Question

absolute value of(x + 3) + absolute value of (3 - x) < 9.

I am not sure how to go about this question. Although, we generally use three cases.

Answer 1

let's rewrite it as
|x+3| < 9 - |3-x|
x+3 < 9 - |3-x| or -x-3 < 9 - |3-x|
|3-x| < 6-x , #1)
or
|3-x| < 12 + x , (#2)

from #1
3-x < 6-x
no solution
or
-3 + x < 6-x
2x < 9
x < 9/2

from #2
3-x < 12+x
-2x < 9
x > -9/2
or
-3+x < 12 +x
no solution

so we have x< 9/2 and x > -9/2

our critical values of -9/2 and +9/2 divide our number line into 3 parts
1. x < -9/s
2. x between -9/2 and 9/2
3. x > 9/2

I usually pick an arbitrary number in each region and test it in the original

1. let x = -10 --> |-7|< 9 -|13| false
2. let x = 0 ---> |3| < 9 - |3| true
3. let x = 10 --> |13| < 9 - |-7| false

so -9/2 < x < 9/2

Answer 2

Thank You!