# How do you solve (16-x^2)/(x^2-9)>=0?

$3 < | x | \le 4$ alternatively $3 < x \le 4$ or $- 4 \le x < - 3$
For the inequality to be two either $16 - {x}^{2} \ge 0 \mathmr{and} {x}^{2} - 9 \ge 0$ or $16 - {x}^{2} \le 0 \mathmr{and} {x}^{2} - 9 \le 0$
Note that $| x | = 3$ is not defined
Now $16 - {x}^{2} \ge 0$ if $| x | \le 4$ and ${x}^{2} - 9 \ge 0$ if $| x | > 3$ so $3 < | x | \le 4$
Now $16 - {x}^{2} \le 0$ if $| x | \ge 4$ and ${x}^{2} - 9 \le 0$ if $| x | < 3$ which has no solution