document.write( "Question 235487: Question is Area = 35sq ft and Perimeter = 27 ft. Solve for Length in the Perimeter. I have P = 2(W) x 2(L) and that would be 27= 2(W)x 2(L) and then move 2(W) to the other side so 27-2(W) = 2(L) then divide both by 2. 27-2(W)/2 = L. But I can't figure it out. Nothing makes 35sq ftof area and 27 feet in a perimeter. My only options as I see it are 35 and 1 or 7 and 5. Someone please help me figure this out. \n" ); document.write( "
Algebra.Com's Answer #173449 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
Area = 35sq ft and Perimeter = 27 ft. Solve for Length in the Perimeter.
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\n" ); document.write( "Area = LW
\n" ); document.write( "Perimeter = 2(L+W)
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\n" ); document.write( "Substitute and solve:
\n" ); document.write( "35 = LW
\n" ); document.write( "27 = 2(L+W)
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\n" ); document.write( "L = 35/W
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\n" ); document.write( "Substitute into the Perimeter equation and solve for W:
\n" ); document.write( "13.5 = (35/W)+ W
\n" ); document.write( "Multiply thru by W and solve:
\n" ); document.write( "W^2 -13.5W + 35 = 0
\n" ); document.write( "W = [13.5 +- sqrt(13.5^2-4*1*35)]/2
\n" ); document.write( "W = [13.5 +- sqrt(42.25)]/2
\n" ); document.write( "Positive solution:
\n" ); document.write( "W = 20/2 = 10 ft. (the width of the rectangle)
\n" ); document.write( "L+W = 13.5 so L = 3.5 ft. (length of the rectangle)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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