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March 25, 2017

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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.

  • math - ,

    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

  • math - ,

    Thank You!

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