SOLUTION: May I please have some help? A soccer field is 336yd^2. The width is 5yd longer than its length. What is the length and width of the soccer field. Thanks for your help.

Algebra ->  Length-and-distance -> SOLUTION: May I please have some help? A soccer field is 336yd^2. The width is 5yd longer than its length. What is the length and width of the soccer field. Thanks for your help.      Log On


   

Question 41579: May I please have some help? A soccer field is 336yd^2. The width is 5yd longer than its length. What is the length and width of the soccer field. Thanks for your help.
Answer by mszlmb(115) About Me  (Show Source):
You can put this solution on YOUR website!
Given soccer fields are rectangular:

2 variables, the width, and the length of the soccer field.

The width is 5 longer than the length.

The width times the length is 336.

The unit is yards.

W=L+5

WL=336

substitute for W in "WL=336".

(L+5)L=336  distribute

L2+5L=336  subtract 336 from both sides

L2+5L-336=0  Factor (or use quadradic formula)



2 numbers which multiply to -336 and add to 5, so 1 number must be negative.

336=112*3=56*6=28*12=14*24=7*48, but you'll notice none of those have a difference of 5..let's try 2's

336=168*2=84*4=42*8=21*16 aha!  difference of 5. so we alter the equation:



L2+5L-336=L2-16L+21L-336 

Now we factor:

L2-16L+21L-336  take out like terms:

L(L-16)+21(L-16)  put like terms 2gether

(L+21)(L-16)=0  so 1 of thoz = 0

L=-21, or it equals 16. A side is hardly negative; but definitely not for a football field, therefore L=16.

W=L+5 so W=16+5, or 21.

W=21yards

L=15yards



..I'm beginning to think there was a simpler way of figuring this out, I cannot put my finger on it.