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Math Help - (simple) Inequalities.

  1. #1
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    Smile (simple) Inequalities.

    0 <= e^(-x) < 1

    I have to find the solution set so i got:
    minus infinity > x > 0
    correct?

    and for
    |2-3x|<3
    i get
    x > 1/3
    and
    x < 5/3

    now my question is:
    on the second inequality, the solution set is the INTERSECTION of both solutions? (is it an OR or and AND connection of both sets?)
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  2. #2
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    Re: (simple) Inequalities.

    Quote Originally Posted by nappysnake View Post
    0 <= e^(-x) < 1

    I have to find the solution set so i got:
    minus infinity > x > 0
    correct?

    and for
    |2-3x|<3
    i get
    x > 1/3
    and
    x < 5/3

    now my question is:
    on the second inequality, the solution set is the INTERSECTION of both solutions? (is it an OR or and AND connection of both sets?)
    1.

    e^x > 0  \land e^x>1

    2.

    |2-3x| =\begin{cases}3x-2, & \text{if }x>\frac{2}{3} \\2-3x, & \text{if }x \leq \frac{2}{3}\end{cases}

    2a.

    x \in (\frac{2}{3},+\infty)

    3x-2<3

    2b.

    x \in \left(-\infty,\frac{2}{3}\right]

    2-3x<3

    SolutionOf(2)=SolutionOf(2a) \lor SolutionOf(2b)
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  3. #3
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    Re: (simple) Inequalities.

    For the second, I think it's really better to do it simultaneously. So

    |2-3x| < 3

    or

    -3 < 2-3x < 3.

    So -5 < -3x < 1

    then 5 > 3x > 1

    then \frac{5}{3} > x > \frac{1}{3}

    or

     \frac{1}{3} < x <\frac{5}{3}
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