Can someone explain the math involved in parallel/series/combined circuits?

The trick is to pay attention to the potentials (volts) across each segment marked by a connection.

For a series, for example: x--------x--------x where the x's are connections and the ------- is the wire with some resistance R.  Assume a battery plus is attached to the LH x and its minus is attached to the RH x.  The battery has a V volts rating.  So we know V = iR + iR; where i is the current through each of the two segments.  We know i is the same in both segments because a wire of a given resitance is like a hose of a given size, that is, the current in both has to be the same throughout their respective lengths.

Further, as IR = v, a voltage drop, we also know that V = v + v and that the voltage across the inner and two outer contacts is V/2 = v for each of the two segments.  The equivalent resistance rho for this series line of two segments is simply rho = 2R because i2R = 2v = V.

Now let's add a parallel line.

../----X------\
x------x-------x

The resistance in the parallel line through X is r < R in each segment.  The voltage across the two outer x's is still V from the battery.  Thus, the voltage across the parallel line with the X is the same voltage as across the original line with the x's.  

Recognizing that both lines, through x and X, have the same potential V across them is key to finding the equivalent resistance rho of the parallel circuit.  Because now we can write V = Ir + Ir = iR + iR; then 2Ir = 2iR or I = i(R/r), which, since R > r shows that the current through the parallel line will be greater than that of the original line.  Makes sense because the r < R means the "hose" is larger for the parallel line; so the current will be greater.

The equivalent resitance of our parallel circuit example is found by again noting that V is the potential across both lines (with the X and with the x).  Let's call rho the equivalent resistance, then V = iota rho; where iota is the equivalent current needed to give the same voltage as the battery over the equivalent resistance rho.  

By I = i(R/r) and looking at the parallel circuit diagram, such as it is, we see the total current between the LH x and the RH x is the sum of the two currents through the two lines.  That is, iota = I + i = i(R/r) + i = i(1 + R/r) = i((r + R)/r)  Then from V = iota*rho we have V/iota = rho = (V/i)(r/(r + R))  As V = i(2R) because the voltage is across two segments of R resitance each and each segment in the orginal line is carrying i current, we have V/i = 2R.

Put this all together and rho = (V/i)(r/(r + R)) = 2Rr/(r + R), which is the equivalent resistance of the parallel circuit example.  And we got to this point by keeping in mind that the voltage V is across both lines of the parallel circuit.  As I said ealier, the trick to solving these is in the voltages.

dude... don't flake out on physics class.

A circuit that is made up of only resistors and current/voltage sources can be solved by only three laws.

ohms law: v=ir

Kirchhoff's current law: net current entering any point is equal to the current leaving any point on the circuit.

Kirchhoff' voltage law: the sum of all voltages in a loop in one direction is zero.

Now you can determine all the voltages and currents in a circuit. Since resistors in series are share the same point or node on the circuit they have to have the same current (current law). And because resistors in parallel have to have a voltage that sums up to zero within the parallel loop the voltages have to equal.  

I think the second part of the question has to do with equivalent circuits? Generally any circuit with only sources and resistors can be expressed as a "non-ideal" source. (a voltage source with a resistor in series.) All can be expressed by the linear equation:

v = v_eq + R_eq * i

where v_eq is the equivalent voltage and R_eq is the equivalent resistance.

Hope this helps.