How do you solve #(x+3)/(x+4)<0#?

1 Answer
Aug 7, 2016

Answer:

#x in (-4, -3)#

Explanation:

There are three intervals within which the sign of the quotient does not change:

#(-oo, -4)#

#(-4, -3)#

#(-3, oo)#

We do not need to consider the points #x=-4# and #x=-3# as the former gives an undefined quotient and the latter a zero quotient. So they are not in the solution set.

#color(white)()#
Case #bb(x in (-oo, -4))#

Both #(x+3) < 0# and #(x+4) < 0#, so the quotient is positive.

So this interval is not part of the solution set.

#color(white)()#
Case #bb(x in (-4, -3))#

In this interval, #(x+3) < 0# but #(x+4) > 0#, so the quotient is negative.

So this interval is part of the solution set.

#color(white)()#
Case #bb(x in (-3, oo))#

In this interval, #(x+3) > 0# and #(x+4) > 0#, so the quotient is positive.

So this interval is not part of the solution set.

#color(white)()#
Conclusion

The solution set is the interval #(-4, -3)#