document.write( "Question 143876: The perimeter of a rectangle is 178 inches. The length exceeds the width by 15 inches. Find the width and the legnth.
\n" ); document.write( " The length is ? inches?
\n" ); document.write( "The width is ? inches?
\n" ); document.write( "Write in an integer or a decimal.
\n" ); document.write( "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.
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Algebra.Com's Answer #104693 by ankor@dixie-net.com(22740)\"\" \"About 
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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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\n" ); document.write( "Let x = the width
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\n" ); document.write( "It says,\"The length exceeds the width by 15 inches.\" so we can say:
\n" ); document.write( "(x+15) = the length
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\n" ); document.write( "we know,\"The perimeter of a rectangle is 178 inches.\"
\n" ); document.write( "2L + 2W = 178
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\n" ); document.write( "Substitute (x+15) for L and x for w, and find x (the width)
\n" ); document.write( "2(x+15) + 2x = 178
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\n" ); document.write( "2x + 30 + 2x = 178
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\n" ); document.write( "2x + 2x = 178 - 30
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\n" ); document.write( "4x = 148
\n" ); document.write( "x = \"148%2F4\"
\n" ); document.write( "x = 37 inches is the width
\n" ); document.write( "then
\n" ); document.write( "37 + 15 = 52 inches the length
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\n" ); document.write( "Check solution by finding the perimeter with these values:
\n" ); document.write( "2(52) + 2(37) = 178
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