rectangle perim. 58 in., length is 19 in more than width

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phansen05
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rectangle perim. 58 in., length is 19 in more than width

Postby phansen05 » Mon Jan 17, 2011 6:22 pm

Please help having a difficult time solving this problem! The perimeter of a rectangle is 58 inches. The length exceeds the width by 19 inches. Find the length and width.

snowdrifter
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Re: rectangle perim. 58 in., length is 19 in more than width

Postby snowdrifter » Tue Jan 18, 2011 3:50 am

phansen05 wrote:Please help having a difficult time solving this problem! The perimeter of a rectangle is 58 inches. The length exceeds the width by 19 inches. Find the length and width.


The formula for perimeter is P = 2L + 2W
You know that P = 58 and W + 19 = L
Replace the L in the first equation with (W + 19)
Solve for W and then add 19 to get L.


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