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Math Help - inequality and an interval

  1. #1
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    inequality and an interval

    I have the following inequalities:

    -2 < h*2/x < 0

    -2.78 <h*2/x < 0


    where x is 1 < x < 2 ?

    Are the answers:

    x > h > 0 and 1.39x > h > 0

    or do i have problems with the way I have placed the signs?

    and How does the interval affect the outcome?




    Last edited by iwishiunderstood; September 23rd 2008 at 11:40 AM.
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  2. #2
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    Quote Originally Posted by iwishiunderstood View Post
    I have the following inequalities:

    -2 < h*2/x < 0

    -2.78 <h*2/x < 0


    where x is 1 < x < 2 ?

    Are the answers:

    x > h > 0 and 1.39x > h > 0

    or do i have problems with the way I have placed the signs?

    and How does the interval affect the outcome?

    There are only few rules which you have to observe when dealing with unequalities:

    a) adding or subtracting a term doesn't change the relation-sign

    b) dividing or multiplying by a positive term doesn't change the relation-sign

    c) calculationg the reziprocal of fractions could change the relation-sign depending of the sign of the fractions in question. Examples:

    \dfrac13 > \dfrac14~\implies~3<4

    -\dfrac13 < -\dfrac14~\implies~-3>-4

    BUT:

    -\dfrac13 < \dfrac14~\implies~-3<4

    To your questions:

    -2<h\cdot \frac2x <0~\wedge~x>0 that means x is positive

    -2x<2h<0~\wedge~x>0 divide through by -2 (the divisor is negative). The relation-signs have to be changed. Therefore your answer is correct.

    -2<h\cdot \frac2x <0~\wedge~x>0~\implies~-2x<2h<0 . Dividing thriough by +2 will yield:

    -1.39x<h<0
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  3. #3
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    just notice in both my inequalities h * 2/x should be negative.


    is this just a case of changing the signs round in my answers?
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  4. #4
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    Quote Originally Posted by iwishiunderstood View Post
    just notice in both my inequalities h * 2/x should be negative.


    is this just a case of changing the signs round in my answers?
    You probably have noticed that I've made a silly mistake. Here comes the corrected calculation:

    To your questions:

    -2<h\cdot \frac2x <0~\wedge~x>0 that means x is positive and h must be negative

    -2x<2h<0~\wedge~x>0 divide through by -2 (the divisor is negative). The relation-signs have to be changed. Therefore the answer is:

    x>-h>0
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