The question says that the length of the floor of a room is 'y' meters.
Its width is 5 meters shorter than the length,
If the width is 5 meters shorter than the actual width should be:
[tex](y-5)[/tex] meters
Now the room is in the shape of a rectangle, so we will use the formula of the perimeter of a rectangle:
Perimeter = [tex]2(length +width)[/tex]
The length of the floor is 'y' meters
The width of the floor is 'y-5' meters
Plugging the values of length and width we get,
[tex]2(y+ (y-5))[/tex]
Question says that the perimeter of the rectangle is [tex](4y+1)[/tex] meters.
So,
[tex]2(y+(y-5))=4y+1[/tex]
We will solve for 'y'.
[tex]2(y+y-5)=4y+1[/tex]
[tex]2(2y-5)=4y+1[/tex]
[tex]4y-10=4y+1[/tex]
Since,
[tex]-10\neq 1[/tex]
The system of equation seems to have no solution.
Hence, no such floor exists.
