Thanks!
hello,
$\displaystyle \frac{|x-3|}{x+4}\leq 5$ transform into:
$\displaystyle \frac{x-3}{x+4}\leq 5 ~ \wedge ~ x\geq 3~~\vee~~ \frac{-x+3}{x+4}\leq 5 ~ \wedge ~ x < 3$
Solve both inequalities separately. You should get as a final result:
$\displaystyle x < -4 ~~\vee~~x \geq -\frac{17}{6}$
I won't give the details, but another way would be to simply square the expression. $\displaystyle (|x|)^2 = x^2$. The downside is that you may have extra solutions, and instead of one inequality you have two:
$\displaystyle -25 \leq \frac{(x - 3)^2}{(x + 4)^2} \leq 25$.
However, despite the downsides I like the logic inherent in this method slightly better, even if it makes the problem more difficult to solve. And, of course, when you are done you get earboth's answer.
-Dan