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# inequality help pls!

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

Jun 21, 2020

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

3 - z >= z + 1

2z <= 2

z <= 1

The solution is (-inf,1].

Jun 21, 2020
#2
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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)

Jun 21, 2020
#3
+1

hmmm the answer is actually $$(-1,1]$$

thanks for trying to help though

#4
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$$\dfrac{3-z}{z+1} \ge 1\\~\\ z\ne-1\\ If \;\;z<-1\;\;then\\ 3-z\le z+1\\ -2z\le-2\\ z\ge1\\$$

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]

Jun 22, 2020
edited by Melody  Jun 22, 2020
edited by Melody  Jun 22, 2020
#5
+1

thank you so much Melody!!! I really appreciate it! :)