The inequality is

#(5x-8)/(x-5)>=2#

We cannot do crossing over, so

#(5x-8)/(x-5)-2>=0#

Putting on the same denominator

#(5x-8-2(x-5))/(x-5)>=0#

#(5x-8-2x+10)/(x-5)>=0#

#(3x+2)/(x-5)>=0#

Let #f(x)=(3x+2)/(x-5)#

Build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaa)##-2/3##color(white)(aaaaaaa)##5##color(white)(aaaa)##+oo#

#color(white)(aaaa)##3x+2##color(white)(aaaaa)##-##color(white)(aaa)##0##color(white)(aaa)##+##color(white)(aaaaa)##+#

#color(white)(aaaa)##x-5##color(white)(aaaaaa)##-##color(white)(aaa)####color(white)(aaaa)##-##color(white)(aa)##||##color(white)(aa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##+##color(white)(aaa)##0##color(white)(aaa)##-##color(white)(aa)##||##color(white)(aa)##+#

Therefore,

#f(x)>=0# when #x in (-oo,-2/3] uu(5, +oo)# in interval notation

and #x<=-2/3# and #x>5# as an inequality

graph{(5x-8)/(x-5)-2 [-29.8, 35.14, -12.44, 20.05]}