How to combine two resistors in a circuit reduction
Simplifying series/parallel resistive circuits you use two formulas.  

For N resistors in series, 

\$R_T=R_1+R_2+...+R_N\$ 

and for N resistors in parallel,

\$1/R_T=1/R_1+1/R_2+...+1/R_N\$

or transpose to get:

\$R_T=1/(1/R_1+1/R_2+...+1/R_N)\$

You can solve these circuits by identifying series connected resistors and combining them, then parallel connected resistors and combining them, and repeating until only one resistor is left.  In the simple circuit you provide:







simulate this circuit &ndash; Schematic created using CircuitLab

There are no simple series connected resistors, so check parallel.  R2 and R3 are connected in parallel, so we combine them using the formula:
\$R_{2+3}=1/(1/100\Omega+1/100\Omega)=1/(2/100\Omega)=50\Omega\$







simulate this circuit

And now we check for series connections again and see we can combine the remaining two resistors.

\$R_T=R_1+R_{2+3}=150\Omega\$

You can invert the formulas and resistances to figure it out based on conductance, but usually this type of problem is figured out in terms of resistance.