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Math Help - Equation involving modulus -- help..

  1. #1
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    Equation involving modulus -- help..

    I need help with solving this equation:

    |x - 1| = |x| + 1

    I know that |x| = x when x >= 0 and |x| = -x when x < 0 but I am not sure how to apply it here.

    When |x - 1| >= 0 , i.e. when x > 1, x - 1 = x + 1 => 0

    Probably my reasoning is wrong thus far so I need help...
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  2. #2
    MHF Contributor
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    Hi

    I can see 2 possibilities to answer
    1) The equation is equivalent to |x-1| - |x| = 1
    You can build a table like this
    --------------------oo------------0-----------1------------+oo
    -------|x-1|-------------1-x----------1-x----------x-1
    --------|x|----------------x------------x-----------x
    ----|x-1|-|x|------------1------------1-2x----------1

    Now you can easily conclude

    2) An alternative way is to square the equation
    |x-1| = |x| + 1
    |x-1| = (|x| + 1)
    x-2x+1 = x+2|x|+1 because |x|=x
    -x = |x|
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  3. #3
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    Hello, struck!

    Solve: . |x-1| \:=\:|x| + 1
    I'll do it the Long Way . . .


    Since both x-1 and x can be positive or negative,
    . . there are four cases to consider:


    [1] Both positive: . x-1 > 0,\;\;x> 0 \quad\Rightarrow\quad x > 1
    . . .Then we have: . x - 1 \:=\:x + 1 \quad\Rightarrow\quad 0 \:=\:2 . . . impossible

    [2] Positive-negative: . x - 1 > 0,\;\;x < 0
    . . .But this means: . (x > 1) \:\wedge\:(x < 0) . . . impossible

    [3] Negative-positive: . x-1 < 0,\;\;x > 0
    . . .Then we have: . 1 - x \:=\:x + 1 \quad\Rightarrow\quad x = 0

    [4] Negative-negative: . x-1 < 0,\;\;x < 0 \quad\Rightarrow\quad x < 0
    . . .Then we have: . 1 - x \:=\:-x + 1 \quad\Rightarrow\quad 0 \:=\:0 . . . always true.


    The inequality is satisfied in cases [3] and [4]

    The solution is: .  x \:\leq \:0


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    If we graph the two functions, we can "eyeball" the solution.


    The graph of y = |x| is a \vee, vertex at (0,0). .[1]


    The graph of . y \:=\:|x-1| . is [1], moved one unit to the right.
    Code:
                 \|
                 1*           /
                  |\         /
                  | \       /
                  |  \     /
                  |   \   /
                  |    \ /
          - - - - + - - * - - - -
                  |     1

    The graph of . y \:=\:|x|+1 .is [1], moved one unit upward.
    Code:
                  |
            \     |     /
             \    |    /
              \   |   /
               \  |  /
                \ | /
                 \|/
                 1*
                  |
                  |
                  |
                  |
                  |
          - - - - + - - - - - - -
                  |

    Sketch them on the same coordinate system
    . . and we can see the solotion: . x \:\leq\:0

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