SOLUTION: mary needs to enclose a rectangular section of his yard. The area is 35sq.ft and the perimeter is 27ft. Find the length and the width of the section.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: mary needs to enclose a rectangular section of his yard. The area is 35sq.ft and the perimeter is 27ft. Find the length and the width of the section.      Log On


   

Question 166995: mary needs to enclose a rectangular section of his yard. The area is 35sq.ft and the perimeter is 27ft. Find the length and the width of the section.
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
xy=35 or x=35/y
2x+2y=27
2(35/y)+2y=27
70/y+2y=27
(70+2y^2)/y=27
70+2y^2=27*y
2y^2+70=27y
2y^2-27y+70=0
(2y-7)(y-10)=0
2y-7=0
2y=7
y=7/2
y=3.5 answer.
x*3.5=35
3.5x=35
x=35/3.5
x=10 answer.
Proof:
2*3.5+2*10=27
7+20=27
27=27