If #x=-5# then the denominator of the rational expression is zero and the quotient is undefined. So #-5# is not part of the solution set.

#color(white)()#

**Case #bb(x < -5)#**

If #x < -5# then #(x+5) < 0#

Multiply both sides of the inequality by #(x+5)# and reverse the inequality (since #(x+5) < 0#) to get:

#x+26 <= 0#

Subtract #26# from both sides to get:

#x <= -26#

So #(-oo, -26]# is part of the solution set.

#color(white)()#

**Case #bb(x > -5)#**

If #x > -5# then #(x+5) > 0#

Multiply both sides of the inequality by #(x+5)# to get:

#x+26 >= 0#

Subtract #26# from both sides to get:

#x >= -26#

Since #x > -5# this is already true.

So #(-5, oo)# is part of the solution set.

#color(white)()#

**Conclusion**

The solution set is:

#(-oo, -26] uu (-5, oo)#