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Math Help - inequality help

  1. #1
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    inequality help

    lx+10l + 3 > 6

    do i subtract 3 to both sides or subtract 10? thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rj2001 View Post
    lx+10l + 3 > 6

    do i subtract 3 to both sides or subtract 10? thanks
    subtract 3 first to get |x + 10| > 3

    now we must have x + 10 > 3 or -(x + 10) > 3 depending on the sign of (x + 10) with the absolute values
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  3. #3
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    Hi rj2001,

    We have |x+10|+3>6

    Firstly as you suggested subtract 3 from both sides which gives,

    |x+10|>3

    Since the modulus sign implies that the "stuff" inside it is "turned" positive when negative and remains positive when positive this equation means two things. The first thing it means is that,

    x+10>3

    Subtracting 10 from both sides gives

    x>-7

    The second thing it means is

    -(x+10)>3

    Now divide through by -1 . When you divide through by -1 in an inequality the inequality sign switches round (e.g. if -x>3 then x<-3). Since this is the case we are left with,

     <br />
x+10<-3 \implies x<-13<br />

    Thus we are left with

    x>-7~~,~~x<-13
    Last edited by Sean12345; September 20th 2008 at 10:08 AM. Reason: mistake
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