document.write( "Question 612910: The width of a rectangle is 3 m less than twice its length. (By m I think it means meter, not a variable.) If the ratio of the length to the width is 7/5, find the dimensions of the rectangle. \n" ); document.write( "
Algebra.Com's Answer #385787 by jim_thompson5910(35256)\"\" \"About 
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Length/Width = 7/5\r
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\n" ); document.write( "\n" ); document.write( "L/(2L-3) = 7/5\r
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\n" ); document.write( "\n" ); document.write( "5L = 7(2L-3)\r
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\n" ); document.write( "\n" ); document.write( "5L = 7(2L-3)\r
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\n" ); document.write( "\n" ); document.write( "5L = 14L-21\r
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\n" ); document.write( "\n" ); document.write( "5L-14L = -21\r
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\n" ); document.write( "\n" ); document.write( "-9L = -21\r
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\n" ); document.write( "\n" ); document.write( "L = -21/(-9)\r
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\n" ); document.write( "\n" ); document.write( "L = 7/3\r
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\n" ); document.write( "\n" ); document.write( "W = 2L - 3\r
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\n" ); document.write( "\n" ); document.write( "W = 2(7/3) - 3\r
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\n" ); document.write( "\n" ); document.write( "W = 14/3 - 3\r
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\n" ); document.write( "\n" ); document.write( "W = 14/3 - 9/3\r
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\n" ); document.write( "\n" ); document.write( "W = 5/3\r
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\n" ); document.write( "\n" ); document.write( "So the width is 5/3 m and the length is 7/3 m
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