Absolute value cases?

I don't understand the whole "case" thing with absolute value inequalities.

(Pretend that [ and ] are the absolute value thingies)

2x-1-[9-3x]<[3x]

All I know is that Case I: x<0, Case II: 0<x<3, and Case III: x>3.

Can someone show me step by step how to solve this?

2 Answers

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  • 1 decade ago
    Favorite Answer

    Absolute value is making every number into a positive one. For example when you have [x]=3 x can be -3 or +3.

    In this case you don't know what sign did the 3x and 9-3x had, so you have to consider all options:

    3x is positive when x is positive(bigger then 0)

    9-3x is positive when x is <3.

    If a number was positive before it entered the absolute value, you can remove the [], without changing anything. If it was negative, you have to change it's sign to remove the absolute value (to anull it's effect).

    so in Case I

    2x -1-[9-3x]<-3x is the same as 2x-1-(9-3x)<-3x

    2x-1-9+3x<-3x

    8x-10<0

    8x<10

    x<5/4

    So in Case I we have x<5/4 and x<0... So for any x if x<0 (x is then less then 5/4) it will be correct.

    Case II:

    9-3x is still positive, and 3x is positive now

    so 2x-1-9+3x<3x

    2x-10<0

    2x<10

    x<5

    So we have: 0<x<3 and x<5. But if x is less the three, it is also less then 5 so for 0<x<3, it will be correct.

    Case III

    (9-3x) is negative now, which means that we have to change it's sign in order to remove absolute value. 3x>0

    2x-1-(-(9-3x))<3x

    2x-1+9-3x<3x

    8<4x

    x>2

    So, x has to be greater than 3 and greater than 2, so for x>3 it will be correct.

    Now we unite 3 cases:

    For: x<0, 0<x<3 and for x>3, it will be correct... Checking the boundaries: For x=0 we have -10<0, true

    For x=3 we have 5<9, true

    So 2x-1-[9-3x]<[3x] is true for all x

  • locuaz
    Lv 7
    1 decade ago

    case 1: x>=0 and 9-3x>=0,

    x>=0, 9>=3x, then 3>=x

    so this case is: 0<=x<=3

    2x-1-9+3x< 3x

    2x<10, x<1/5

    so the solution in this case is:

    0<=x<1/5

    case 2: x>=0, 9-3x<0,

    9<3x

    3<x, together with x>=0, simply means that

    x>3,

    in the equation:

    2x-1-(3x-9)<3x

    2x-1-3x+9<3x

    8< 4x, 2<x,

    so the solution is: x>3.

    case 3: x<0, 9-3x>=0

    9>=3x

    3>=x, so togegher with x<0 is simply

    x<0

    in the equation:

    2x-1-9+3x<-3x

    8x<10

    x<5/4

    together with x<0,

    the solution is: x<0

    case 4: x<0, 9-3x<0

    then, 3<x,

    but this cannot happen together with x<0, so in this case there are no solutions

    Source(s): h
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