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January 31, 2015

January 31, 2015

Posted by **Jenni** on Tuesday, February 22, 2011 at 10:33pm.

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

- math -
**Reiny**, Tuesday, February 22, 2011 at 11:36pmlet'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

- math -
**Jenni**, Wednesday, February 23, 2011 at 1:12amThank You!

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