NPN transistor numerical
Your "Vx = VIn×(RB/RB+RA)" is incorrect as that is only valid if Ib = 0, which it is not.

You're applying the voltage divider formula to this but it cannot be used here as it ignores that there is current flowing from the output of that voltage divider, that current is Ib.

How I would solve this:

I analyze the circuit of Ra, Rb, Re and the NPN as if it is:







simulate this circuit &ndash; Schematic created using CircuitLab

I prefer to use Thevenin for that, once for \$V_{in}\$ and \$R_a\$ and once for \$V_{BE}\$ and \$R_e\$ which would give:







simulate this circuit

Now I can combine the currents \$I_{tot} = V_{in}/R_a + V_{be}/R_e\$

and all resistors in parallel: \$R_{tot} = R_a ||R_b||R_e\$

Then \$V_1 = I_{tot} * R_{tot}\$

Then the current through Re: \$I(R_e) = V_1 / R_e\$

That current is the \$I_b\$ I need to know.

Multiply that by \$\beta\$ and you have \$I_c\$

Then it is easy to determine \$V_c\$ which is probably the collector potential (the question should have made that more clear).

Assuming that \$V_b\$ is the base potential then that's easy as the emitter is grounded and \$V_{BE}\$ is given.