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Math Help - Absolute Value algebra manipulation

  1. #1
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    Absolute Value algebra manipulation

    Hi all,
    I have arrived at a question involving absolute value I am unsure of. It involves manipulation of a simple expression, the only problem being the absolute value segment, which I havent come across in this sort of context. See below:

    (7x+21) / ((x+5)|x-1|)

    if the denominator were (x+5)(x-1) it would be easy (x^2 + 4x - 5). Can I treat |x-1| the same?

    I have tried simplifying |x-1| to sqrt((x-1)^2) at which stage the sqrt and '^2' would normally cancel, but isn't this taking away that which makes absolute value unique?

    Or am I over thinking this all?
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  2. #2
    MHF Contributor red_dog's Avatar
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    |x-1|=\left\{\begin{array}{ll}x-1, & x\geq 1\\1-x, & x<1\end{array}\right.

    Then,

    \frac{7x+21}{(x+5)|x-1|}=\left\{\begin{array}{ll}\displaystyle\frac{7x+  21}{(x+5)(x-1)}, & x>1\\\displaystyle\frac{7x+21}{(x+5)(1-x)}, & x<1,x\neq -5\end{array}\right.
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  3. #3
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    right, so first of all you isolate the absolute value portion of the equation, then (in the case of the example I gave where I had to find the Horizontal Asymptote) I'd calculate the HA for both denominators listed above? (meaning (x+5)(x-1) and (x+5)(1-x))
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