document.write( "Question 218159: The length of a rectangle is 5 more than twice its width. If the width is decreased by 3 and the length by 1, the perimeter becomes four-fifths of the original. Find the original dimensions of the rectangle.\r
\n" ); document.write( "\n" ); document.write( "Please send me the solution as soon as possible. Thanks MATH Masters~ \n" ); document.write( "
Algebra.Com's Answer #164327 by alicealc(293)\"\" \"About 
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length = l
\n" ); document.write( "width = w\r
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\n" ); document.write( "\n" ); document.write( "l = 5 + 2*w
\n" ); document.write( "2*(w - 3) + 2*(l - 1) = 4/5 * (2*l + 2*w)\r
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\n" ); document.write( "\n" ); document.write( "substitute the first equation into the second equation:
\n" ); document.write( "2*(w - 3) + 2*(5 + 2w - 1) = 4/5 * (2*(5 + 2*w) + 2*w)
\n" ); document.write( "2w - 6 + 10 + 4w - 2 = 4/5 * (10 + 4w + 2w)
\n" ); document.write( "6w + 2 = 40/5 + 16/5 w + 8/5 w
\n" ); document.write( "6w + 2 = 8 + 24/5 w
\n" ); document.write( "6w - 24/5 w = 8 - 2
\n" ); document.write( "30/5 w - 24/5 w = 6
\n" ); document.write( "6/5 w = 6
\n" ); document.write( "w = 6 * 5/6 = 30/6 = 5
\n" ); document.write( "l = 5 + 2*w = 5 + 2*5 = 5 + 10 = 15\r
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\n" ); document.write( "\n" ); document.write( "so, the original dimension of the rectangle is:
\n" ); document.write( "length = 15, and width = 5 \n" ); document.write( "
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