document.write( "Question 86933: The perimeter of a rectangle is 34 feet and its area is 60 square feet. Find the length and width of the rectangle. \n" ); document.write( "
Algebra.Com's Answer #62880 by checkley75(3666)\"\" \"About 
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xy=60
\n" ); document.write( "x=60/y
\n" ); document.write( "2x+2y=34
\n" ); document.write( "2(60/y)+2y=34
\n" ); document.write( "120/y+2y=34
\n" ); document.write( "(120+2y^2)/y=34
\n" ); document.write( "120+2y^2=34y
\n" ); document.write( "2y^2-34y+120=0
\n" ); document.write( "2(y^2-17y+60)=0
\n" ); document.write( "2(y-12)(y-5)=0
\n" ); document.write( "y-12=0
\n" ); document.write( "y=12 answer. the x=60/12=5
\n" ); document.write( "y-5=0
\n" ); document.write( "y=5 answer. the y=60/5=12
\n" ); document.write( "proof
\n" ); document.write( "5*12=60
\n" ); document.write( "60=60
\n" ); document.write( "2*5+2*12=34
\n" ); document.write( "10+24=34
\n" ); document.write( "34=34\r
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