How do you solve #(-x-3)/(x+2)<=0#?

1 Answer
Dec 3, 2016

Answer:

The answer is #x in] -oo,-3 ] uu] -2, oo[#

Explanation:

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

As we cannot divide by #0# , the domain of #f(x)# is #D_f(x)=RR-{-2}#

To solve the inequalty, we do a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-3##color(white)(aaaa)##-2##color(white)(aaaa)##+oo#

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

#color(white)(aaaaa)##x+2##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

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

So,

#f(x<=0)# when

#x in] -oo,-3 ] uu] -2, oo[ #