Question

Find all values of $z$ such that $\dfrac{3-z}{z+1} \ge 1$. Answer in interval notation.

Answer 1

Answer:

) 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 <========