Following up the exact analysis earlier, this CE bjt amplifier can be analysed with the approximation quickly. Still starting from this:
Removing all AC components such as AC signal input and capacitors to have this circuit:
Now the circuit is ready to be analysed by without Thevenizing using voltage divider rule for the base voltage and then obtain the amplfied current IE.
VB=VCC*R2/(R1+R2)=15V*30k/(130k+30k)=2.8125V approximately without considering the drop due to the base current loading effect. Following the diode BE drop of 0.7V,
Since IB is small or insignificant, IE=VRE/RE=2.1125V/1.2K=1.7604mA
Compare the IE=1.4mA using the exact analysis, there is a slight difference. Can you tolerate this error?
Everything else can be solved after this: VRC=IC*RC, VRE=IE*RE and IE=IC approximately and using KVL to find VCE=VCC-VRC-VRE. VCE found to be 7.07V which is a little different from the exact analysis.