Absolute value cases?
I don't understand the whole "case" thing with absolute value inequalities.
(Pretend that [ and ] are the absolute value thingies)
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?
- 1 decade agoFavorite 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
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.
9-3x is still positive, and 3x is positive now
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.
(9-3x) is negative now, which means that we have to change it's sign in order to remove absolute value. 3x>0
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
- locuazLv 71 decade ago
case 1: x>=0 and 9-3x>=0,
x>=0, 9>=3x, then 3>=x
so this case is: 0<=x<=3
so the solution in this case is:
case 2: x>=0, 9-3x<0,
3<x, together with x>=0, simply means that
in the equation:
8< 4x, 2<x,
so the solution is: x>3.
case 3: x<0, 9-3x>=0
3>=x, so togegher with x<0 is simply
in the equation:
together with x<0,
the solution is: x<0
case 4: x<0, 9-3x<0
but this cannot happen together with x<0, so in this case there are no solutionsSource(s): h