document.write( "Question 757671: The 1st problem is: The perimeter of a rectangle is 860 cm. The length is 4 times the width. Find the length.\r
\n" ); document.write( "\n" ); document.write( "The 2nd problems is: The perimeter of a rectangle is 146 cm. The width is 10 cm longer than TWICE the length. Find the width.\r
\n" ); document.write( "\n" ); document.write( "I am totally stumped on these problems it is just confusing.\r
\n" ); document.write( "\n" ); document.write( "Thank you for your help.
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Algebra.Com's Answer #460977 by nerdybill(7384)\"\" \"About 
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The 1st problem is: The perimeter of a rectangle is 860 cm. The length is 4 times the width. Find the length.
\n" ); document.write( "Let x = length of rectangle
\n" ); document.write( "then
\n" ); document.write( "4x = width of rectangle
\n" ); document.write( ".
\n" ); document.write( "from: \"The perimeter of a rectangle is 860 cm\" we get our equation:
\n" ); document.write( "2(4x + x) = 860
\n" ); document.write( "2(5x) = 860
\n" ); document.write( "10x = 860
\n" ); document.write( "x = 86 cm (length)
\n" ); document.write( ".
\n" ); document.write( "The 2nd problems is: The perimeter of a rectangle is 146 cm. The width is 10 cm longer than TWICE the length. Find the width.
\n" ); document.write( "Let x = length
\n" ); document.write( "then from \" width is 10 cm longer than TWICE the length\"
\n" ); document.write( "2x+10 = width
\n" ); document.write( ".
\n" ); document.write( "from:\" The perimeter of a rectangle is 146 cm\"
\n" ); document.write( "2(2x+10 + x) = 146
\n" ); document.write( "2(3x+10) = 146
\n" ); document.write( "6x+20 = 146
\n" ); document.write( "6x = 126
\n" ); document.write( "x = 21 cm (length)
\n" ); document.write( "width:
\n" ); document.write( "2x+10 = 2(21)+10 = 42+10 = 52 cm (width)
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