Q:

How to derive partial gas equation?

If the following gasses are to mix, what is the partial pressure of neon?

The correct answer is 0.1. I cannot figure out how my professor derived this equation. From the ideal gas law, all I can arrive to is this:
$P_x = \frac{P_tV_t}{V_x}$
This gives me an answer of 4.4, which is obviously incorrect.
I've determined the equation he used is as follows:
$P_x = \frac{P_xV_x}{V_t}$
which gives the correct answer of 0.1. How did he arrive to this answer?

A:

The total pressure of the system is:
$\frac{1.0\times0.5+1.0\times0.4+2.0\times0.2}{1.0+1.0+2.0}=1.3/4=0.325$
i.e. just the sum of the $PV$s divided by the sum of the $V$s.  Then $0.325$ is the sum of the partial pressures by the Law of Partial Pressures.  So this at least suggest that the individual partial pressures are near $0.1$.
Can you figure out how to parcel out the individual pressures from here?

A:

Prepare for an amazing feat of algebra! I sincerely hope there is a shorter way to the endpoint. (there is! See the end of my answer.)
To calculate partial pressure, $P_{\ce{He}}$, we need the total pressure $P_t$ and the fraction of the gas that is He $X_{\ce{He}}$: $$P_{\ce{He}}=X_{\ce{He}} P_t$$
You have $PV=nRT$. $T$ is not given, so assume it is constant. $n$ is unknown, but the system is closed, so $n_x$ for each gas is constant. Thus we have: 
$$PV=\text{ constant}$$

Total Pressure

It is tempting to write variations of $$P_1 V_1 = P_2 V_2$$
However, for each gas, both $P$ and $V$ are changing, so we need to consider their products: $$(PV)_1 = (PV)_2$$
Thus, as trb456 suggests, $$P_t V_t = (PV)_t \implies P_t =\frac{\sum{(PV)_i}}{\sum{V_i}}$$

Fraction of He

The fraction of the mixture that is helium is determined by the ratio $\dfrac{n_{\ce{He}} }{n_t}$. Since $n=PV/RT$, and $R$ and $T$ are constant, we can write: $$X_{\ce{He}}=\frac{n_{\ce{He}}}{\sum{n_i}}=\frac{(PV)_{\ce{He}}}{\sum{(PV)_i}}$$

At last!

$$P_{\ce{He}}=X_{\text{He}} P_t = \left( \frac{(PV)_{\ce{He}}}{\sum{(PV)_i}}\right) \left( \frac{\sum{(PV)_i}}{\sum{V_i}}\right)$$
$$P_{\ce{He}_2}=\frac{(PV)_{\ce{He}_1}}{V_t}$$

Or, as it struck me as I finished:

Helium expands to fill the total volume. Now we can use $P_1 V_1 = P_2 V_2$ and completely ignore the other gasses (there is a lot empty space for the He atoms to fit in).
$$P_{\ce{He}_2}=\frac{P_{\ce{He}_1}V_{\ce{He}_1}}{V_t}$$
This equation is very similar to your second equation, except the subscripts are added to denote initial and final pressure of helium. 

