Finding the perimeter of the room If the length and breadth of a room are increased by $1$ $m$, the area is increased by $21$ $m^2$. If the length is increased by $1$ $m$ and breadth is decreased by $1$ $m$ the area is decreased by $5$ $m^2$. Find the perimeter of the room.
Let the length be $x$ and the breadth be $y$
Therefore, Area$=$$xy$ $m^2$
Accordingly, $(x+1) \cdot (y+1)  \ = \ xy+21$ $m^2$
What should I do now? How should I find the second equation?
Should the second equation look like:
$(x+1) \ \cdot \ (y-1) \ = \ xy -5 \ $ $m^2$

$A=xy$
$(x+1)(y+1)=xy+21$
$(x+1)(y-1)=xy-5$
Foil out both equations to get:
$xy+x+y+1=xy+21 \quad \to \quad x+y=20 \quad \to \quad y=20-x$
$xy-x+y-1=xy-5 \quad \to \quad -x+y=-4 \quad \to \quad y=-4+x$
Set them equal to each other:
$20-x=-4+x$
$2x=24 \to x=12$
Since we know $x+y=20, y=8$.
You can verify this solution by checking the conditions given.
$A=12\cdot 8=96$ 
Adding $1$ to both the length and width:
$A=13\cdot 9=117 \to 96+21=117$
Also, if you add $1$ to the length and subtract $1$ from the width: $A=13\cdot 7=91 \to 96-5=91$
