Solving $\frac{4x}{x+7}I know how to solve the problem. The reason I post the problem here is to see whether there is a quick approach, rather than a traditional method, to solve the problem.  

The problem is: find $x$ that satisfy $\frac{4x}{x+7}<x$
  I considered two cases: $x+7>0$ and $x+7<0$, and then went through details to find x. The process took me a few minutes. I wonder whether there is a simple way to find the answer.


First of all, note that $x=0$ is not a solution. Thus we can consider two cases: $x$ is positive, or $x$ is negative.
If $x$ is positive, then we can divide by $x$ to obtain $\frac{4}{x+7}<1$. Since $x$ is positive, this is always true. Thus, $(0,\infty)$ is in our solution set.
If $x$ is negative, we get $\frac{4}{x+7}>1$. This works if $x$ is greater than $-7$, but less than $-3$. Thus, $(-7,-3)$ is in our solution set.
