Solve the inequality \frac{3-z}{z+1} \ge 2(z + 4). Write your answer in interval notation.
Multiply by (z + 1)^2 on both sides. Then,
(3−z)(z+1)≥2(z+4)(z+1)2 and z+1≠0(z+1)(2(z+4)(z+1)−(3−z))≤0 and z+1≠0(z+1)(2z2+11z+5)≤0 and z+1≠0(z+1)(2z+1)(z+5)≤0 and z+1≠0z≤−5 or −1<z≤−12
In interval notation, z∈(−∞,−5]∪(−1,−12]
+1000 
