SOLUTION: The perimeter of a rectangle is 32 inches and the area is 60 square inches. Find the length and the width of the rectangle. I drew a rectangle and labled the sides x and y. I w

Algebra ->  Human-and-algebraic-language -> SOLUTION: The perimeter of a rectangle is 32 inches and the area is 60 square inches. Find the length and the width of the rectangle. I drew a rectangle and labled the sides x and y. I w      Log On


   

Question 11682: The perimeter of a rectangle is 32 inches and the area is 60 square inches. Find the length and the width of the rectangle.
I drew a rectangle and labled the sides x and y.
I wrote the following two exuations:
2x+2y=32 and xy=60.
I am now stuck. Can you help, please?
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
perfect so far....

2x+2y=32 and xy=60
--> x+y=16 and xy=60
--> y = 16-x
--> x(16-x)=60
16x-x%5E2+=+60
x%5E2+-+16x+%2B+60+=+0

so we have a quadratic, which needs to be solved. Hopefully by factorisation. If not..the quadratic formula.

1x60
2x30
3x20
4x15
5x12
6x10

looking at these, 6 and 10 can add to make 16, so

(x-10)(x-6) = 0
so x-10=0 OR x-6=0
so x=10 OR x=6

so, basically we have x=10, so y=6
or we have the opposite, x=6 and y=10

jon.