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 \ $ $m^2$

$(x+1)(y+1)=xy+21$, $(x+1)(y-1)=xy-5$, so by subtraction $(x+1)(y+1-y+1)=2x+2=26$. Solving gives $x=12$ and then using this to solve the first equation 13(y+1)=12y+21, so $y=8$.
