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Mathematics 75 Online
OpenStudy (anonymous):

The perimeter of a rectangle is 158 inches. The length exceeds the width by 57 inches. Find the length and the width.

myininaya (myininaya):

2l+2w=158 is given in the first sentence

myininaya (myininaya):

l=w+57 is given by second sentence

OpenStudy (watchmath):

L=length, W=width Then 2L+2W=158. Also L=W+57 Plug in that to the first equation we have 2W+114+2W=158 4W=44 W=11 Hence L=68

myininaya (myininaya):

so 2(w+57)+2w=158 2w+2(57)+2w=158 4w+2(57)=158 4w=158-2(57)=158-114 4w=44 w=44/4=11 so l=11+57=68

OpenStudy (anonymous):

Thank you :)

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