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
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.
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
:)
The infinity symbol at this forum can be typed as follows:
& i n f i n ;
but without the spaces between the characters.
To solve the inequality |x|/|x+2|<2, you have used the case method, which is a valid approach. Let's go through each case and check your work:
Case 1: x >= 0 and x+2 > 0
You correctly simplified the inequality step by step:
x/(x+2) < 2
x < 2(x+2)
x < 2x+4
-x < 4
x > -4
In this case, since the assumptions were x >= 0 and x+2 > 0, we can conclude that x must be greater than or equal to zero. So, x >= 0 is the solution for this case. You've done this part correctly.
Case 2: x >= 0 and x+2 < 0
You noticed that this case is not possible because there are no numbers that are both greater than or equal to zero and less than -2. Hence, there is no solution in this case.
Case 3: x < 0 and x+2 > 0
You correctly simplified the inequality:
-x/(x+2) < 2
-x < 2(x+2)
-x < 2x+4
-3x < 4
x > -4/3
In this case, the assumption was x < 0 and x+2 > 0, which implies x > -2. Comparing the constraints, we find that x must be greater than -4/3 but less than 0. Your solution for this case is correct.
Case 4: x < 0 and x+2 < 0
You correctly simplified the inequality:
-x/-(x+2) < 2
-x < -2(x+2)
-x < -2x-4
x < -4
In this case, the assumption was x < 0 and x+2 < 0, which implies x < -2. Comparing the constraints, we find that x must be less than -4. Your solution for this case is correct.
Combining all the solutions from all the cases, we have x > -4/3 and x < -4. This can also be written as the solution set {x | x < -4 or x > -4/3}.
Overall, your work is correct, and you have correctly solved the inequality using the case method.