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Math Help - Absolute Values

  1. #1
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    Post Absolute Values

    Can anyone please hammer out this idea of absolute value for me. Please.

    y = |2x - 3|

    using

    |x| = \begin{cases}x \ldots & x\ge 0 \\\\<br />
 -x \ldots & x \le 0\end{cases}

    I know its relatively simple but I just cant seem to get the ideas into my head. I know what it looks like graphically using a graphing utility but once I come to actually solve it by hand, I have no idea what to do

    Please help
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  2. #2
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    To start, first just substitute 2x - 3 for each occurance of "x" in the definition. Then solve for x. I.e, |2x-3| = 2x-3 when 2x-3 > 0, |2x-3| = -(2x - 3) when 2x - 3 < 0.
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  3. #3
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    I love this question. It is at the very base of a great deal of mathematics.

    The absolute value of x, \left| x \right| is the distance x is from 0.

    So we can see -3 is three units from zero so  \left| -3 \right|=3 .

    So look at this question: Solve \left| {x + 6} \right| = 2.
    O.K. What number is two units from zero? Well of course that answer is \pm 2.
    So solve this system x+6=2\text{ or }x+6=-2.
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  4. #4
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    Ok is this right?

    2x - 3 \ge 0

    x \ge \frac{3}{2}

    and

    -2x - 3 \le 0

    x \ge \frac{-3}{2}

    I've done something wrong haven't I, because that is not the same output my graphing utility is showing me...
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  5. #5
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    Quote Originally Posted by Plato View Post
    I love this question. It is at the very base of a great deal of mathematics.

    The absolute value of x, \left| x \right| is the distance x is from 0.

    So we can see -3 is three units from zero so  \left| -3 \right|=3 .

    So look at this question: Solve \left| {x + 6} \right| = 2.
    O.K. What number is two units from zero? Well of course that answer is \pm 2.
    So solve this system x+6=2\text{ or }x+6=-2.
    Oh my, that's an explanation of a true teacher hehe, okay so looking at the example you have given me,

    does that mean my equation should be

    2x = 3

    and

    2x = -3

    so  x = \pm \Big(\frac{3}{2}\Big)

    so how do I interpret that as:

    x \le \frac{3}{2}

    x\ge \frac{3}{2}
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  6. #6
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    Quote Originally Posted by dwat View Post
    Ok is this right?

    2x - 3 \ge 0

    x \ge \frac{3}{2}

    and

    -2x - 3 \le 0

    x \ge \frac{-3}{2}

    I've done something wrong haven't I, because that is not the same output my graphing utility is showing me...
    Well, -(2x - 3) is not -2x - 3. You forgot to change the sign of -3.

    Also, getting back to your original question, there is nothing to "solve" in this equation. Is this just a question about "how do I graph this equation"?
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  7. #7
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    Quote Originally Posted by dwat View Post

    so how do I interpret that as:

    x \le \frac{3}{2}

    x\ge \frac{3}{2}

    2x - 3 will be positive when x > 3/2
    2x - 3 will be negative when x < 3/2
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  8. #8
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    Quote Originally Posted by QM deFuturo View Post
    Well, -(2x - 3) is not -2x - 3. You forgot to change the sign of -3.

    Also, getting back to your original question, there is nothing to "solve" in this equation. Is this just a question about "how do I graph this equation"?
    ohh thanks for correcting my errors.
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  9. #9
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    I think I've got the idea of this now....

    thank you all for your help

    All deserve plus rep from me
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