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Solve the inequality
\[\frac{3-z}{z+1} \ge 2 + 7z.\]
Write your answer in interval notation.

 Dec 12, 2023
 #1
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Multiplying both sides by z+1, we get 3−z≥2(z+1)+7z2+7z.

 

Simplifying, we get −8z2−9z−1≥0. Dividing both sides by −8, we get [z^2 + \frac{9}{8}z + \frac{1}{8} \le 0.]

 

Completing the square, we get [(z + \frac{9}{16})^2 \le \frac{1}{64},] so ∣z+169​∣≤81​

 

This means −81​≤z+169​≤81​, so [-\frac{25}{16} \le z \le -\frac{13}{16}.]

 Dec 12, 2023

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