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# 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

- 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

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

- 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

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