SOLUTION: a rectangle is 5 meters longer than its width. If thee length is shortened by 2 meters, and the width is increased by one meter, the area will remain the same. Find the length and

Algebra ->  Rectangles -> SOLUTION: a rectangle is 5 meters longer than its width. If thee length is shortened by 2 meters, and the width is increased by one meter, the area will remain the same. Find the length and       Log On


   

Question 242342: a rectangle is 5 meters longer than its width. If thee length is shortened by 2 meters, and the width is increased by one meter, the area will remain the same. Find the length and the area.
Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
Let w=width of original rectangle
Then w+5=length of original rectangle
(w+5)-2=length shortened by 2 meters
And w+1=width increased by 1 meter
Now we know that the area of a rectangle=Length times Width or A=lw
Area of original rectangle=w*(w+5)
Area of modified rectangle=(w+1)*(w+5-2)=(w+1)*(w+3)
We are told that the above two areas are equal, so our equation to solve is:
w(w+5)=(w+1)(w+3) get rid of parens by expanding each side
w^2+5w=w^2+4w+3 subtract w^2 and also 5w from each side and we get:
w^2-w^2+5w-5w=w^2-w^2+4w-5w+3 collect like terms
-w+3=0 subtract 3 from each side
-w=-3 or
w=3 meters width of original rectangle
w+5=3+5=8 length of original rectangle
Area of original rectangle=8*3=24 sq meters
CK
Area of modified rectangle=(3+1)(3+3)=4*6=24 sq meters
Hope this helps---ptaylor