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    mix786's Avatar
    mix786 Posts: 21, Reputation: 4
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    #1

    Oct 1, 2007, 06:51 AM
    algebra 2
    Graph on the number line :
    -|x| - 2 > -5; D = [ integers ]
    CellistJames's Avatar
    CellistJames Posts: 19, Reputation: 2
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    #2

    Oct 1, 2007, 03:01 PM
    OK so for starters you need to get the absolute value alone with out any thing on its side

    so...

    -|x| - 2 > -5

    add a 2 to both sides

    -|x| > -3

    multiply through with a -1 (dont forget that the > switches to a < when you multiply or divided with a negative, vice versa)

    |x| < 3
    so x is either equal to a negative or a positive so you have two equations to set up past this point.

    x < 3

    or

    x > -3

    3 and -3 are your critical values you need to to put those on a number line and test a point in each region.

    so pick a point between -3 and 3
    lets say 0 and put it in the original equation
    -|0| -2 > -5
    -2 > -5 True

    and a number between -infinity and -3
    say -4 plug that into your original equation
    -|-4| -2 > -5
    -4 -2 > -5
    -6 > -5 False

    and a number between 3 and infinity
    say 4 and plug that into you equation
    -|4| -2 > -5
    -4 -2 > -5
    -6 > -5 False

    so that being said you know that x can be any value between -3 and 3 but is not: x<-3 or x>3 so -3<x<3 would be your answer and graph it however you were taught.

    Sorry if I confused you more..

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