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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?

    Hi there, 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 angel.white's Avatar
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    Quote Originally Posted by Lisa87654 View Post
    Hi there, 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?
    \mid x-3 \mid = x + 2

    You can see that the two sides are going to have alternative signs, meaning either (x-3) will be negative or (x+2) will be negative so cancel the absolute value, and make either side negative:
    -(x-3)= x + 2

    Distribute the negative sign
    -x+3= x + 2

    Subtract (x+3) from both sides
    -x+3 -(x+3)= x + 2 -(x+3)

    Distribute the negatives
    -x+3 -x -3= x + 2 -x-3

    Simplify
    -2x= -1

    Divide both sides by -2
    x= \frac {-1}{-2}

    Simplify
    x= \frac 12
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  3. #3
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    Quote Originally Posted by Lisa87654 View Post
    Hi there, 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?
    Use the definition of the absolute value:

    |x-3| = \left\{\begin{array}{l}x-3 ~,~ x-3 \geq 0~\implies~x \geq 3\\ -(x-3) ~,~ x-3 < 0~\implies~x < 3 \end{array} \right.

    That means you get 2 different equations:

    x-3 = x+2~,~x \geq 3 This equation has no real roots.

    -(x-3) = x+2~,~x < 3~\implies~ x = \frac12
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