[SPLIT] length of a rectangle is 3 more than twice its width  TOPIC_SOLVED

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[SPLIT] length of a rectangle is 3 more than twice its width

Postby Xo Exia oX on Thu Jul 30, 2009 11:25 pm

I have another problem I can't solve.
The length of a rectangle is 3 more than twice its width. The perimeter is 48 feet. Find the width.


I came up with an equation 2w + 3 + w = 48.

When I solved, W = 15. There is no choice for 15.
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Postby stapel_eliz on Fri Jul 31, 2009 12:15 am

Check the formula for "perimeter". I'm fairly certain it's P = 2L + 2w, not P = L + w. :wink:
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Re: [SPLIT] length of a rectangle is 3 more than twice its width

Postby Xo Exia oX on Sat Aug 01, 2009 6:52 pm

I tried again, and came up with 2w + 3 + 2w = 48

Solve for W = 11.25
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Postby stapel_eliz on Sat Aug 01, 2009 7:05 pm

You don't show much work, so I can only guess that you're trying to do too much at once. Instead, trying working step-by-step and clearly.

Write down your explicit variable for the width.

Write down your explicit expression for the length, in terms of the variable for the width.

Plug the variable in for "w" in the "perimeter" formula.

Plug the expression in for "L" in the "perimeter" formula.

Plug the given value in for "P" in the "perimeter" formula.

Solve this equation for the variable.
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Re: [SPLIT] length of a rectangle is 3 more than twice its width

Postby Xo Exia oX on Sun Aug 02, 2009 7:27 pm

1. W

2. 2w + 3

3. P = 2(w) + 2L

4. P = 2(w) + 2(2w + 3)

5. 2W + 4W + 3 = 48

6W + 3 = 48

48 - 3 = 45

6w = 45
/6 /6

w = 7.5
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Postby stapel_eliz on Sun Aug 02, 2009 8:44 pm

Xo Exia oX wrote:4. P = 2(w) + 2(2w + 3)

5. 2W + 4W + 3 = 48

2(2w + 3) = 2(2w) + 2(3) = ...?

:wink:
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Re: [SPLIT] length of a rectangle is 3 more than twice its width  TOPIC_SOLVED

Postby Xo Exia oX on Mon Aug 03, 2009 3:38 am

W = 7

Thanks!! :D
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