Beginning Algebra-Applications

Algebra
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The length of a rectangular piece of steel in a bridge is 3 meters less than double the width. The perimeter of the piece of steel is 42 meters. Find the length of the piece of steel. Find the width of the piece of steel.

Sep 30th, 2015

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Let the length = L

Let the width = W

Now

length of a rectangular piece of steel in a bridge is 3 meters less than double the width

Meaning

L=2W-3

Perimeter = 2(Length + width)

                 =2(2W-3+W)

                24  =2(3W-3)

             or

             6W-6=24

              6W=30

              W=30

Width is 30 meters

Length = 2*30-3 = 60-3 = 57 meters.

Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 30th, 2015

The question is asking for a length and width only. The answer in back of the book came up with the following answer:

L=13 meters

W=8 meters

I'm having problems figuring out how to write the equation...

Sep 30th, 2015

I tried

x+x-3=42

2x-3=42

2x+45

x=22.5

and lost it from there....


Sep 30th, 2015

Sorry my mistake, i used perimeter to be 24 meters instead of 42 meters. It is now correct here.

Let the length = L

Let the width = W

Now

length of a rectangular piece of steel in a bridge is 3 meters less than double the width

Meaning

L=2W-3

Perimeter = 2(Length + width)

  =2(2W-3+W)

  42  =2(3W-3)

  or

  6W-6=42

  6W=48

  W=8

Width is 8 meters

L=2*8-3 = 16-3=13


Sep 30th, 2015

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Sep 30th, 2015
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Sep 30th, 2015
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