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Math Help - Solving for x when x's cancel out?

  1. #1
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    Solving for x when x's cancel out?

    (I also posted this is the Algebra forum, but got a lot of views, but no replies, so I'm trying here instead...)

    I'm sure the answer to this is incredibly obvious, but something is slipping my mind:

    │x-3│= x + 2

    if you move the x out of the absolute value braket, the x's cancel... so how do you solve it?
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  2. #2
    Super Member wingless's Avatar
    Joined
    Dec 2007
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    Istanbul
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    |x-3|=x+2

    There are 3 possible cases.

    Case 1:
    x-3 > 0
    x > 3
    If it's positive, we can say that x-3 = x+2 which doesn't have a root.

    Case 2:
    x-3<0
    x<3
    It's negative, we can rewrite it as x-3 = -(x+2)
    2x = 1
    \boxed{x = \frac{1}{2}} --! We found a root

    Case 3:
    x-3 = 0
    x=3
    For x=3, |x-3|\neq x+2 so we don't have roots in this region.

    Conclusion: The equation has only the root x=\frac{1}{2}.


    Graph:
    We can rewrite the equation as |x-3|-x-2=0 and plot this function.

    y=|x-3|-x-2
    See where this graph crosses the x axis.

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