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| Question 572611:  The length of a rectangle is 2 feet more than 3 times the width. If the area is 16 square feet, find the width and the length.
 Answer by mathsmiles(68)
      (Show Source): 
You can put this solution on YOUR website! Let's write down everything we know:
 L = 2 + 3W  (length is 2 feet more than 3 times the width)
 A is 16 sq ft
 A = L x W
 
 If we substitute in the area formula so we only have one variable, W:
 L x W = 16
 (2+3W) x W = 16  Multiplying out the paren
 2W + 3W^2 = 16
 Let's rearrange the terms:
 3W^2 + 2W = 16
 Subtracting 16 from each side:
 3W^2 + 2W - 16 = 0
 
 Now we factor this:
 (3W _ _) (W _ _) = 0  Where we have to figure out the operators and terms
 What are factors of 16 that we can try?
 16 = 4x4
 16 = 8x2 (oh good, there's only 2)
 
 Let's try the 2 4s:
 (3W _ 4)(W _ 4) = 0
 Since 16 is negative, one must be positive, the other negative:
 Try:
 (3W - 4)(W + 4) = 3W^2 + 12W - 4W -16    (12-4 is not 2 so this isn't it)
 Try:
 (3W - 2)(W + 8) = 3W^2 + 24W - 2W -16    (24-2 is not 2 so let's switch the 2 & 8 and try this one again)
 (3W - 8)(W + 2) = 3W^2 + 6W - 8W - 16    (6-8 is neg 2 so let's switch the signs)
 (3W + 8)(W - 2) = 3W^2 -6W + 8W -16
 = 3W^2 + 2W -16  Yay!
 Writing it again so we can finish:
 (3W + 8)(W-2) = 0
 3W + 8 = 0 Subtract 8 from both sides:
 3W = -8  Divide both sides by 3
 W = -8/3  (not likely since this is a rectangle that the width is negative)
 W - 2 = 0  Add 2 to both sides
 W = 2 Aha!
 
 Now we need to figure the length:
 L = 2+3xW
 L = 2+3(2)
 L = 2+6
 L = 8
 
 Checking:
 A = L x W
 A = 8 x 2
 16 = 16 Correct!
 
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