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Math Help - absolute values

  1. #1
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    absolute values

    In solving an equation for e.g : 2x-7=9 we start by using some axioms or thoerem ,like adding or subtracting or multiplying both sides of the equation the equation will not change e.t.c

    But in solving the equation : |x+2|+|x-3| =2 what axioms or definitions e.t.c can we use to start solving the equation??
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: absolute values

    Use the definition of the absolut value. You should consider 4 cases.
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    MHF Contributor MarkFL's Avatar
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    Re: absolute values

    I only find 3 cases to consider:

    x<-2

    -2\le x<3

    3\le x

    What am I missing?
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    MHF Contributor Siron's Avatar
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    Re: absolute values

    Quote Originally Posted by MarkFL2 View Post
    I only find 3 cases to consider:

    x<-2

    -2\le x<3

    3\le x

    What am I missing?
    Yes, you're right because one case isn't possible, but my intention was that psolaki would notice that.
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    MHF Contributor MarkFL's Avatar
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    Re: absolute values

    Ah, now I get what you mean...I thought I was neglecting something...my apologies for being obtuse and spoiling your intention.
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    Re: absolute values

    Since x+2\geq \vee x+2<0 and x-3\geq 0\vee x-3<0by the axiom of trichotomy,the 4 cases we should consider are the following:

    1) x+2\geq 0\wedge x-3\geq 0

    2) x+2\geq 0\wedge x-3<0

    3) x+2<0\wedge x-3\geq 0

    4) x+2<0\wedge x-3<0

    Correct??
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  7. #7
    MHF Contributor Siron's Avatar
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    Re: absolute values

    Yes, indeed! But there's one case which is not possible. Which one? Can you solve the equation now? What's your conclusion?
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    Re: absolute values

    Quote Originally Posted by psolaki View Post
    But in solving the equation : |x+2|+|x-3| =2 what axioms or definitions e.t.c can we use to start solving the equation??
    The equation |x+2|+|x-3| =2 has no solution.
    It is a very good example of showing that knowing the basics is the best way to do mathematics.

    The expression |x-y| is the distance from x\text{ to }y.

    So look at that equation again.
    The distance from x to -2 plus the distance from x to 3 equals 2.

    Now draw a number line and mark off -2~\&~3.
    They are five units apart: |(-2)-(3)|=5.
    So can there be an x such that |x-(-2)|+|x-(3)|=2~?
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    Re: absolute values

    o.k

    1) x+2\geq\wedge x-3\geq 0\Longrightarrow x+2+x-3=2\Longrightarrow x=\frac{3}{2}

    2) x+2\geq 0\wedge x-3<0\Longrightarrow x+2-x+3=2\Longrightarrow 5=2

    3) x+2<0\wedge x-3\geq 0\Longrightarrow -x-2+x-3=2\Longrightarrow -5=2

    4) x+2<0\wedge x-3<0\Longrightarrow -x-2-x+3=2\Longrightarrow x=-\frac{1}{2}

    So the solutions are: x=3/2 or x= -1/2

    Correct??
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  10. #10
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    Re: absolute values

    Quote Originally Posted by psolaki View Post
    o.k

    1) x+2\geq\wedge x-3\geq 0\Longrightarrow x+2+x-3=2\Longrightarrow x=\frac{3}{2}

    2) x+2\geq 0\wedge x-3<0\Longrightarrow x+2-x+3=2\Longrightarrow 5=2

    3) x+2<0\wedge x-3\geq 0\Longrightarrow -x-2+x-3=2\Longrightarrow -5=2

    4) x+2<0\wedge x-3<0\Longrightarrow -x-2-x+3=2\Longrightarrow x=-\frac{1}{2}

    So the solutions are: x=3/2 or x= -1/2

    Correct??
    Did you read Plato's post directly above yours?
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    Re: absolute values

    Siron in guiding me thru the solution of the equation asked me to show calculations,or must accept platos way of solving the problem as a general way of solving equations of this type??

    I am confused
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  12. #12
    MHF Contributor MarkFL's Avatar
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    Re: absolute values

    I believe Plato wanted you to recognize that the sum of the distances from any point x on the number line to the points -2 and 3 is a minimum when:

    -2\le x\le 3 and that sum is 5.

    Look at the 2 solutions you obtained...do they violate your stated conditions?
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  13. #13
    Super Member ILikeSerena's Avatar
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    Re: absolute values

    Let's start with:
    Quote Originally Posted by psolaki View Post
    1) x+2\geq 0 \wedge x-3\geq 0\Longrightarrow x+2+x-3=2\Longrightarrow x=\frac{3}{2}
    You found x=\frac{3}{2} which is good.
    However, this can only be a solution if it satisfies the conditions x+2\geq 0 \wedge x-3\geq 0.
    Does it?
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  14. #14
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    Re: absolute values

    Quote Originally Posted by psolaki View Post
    Siron in guiding me thru the solution of the equation asked me to show calculations,or must accept platos way of solving the problem as a general way of solving equations of this type??

    I am confused
    your calculations are ok, but you haven't gone far enough.

    x+2 ≥ 0 and x-3 ≥ 0, means that x ≥ -2 and x ≥ 3, that is, x ≥ 3. however, if x ≥ 3, then that is inconsistent with x = 3/2. so that's a contradiction.

    now examine your other three cases.
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    Re: absolute values

    Quote Originally Posted by psolaki View Post
    o.k

    1) x+2\geq\wedge x-3\geq 0\Longrightarrow x+2+x-3=2\Longrightarrow x=\frac{3}{2}
    If x= 3/2, then x- 3= -3/2 which is NOT greater than or equal to 0

    2) x+2\geq 0\wedge x-3<0\Longrightarrow x+2-x+3=2\Longrightarrow 5=2

    3) x+2<0\wedge x-3\geq 0\Longrightarrow -x-2+x-3=2\Longrightarrow -5=2

    4) x+2<0\wedge x-3<0\Longrightarrow -x-2-x+3=2\Longrightarrow x=-\frac{1}{2}
    If x=-1/2 then x+ 2= 3/2 which is NOT less than 0.

    So the solutions are: x=3/2 or x= -1/2

    Correct??
    No. You did not check to see if your values for x met the required conditions.
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