Simple transistor resistor amplifier circuit question
Well, we have the following circuit:







simulate this circuit &ndash; Schematic created using CircuitLab

When analyzing a transistor we need to use the following relations:


$$\text{I}_\text{E}=\text{I}_\text{B}+\text{I}_\text{C}\tag1$$
Transistor gain \$\beta\$:
$$\beta=\frac{\text{I}_\text{C}}{\text{I}_\text{B}}\tag2$$


Using KCL, we can write:

$$
\begin{cases}
\text{I}_\text{x}=\text{I}_1+\text{I}_3\\
\\
\text{I}_3=\text{I}_\text{T}+\text{I}_4\\
\\
\text{I}_\text{x}=\text{I}_2+\text{I}_4\\
\\
\text{I}_2=\text{I}_\text{T}+\text{I}_1\\
\\
\beta=\frac{\text{I}_1}{\text{I}_\text{T}}
\end{cases}\tag3
$$

Using KVL, we can write:

$$
\begin{cases}
\text{I}_1=\frac{\text{V}_\text{x}-\text{V}_1}{\text{R}_1}\\
\\
\text{I}_2=\frac{\text{V}_3}{\text{R}_2}\\
\\
\text{I}_3=\frac{\text{V}_\text{x}-\text{V}_2}{\text{R}_3}\\
\\
\text{I}_4=\frac{\text{V}_2}{\text{R}_4}\\
\\
\text{V}_2-\text{V}_3=\alpha
\end{cases}\tag4
$$

Now, we can solve for \$\text{R}_2,\text{R}_3\$ and \$\text{R}_4\$.



Using your values we get:

FullSimplify[
 Solve[{Ix == 2*10^(-3) + I3, I3 == It + I4, Ix == I2 + I4, 
   I2 == It + 2*10^(-3), 100 == 2*10^(-3)/It, 
   2*10^(-3) == (9 - V1)/1200, I2 == (V3)/R2, I3 == (9 - V2)/R3, 
   I4 == (V2)/R4, V2 - V3 == 3/10, V1 - V3 == 3, I3 == 10*It}, {Ix, 
   It, I2, I3, I4, V1, V2, V3, R2, R3, R4}]]


Which gives:

{{Ix -&gt; 11/5000, It -&gt; 1/50000, I2 -&gt; 101/50000, I3 -&gt; 1/5000, 
  I4 -&gt; 9/50000, V1 -&gt; 33/5, V2 -&gt; 39/10, V3 -&gt; 18/5, 
  R2 -&gt; 180000/101, R3 -&gt; 25500, R4 -&gt; 65000/3}}


And that is the same as:

{{Ix -&gt; 0.0022, It -&gt; 0.00002, I2 -&gt; 0.00202, I3 -&gt; 0.0002, 
  I4 -&gt; 0.00018, V1 -&gt; 6.6, V2 -&gt; 3.9, V3 -&gt; 3.6, R2 -&gt; 1782.18, 
  R3 -&gt; 25500., R4 -&gt; 21666.7}}