SOLUTION: The perimeter of a rectangle is 178 inches. The length exceeds the width by 15 inches. Find the width and the legnth.
The length is ? inches?
The width is ? inches?
Write in
Question 143876: The perimeter of a rectangle is 178 inches. The length exceeds the width by 15 inches. Find the width and the legnth.
The length is ? inches?
The width is ? inches?
Write in an integer or a decimal.
Will someone please help me with this one. I thought last week would be all the questions I would have but NO I still have questions that I plainly do not understand. Help please. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! The perimeter of a rectangle is 178 inches. The length exceeds the width by 15 inches. Find the width and the length.
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Let x = the width
:
It says,"The length exceeds the width by 15 inches." so we can say:
(x+15) = the length
:
we know,"The perimeter of a rectangle is 178 inches."
2L + 2W = 178
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Substitute (x+15) for L and x for w, and find x (the width)
2(x+15) + 2x = 178
:
2x + 30 + 2x = 178
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2x + 2x = 178 - 30
:
4x = 148
x =
x = 37 inches is the width
then
37 + 15 = 52 inches the length
:
:
Check solution by finding the perimeter with these values:
2(52) + 2(37) = 178
:
:
Do you understand it now?
