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Thread: Absolute Values

  1. #1
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    Post Absolute Values

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

    $\displaystyle y = |2x - 3|$

    using

    $\displaystyle |x| = \begin{cases}x \ldots & x\ge 0 \\\\
    -x \ldots & x \le 0\end{cases}$

    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, $\displaystyle \left| x \right|$ is the distance $\displaystyle x$ is from $\displaystyle 0$.

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

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

    $\displaystyle 2x - 3 \ge 0$

    $\displaystyle x \ge \frac{3}{2}$

    and

    $\displaystyle -2x - 3 \le 0$

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

    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, $\displaystyle \left| x \right|$ is the distance $\displaystyle x$ is from $\displaystyle 0$.

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

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

    $\displaystyle 2x = 3$

    and

    $\displaystyle 2x = -3$

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

    so how do I interpret that as:

    $\displaystyle x \le \frac{3}{2}$

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

    $\displaystyle 2x - 3 \ge 0$

    $\displaystyle x \ge \frac{3}{2}$

    and

    $\displaystyle -2x - 3 \le 0$

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

    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:

    $\displaystyle x \le \frac{3}{2}$

    $\displaystyle x\ge \frac{3}{2}$

    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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