Answer
There is no solution for $\frac{z^2+z}{z^2-1}\geq3$
Work Step by Step
$\frac{z^2+z}{z^2-1}\geq3$
Distribute the 3 into the brackets
$3(z^2-1)\geq z^2+z$
$3z^2-3-z^2-z\geq0$
$2z^2-z-3\geq0$
$(z+1)(z-\frac{3}{2})\geq0$
$p\lt-1$or $p\gt\frac{3}{2}$
but z can not be in $(-\infty,-1) \cup (1,\infty)$
Thus, there is no solution for $\frac{z^2+z}{z^2-1}\geq3$
