First divide both sides by #3# to get:
#1/(x+3) > 1/(x-2)#
We can ignore the values #x = -3# and #x = 2# since one or the other side of the inequality is undefined for these values. That leaves #(-oo, -3) uu (-3, 2) uu (2, oo)#
It is probably easiest to split this into cases as follows:
Case #bb(x in (-oo, -3))#
#(x+3) < 0# and #(x-2) < 0# so #(x+3)(x-2) > 0#
Multiply both sides of the inequality by #(x+3)(x-2)# to get:
#x-2 > x+3#
Subtract #x# from both sides to get #-2 > 3#, which is false.
So there is no solution for #x in (-oo, -3)#
Case #bb(x in (-3, 2))#
#(x+3) > 0# and #(x-2) < 0# so #(x+3)(x-2) < 0#
Multiply both sides of the inequality by #(x+3)(x-2)# and reverse the inequality (since we're multiplying by a negative value) to get:
#x-2 < x+3#
Subtract #x# from both sides to get #-2 < 3#, which is true.
So the inequality is true for all #x in (-3, 2)#
Case #bb(x in (2, oo))#
#(x+3) > 0# and #(x-2) > 0# so #(x+3)(x-2) > 0#
Multiply both sides of the inequality by #(x+3)(x-2)# to get:
#x-2 > x+3#
Subtract #x# from both sides to get #-2 > 3#, which is false.
So there is no solution with #x in (2, oo)#
