Amplifier design using BJT
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\$:
$$\beta=\frac{\text{I}_\text{C}}{\text{I}_\text{B}}\tag2$$
Emitter voltage:
$$\text{V}_\text{BE}=\text{V}_2-\text{V}_3\tag3$$


When we use and apply KCL, we can write the following set of equations:

$$
\begin{cases}
\text{I}_\text{x}=\text{I}_\text{C}+\text{I}_3\\
\\
\text{I}_3=\text{I}_\text{B}+\text{I}_4\\
\\
\text{I}_\text{x}=\text{I}_\text{E}+\text{I}_4
\end{cases}\tag4
$$

When we use and apply Ohm's law, we can write the following set of equations:

$$
\begin{cases}
\text{I}_\text{C}=\frac{\text{V}_\text{x}-\text{V}_1}{\text{R}_1}\\
\\
\text{I}_\text{E}=\frac{\text{V}_3}{\text{R}_2}\\
\\
\text{I}_3=\frac{\text{V}_\text{x}-\text{V}_2}{\text{R}_3}\\
\\
\text{I}_4=\frac{\text{V}_2}{\text{R}_4}
\end{cases}\tag5
$$

Now, I use Mathematica to solve your problem using \$\$:

In[1]:=VBE = -6/10;
\[Beta] = 200;
Vx = 24;
V1 = 16;
V2 = 25*10^(-3);
FullSimplify[
 Solve[{IE == IB + IC, \[Beta] == IC/IB, VBE == V2 - V3, 
   Ix == IC + I3, I3 == IB + I4, Ix == IE + I4, IC == (Vx - V1)/R1, 
   IE == V3/R2, I3 == (Vx - V2)/R3, I4 == V2/R4, 
   I3 == (1/10)*IC, (Vx - V2)*I3 == 10*10^(-3)}, {IE, IB, IC, I3, I4, 
   V3, R1, R2, R3, R4, Ix}]]

Out[1]={{IE -&gt; 201/47950, IB -&gt; 1/47950, IC -&gt; 4/959, I3 -&gt; 2/4795, 
  I4 -&gt; 19/47950, V3 -&gt; 5/8, R1 -&gt; 1918, R2 -&gt; 119875/804, 
  R3 -&gt; 919681/16, R4 -&gt; 4795/76, Ix -&gt; 22/4795}}

In[2]:=N[%1]

Out[2]={{IE -&gt; 0.00419187, IB -&gt; 0.0000208551, IC -&gt; 0.00417101, 
  I3 -&gt; 0.000417101, I4 -&gt; 0.000396246, V3 -&gt; 0.625, R1 -&gt; 1918., 
  R2 -&gt; 149.098, R3 -&gt; 57480.1, R4 -&gt; 63.0921, Ix -&gt; 0.00458811}}