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.

 A: 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 $(-7,-3) \cup (0, +\infty)$.
A: Here is another way--whether or not it is simpler depends on the problem and your previous experience.
First solve the equality
$$\frac{4x}{x+7}=x$$
I'm sure you can do that fairly quickly, getting $x=0$ or $x=-3$. Then find the values of $x$ where one or both of the sides of the equation are undefined. In your problem, that is $x=-7$.
Those finitely many values of $x$ break up the real number line into a finite number of intervals, two of them (the left-most and the right-most) infinitely large. Check each of those intervals to see if they make your initial equality true or false. You are guaranteed that any point in an interval will give the same answer as any other point in that interval.
Your final answer is the union of the intervals that made the equality true.
In your case examine the intervals:
$(-\infty,-7)$: Inequality is False
$(-7, -3)$: Inequality is True
$(-3, 0)$: Inequality is False
$(0, \infty)$: Inequality is True
So your final answer is
$x\in (-7, -3) \cup (0, \infty)$
NOTE: My first version of the answer had the Trues and Falses reversed so the answer was wrong. I now realize just what I did wrong--thanks for the corrections in the comments!
A: 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.
A: Reducing both sides to the same denominator, the inequality is equivalent to
$$\frac{x(x+3)}{x+7}>0\iff x(x+3)(x+7)>0\;\text{ AND }\; x\ne -7.$$
Now, by the I.V.T., the  polynomial has a constant sign between its roots, which determine 4 intervals and signs alternate when stepping from one interval to the next (or the previous) interval. As $\;\lim\limits_{x\to\infty}x(x+3)(x+7)=+\infty$, the successive signs are as follows:
$$\begin{array}{r}x\\\hline x(x+3)/(x+7)\end{array}\quad\begin{array}{rccccccc}
-\infty&&-7&&-3&&0&&+\infty\\\hline
&-&0&+&||&-&0&+
\end{array}$$
and we see at once the set of solutions is $\;(-7,-3)\cup(0,+\infty)$.
A: This is not a faster way, it is an indication of how to get better at this. Learn how to sketch a graph of the hyperbola $y = 4x /(x+7).$ Draw $y=x$ on the same graph paper. Then explain what you see. You can get a pdf of graph paper, then print out as needed, from https://www.printablepaper.net/category/graph

