TrueBudSmoker
2008-09-12, 04:05
This is fucking pissing me off. The teacher didnt say how to to this type of question and the normal method of solving the positive and negative case isnt working. When I solve the positive case there is no solution, and the book says the answer to the negative case in the answer to the full question. But the teacher said you have to take the intersection between the 2 casses for "x is less then" questions. I'm starting to think my teacher doesn't know what he is talking about.
|x/(2+x)| < 1
(absolute value less than 1)
answer to negative case:
-x/(2+x) < 0
-x < 2 + x
-2x < 2
x > -1
Why would the answer to the negative case be the full answer to the question when he said you need to take the intersection? That fucking cunt I'm going to call him out in front of the class tomorrow.
Edit: Never mind I figured it out, that son of a bitch shouldn't generalize like that though. In fact, he is a fucking lier.
Optionryder420
2008-09-13, 16:49
I thought this was going to be a hard problem...
EDIT: The way you did that was kind of strange to me... usually I'll just take an absolute value with a greater than or less than symbol, whichever it is, and make both.
For instance, here's how I solved your problem.
|x/(2+x)| < 1
-1<x/(2+x)<1
-1(2+x)<x<1(2+x)
-2+-x<x<2+x
Then, I split them into two different problems...
-2+-x<x
2+x>x
Since 2>0 obviously, the positive problem I just get rid of.
Then I get -2<2x
Divide by two.
-1<x
There's your solution.
Me, I'm good at math, so typically I can see problems and understand them just from seeing them. Looking at that problem just for a second I had the answer at basically any number as long as it wasn't -2 or greater since that would equal >1.
Why not, before you attempt to do a problem, throw some random numbers in there and see if you can get a few answers that make sense, that way once you work out the problem you'll know if it's correct or not.
Basically what I'm saying is, check your problem before solving it. At least, I always do. Write it out if you absolutely have to and there are many numbers which work well for checking problems.
One is always a good start to see how the problem works out. For instance, put one into that problem.
1/3<1 is correct... at that point I'll throw a few more numbers into there closer in range to one. Usually I'll go both ways too, so I'll put a 2 and a -1 in there (I usually don't test zero, and honestly one isn't a good way to check larger problems, this is a smaller number so smaller numbers will work well enough)
Using 2, you're going to get 1/2.
Using -1 you get 1.
So it obviously isn't -1, using -2 is undefined, and using -3 would equal 3, so basically you can see from this pattern that using larger negative numbers isn't going to work.
Sorry for ranting, I just strangely can see problems in my head and this is how I solve them.