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Math Help - solution set/ interval notation

  1. #1
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    solution set/ interval notation

    I tried the latex thing and it still won't work...

    [tex]\frac{\pi^2}{6}[tex]

    The problem I need help with is this, all steps please so I can follow along and see how you got to the answer.

    1)
    6-x/12 > 1/4 or 6-x/12 < -1/6

    2) Absolute Problem

    lwl+8=2

    For this I am thinking it has one answer because isn't it true if the number on the right side of the equal sign is negative the negative equation is false?

    w+8=2 w+8=-2
    w=6 w=-10

    Now seems both do not work... is this correct?
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  2. #2
    Junior Member Freaky-Person's Avatar
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    1) the "math" in the brackets should be capitol and you need a space after pi so it can register that that's a separate call.

    and you need to close it with [\MATH ] (minus the space)

    2) I don't know what you mean with the "or" but here goes:

    6 - \frac{x}{12} > \frac{1}{4}<br />
\Rightarrow 72 - x > 3<br />
\Rightarrow -x > -69<br />
\Rightarrow x < 69


    6 - \frac{x}{12} > -\frac{1}{6}<br />
\Rightarrow 72 - x > -2<br />
\Rightarrow -x > -74<br />
\Rightarrow x < 74

    For the last question there is no way for that to be true. Because if you move things around you get |w| = -6 and an absolute value's not going to be negative.

    But.. hmm..

    (case 1: w > 0 : fails, by above
    case 2: w < 0 : -w + 8 = 2 \Rightarrow w = 6)

    I'm not sure about what I wrote in there, I think that was from graphing and when you had to find the intervals for which the equation worked or something. I don't know if it still applies but you can take it and show it to someone who knows what they're doing, unlike me obviously. :P
    Last edited by Freaky-Person; September 26th 2009 at 04:20 PM.
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  3. #3
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    The "Or" meant that it is a solution set and either equality is the answer of both equalities.
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  4. #4
    Junior Member Freaky-Person's Avatar
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    Quote Originally Posted by pippers View Post
    The "Or" meant that it is a solution set and either equality is the answer of both equalities.
    oh, so.. then... x < 69, because all numbers < 69 are in both sets (as long as it's less than 69 then both equations work)?
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