# Inequality with absolute value

I am unsure if have solved the following inequality correctly:

$\dfrac{2x+3}{x+5} \leq \dfrac{x+1}{|x-1|} \tag{1}$

I've proceeded as follows.

If $x>1$ then $|x-1|=(x-1)$

If $x<1$ then $|x-1|=-x-1$

I've then solved for those seperate inequalities,

$\dfrac{2x+3}{x+5} \leq \dfrac{x+1}{x-1}$

$\dfrac{2x+3}{x+5} \leq \dfrac{x+1}{-x-1}$

The problem is that the union of their solution intervals yields a different result from the inequality (1) when I enter it into Wolfram Alpha. I am afraid I have not solved it correctly.

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## 1 Answer

When $a <0$, $|a|=-a$. Thus, when $x<1$, $$|x-1|=-(x-1)=1-x,$$ or, if you prefer, $-x+1$ (I don't). With this correction, the rest of the work should not be difficult. But if you "cross-multiply," there is the need to remember that the inequality is reversed when we multiply through by a negative number. That issue arises when $x<-5$.

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(Neither do I.) –  Brian M. Scott Sep 22 '11 at 2:27