SOLUTION: A rectangle is three times as long as it is wide. If the length and the width are each increased by 4, the area is increased by 176. Find the dimensions of the orginial rectangle.

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Question 168379: A rectangle is three times as long as it is wide. If the length and the width are each increased by 4, the area is increased by 176. Find the dimensions of the orginial rectangle.
Found 2 solutions by checkley77, stanbon:
Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website!
L=3W
3W*W=(L+4)(W+4)-176
3W^2=(3W+4)(W+4)-176
3W^2=3W^2+12W+4W+16-176
0=16W+16-176
16W=-16+176
16W=160
W=160/16
W=10 ANS. FOR THE ORIGINAL WIDTH .
L=3*10=30 ANS. FOR THE ORIGINAL LENGTH.
PROOF:
10*30=(10+4)(30+4)-176
300=14*34-176
300=476-176
300=300

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
A rectangle is three times as long as it is wide. If the length and the width are each increased by 4, the area is increased by 176. Find the dimensions of the orginial rectangle.
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Let the width be "x".
Then the length is "3x".
Area based on this data = x(3x) = 3x^2
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New conditions:
New width: "x+4"
New length: "3x+4"
Area based on this data = (x+4)(3x+4)
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EQUATION:
New area - Old area = 176
(x+4)(3x+4) - 3x^2 = 176
3x^2+16x+16 - 3x^2 = 176
16x + 16 = 176
16x = 160
x = 10 (original width)
3x= 30 (original length)
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Cheers,
Stan H.