Geometry Help!!!!!!!!!!

If the sides of a triangle have measurements 3x + 4 , 6x - 1, and 8x + 2, find all possible values of x.
I also need help with this question please

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  1. All sides must be greater of zero.

    3x+4 always>0

    8x+2 always>0

    Only side: 6x-1 can be negative.

    That's why:

    6x-1>0

    6x>1 Divide both sides with 6

    x>1/6

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  2. Thanks anyway but that isn't one of the choices the answer is x> 3/11 i figured it out thanks for trying :)

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  3. Anonymus solution not completely.

    All sides must be greater of zero:

    3x+4>0

    3x> -4 Divide with 3

    x> -4/3

    6x-1>0

    6x>1 Divide with 6

    x>1/6

    8x+2>0

    8x> -2 Divide with 8

    x> -2/8

    x> -1/4

    Least of that numbers is -4/3= -1.3333

    x> -4/3 is solution

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  4. x>3/11 is also > -4/3

    All x> -4/3 is choices

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  5. The condition that all sides have to be positive, as Anonymous used, is not sufficient.
    In any triangle the sum of 2 sides must be greater than the third side, so
    3x+4 + 6x-1 > 8x + 2 ----> x > -1
    AND
    3x+4 + 8x + 2 > 6x-1 ---> x > -7/5
    AND
    8x + 2 + 6x - 1 > 3x+4 --> x > 3/11

    the intersection of all three conditions is
    x > 3/11

    To show that Anonymous is incorrect, pick a value of x between his/her answer of 1/6 and mine of 3/11
    e.g. x = 11/50
    3x+4 --> 4.66
    6x-1 --> .32
    8x+2 --> -.24 , contradiction
    I have a value of x > 1/6 which did not work.

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