simple circuit analysis kcl kvl
Thevenin
Let's start with a simple voltage divider. I'll use the name \$V_\text{CC}\$ for the name of a given voltage source for the divider, for lack of a better name. The point of the exercize is to replace the circuit on the left with an equivalent circuit on the right. This is the process of creating a Thevenin equivalent.



simulate this circuit &ndash; Schematic created using CircuitLab
For the above case, the right-hand side Thevenin equivalent is \$V_\text{TH}=V_\text{CC}\cdot\frac{R_2}{R_1+R_2}\$ and \$R_\text{TH}=\frac{R_1\cdot R_2}{R_1+R_2}\$.
So, now your circuit can be transformed using the above Thevenin idea (if you need a proof, there are plenty available on the web.)



simulate this circuit
From the right side, you can compute the current as \$I=\frac{6\:\text{V}-700\:\text{mV}}{6.667\:\text{k}\Omega+20\:\text{k}\Omega}=198.75\:\mu\text{A}\$ (the diode voltage subtracts from the total voltage available, prior to dividing by the total series resistance) and the voltage at \$V\$ as being \$V=I\cdot 20\:\text{k}\Omega=3.975\:\text{V}\$. (\$V_\text{X}\$ would be \$700\:\text{mV}\$ above \$V\$, of course.)
As you can see, this appears to match the answer you presented at the outset.
Nodal
Using nodal analysis I get:
$$\begin{align*}
\frac{V}{R_3}&amp;= I_{D_1}\\\\
\frac{V_\text{X}}{R_1}+\frac{V_\text{X}}{R_2}+I_{D_1}&amp;=\frac{9\:\text{V}}{R_1}+\frac{0\:\text{V}}{R_2}\\\\
V_\text{X}&amp;=V+V_{D_1}
\end{align*}$$
This is three equations and three unknowns. Given \$V_\text{CC}=9\:\text{V}\$ and \$V_{D_1}=700\:\text{mV}\$ and your resistor values, this provides:
$$\begin{align*}
I_{D_1}&amp;=198.75\:\mu\text{A}\\\\
V&amp;=3.975\:\text{V}\\\\
V_\text{X}&amp;=4.675\:\text{V}
\end{align*}$$
Same results as before.