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

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
Xo Exia oX
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### [SPLIT] length of a rectangle is 3 more than twice its width

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.

stapel_eliz
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Check the formula for "perimeter". I'm fairly certain it's P = 2L + 2w, not P = L + w.

Xo Exia oX
Posts: 18
Joined: Wed Jul 08, 2009 12:59 am
Contact:

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

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

Solve for W = 11.25

stapel_eliz
Posts: 1687
Joined: Mon Dec 08, 2008 4:22 pm
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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.

Xo Exia oX
Posts: 18
Joined: Wed Jul 08, 2009 12:59 am
Contact:

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

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

stapel_eliz
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Joined: Mon Dec 08, 2008 4:22 pm
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Xo Exia oX wrote:4. P = 2(w) + 2(2w + 3)

5. 2W + 4W + 3 = 48

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

Xo Exia oX
Posts: 18
Joined: Wed Jul 08, 2009 12:59 am
Contact:

W = 7

Thanks!!