# Solving rational inequalities with an unknown using cases

#### hanniball

##### New member
So the question is,
For what values of x is 1/(1 + x) > -1?
Now, one way to do this is to write the inequality relative to zero, and then find the behavior of the graph relative to the zeroes of the numerator and denominator:

1/(1 + x) > -1 => 1/(1+x) + 1> 0 => 1/(1+x) + (1+x)/(1+x) > 0
=> (2+x)(1+x) > 0, and we simply find what happens when x < -2, -2 < x < -1, and x > -1.

What I don't understand is why I can't answer this by multiplying through by 1 +x if I consider both a) 1 +x >0 and b) 1+x < 0.

Here's the attempt:

Case a): 1 + x > 0 => x > -1, so
1/(1+x) > -1
=> (1+x) (1)/(1+x) > -1 (1+x) (we can multiply both sides by 1+x because we assume 1+x is positive)
=> x > -2.

Hence, the inequality is true when x > -2 and x > -1--so x > -1.

Case b): 1+x < 0 => x < -1, so
1/(1+x) > -1
=> (1+x) 1/(1+x) < -1 (1+x) (we assume 1+x is negative, so we reverse the sign)
=> 1 < 1 + x
=> x > 0.

Hence, the inequality is true when x < -1 and x > 0--this is never true.

So the real answer is x > -1 or x <-2, and I think the reason my attempt at multiplying through by an unknown doesn't work is that, in my assumption x in (-1, infinity) clearly includes both negative and positive values, so we can't say what the value of x will be when 1 + x > 0, and that leads the answer to be wrong.

But I'm not 100 percent certain it's impossible, and would like some help!

#### tkhunny

##### Moderator
Staff member
You'redoing fine. You just need a little fine tuning in your conclusions.

You CAN do what you are doing. It must be done with care. Watch the"AND" and "OR" very carefully.

You have correctly concluded that EITHER (1) 1+x > 0 OR (2) 1+x < 0

(1) When you say (1+x) > 0, or x > -1, you commit to taking NO othersolution as long as that assumption holds.
When you subsequently conclude that x > -2 you must IMMEDIATELY discardanything that violates you first promise.
Mathematically, your solution set is x > -1 AND x > -2 which is validonly for x > -1

(2) When yousay (1+x) < 0, or x < -1, you commit to taking NO other solution as longas that assumption holds.
When you subsequently conclude that x < -2 you must IMMEDIATELY discard anythingthat violates you first promise.
Mathematically, your solution set is x < -1 AND x < -2 which is valid onlyfor x < -2

Having made conclusions in (1) and (2), we new accumulate the result, x > -1OR x < -2

Just for fun, let's try it out.

x = -3 should work. ==> 1/(1-3) =-1/2 > -1 CHECK

x =-3/2 should not work. ==> 1/(1-3/2) = 1/(-1/2) = -2 < -1 AS EXPECTED.

x = 0 should work ==> 1/(1-0) 1 > -1 CHECK

Is it AND or is it OR. You must be VERY CLEAR with this.

#### stapel

##### Super Moderator
Staff member
So the question is,

For what values of x is 1/(1 + x) > -1?
Now, one way to do this is to write the inequality relative to zero, and then find the behavior of the graph relative to the zeroes of the numerator and denominator...

What I don't understand is why I can't answer this by multiplying through by 1 +x if I consider both a) 1 +x >0 and b) 1+x < 0....
By converting the inequality to having zero on one side, you're kind of taking care of those "cases", but more clearly. Also, things are quickly going to become more complicated, with many more than just one factor in the denominator. They start you off with the easy case, so you learn the method and can adapt it to messier things.