Solve The Inequality \frac{3-z}{z+1} \geq 1

Multiply both sides by z + 1: 3 - z >= z + 1

Then 2z <= 2, so z <= 1.

Solution: (-inf,1].

$\displaystyle \frac{3-z}{z+1} \geq{1}$

Ignoring the toxic "discussion" above and actually solving the problem:

$\dfrac{3 - z}{z + 1} \geq 1\\ \dfrac{3 - z}{z + 1} (z + 1)^2 \geq (z + 1)^2\text{ and }z +1 \neq 0\\ (3 - z)(z + 1) \geq (z + 1)^2\text{ and }z +1 \neq 0\\ -z^2 + 2z + 3\geq z^2 + 2z + 1 \text{ and }z +1 \neq 0\\ 2z^2 - 2 \leq 0 \text{ and }z +1 \neq 0\\ z^2 - 1 \leq 0 \text{ and }z +1 \neq 0\\ -1 \leq z \leq 1 \text{ and }z \neq -1\\ -1 < z \leq 1$

Sometimes I just gotta laugh.   

\(\frac{3-z}{z+1} \geq 1 \\ \text{ Assuming that z is a real number}\\ z\ne-1\\ \frac{3-z}{z+1} \geq 1 \\~\\ If\;\;z>-1\;\;then\\ 3-z \geq z+1 \\ -2z\ge -2\\ z\le1\\ \therefore -1

or

$If\;\;z<-1\;\;then\\ 3-z \leq z+1 \\ -2z\le -2\\ z\ge1\\ \therefore no\;\;soln$

Hence

\(-1

Sorry Max, I didn't see your answer there. :)