For what x is #1 / ( x + 3 ) > 2/(x-3)#?

1 Answer
Oct 28, 2015

Answer:

#x in (-oo, -9) uu (-3, 3)#

Explanation:

Note that #x = -3# and #x = 3# are not solutions since one or the other side of the inequality is undefined for these values of #x#.

Split the remaining possibilities into cases:

Case #bb (x in (-oo, -3) uu (3, oo))#

#x+3# and #x-3# are both negative or both positive.

In either case #(x+3)(x-3) > 0# so multiply both sides of the inequality by that to get:

#x-3 > 2x+6#

Subtract #x+6# from both sides to get:

#-9 > x#

Combining with the conditions of the case, that gives us the solution:

#x in (-oo, -9)#

Case #bb (x in (-3, 3))#

#x+3 > 0# and #x-3 < 0#, so #(x+3)(x-3) < 0#, so multiply both sides of the inequality by this and reverse the inequality to get:

#x-3 < 2x+6#

Subtract #x+6# from both sides to get:

#-9 < x#

Combining with the conditions of the case, that gives us the solution:

#x in (-3, 3)#

Conclusion

Putting the two cases together, the original inequality holds for all

#x in (-oo, -9) uu (-3, 3)#