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

Posted: **Thu Jul 30, 2009 11:25 pm**

by **Xo Exia oX**

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.

Posted: **Fri Jul 31, 2009 12:15 am**

by **stapel_eliz**

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

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

Posted: **Sat Aug 01, 2009 6:52 pm**

by **Xo Exia oX**

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

Solve for W = 11.25

Posted: **Sat Aug 01, 2009 7:05 pm**

by **stapel_eliz**

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.

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

Posted: **Sun Aug 02, 2009 7:27 pm**

by **Xo Exia oX**

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

Posted: **Sun Aug 02, 2009 8:44 pm**

by **stapel_eliz**

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

5. 2W + 4W + 3 = 48

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

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

Posted: **Mon Aug 03, 2009 3:38 am**

by **Xo Exia oX**

W = 7

Thanks!!