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Math Help - Absolute Values

  1. #1
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    Absolute Values

    all x ∈ R that satisfy |x+1|-2|x-4|= 3

    help?
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  2. #2
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    In this problem, use cases.
    \begin{array}{l}<br />
 x \in ( - \infty , - 1]\quad  \Rightarrow \quad \left( { - x - 1} \right) - 2\left( { - x + 4} \right) = 3 \\ <br />
 x \in ( - 1,4)\quad  \Rightarrow \quad \left( {x + 1} \right) - 2\left( { - x + 4} \right) = 3 \\ <br />
 x \in [4,\infty )\quad  \Rightarrow \quad (x + 1) - 2(x - 4) = 3 \\ <br />
 \end{array}

    Now solve each of the three cases.
    But make sure the solution fits the case.
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  3. #3
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    Quote Originally Posted by Plato View Post
    In this problem, use cases.
    \begin{array}{l}<br />
 x \in ( - \infty , - 1]\quad  \Rightarrow \quad \left( { - x - 1} \right) - 2\left( { - x + 4} \right) = 3 \\ <br />
 x \in ( - 1,4)\quad  \Rightarrow \quad \left( {x + 1} \right) - 2\left( { - x + 4} \right) = 3 \\ <br />
 x \in [4,\infty )\quad  \Rightarrow \quad (x + 1) - 2(x - 4) = 3 \\ <br />
 \end{array}

    Now solve each of the three cases.
    But make sure the solution fits the case.

    Thanks, but how do we know this? I mean, how do we know that it's -infintiy, -1, and -1,4 etc?
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  4. #4
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    Quote Originally Posted by weasley74 View Post
    Thanks, but how do we know this? I mean, how do we know that it's -infintiy, -1, and -1,4 etc?
    Well you are studying absolute value.
    Are you not expected to understand how it all works?
    Go look at your text material. Go over your notes.
    You must have been given examples.
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  5. #5
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    Quote Originally Posted by Plato View Post
    \begin{array}{l}<br />
 x \in ( - \infty , - 1]\\ <br />
 x \in ( - 1,4)\\<br />
 x \in [4,\infty )\\\ <br />
 \end{array}
    is it supposed to be ( and not ] ?
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  6. #6
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    Quote Originally Posted by weasley74 View Post
    is it supposed to be ( and not ] ?
    Because  <br />
\Re  = ( - \infty , - 1] \cup ( - 1,4) \cup [4,\infty ) we have considered every real number.
    That is the whole point. But there are other ways this can be done.
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  7. #7
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    Quote Originally Posted by Plato View Post
    In this problem, use cases.
    \begin{array}{l}<br />
 x \in ( - \infty , - 1]\quad  \Rightarrow \quad \left( { - x - 1} \right) - 2\left( { - x + 4} \right) = 3 \\ <br />
 x \in ( - 1,4)\quad  \Rightarrow \quad \left( {x + 1} \right) - 2\left( { - x + 4} \right) = 3 \\ <br />
 x \in [4,\infty )\quad  \Rightarrow \quad (x + 1) - 2(x - 4) = 3 \\ <br />
 \end{array}

    Now solve each of the three cases.
    But make sure the solution fits the case.
    so for the first one, for which x=12, the solution doesn't fit the case?
    the second one, x=10/3, fits and the same for the third case when x=6 ?
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  8. #8
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    Quote Originally Posted by weasley74 View Post
    so for the first one, for which x=12, the solution doesn't fit the case?
    the second one, x=10/3, fits and the same for the third case when x=6 ?
    Seems correct

    Here's the graph for y=|x+1| - 2 |x-4|-3, do you see where the lines crosses the x axis?

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