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Author Topic: Modulus Inequalities Question  (Read 2142 times)

Neko Desu

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Modulus Inequalities Question
« on: February 06, 2012, 10:29:38 pm »

What do you do when all the conditions you arrive at in a modulus inequality equation contradict each other? This is the question I'm talking about:

|3x - 2| >= |2x + 3|

The solutions to this inequality are:

1. (Take positive on both sides) x >= 5
2. (Left positive, right negative) x >= -1/5
3. (Left negative, right positive) x <= -1/5
4. (Both negative) x <= 5

Yeah, I'm kind of stuck

inb4 Kwiftee?
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Venser

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Re: Modulus Inequalities Question
« Reply #1 on: February 06, 2012, 11:00:04 pm »

I'm not bothering to check your work, but assuming it's right, then the statement is simply false. Nothing else can be done with it.
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Pokota

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Re: Modulus Inequalities Question
« Reply #2 on: February 06, 2012, 11:29:39 pm »

Modulus inequalities?

I see no modulus here. Perhaps you should start with the starting conditions rather than just showing us what your results are (you neoscientist, you :) )

Anyway, for the absolute value of 3x - 2 to be greater than or equal to the absolute value of 2x + 3 requires some funky shit. Of course, graphically an absolute value of a linear equation can have up to two solutions, and inequalities weren't really my forte, so you're probably looking for either all real numbers or no solution in this instance - which really wouldn't surprise me.

EDIT: check your negatives, I got a completely different set of results when I worked out the four scenarios. Disregard what's crossed out, I'm an idiot. What would this look like if you tried to graph it? I always handle this sort of crap better with visual aids.
« Last Edit: February 06, 2012, 11:42:44 pm by Ponchichi »
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Neko Desu

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Re: Modulus Inequalities Question
« Reply #3 on: February 07, 2012, 12:24:25 am »

This is pretty messed up. A graph shows that only solutions 1. and 3. are correct (x >= 5 and x <= -1/5). This doesn't explain why 2. and 4. are wrong, though, and I have indeed checked over my math multiple times.
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Kwiftee

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Re: Modulus Inequalities Question
« Reply #4 on: February 07, 2012, 06:24:45 am »

The issue is that your final solutions don't actually contradict each other; you're just getting mixed up between AND and OR statements. You're assuming that all the statements are 'and's, but they aren't necessarily.

Firstly, assume (as you did) that both sides are positive (or zero). Then, you have:
3x-2 >= 0 AND 2x+3 >= 0 AND 3x - 2 >= 2x + 3
simplifying to x>=2/3 AND x>=-3/2 AND x>=5.
which simplifies to x >= 5.

But, as you know, that assumption we made isn't the only case. Another case is that left hand side is positive, RHS negative.
3x-2 >= 0 AND 2x+3 <= 0 AND 3x - 2 >= -2x - 3
x >= 2/3 AND x <= -3/2 AND x >= -1/5
Note that this leads to a contradiction; there's no values of x which satisfy all three of those statements. So, the second assumption was just never true.

as for the other two assumptions:
LHS<=0, RHS>=0
gives 3x-2 <= 0 AND 2x+3 >= 0 AND 2 - 3x >= 2x + 3
x <= 2/3 AND x >= -3/2 AND x <= -1/5
hence, -3/2 <= x <= -1/5

LHS<=0, RHS<=0
gives 3x-2 <= 0 AND 2x+3 <= 0 AND 2 - 3x >= -2x - 3
x <= 2/3 AND x <= -3/2 AND x <= 5.
hence, x <= -3/2

Now, since all of these were assumptions, any one of them could lead to a solution. So now, we use OR statements, giving us our answer of  x >= 5 or -3/2 <= x <= -1/5 or x <= -3/2

...simplifying to x >= 5 or x <= -1/5

When to use AND and when to use OR requires a bit of intuition sometimes; I'm not quite sure how to explain it. Plus I remember there was a problem I once did where this way of thinking didn't work out; something involving the range of arccos(x2) from memory, but I don't remember.

-

At any rate this isn't the simplest way of doing the above problem; it'd be much easier to just graph both of them and do it visually. Or failing that, you can recognise that both y = |3x - 2| and y = |2x + 3| are continuous functions, solve them equal to each other which tells you at what points they cross and just use intuition once again to work out which one is on top at a given value of x.
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