document.write( "Question 51268: (20x)/(102-x) for 0<=x<=100 \n" ); document.write( "
Algebra.Com's Answer #34228 by THANApHD(104)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "let
\n" ); document.write( " 20x/102-x <= y
\n" ); document.write( " 20x <= 102y-xy
\n" ); document.write( " 20x+xy <= 102y
\n" ); document.write( " X(20+y) <= 102y
\n" ); document.write( " x <= 102y/20+y
\n" ); document.write( "
\n" ); document.write( " but x <=100\r
\n" ); document.write( "\n" ); document.write( "so its obvious 100 = 102y/20+y
\n" ); document.write( " 2000+100y =102y
\n" ); document.write( " 2y =2000
\n" ); document.write( " y=1000
\n" ); document.write( " and other case the least value of the 20x/102-x is 0
\n" ); document.write( " so
\n" ); document.write( " 0<=20x/102-x<=1000
\n" ); document.write( "
\n" );