + 4$.

3. Modified Condition:

   • Both the length and breadth are increased by 3 units. So, the new length $l' = (3b + 2) + 3 = 3b + 5$ and the new breadth $b' = b + 3$.
   • The new perimeter $P' = 2(l' + b') = 2((3b + 5) + (b + 3)) = 2(4b + 8) = 8b + 16$.

4. Condition on Perimeters:

   • The new perimeter is twice the original perimeter. So, $P' = 2P$.
   • Substituting for $P'$ and $P$: $8b + 16 = 2(8b + 4)$ Simplify this equation: $8b + 16 = 16b + 8$ $16 - 8 = 16b - 8b$ [ b =