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Math Help - inequalities

  1. #1
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    inequalities

    i dont know but i just cant seem to understand how the inequalities have been obtained

    given that 0<lx-2l<1,
    i get that it becomes -1<x-2<1

    but the book went on to say that
    -1<x-2<3
    2<3x-1<8
    thus l 3x-1 l>2

    how is the LHS of the inequalities obtained? namely that it is -1 and 2..
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  2. #2
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    Quote Originally Posted by alexandrabel90 View Post
    i dont know but i just cant seem to understand how the inequalities have been obtained

    given that 0<lx-2l<1,
    i get that it becomes -1<x-2<1

    but the book went on to say that
    -1<x-2<3
    2<3x-1<8
    thus l 3x-1 l>2

    how is the LHS of the inequalities obtained? namely that it is -1 and 2..
    Dear alexandrabel90,

    -1<x-2<1\mbox{~and~}1<3\Rightarrow -1<x-2<3 --------(1)

    -1<x-2<1

    \Rightarrow -3<3x-6<3

    \Rightarrow -3+5<3x-6+5<3+5

    \Rightarrow 2<3x-1<8 --------(2)

    Since, 3x-2>2; 3x-2 is a positive number. Therefore, \mid 3x-2\mid>2 ------------(3)

    Hope you understood.
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  3. #3
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    A slightly different way to do it: from 0< |x- 2|< 1 is the same as -1< x- 2< 1 (with the stipulation that x- 2\ne 0! That is -1< x-2< 0 or 0< x-2< 1) Adding 2 to each part, 1< x< 3 (with the added stipulation that x is not 2: 1< x< 2 or 2< x< 3). Multiply by 3: 3< 3x< 9 (with the added stipulation that 3x is not 6). Subtracting 1, 2< 3x-1< 8 (with the added stipulation that 3x-1 is not 5: 2< 3x- 1< 5 or 5< 3x- 1< 8. But since all possible values of 3x-1 are positive and larger than 2, it is true that, for all x satisfying 0< |x- 2|< 1, 2< |3x-1|
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