How to solve a LED-resistor parallel with a resistor?
I'll try and analyze your circuit in a way that requires you to learn just a few ideas.
Thevenin Equivalent of a Resistor Voltage Divider
A resistor voltage divider looks like this:



simulate this circuit &ndash; Schematic created using CircuitLab
On the left side we have two resistors in series between a power supply. What I'd like to know is what is the voltage at \$+V_\text{TH}\$. But when I ask that question, I have to say, &quot;Relative to what other location in the circuit?&quot; So I've labeled another spot (node) called \$-V_\text{TH}\$, which identifies the location I've picked as the &quot;relative to&quot; answer. I'm asking, &quot;What is the voltage at \$+V_\text{TH}\$ with respect to the voltage at \$-V_\text{TH}\$?&quot;
On the right side, I'm showing you a &quot;cheat&quot; that you will often find in circuits. That is that one of the nodes is named &quot;GND.&quot; This becomes the &quot;default reference point&quot; whenever anyone is talking about the voltage at some other place in the circuit. We just &quot;assume&quot; that's the &quot;relative to&quot; location. So now, I can just ask &quot;What the voltage at \$+V_\text{TH}\$?&quot; and you are then supposed to insert in your head &quot;with respect to GND&quot; in your own head. It's just a &quot;common&quot; that is always inferred whenever anyone talks about voltages at any one point. (Voltages are always a &quot;voltage here with respect to a voltage there,&quot; as they are always relative measurements and have no absolute meaning.)
With that in mind, we can work out the voltage at \$+V_\text{TH}\$. We know the current through the series circuit is \$\frac{5\:\text{V}}{1\:\text{k}\Omega+4\:\text{k}\Omega}=1\:\text{mA}\$. But \$1\:\text{mA}\$ through \$R_2\$ will cause a voltage difference from one end to the other end of the resistor of \$1\:\text{mA}\cdot 4\:\text{k}\Omega=4\:\text{V}\$. So it must be the case that \$+V_\text{TH}=4\:\text{V}\$ with respect to GND (or \$-V_\text{TH}\$.)
This is the often called the Thevenin voltage at \$+V_\text{TH}\$ (with that implied reference to GND, of course.)
There is also a Thevenin resistance. This is a little trickier to gather up, at first. But in this case it will, in effect, just be the resistance of the two resistors take &quot;in parallel&quot; to each other. There's a reason for this. But for now, just believe me. So the Thevenin resistance is \$1\:\text{k}\Omega\mid\mid 4\:\text{k}\Omega=800\:\Omega\$. This allows us to write:



simulate this circuit
Note that the voltage has changed. But we've also simplified the schematic so that there is only one resistor instead of two. This usually makes further analysis easier to do. But before proceeding, let's test this.
Let's consider two different resistor values that we'll place between the \$+V_\text{TH}\$ output wire and the GND wire. Suppose we use \$R_\text{LOAD}=800\:\Omega\$ and \$R_\text{LOAD}=1200\:\Omega\$. We'll analyze the first circuit and then we'll analyze the &quot;Thevenin equivalent&quot; circuit for both cases. So we'll have four results and we'll compare them.



simulate this circuit
On the upper-left, we have \$800\:\Omega\mid\mid 4\:\text{k}\Omega=\frac23\:\text{k}\Omega\$ that is in-series with \$1\:\text{k}\Omega\$. So the total current from the power supply will be \$\frac{5\:\text{V}}{1\:\text{k}\Omega+\frac23\:\text{k}\Omega}=3\:\text{mA}\$. This means that \$R_1\$ will drop \$1\:\text{k}\Omega\cdot 3\:\text{mA}=3\:\text{V}\$, leaving \$+V_\text{TH}=5\:\text{V}-3\:\text{V}=2\:\text{V}\$. From this, we find that \$I_\text{LOAD}=\frac{2\:\text{V}}{800\:\Omega}=2.5\:\text{mA}\$.
On the upper-right, we have a total current of \$\frac{4\:\text{V}}{800\:\Omega+800\:\Omega}=2.5\:\text{mA}\$. Note that all of the total current is flowing through \$R_\text{LOAD}\$. So this matches up with what we just calculated for the upper-left circuit.
On the lower-right, we have \$1.2\:\text{k}\Omega\mid\mid 4\:\text{k}\Omega=923 \frac1{13}\:\Omega\$ that is in-series with \$1\:\text{k}\Omega\$. So the total current from the power supply will be \$\frac{5\:\text{V}}{1\:\text{k}\Omega+923 \frac1{13}\:\Omega}=2.6\:\text{mA}\$. This means that \$R_1\$ will drop \$1\:\text{k}\Omega\cdot 2.6\:\text{mA}=2.6\:\text{V}\$, leaving \$+V_\text{TH}=5\:\text{V}-2.6\:\text{V}=2.4\:\text{V}\$. From this, we find that \$I_\text{LOAD}=\frac{2.4\:\text{V}}{1.2\:\text{k}\Omega}=2\:\text{mA}\$.
On the lower-right, we have a total current of \$\frac{4\:\text{V}}{800\:\Omega+1.2\:\text{k}\Omega}=2\:\text{mA}\$. Note that all of the total current is flowing through \$R_\text{LOAD}\$. So this matches up with what we just calculated for the lower-left circuit.
I think you can see, at least from these examples anyway, that it appears this &quot;trick&quot; works. There exists a Thevenin equivalent for a resistor divider and you can work it out using the above rules.
Hopefully, the above convinces you. But to be sure, you should repeat the above process any number of ways and verify that it &quot;just works right.&quot; It will, I assure you. But you should repeat this process many times with varying values in order to be more fully convinced. It's worth the effort.
Applying the Thevenin Equivalent to your Circuit
If you think closely about your circuit, you'll see the following as true:



simulate this circuit
If we assume that \$V_\text{LED}\approx 2\:\text{V}\$ here, then we would find that \$I_\text{LED}=\frac{4\frac1{26}\:\text{V}-2\:\text{V}}{40\frac5{13}\:\Omega+200\:\Omega}=8.48\:\text{mA}\$.
This is pretty close to your simulation value. However, take note that your simulation yielded \$V_\text{LED}=1.91\:\text{V}\$. If we use that value instead, we find that \$I_\text{LED}=\frac{4\frac1{26}\:\text{V}-1.91\:\text{V}}{40\frac5{13}\:\Omega+200\:\Omega}=8.85\:\text{mA}\$. And that's almost an exact match with the simulation you tried.
Summary
If you work with this simple resistor divider's Thevenin equivalent for a while, you'll find it works well in every situation. So spend some time with it until you are comfortable with how it works for you.
There is much more to Thevenin equivalent circuits. But the voltage divider is a good way to start learning about Thevenin equivalents. I hope you will practice this, over and over, for a while to make sure you accept how well it works for you. Eventually, you can learn more about why it works like this. But for now, it's enough to see that it does work.