inequality help pls!

(3 - z)/(z + 1) >= 1

3 - z >= z + 1

2z <= 2

z <= 1

The solution is (-inf,1].

There are two cases (Guest answered the first one):

Case 1:  z + 1  >=  0:     --->     3 - z  >=  (z + 1)·1

                                                 3 - z  > =  z + 1

                                                      2  >=  2z

                                                      1  >=  z     --->     z  <=  1     (-inf, 1]

 Case 2:  z + 1  <= 0:     --->     3 - z  <=  (z + 1)·1                    (change the direction of the inequality

                                                 3 - z  <=  z + 1                            because you are multiplying by a

                                                      2  <=  2z                                negative)

                                                      1  <=  z     --->     z  >= 1     [1, inf)

The final answer is the union of the two above answers.

\(\dfrac{3-z}{z+1} \ge 1\\~\\ z\ne-1\\ If \;\;z<-1\;\;then\\ 3-z\le z+1\\ -2z\le-2\\ z\ge1\\ \)

z cannot be less than -1 and also greater than +1 so no solution here

\( If \;\;z>-1\;\;then\\ 3-z\ge z+1\\ -2z\ge-2\\ z\le1\\ -1

My latex is not rendering. Maybe it will is itself later???

Here is an image of it:

LaTex

\dfrac{3-z}{z+1} \ge 1\\~\\ z\ne-1\\ If \;\;z<-1\;\;then\\ 3-z\le z+1\\ -2z\le-2\\ z\ge1\\ 
\text{z cannot be less than -1 and also greater than +1 so no solution here}\\~\\ 
If \;\;z>-1\;\;then\\ 3-z\ge z+1\\ -2z\ge-2\\ 
z\le1\\ 
-1  (-1,1]