SOLUTION: Find the length of each side of the rectangle so that it has the given perimeter. Note: Perimeter = 2*length + 2*width Length= 2x-5 Width= x p=50 in.

Algebra ->  Rectangles -> SOLUTION: Find the length of each side of the rectangle so that it has the given perimeter. Note: Perimeter = 2*length + 2*width Length= 2x-5 Width= x p=50 in.      Log On


   

Question 76678: Find the length of each side of the rectangle so that it has the given perimeter.
Note: Perimeter = 2*length + 2*width
Length= 2x-5
Width= x

p=50 in.
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
Find the length of each side of the rectangle so that it has the given perimeter.
Note: Perimeter = 2*length + 2*width
Length= 2x-5
Width= x

p=50 in.
50=2%282x-5%29%2B2%28x%29
50=4x-10%2B2x
50=6x-10
50%2B10=6x-10%2B10
60=6x
60%2F6=6x%2F6
10=x
Length=2x-5= 2(10)-5=20-5=15 in.
Width=x=10 in
Sanity check, does a rectangle with a length of 15 in and a width of 10 in have a peimeter that is 50 in?
50=2(15)+2(10)
50=30+20
50=50
We're not off our rocker!!!!
Happy Calculating!!!!