It requires an iterative solution (or solving a set of equations simultaneously) to find all of the node voltages and device currents.
Refer back to post #8. To solve for a consistent set of node voltages and branch currents with pencil and paper, you would have to guess a value of V(b), which determines Ib(Q2) =(18-V(b))/3.3meg, which makes Ib(Q1)=β2*Ib(q2), which makes V(e) = β1*Ib(Q1)*390. Now, we know that V(b) is V(e) + two Vbe drops, so V(b)= V(e)+2*0.65= V(e)+1.3.
You will have to keep iterating around that loop until you find a voltage at V(e) that is consistent with the steps above. I cannot see how you can do this without an iterative solution, or using an assumption as to what the Vbe of the two transistors is and then setting up a solution of two equations in two unknowns (V(b) and V(e)).
That is what the simulator does intrinsically. It effectively iterates the expression(s) above until it reaches a consistent set of branch currents and node voltages, all based on the transistor models. In post #8, I showed how V(e), and by implication V(b), is effected as β1 = β2 ranges from 50 to 300.
Here, I repeat the simulation asking LTSpice to do just the DC (.op) solution as a function of β (the x-axis). I plot V(b), V(e), the difference V(b)-V(e), and the Q2 base current Ib(Q2), all vs β of both transistors.
Les Jones and
aliraza54, compare your pencil and paper solutions to these plots: