Finding the perimeter of the room

Question

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 \ $ $m^2$

Answer 1

$A=xy$

$(x+1)(y+1)=xy+21$

$(x+1)(y-1)=xy-5$

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: $A=13\cdot 7=91 \to 96-5=91$

Answer 2

The area is $xy$. You know that $$ (x+1)(y+1)=xy+21 $$ and that $$ (x+1)(y-1)=xy-5 $$ This leaves two linear equations: the first can be developed as $$ xy+x+y+1=xy+21 $$ that is, $$ x+y=20 $$ Do the same to the second relation and find $x$ and $y$ from the two equations.

Answer 3

$(x+1)(y+1)=xy+21$, $(x+1)(y-1)=xy-5$, 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 $y=8$.