I don't understand this: On an LED Voltage-Current graph, does the LED "pull" a certain current like it has resistance?
Yes, to understand intuitively this phenomenon known as "current steering", you can think of the LED of as a resistor... but having nonlinear resistance with the property to keep up the voltage across it constant. It is interesting to see how it does this "magic". Here is a simple but very intuitive explanation.
If the LED was an ordinary (ohmic) resistor, the voltage across it would be V = I.R. But it behaves as a "dynamic" resistor that decreases its resistance R when the current I increases and v.v. so their product - the voltage V, does not change.
When you connect two such voltage stabilizing elements in parallel, they form a dynamic current divider. Each of them tries to set its voltage across this network... and if there is at least a small difference between their voltage thresholds, they vigorously change their resistances to divert the current ("to 'pull' a certain current").
The advantage of this simple explanation is that you can emulate this arrangement by two variable resistors (rheostats) in parallel.
EDIT - answering the OP's questions in the comments below
Not because I want to do so (could save some time), but just to understand concepts.
Exactly... all we need to understand concepts... "to see the forest for the trees". In this way, we can make a connection between seemingly different circuit phenomena.
Do you think you could explain another situation? If I had 5V with a 300 ohm resistor, and after it split into parallel, each side had an LED. One's datasheet showed 1.9V at 12mA, the other 1.9V at 30mA (a large difference). This seems like it would "pull" 42mA, but the resistor would only allow 10.3 mA to come through.
First of all, we have to create a notion about the non-linear resistance. Most people do not feel a need for this; they simply accept that it is what its IV curve represents... they take it for granted. But we are both different from them and we need an even deeper explanation. We want to know not only that the curve is such (nonlinear) but also to somehow imagine how such a shape is obtained in principle.
1. Presenting the red LED1 as a "dynamic resistor". The most natural and intuitive way to answer this question is to imagine LED as a varying resistor (rheostat) that changes its own resistance. This is illustrated by the graphical interpretation in Fig. 1 where the supply voltage varies in the range 0 - 5 V. The resistor R can be considered as internal resistance of this real source. As its voltage increases, its IV curve moves (translates) to the right. At the same time, the IV curve of the static resistance R1 of the nonlinear "resistor" (LED1) does not stay immovable but rotates counterclockwise. As a result, the intersection (operating) point OP1 moves upwards along and pictures the LED1 nonlinear characteristic. So, the idea is to dynamically change the current resistance of the element.
Fig. 1. LEDs as dynamic resistors (graphical representation: top - red LED1; bottom - green LED2
2. Presenting the green LED2 as a "dynamic resistor".
In a similar way, we can explain how the non-linear IV curve of the second "resistor" (the green LED2) is obtained. The only difference is that it is more sloping.
Could it be that both of the LED's I could count as resistance, then solving for the exact current per LED?
Exactly... This is what I will show with the following drawing...
3. Presenting the two LEDs in parallel by an equivalent "dynamic resistor". I fully accept your idea to replace the combination of two nonlinear resistors (red and green LEDs) in parallel with only one "dynamic resistor" (represented by a blue IV curve).
Fig. 2. LEDs in parallel as equivalent resistor: top - nominal R (74 ohm); bottom - higher R (300 ohm)
If we supply the LED network through a resistor R with a nominal value of 74 ohm (3.1V/42 mA) - the upper Fig. 2a, everything will be fine and will correspond to the datasheet. But if you supply it through R = 300 ohm - the lower Fig. 2b, the IV curve of the real voltage source (aka "load line"), will tilt significantly... and both voltage and common current will decrease.
Is there some resistance that I am not thinking about?
No, there are no other resistance; only the LEDs have increased their static resistances.