A rectangle is three times as long as it is wide. If...

sandyk109

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May 7, 2007
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A rectangle is three times as long as it is wide. If its lenght and width are both decreaded by 2 cm, its area is decreased by 36 cm2. Find it orginal dimensions.

(3x-40(x-4) + 3x2(x) = x + 36

-12x -4x + 16 = x+ 36

-16x = X + 20

Please explain and show work.

Thanks,

Sandy
 

TchrWill

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Jul 7, 2005
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Re: Beginning Algebra

A rectangle is three times as long as it is wide. If its lenght and width are both decreased by 2 cm, its area is decreased by 36 cm2. Find it orginal dimensions.

(3x-40(x-4) + 3x2(x) = x + 36

-12x -4x + 16 = x+ 36

-16x = X + 20

Please explain and show work.

Thanks,

X = width
3x = length

Then, x(3x) - (x - 2)(3x - 2) = 36.

3x^2 - 3x^2 + 8x - 4 = 36

8x = 40

I bet you can take it from here.
 

Subhotosh Khan

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Re: Beginning Algebra

sandyk109 said:
A rectangle is three times as long as it is wide. If its lenght and width are both decreaded by 2 cm, its area is decreased by 36 cm2. Find it orginal dimension.

Original width = W

Original length = 3*W

Original area = 3*W^2

Decreased width = W-2

Decreased length = 3W-2

Decresed area = (W-2)(3W-2) = 3W^2 - 8W + 4

so

3*W^2 - (3W^2 - 8W + 4) = 36

8W - 4 = 36

Now solve W(idth) and Length and the area.
 
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