Rewrite the equation
#(2t+7)/(t-4)>=3#
#(2t+7)/(t-4)-3>=0#
#((2t+7)-3(t-4))/(t-4)>=0#
#(2t+7-3t+12)/(t-4)>=0#
#(-t+19)/(t-4)>=0#
Let #f(t)=(-t+19)/(t-4)#
We can construct the sign chart
#color(white)(aaaa)##t##color(white)(aaaa)##-oo##color(white)(aaaaaaa)##4##color(white)(aaaaaaaa)##19##color(white)(aaaaaa)##+oo#
#color(white)(aaaa)##t-4##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaaa)##+#
#color(white)(aaaa)##19-t##color(white)(aaaa)##+##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaaa)##-#
#color(white)(aaaa)##f(t)##color(white)(aaaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaaa)##-#
Therefore,
#f(t)>=0# when #t in ]4, 19]#