You cannot do crossing over in inequalities
#(2x)/(x-5)>=3#
#<=>#, #(2x)/(x-5)-3>=0#
#<=>#, #(2x-3(x-5))/(x-5)>=0#
#<=>#, #(2x-3x+15)/(x-5)>=0#
#<=>#, #(-x+15)/(x-5)>=0#
Let #f(x)=(-x+15)/(x-5)#
Build a sign chart
#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaaa)##5##color(white)(aaaaaaa)##15##color(white)(aaaa)##+oo#
#color(white)(aaaa)##x-5##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aa)##+##color(white)(aaaaaa)##+#
#color(white)(aaaa)##-x+15##color(white)(aaa)##+##color(white)(aaaaa)####color(white)(aa)##+##color(white)(aaa)##0##color(white)(aa)##-#
#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##-##color(white)(aaa)##||##color(white)(aaa)##+##color(white)(aaa)##0##color(white)(aa)##-#
Therefore,
#f(x)>=0# when #x in (5,15]#
graph{(2x)/(x-5)-3 [-25.56, 32.17, -14.65, 14.22]}