How do solve #(3-x)/(x+5)<=0# and write the answer as a inequality and interval notation?

1 Answer
May 30, 2017

Answer:

The solution is #x in (-oo,-5) uu [3, +oo)# or
#x<-5# and #x>=3#

Explanation:

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

We build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaa)##-5##color(white)(aaaaaaa)##3##color(white)(aaaaaa)##+oo#

#color(white)(aaaa)##x+5##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##3-x##color(white)(aaaaa)##+##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##-#

Therefore,

#f(x)<=0# when #x in (-oo,-5) uu [3, +oo)# in interval notation or

#x<-5# and #x>=3# as inequality