Algebra

$\frac{3-z}{z+1} \ge 2(z + 4)$

 

Multiply both sides by $z+1$: $3-z \ge 2(z+4)(z+1)$

 

When we simplify, we get: $3-z \ge 2z^2+10z+8$

 

Take everything to one side: $2z^2+11z+5 \le 0$

 

Now, we can factor the quadratic: $(2z+1)(z+5) \le 0$. Clearly, the critical points are $z=-1/2,-5,$ and $-1$ (because that is what is undefined in the original equation)

 

We can now test values below $-5$, between $-5$ and $-1$, between $-1$ and $-1/2$ and values greater than $-1/2$, and see which intervals satisfy the inequality.

 

Substiuting $-6$ in (interval below $-5$): $-9/5 \ge -4$ - True

Substituting $-2$ in (interval between $-5$ and $-1$): $-5 \ge 8$ - Not True

Substituting $-3/4$ in (interval between $-1$ and $-1/2$): $15 \ge 13/2$ - True

Substituting $0$ in (interval greater than $-1/2$): $3 \ge 8$ - Not True

 

There is two intervals which satisfy the original inequality, so our answer In interval notation is $(-\infty,-5)\cup(-1,-1/2)$

 

  

Ahh...Owinner. the sign is greater and EQUAL to, so -5 and -1/2 are actually valid solutions there.

the answer is

$(-\infty, -5]\cup(-1, -\frac{1}{2}]$