Trouble analyzing transistor's circuit
Normally, a circuit like this really would take careful consideration. But less so, provided some of the givens. I'm a little concerned about your value for \$R_4\$, only because the schematic editor uses that value as a default and I'm not sure you intended that value. \$R_3\$ seems a bit odd, as well. But let's go with it:







simulate this circuit &ndash; Schematic created using CircuitLab

I think you can see that pretty much everything turns on the value of \$V_x\$. So let's just do nodal analysis and shoot for its value:

$$\begin{align*}
\frac{V_x}{R_2} + \frac{V_x}{R_3} + \frac{V_x}{R_4} + I_{B_1} &amp;= \frac{600\:\textrm{mV}}{R_2} + \frac{20\:\textrm{V}}{R_3} + \frac{V_x-1\:\textrm{V}}{R_4} \\
  \\
V_x\cdot\left(\frac{1}{R_2} + \frac{1}{R_3}\right) + I_{B_1} &amp;= \frac{600\:\textrm{mV}}{R_2} + \frac{20\:\textrm{V}}{R_3} - \frac{1\:\textrm{V}}{R_4} \\
  \\
V_x &amp;\approx 12\:\textrm{V} - 755\cdot I_{B_1}
\end{align*}$$

Hmm. We don't know \$I_{B_1}\$. So let's re-group a bit.

You know that \$Q_3\$ has \$\beta=50\$ and that it's collector current must include \$I_{R_4}\$ and the base current of \$Q_2\$. This must be more than \$10\:\textrm{mA}\$. It follows that \$I_{B_3}\ge 200\:\mu\textrm{A}\$. So you know that \$I_{R_2}\ge 260\:\mu\textrm{A}\$ and therefore also that \$V_x\ge 9.18\:\textrm{V}\$.

You also know that \$R_3\$ must include \$I_{R_2}\$, \$I_{R_4}\$, and \$Q_1\$'s base current. So \$I_{R_3}\ge 10.26\:\textrm{mA}\$ and therefore \$V_x\le 12.07\:\textrm{V}\$.

So we now can at least say this:

$$\begin{align*}
9.18\:\textrm{V} \le \left(V_x \approx 12\:\textrm{V} - 755\cdot I_x\right) \le 12.07\:\textrm{V}
\end{align*}$$



Thanks for asking questions about this. Here's my additions. I've added the current \$I_{B_1}=I_x\$ to the schematic. (This must be returned to the collector of \$Q_3\$ via the base of \$Q_2\$, since both \$Q_1\$ and \$Q_2\$ share the same \$\beta=100\$ value.)

So we have the above equation. But what is missing is the value for \$I_x\$.

\$I_x\$ adds to the \$10\:\textrm{mA}\$ from \$R_4\$, so that the collector current of \$Q_3\$ increases by that amount. This means that the base current for \$Q_3\$ increases by \$\tfrac{1}{50}\$ of that. So we can set this up:

$$\begin{align*}
V_x = 12 - 755\cdot I_x &amp;= 600\:\textrm{mV}+\left(60\:\mu\textrm{A}+\frac{10\:\textrm{mA}}{50}+\frac{I_x}{50}\right)\cdot R_2 \\
 \\
12 - 755\cdot I_x &amp;= 600\:\textrm{mV}+\left(260\:\mu\textrm{A}+\frac{I_x}{50}\right)\cdot R_2 \\
 \\
I_x &amp;\approx 2\:\textrm{mA}
\end{align*}$$

From this, we can now estimate \$V_x\approx 10.5\:\textrm{V}\$ and then that \$V\approx 10\:\textrm{V}\$.

Thanks for asking additional questions. It helped me to add my own additional thinking to this and to provide what I think is now a more complete answer to your question.