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Math Help - solve the equatio

  1. #1
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    solve the equatio

    abs(2x-1 -abs(x+5=3


    thank you
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  2. #2
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    Quote Originally Posted by hoger View Post
    abs(2x-1 -abs(x+5=3


    thank you
    Re-write

    |2x-1|=\left\{\begin{array}{lcr}2x-1&if&x\geq\dfrac12 \\ -(2x-1)&if& x<\dfrac12\end{array}\right.

    |x+5|=\left\{\begin{array}{lcr}x+5&if&x\geq-5\\ -(x+5)&if& x<-5\end{array}\right.

    Combine the domains (x<-5; -5<x<1/2; x>1/2). You'll get 3 different linear equations.

    The final solution is x=-\dfrac73~\vee~x=9
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  3. #3
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    As I've told you there are three distinct intervals. To each interval belongs one equation without any absolute values:

    A) x<-5:~~~~~~~~-(2x-1)-(-(x+5))=3

    B) -5\leq x < \frac12:~-(2x-1)-(x+5)=3

    C) x\geq \frac12:~~~~~~~~~~~(2x-1)-(x+5)=3

    Expand the brackets and solve for x. Afterwards you have to check if the solution is possible:

    A) -(2x-1)-(-(x+5))=3~\iff~-x+6=3~\iff~x=3.......BUT: 3 is not smaller than -5

    B) -(2x-1)-(x+5)=3~\iff~-3x-4=3~\iff~-3x=7~\iff~x=-\frac73

    I'll leave the last one for you.

    Btw: If you need some additional informations ask in the original thread, please.
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