How do you solve and write the following in interval notation: #12/ (x-5) > 10 /(x+1)#?

We cannot do crossing over.

Rearrange the inequality

#12/(x-5)>10/(x+1)#

#12/(x-5)-10/(x+1)>0#

#(12(x+1)-10(x-5))/((x-5)(x+1))>0#

#(12x+12-10x+50)/((x-5)(x+1))>0#

#(2x+62)/((x-5)(x+1))>0#

#(2(x+31))/((x-5)(x+1))>0#

Let #f(x)=(2(x+31))/((x-5)(x+1))#

We can build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-31##color(white)(aaaaaa)##-1##color(white)(aaaaaaaa)##5##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+31##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aa)##||##color(white)(aa)##+#

#color(white)(aaaa)##x+1##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aa)##||##color(white)(aa)##+#

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

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

Therefore,

#f(x)>0#, when #x in (-31,-1) uu (5, +oo)#