Question

Here's a question I'm having trouble with:

|x|/|x+2|<2

Solve for x.

I'm trying to use the case method to solve this problem. Basically I just need confirmation that I did this correctly. I'm doing this through distance learning so I can't ask a teacher until Monday.

Case 1:

x is greater than or equal to zero, and x+2 is greater than zero.

x/(x+2)<2
x<2(x+2)
x<2x+4
-x<4
x>-4

Now here's where I get confused... since the assumptions for this case were x>=0 and x+2>0, x must be greater than or equal to zero. When I solved for x in this case, I got x>-4. Reconciling the constraints leaves me back at x>=0. Did I do that correctly?

Case 2:

x is greater than or equal to zero, x+2 is less than xero (x is less than -2.) This isn't possible since there are no numbers that are both greater than or equal to zero and less than -2.

Case 3:

x is less than zero, and x+2 is greater than zero (x>-2)

-x/(x+2)<2
-x<2(x+2)
-x<2x+4
-3x<4
x>-4/3

When I reconcile the constraints, I end up with x is greater than -4/3 but less than zero. Again I am not sure if I did that correctly.

Case 4:

x is less than zero, and x+2 is less than zero (x<-2.)

-x/-(x+2)<2
-x<-2(x+2)
-x<-2x-4
x<-4

Reconciling the constraints leaves me with x<-4.

When I combine all of the solutions from all of the cases, I get a solution set of {x|x<-4 or x>-4/3}

This seems to be correct when I enter test points, but I was wondering if someone could check my work and confirm that I've done it correctly.

Thanks
Randy

Answer 1

Your approach is impeccable.

There is only one little glitch for which you have to watch out.

In case one, you assumed x>0, and x+2>0, and you obtained x>-4, which is inconsistent with your assumption of x>0.
So it should have been revised to x>0.

If you submitted your answer as
(-∞,-4)∪(-4/3,∞)
you would have the correct answer, since [0,∞) from case 1 is absorbed in case 3.

If you had included x>-4 (from case 1), the answer would have been (-∞,∞) which is incorrect.

Answer 2

Thanks for you help. I would have expressed my answer in the way you suggested, but I wasn't sure how to type the infinity symbol. I will watch out for the glitch you mentioned.

Thanks again,
Randy

Answer 3

:)
The infinity symbol at this forum can be typed as follows:
& i n f i n ;
but without the spaces between the characters.