Multiply both sides by z + 1: 3 - z >= z + 1
Then 2z <= 2, so z <= 1.
Solution: (-inf,1].
You're very new to math, in this case, if you include -1 in the denominator, the answer is undefined, so you're WRONG. The answer is $(-\infty, -1)U(-1, 1)$
You're very new to math too, as if the answer is negative and lower than -1 then it will be less than 1. so the answer then, is obviously (-1, 1]
+9673 Ignoring the toxic "discussion" above and actually solving the problem:
\(\dfrac{3 - z}{z + 1} \geq 1\\ \dfrac{3 - z}{z + 1} (z + 1)^2 \geq (z + 1)^2\text{ and }z +1 \neq 0\\ (3 - z)(z + 1) \geq (z + 1)^2\text{ and }z +1 \neq 0\\ -z^2 + 2z + 3\geq z^2 + 2z + 1 \text{ and }z +1 \neq 0\\ 2z^2 - 2 \leq 0 \text{ and }z +1 \neq 0\\ z^2 - 1 \leq 0 \text{ and }z +1 \neq 0\\ -1 \leq z \leq 1 \text{ and }z \neq -1\\ -1 < z \leq 1\)
+118687 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
