- #1
Plutonium88
- 174
- 0
So I broke it into a number line and calculated when
|3x-7|
(3x-7) x>orequAl 7/3
-(3x-7) x < 7/3 (strict)
|x-8|
-(x-8) x<8
(x-8) x>orequal 8
So for x<7/3
-(3x-7) - [-(x-8)] > 4
Simplified to x < -5/2
For the interval 7/3<x<8
(3x-7)-[-(x-8)]
x>19/4
This is part of solution be ause it lies within the interval 7/3<x<8
Now for x>8
(3x-7)-(x-8)
X<3/2
Now I'm not sure how to adress this but I want to say this is not part of the solution cause it doesn't lie within the interval x>8
And also in terms of writing my Ana after in interval notation how can I do this with the info attained. Do I just plot my newly found restrictions on the number line?
|3x-7|
(3x-7) x>orequAl 7/3
-(3x-7) x < 7/3 (strict)
|x-8|
-(x-8) x<8
(x-8) x>orequal 8
So for x<7/3
-(3x-7) - [-(x-8)] > 4
Simplified to x < -5/2
For the interval 7/3<x<8
(3x-7)-[-(x-8)]
x>19/4
This is part of solution be ause it lies within the interval 7/3<x<8
Now for x>8
(3x-7)-(x-8)
X<3/2
Now I'm not sure how to adress this but I want to say this is not part of the solution cause it doesn't lie within the interval x>8
And also in terms of writing my Ana after in interval notation how can I do this with the info attained. Do I just plot my newly found restrictions on the number line?