Playing with it and learn is a great way to approach... Besides playing around with voltage regulators, allow me to give you a couple more ideas to play and learn...
You could use a zener diodes. 6V zener diode is available. LED's also have a "constant voltage" property up to a point.
If you use two 6V zener diodes instead of two 10K ohms... Each zener will try to hold the voltage to 6V giving you a much nicer divider. Once when a zener exceeds it designed voltage, it will "break down" and let current just pass on. Look at this link to see how zener diodes can be used to regulate voltage.
https://www.electronics-tutorials.ws/diode/diode_7.htmlNow for something more fun...
I said LEDs has a "constant voltage" property up to a point as well. When the current is high enough for an LED to light, increasing the current doesn't increase the voltage linearly like a resister would. It would increase voltage by some but not as much - till current is so high that it burns out. So, a regular LED can be use as an
improvise voltage regulator - not well, but to some degree.
Say you find yourself four 3V LEDs. Connect the 4 in a series, top 2 adds up to 6V and bottom two add up to be 6v. So you would have about 6V on the top two LEDs and about 6V on the bottom two LEDs.
If your loads are not drawing a lot of current and they are rather close in current needs, that could work a lot better than dividing resistors. Typical blue and white LEDs are in the 3.2-3.7 volt range, they will serve well in this experiment.
Now the fun part in making it work...
If you get some of those 700mA+ LEDs (over a dollar each instead of 5 cents), you wont need by-pass as all, but for the cheap 5-cents LEDs, you will need the by-pass. Read on, you will understand.
To reuse my earlier example of 0.5A top load and 0.1A for bottom load, you have a delta of 0.4A. So your two LEDs at the bottom must be able to carry that 0.4A that the 0.1A load isn't using, or your bottom LEDs will burn-out and probably cause other issues at well. But, we need certain minimum going via the LEDs on the top load. Figure typical 5-cents LEDs are about 1/4W to 1/8W, at 1/8W, and takes about 20mA to light up (bring it up to voltage), so the top load is about 0.52A. Each of the two bottom LED in series are 3V, and it needs to carry that 0.42A excess current that the bottom load doesn't use. Typical 5 cents LEDs are 1/4 or 1/8 watt, about 40mA max. 420mA excess is way too much for them to carry. So one would need a by-pass resister so the excess current is not all for the LEDs to carry. Bottom two LEDs adds to 6V, bypass current is 0.42A, a 14 or 15ohms resister should do it nicely.
So, your top 10K ohm running the 0.5A load (the larger) is replaced by 2 typical blue (or white) LEDs (3.2 to 3.4 volt
typically) in series. 3.4V+3.4V putting it to
about 6.8Volt max. Your bottom 10K with the smaller load is replaced by 2 blue LEDs in series then parallel by a 15ohm for by-pass... This would work a lot better than a 10K at top and a 10K at bottom voltage divider for 0.5A top load and 0.1A bottom load. Your top and bottom are both regulated to
around 6.8V max, and bottom one use 0.4A less than top. So unlike the bare 10K resister which changes the bottom to > 9Volt (as seen in my example in the earlier reply), this will fix it to
around 6.8V max at the bottom.
Of course 0.5A and 0.1A are my examples and not your actual loads. You will have to redo the math and do some experimentation with your actual load's current range -- that would be the fun part.
Depending on your load's current requirement delta and variation, this may not work, but trying and experimenting is the fun part.Lots of ways to play with that to have fun and learn...
EDIT: I better add this:
If you do try this, make sure that your loads can handle some excessive voltage. During the experimentation, you would also be burning out some of the LED's you use for voltage regulation. So be prepare, you will kill some LEDs and you may damage your load.