document.write( "Question 166995: mary needs to enclose a rectangular section of his yard. The area is 35sq.ft and the perimeter is 27ft. Find the length and the width of the section. \n" ); document.write( "
Algebra.Com's Answer #123017 by checkley77(12844)\"\" \"About 
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xy=35 or x=35/y
\n" ); document.write( "2x+2y=27
\n" ); document.write( "2(35/y)+2y=27
\n" ); document.write( "70/y+2y=27
\n" ); document.write( "(70+2y^2)/y=27
\n" ); document.write( "70+2y^2=27*y
\n" ); document.write( "2y^2+70=27y
\n" ); document.write( "2y^2-27y+70=0
\n" ); document.write( "(2y-7)(y-10)=0
\n" ); document.write( "2y-7=0
\n" ); document.write( "2y=7
\n" ); document.write( "y=7/2
\n" ); document.write( "y=3.5 answer.
\n" ); document.write( "x*3.5=35
\n" ); document.write( "3.5x=35
\n" ); document.write( "x=35/3.5
\n" ); document.write( "x=10 answer.
\n" ); document.write( "Proof:
\n" ); document.write( "2*3.5+2*10=27
\n" ); document.write( "7+20=27
\n" ); document.write( "27=27
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