#1**0 **

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

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

Solution: (-inf,1].

Guest Jun 15, 2022

#3**0 **

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 $(-\infty, -1)U(-1, 1)$

Guest Jun 15, 2022

#4**0 **

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]

Guest Jun 15, 2022

#10**+1 **

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

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

MaxWong Jun 16, 2022

#11**+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

Melody Jun 16, 2022