SOLUTION: Can you help me with this problem. The lenght of a rectangular garden is 20 feet longer than the width. The perimeter must be between 80 and 100feet for the fencing that has been p

Algebra ->  Rectangles -> SOLUTION: Can you help me with this problem. The lenght of a rectangular garden is 20 feet longer than the width. The perimeter must be between 80 and 100feet for the fencing that has been p      Log On


   

Question 151539: Can you help me with this problem. The lenght of a rectangular garden is 20 feet longer than the width. The perimeter must be between 80 and 100feet for the fencing that has been purchased. What are the possible widths for the garden.
I started with p=2L=2W Perimeter
A=LW
I don't where to even start.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Since the "lenght of a rectangular garden is 20 feet longer than the width", this means that L=W%2B20


P=2W%2B2L Start with the perimeter equation.


P=2W%2B2%28W%2B20%29 Plug in L=W%2B20


P=2W%2B2W%2B40 Distribute


P=4W%2B40 Combine like terms.


Since the "perimeter must be between 80 and 100 feet", this means that 80%3C=P%3C=100


80%3C=P%3C=100 Start with the given compound inequality.


80%3C=4W%2B40%3C=100 Plug in P=4W%2B40


%2880%29%2F4%3C=w%3C=%28100%29%2F4 Divide all sides by 4.


20%3C=w%3C=25 Reduce.


80-40%3C=4w%3C=100-40 Subtract 40 from all sides.


40%3C=4w%3C=100-40 Combine like terms on the left side.


40%3C=4w%3C=60 Combine like terms on the right side.


%2840%29%2F4%3C=w%3C=%2860%29%2F4 Divide all sides by 4.


10%3C=w%3C=15 Reduce.


So our answer is 10%3C=w%3C=15 which means that the width must be between 10 and 15 feet.