SOLUTION: 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.

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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