Expert: Paul Klarreich - 12/7/2008

Question
I studied fractional inequalities last year, and no longer remember how to solve for them very well. I looked in your past answered questions and didn't find anything similar to what I am doing, nor was the internet helpful to me.
The problem I was given was to solve for "x" when
(x+5) divided by (x-3) is greater than or equal to zero.

I know that the denominator cannot equal zero, therefore x must be greater than 3, but beyond that I have no idea how to approach the problem, or what steps to take.

If you could please tell me how I should follow through on this problem in order to solve for all values of x I would greatly appreciate it.

Thanks!

Answer
Questioner:   Jen
Category:  Advanced Math
Private:  No
 
Subject:  Precalculus: Fractional Inequalities
Question:  I studied fractional inequalities last year, and no longer remember how to solve for them very well. I looked in your past answered questions and didn't find anything similar to what I am doing, nor was the internet helpful to me.
The problem I was given was to solve for "x" when
(x+5) divided by (x-3) is greater than or equal to zero.

>> Actually it is there, but I don't think I could find it, either.

I know that the denominator cannot equal zero, therefore x must be greater than 3, but beyond that I have no idea how to approach the problem, or what steps to take.

If you could please tell me how I should follow through on this problem in order to solve for all values of x I would greatly appreciate it.

Thanks!
..................................
Hi, Jen,

Your inequality:
x + 5
----- > 0  (yes, it said  >=; but we'll worry about that later)
x - 3

needs this thinking:

This is a fraction; it is positive if both numerator and denominator have the same sign; a fraction is positive if you have +/+ or -/-.

So you need EITHER:

x + 5 > 0  and  x - 3 > 0; which means

   x > -5 and      x > 3, which is just x > 3

OR

x + 5 < 0  and  x - 3 < 0; which means

   x < -5 and      x < 3, which is just x < -5.

SO values from (-infinity, -5) and (3,+infinity) are solutions.

What about  >= 0?  Well, if x = -5, your fraction will be zero, so that is also a solution.  Not x = 3, of course.

Finally;  (-inf,-5] union with (3,+inf) is your solution set.