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Math Help - Inequality

  1. #1
    Math Engineering Student
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    Inequality

    I forgot how to do these things:

    \left|x-2\right|+\left|x+4\right|>6

    Thanks.
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  2. #2
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    Quote Originally Posted by Krizalid View Post
    I forgot how to do these things:

    \left|x-2\right|+\left|x+4\right|>6

    Thanks.
    Consider cases.
    If x\geq 2 \implies |x-2|=x-2\mbox{ and }x<2\implies |x-2|=2-x

    If x\geq -4 \implies |x+4|=x+4 \mbox{ and }x<-4\implies |x+4|=-x-4

    Thus,
    x<-4\implies (2-x)+(-x-4) >6
    -4\leq x < 2 \implies (2-x)+(x-4)>6
    2\leq x\implies (x+2) + (x-4) > 6

    Now do each one ... carefully.
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  3. #3
    Math Engineering Student
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    Now I reviewed these things and the procedure is not complicated.

    I'll solve this one, may be it may help someone else.

    Quote Originally Posted by Krizalid View Post

    \left|x-2\right|+\left|x+4\right|>6
    For x=-4\implies6>6, which is false. For x=2\implies6>6, which is false. Hence, we gotta study the following three cases.
    • Case #1: if x<-4. The inequality becomes (2-x)-(x+4)>6\implies x<-4. First solution it's actually S_1=(-\infty,-4).
    • Case #2: if -4<x<2. The inequality becomes (2-x)+(x+4)>6\implies6>6, which is false, so we have no solution here.
    • Case #3: if x>2. The inequality becomes (x-2)+(x+4)>6\implies x>2. Second solution it's actually S_2=(2,\infty).

    Final solution S will be given by S=S_{1}\cup S_{2}=(-\infty ,-4)\cup (2,\infty ).
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