Answer
The solution is $(-\infty,-1] \cup [1,\infty)$
Work Step by Step
$\frac{r+1}{r-1}\gt0$
First solve the corresponding equation
$\frac{r+1}{r-1} = 0$
$r+1=0$
$r=-1$
For $(-\infty,-1)$, choose $-2: \frac{-2+1}{-2-1}=\frac{1}{3}\gt0$
For $(-1,1)$, choose $0: \frac{0+1}{0-1}=-1\leq0$
For $(1,\infty)$, choose $2: \frac{2+1}{2-1}=3\gt0$
The solution is $(-\infty,-1] \cup [1,\infty)$
