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Solve the inequality
(3-z)/(z+1) ≥ 1

 Jun 25, 2020
 #1
avatar+134 
+2

We know how to compare quotients to 0. We start by subtracting 1 from both sides to get (3-z)/(z+1)-1 ≥0
We can simplify the left-hand side as (3-z)/(z+1)-1=(3-z)/(z+1)-(z+1)/(z+1)=(3-z-(z+1))/z+1=(2-2z)/(z+1)
so now our inequality is (2-2z0/(z+1) ≥ 0.
We can simplify this a bit by dividing both sides by 2 to get (1-z)/(z+1)≤0. Then, we multiply by -1 to turn the numerator on the left into . We must remember that multiplying by a negative means we reverse the inequality, so we have (z-1)/(z+1)≤0.
The quotient of two numbers is negative if and only if one of the numbers is negative and the other is positive, from which we find that (z-1)/(z+1) is negative for -1< z <1. We also could have built a table to analyze the possibilities:

  z+1 z-1 (z-1)/(z+1)
z<-1  -  -  +
z=-1  0  -  Undefined
-1  +  -  -
z=1  +  0  0
z>1  +  +  +


The inequality is nonstrict, so we must also include any values of  for which (z-1)/(z+1)=0. Therefore, the full solution is -1

 Jun 25, 2020
 #2
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0

Oh, ok thanks!

 Jun 25, 2020
 #3
avatar+134 
+1

No problem!

ItzMe  Jun 25, 2020
 #4
avatar+30933 
+2

You could also draw a graph, which often helps in these sort of questions:

 

 Jun 25, 2020

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