How to calculate x in formula $ y =\left( \frac{\frac{17000}{x+400}+8.5}{100}+ 1\right) x $? I am not even sure how it's officially called (so not sure with tag to give it). As an example if you have a math problem $y = x + 1$. You have a $y$ value, but not $x$. To you revese the problem $y - 1 = x$. However I have a rather more complicated problem on with I tried this, but I fear it might not be possible. The formula is. 
$$
y =\left( \frac{\frac{17000}{x+400}+8.5}{100}+ 1\right) x 
$$
This is the orgional formula for calculating a price including it's profit. $x$ is the value without profit. $\frac{17000}{x+400}+8.5$ calculated the profit in procentage and the rest applies the procentage to the value without profit. So $y$ is the value with profit.
Now I want to be able to give $y$ and remove the profit from it. Not sure if this is possible. I tried it and I was not able to do so. I even tried to split the problem in part.
You have $y = \frac{17000}{x+400}+8.5$, $\left(y=\frac{k}{x+h}+a\right)$
then $z = \frac{y}{100} + 1$
then $c = z * x$
Reversing the first two part is not a problem $\frac{17000}{y-8.5} - 400 = x$ and $(z-1)*100 = y$ however the last part would be $c * x = z$. Again we need value $x$ to calculate it. With is the part I want calculated. So I think I have it right that I can not reverse this problem. If not then is there an other way to calculate what I want?
Any help would be great! 

Try to write things with the same denominators and then ollect the $x$ terms. THat way you can get a regular polynomial, which is much easier to solve. Start by using the same denominators for all of it, and multiply in the $x$:
$$
y=x\left(\frac{\frac{1700+108.5(x+400)}{x+400}}{100}\right)=x\left(\frac{1700+108.5(x+400)}{100(x+400)}\right)=\left(\frac{108.5x^2+45100x}{100x+40000}\right)
$$
Multiply both sides with the denominator:
$$
\begin{align}
y(100x+40000)&=108.5x^2+45100x\\
40000y&=108.5x^2+x(45100-100y)\\
108.5x^2+x(45100-100y)-40000y&=0\\
\end{align}
$$
and then you get a second degree polynomial which can be solved using regular methods. 
