# 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 ->  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

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 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)   (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. ```