document.write( "Question 800820: The length of a rectangular piece of carpet is 2 meters less than 5 times its width. Find the dimensions of the carpet if its area is 16 square meters. Solve algebraically \n" ); document.write( "
Algebra.Com's Answer #483277 by algebrahouse.com(1659)\"\" \"About 
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w = width
\n" ); document.write( "5w - 2 = length {length is 2 less than 5 times width}\r
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\n" ); document.write( "\n" ); document.write( "Area of a rectangle = length x width\r
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\n" ); document.write( "\n" ); document.write( "w(5w - 2) = 16 {substituted length, width, and given area into area formula}
\n" ); document.write( "5w² - 2w = 16 {used distributive property}
\n" ); document.write( "5w² - 2w - 16 = 0 {subtracted 16 from each side}
\n" ); document.write( "5w² - 10w + 8w - 16 = 0 {split -2w into -10w and 8w, factoring by grouping}
\n" ); document.write( "5w(w - 2) + 8(w - 2) = 0 {factored 5w out of first two terms and 8 out of last two terms}
\n" ); document.write( "(5w + 8)(w - 2) = 0 {factored out the common factor of the two terms, (w - 2)}
\n" ); document.write( "5w + 8 = 0 or w - 2 = 0 {set each factor equal to 0}
\n" ); document.write( "w = -8/5 ow w = 2 {solved each equation}\r
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\n" ); document.write( "\n" ); document.write( "w = 2 {width cannot be negative}
\n" ); document.write( "5w - 2 = 8 {substituted 2, in for w, into 5w - 2}\r
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\n" ); document.write( "\n" ); document.write( "width = 2m
\n" ); document.write( "length = 8m
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