<s>[INST] 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.

 [/INST] Transform $\frac{4x}{x+7}<x$ into the equivalent $-\frac{x^2+3x}{x+7}=\frac{4x}{x+7}-x<0$ or $\frac{x^2+3x}{x+7}>0$.
This reduces to question to finding the sign of $\frac{x^2+3x}{x+7}$, which can be solved by making a simple table:
$$
\begin{array}{c|ccccc}
&  & -7 &  & -3 &  & 0 & \\
\hline
x & - & - & - & - & -& 0 & + \\
x+3 & - & - & - & 0 & + & + & + \\
x(x+3) & + & + & + & 0 & - & 0 & +\\
x+7 & - & 0 & + & + & + & + & +\\
\dfrac{x^2+3x}{x+7} & - & ! & + & 0 & -& 0 & +
\end{array}
$$
The answer is therefore $(-7,-3) \cup (0, +\infty)$.
</s>