Analyzing a simple circuit with transistor
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\$:
$$\beta=\frac{\text{I}_\text{C}}{\text{I}_\text{B}}\tag2$$


Now, using KCL, we can write:

$$
\begin{cases}
\text{I}_\text{x}=\text{I}_3+\text{I}_4\\
\\
\text{I}_3=\text{I}_1+\text{I}_\text{t}+\text{I}_\text{d}\\
\\
\text{I}_2=\text{I}_\text{x}+\text{I}_5\\
\\
0=\text{I}_\text{d}+\text{I}_1+\text{I}_5\\
\\
\text{I}_2=\text{I}_\text{t}+\text{I}_4\\
\\
\beta=\frac{\text{I}_4}{\text{I}_\text{t}}
\end{cases}\tag3
$$

Using KVL, we get:

$$
\begin{cases}
\text{I}_3=\frac{\text{V}_\text{x}-\text{V}_1}{\text{R}_3}\\
\\
\text{I}_4=\frac{\text{V}_\text{x}-\text{V}_5}{\text{R}_4}\\
\\
\text{I}_2=\frac{\text{V}_3-\text{V}_4}{\text{R}_2}\\
\\
\text{I}_\text{d}=\frac{\text{V}_2}{\text{R}_1}\\
\\
\text{V}_1-\text{V}_3=\alpha\\
\\
\text{V}_1-\text{V}_2=\eta
\end{cases}\tag4
$$

Solving the circuit using your given values, and using \$\beta=10^{20}\$ and \$\alpha=\eta=\frac{7}{10}\$ leads to:

FullSimplify[
 Solve[{Ix == I3 + I4, 
   I3 == (1*10^(-3)) + It + Id, (1*10^(-3)) == Ix + I5, 
   0 == Id + (1*10^(-3)) + I5, (1*10^(-3)) == It + I4, 10^(20) == I4/It, 
   I3 == ((56/10) - V1)/1000, 
   I4 == ((56/10) - V5)/1000, (1*10^(-3)) == (V3 - V4)/1000, 
   Id == V2/1000, V1 - V3 == 7/10, V1 - V2 == 7/10}, {Id, It, Ix, I3, 
   I4, I5, V1, V2, V3, V4, V5}]]


So:

{{Id -&gt; 3900000000000000000029/2000000000000000000020000, 
  It -&gt; 1/100000000000000000001000, 
  Ix -&gt; 7900000000000000000069/2000000000000000000020000, 
  I3 -&gt; 5900000000000000000069/2000000000000000000020000, 
  I4 -&gt; 100000000000000000/100000000000000000001, 
  I5 -&gt; -(5900000000000000000049/2000000000000000000020000), 
  V1 -&gt; 5300000000000000000043/2000000000000000000020, 
  V2 -&gt; 3900000000000000000029/2000000000000000000020, 
  V3 -&gt; 3900000000000000000029/2000000000000000000020, 
  V4 -&gt; 1900000000000000000009/2000000000000000000020, 
  V5 -&gt; 2300000000000000000028/500000000000000000005}}


And:

{{Id -&gt; 0.00195, It -&gt; 1.*10^-23, Ix -&gt; 0.00395, I3 -&gt; 0.00295, 
  I4 -&gt; 0.001, I5 -&gt; -0.00295, V1 -&gt; 2.65, V2 -&gt; 1.95, V3 -&gt; 1.95, 
  V4 -&gt; 0.95, V5 -&gt; 4.6}}




EDIT 1

For \$\beta\to\infty\$:

{{Id -&gt; 39/20000, It -&gt; 0, Ix -&gt; 79/20000, I3 -&gt; 59/20000, 
  I4 -&gt; 1/1000, I5 -&gt; -(59/20000), V1 -&gt; 53/20, V2 -&gt; 39/20, 
  V3 -&gt; 39/20, V4 -&gt; 19/20, V5 -&gt; 23/5}}


Wich is:

{{Id -&gt; 0.00195, It -&gt; 0., Ix -&gt; 0.00395, I3 -&gt; 0.00295, I4 -&gt; 0.001, 
  I5 -&gt; -0.00295, V1 -&gt; 2.65, V2 -&gt; 1.95, V3 -&gt; 1.95, V4 -&gt; 0.95, 
  V5 -&gt; 4.6}}




EDIT 2

FullSimplify[
 Solve[{Ix == I3 + I4, I3 == I1 + It + Id, I2 == Ix + I5, 
   0 == Id + I1 + I5, I2 == It + I4, β == I4/It, 
   I3 == (Vx - V1)/R3, I4 == (Vx - V5)/R4, I2 == (V3 - V4)/R2, 
   Id == V2/R1, V1 - V3 == α, V1 - V2 == η}, {Id, It, Ix, 
   I3, I4, I5, V1, V2, V3, V4, V5}]]


Which gives:

{{Id -&gt; -((I1 R3 - Vx + (I2 R3)/(1 + β) + η)/(R1 + R3)), 
  It -&gt; I2/(1 + β), 
  Ix -&gt; I2 - (I2 R3)/((R1 + R3) (1 + β)) + (
    I1 R1 + Vx - η)/(R1 + R3), 
  I3 -&gt; (I2 R1 + 
    I1 R1 (1 + β) + (1 + β) (Vx - η))/((R1 + 
      R3) (1 + β)), I4 -&gt; (I2 β)/(1 + β), 
  I5 -&gt; (-I1 R1 - Vx + (I2 R3)/(1 + β) + η)/(R1 + R3), 
  V1 -&gt; (-I2 R1 R3 - 
    I1 R1 R3 (1 + β) + (1 + β) (R1 Vx + R3 η))/((R1 +
       R3) (1 + β)), 
  V2 -&gt; (R1 (-I1 R3 + Vx - (I2 R3)/(1 + β) - η))/(R1 + R3),
   V3 -&gt; -α - (I2 R1 R3)/((R1 + R3) (1 + β)) + η - (
    R1 (I1 R3 - Vx + η))/(R1 + R3), 
  V4 -&gt; -I2 R2 - α - (
    I2 R1 R3)/((R1 + R3) (1 + β)) + η - (
    R1 (I1 R3 - Vx + η))/(R1 + R3), 
  V5 -&gt; Vx - (I2 R4 β)/(1 + β)}}