document.write( "Question 171016: The length of a rectangle is 1ft less than twice its width. The area is 55ft^2 find the perimeter. \n" ); document.write( "
Algebra.Com's Answer #126264 by checkley77(12844)\"\" \"About 
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L=2W-1
\n" ); document.write( "L*W=55
\n" ); document.write( "(2W-1)W=55
\n" ); document.write( "2W^2-W-55=0
\n" ); document.write( "(2W-11)(W+5)=0
\n" ); document.write( "2W=11
\n" ); document.write( "W=11/2
\n" ); document.write( "W=5.5 FT. ANS.
\n" ); document.write( "L=2*5.5-1
\n" ); document.write( "L=11-1
\n" ); document.write( "L=10 FT. ANS.
\n" ); document.write( "PROOF:
\n" ); document.write( "10*5.5=55
\n" ); document.write( "55=55
\n" ); document.write( "THUS THE PERIMETERIS:
\n" ); document.write( "2L+2W=PERIMETER.
\n" ); document.write( "2*10+2*5.5=20+11=31 FT. IS THE PERIMETER.
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