Find all values of 
such that 
. Answer in interval notation.
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+23254 Consider three cases:
1) z = -1: this is impossible because the denominator would become zero.
2) z > -1 (This makes the denominator positive; so when you multiply both sides by z + 1, you do not
need to change the direction of the inequality.)
(3 - z) / (z + 1) ≥ 1
Multiply both sides by z + 1:
3 - z ≥ 1(z + 1)
3 - z ≥ z + 1
2 - z ≥ z
2 ≥ 2z
1 ≥ z ---> z ≤ 1 and z > -1 ---> -1 < z ≤ 1
3) z < -1 (This makes the denominator negative; so when you multiply both sides by z + 1, you have
to change the direction of the inequality.)
(3 - z) / (z + 1) ≥ 1
Multiply both sides by z + 1:
3 - z ≤ 1(z + 1)
3 - z ≤ z + 1
2 - z ≥ z
2 ≤ 2z
1 ≤ z ---> z ≥ 1 (This is imposible since the assumption of this section was z < -1.)
Conclusion: the answer is: -1 < z ≤ 1 <==============
+23254 Consider three cases:
1) z = -1: this is impossible because the denominator would become zero.
2) z > -1 (This makes the denominator positive; so when you multiply both sides by z + 1, you do not
need to change the direction of the inequality.)
(3 - z) / (z + 1) ≥ 1
Multiply both sides by z + 1:
3 - z ≥ 1(z + 1)
3 - z ≥ z + 1
2 - z ≥ z
2 ≥ 2z
1 ≥ z ---> z ≤ 1 and z > -1 ---> -1 < z ≤ 1
3) z < -1 (This makes the denominator negative; so when you multiply both sides by z + 1, you have
to change the direction of the inequality.)
(3 - z) / (z + 1) ≥ 1
Multiply both sides by z + 1:
3 - z ≤ 1(z + 1)
3 - z ≤ z + 1
2 - z ≥ z
2 ≤ 2z
1 ≤ z ---> z ≥ 1 (This is imposible since the assumption of this section was z < -1.)
Conclusion: the answer is: -1 < z ≤ 1 <==============
