We cannot do crossing over
Let's simplify the inequality
#3/(x-2)<5/(x+2)#
#3/(x-2)-5/(x+2)<0#
#(3(x+2)-5(x-2))/((x-2)(x+2))<0#
#(3x+6-5x+10)/((x-2)(x+2))<0#
#(16-2x)/((x-2)(x+2))<0#
#(2(8-x))/((x-2)(x+2))<0#
Let #f(x)=(2(8-x))/((x-2)(x+2))#
We, now, build the sign chart
#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaaaaaaa)##2##color(white)(aaaaaa)##8##color(white)(aaaaaaa)##+oo#
#color(white)(aaaa)##x+2##color(white)(aaaa)##-##color(white)(aaa)##||##color(white)(aaaa)##+##color(white)(aaa)##||##color(white)(aaa)##+##color(white)(aaaa)##+#
#color(white)(aaaa)##x-2##color(white)(aaaa)##-##color(white)(aaa)##||##color(white)(aaaa)##-##color(white)(aaa)##||##color(white)(aaa)##+##color(white)(aaaa)##+#
#color(white)(aaaa)##8-x##color(white)(aaaa)##+##color(white)(aaa)##||##color(white)(aaaa)##+##color(white)(aaa)##||##color(white)(aaa)##+##color(white)(aaaa)##-#
#color(white)(aaaa)##f(x)##color(white)(aaaaa)##+##color(white)(aaa)##||##color(white)(aaaa)##-##color(white)(aaa)##||##color(white)(aaa)##+##color(white)(aaaa)##-#
Therefore,
#f(x)<0# when #x in ]-2,2 [ uu ]8, +oo [#