# Absolute Value algebra manipulation

• Aug 4th 2009, 10:59 PM
isp_of_doom
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?
• Aug 4th 2009, 11:28 PM
red_dog
$\displaystyle |x-1|=\left\{\begin{array}{ll}x-1, & x\geq 1\\1-x, & x<1\end{array}\right.$

Then,

$\displaystyle \frac{7x+21}{(x+5)|x-1|}=\left\{\begin{array}{ll}\displaystyle\frac{7x+ 21}{(x+5)(x-1)}, & x>1\\\displaystyle\frac{7x+21}{(x+5)(1-x)}, & x<1,x\neq -5\end{array}\right.$
• Aug 5th 2009, 01:21 AM
isp_of_doom
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))