Solve the inequality \(\dfrac{3-z}{z+1} \ge 1.\)

thank you and please help! :)

lokiisnotdead Jun 21, 2020

#2**0 **

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.

geno3141 Jun 21, 2020

#3**+1 **

hmmm the answer is actually \((-1,1]\)

thanks for trying to help though

lokiisnotdead
Jun 22, 2020

#4**+3 **

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

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]

Melody Jun 22, 2020