Absolute value equation problem...can you check my work?

Discussion in 'Undergraduate Math' started by Will H, Apr 19, 2007.

  1. Will H

    Will H Guest

    |1/2x - 1/3| = 1/5|-x|

    Solve the equation.

    For real numbers y and c such that c> 0.
    |y| =c iff y = c OR y = -c

    Therefore...

    1/2x -1/3 = 1/5|-x| OR 1/2x -1/3 = -1/5|-x|
    1/2x-1/5|-x|=1/3 1/2x+1/5|-x|=1/3
    15/30x - 6/30|-x|=10/30 15/30x + 6/30|-x| = 10/30
    9/30x = 10/30 21/30x =10/30
    x=10/9 x= 10/21

    Is this correct?
     
    Will H, Apr 19, 2007
    #1
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  2. Will H

    Virgil Guest

    If you substitute 10/9 or 10/21 back in the original equation in place
    of x, does it balance?
     
    Virgil, Apr 19, 2007
    #2
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  3. I assume "1/2x" means "(1/2)x" and not 1/(2x), given what you write
    later. Please try to be unambiguous.
    You can do better here. See below.

    Of course, this step in isolation is incorrect. You cannot go from the
    line immediately previous to the bottom, because you have to consider
    two cases again: whether x>=0 or x<0. As it happens, though, the four
    cases you get (one from each column) only amount to the two cases you
    describe.
    Looks like. But you can do better:

    If (1/2)x - (1/3) = (1/5)|-x|, then you know that (1/2)x-(1/3) is
    positive. That means that (1/2)x - (1/3) > 0, so x>2/3. In particular,
    |-x|=x, so you can immediately replace the equation with

    (1/2)x - (1/3) = (1/5)x

    and solve this equaltion, which gives

    (1/2)x - (1/5)x = (1/3)

    (3/10)x = (1/3)
    x= 10/9.

    Now, suppose that (1/2)x - (1/3) = -(1/5)|-x|. That means that
    (1/2)x - (1/3) is negative. Hence

    (1/2)x < (1/3)
    x < (2/3).

    Now, you cannot figure out from this whether |-x|=x or |-x|=-x.

    If 0<x<2/3, then |-x|=x, and you get

    (1/2)x - (1/3) = -(1/5) x
    (1/2)x + (1/5)x = (1/3)
    (7/10)x = (1/3)
    x = 10/21.

    On the other hand, if x<0, then |-x|=-x, so you get

    (1/2)x - (1/3) = -(1/5)(-x)

    however, this leads to x=10/9, which is incompatible with the
    assumption that x<0, so this case cannot occur. So you are left with
    only the two solutions you had.

    --
    ======================================================================
    "It's not denial. I'm just very selective about
    what I accept as reality."
    --- Calvin ("Calvin and Hobbes" by Bill Watterson)
    ======================================================================

    Arturo Magidin
    magidin-at-member-ams-org
     
    Arturo Magidin, Apr 19, 2007
    #3
  4. That would show Will that the answers he got were correct. But it
    would not ensure him that he has managed to find ALL solutions.

    --
    ======================================================================
    "It's not denial. I'm just very selective about
    what I accept as reality."
    --- Calvin ("Calvin and Hobbes" by Bill Watterson)
    ======================================================================

    Arturo Magidin
    magidin-at-member-ams-org
     
    Arturo Magidin, Apr 19, 2007
    #4
  5. Will H

    Greg Neill Guest

    Rather than struggling with the absolute values,
    why not recognize that

    |x/2 - 1/3| = (1/5)|-x| assuming x is real

    implies (x/2 - 1/3)^2 = (1/25)x^2

    Solve in the usual manner for x.
     
    Greg Neill, Apr 19, 2007
    #5
  6. Will H

    Mike Terry Guest

    This follows from your statement that I've labeled [1],
    since 1/5|-x| > 0.

    However, you've still got a mix of x and |x| in each column, so this is not
    best way to proceed. See suggestion below...
    This is wrong, because you have just replaced |-x| with x, which is only
    valid if x<0. You could proceed by examining the separate cases x<0, x>=0,
    so you would then have 4 columns to consider, but you could have avoided
    this pickle with a better starting direction...
    A better approach would be to try and eliminate all the absolute values in
    one go. You can use either of the following:

    For real numbers y and z
    |y| = |z| iff y = z OR y = -z [2] or
    |y| = |z| iff y^2 = z^2 [3]

    Your initial equation can very easily be put in this form, and using either
    of [2] or [3] will set you off down a path where the equations you are
    trying to solve do not have any remaining absolute value functions.

    If you use [2] you will have two columns to follow through. (In fact this
    will be in practice the steps you used above.) This will lead you to two
    (possible) solutions by solving two linear equations.

    Alternatively, using [3] you will have one column, but you will need to
    solve a quadratic equation, which will again lead to two possible solutions.
    Is one quadratic easier than two linear equations? (You decide! :)

    In either case, you will then only have succeeded in proving that a solution
    to the original equation must be one of the two values you have obtained.
    Next you will need to determine whether either of these two values *in fact*
    satisfy the original equation. You do this simply by substituting the
    values back into the original equation and checking if they work. (This
    step was missing from your solution above!)

    Regards,
    Mike.
     
    Mike Terry, Apr 19, 2007
    #6
  7. Will H

    Mike Terry Guest

    Upon reflection, all the steps in the solution are reversible, due to the
    "iff"s appearing in [1], [2], and [3]. So solutions to the final equation
    must also be solutions to the original equation, and there is no requirement
    to confirm these solutions by substituting back into the original equation.

    However, this is good practice in any case just to confirm you've not
    slipped up somewhere with a silly mistake...
     
    Mike Terry, Apr 19, 2007
    #7
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