document.write( "Question 516955: a length of a rectangle is 5 more than the width and the perimeter of a rectangle is 6 times the width.find the width length and perimeter of the rectangle.
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Algebra.Com's Answer #344579 by jessica43(140)\"\" \"About 
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\n" ); document.write( "To solve this problem you need to set up two equations using what you know. First, you know the length of the rectangle is 5 more than the width: L = 5 + W. Second, you know the perimeter of the rectangle is 6 times the width: L + L + W + W = 6W, or 2L + 2W = 6W.
\n" ); document.write( "Now plug in the first equation into the second equation, substituting the L with the (5 + w):
\n" ); document.write( "2L + 2W = 6W
\n" ); document.write( "2(5 + W) + 2W = 6W
\n" ); document.write( "10 + 2W + 2W = 6W
\n" ); document.write( "10 + 4W = 6W
\n" ); document.write( "10 = 2W
\n" ); document.write( "5 = W
\n" ); document.write( "So the width is 5. Now plug that into the first equation:
\n" ); document.write( "L = 5 + W
\n" ); document.write( "L = 5 + 5
\n" ); document.write( "L = 10. So using the length and width you can find the perimeter:
\n" ); document.write( "2L + 2W = P
\n" ); document.write( "2(10) + 2(5) = P
\n" ); document.write( "20 + 10 = P
\n" ); document.write( "30 = P. The perimeter is 30. \n" ); document.write( "
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