Help with Thevenin equivalent circuits

[u/throwaway8490247]

In advance, I appreciate the help. I know this sub has a "do your own homework" attitude, but I've been staring at this problem for hours and can't figure out how to proceed. I understand TEC circuits pretty well, and had minimal issues with every other problem, but this specific one has stumped me. I've reread my chapter and watched TEC lessons on Youtube, but can't find anything that fits this specific example.

I've got half of this solved. Finding the Thevenin Resistance is simple enough, I shorted the Voltage Sources and opened the Current Source. 1200 +3300 + ((2200*5600)/(2200+5600)) is ~6080 ohms.

Where I'm running into an issue is finding the Thevenin Voltage. If the Load Resistor is opened, then it seems to me that the entire circuit is open. If the circuit is open, then the current source has nowhere to flow, and I can't find the solution. I would convert the current source to a voltage source if I could, but I don't think that can be done, because there is no parallel resistor.

The best solution I've come up with is to find the voltage at the junction between the two voltage sources and their resistors, while treating the current source as non existent because it doesn't have anywhere to flow to. However when testing this result (12.41 V) in a simulator it doesn't seem to work. What am I doing wrong?

Edit: Solved, Thank you everyone for your help. If anyone is having trouble with a problem like this themselves, I worked it out and posted the steps in the comments below.

[u/South-Mine7572]

If the load is open then all the current will flow through the 3.3k resistor find the voltage at the tip of the current source and since the 1.2k resistor is connected to open circuit and no current is flowing the voltage difference across it is zero meaning it has the same voltage on both terminals (the voltage of the other terminal of the current source) A KCL equation and KVL equation and your circuit is solved

[u/throwaway8490247 → u/South-Mine7572]

Thank you! This made it click for me. I think I had some kind of mental block because the current is flowing into a negative voltage source rather than ground. Obvious in hindsight lol

For anyone who may come across this in the future and is having trouble as well here is what I did;

I1 + I2 + I3 = 0 (V will stand in for the voltage at the junction between the 3 resistors)

(22-V)/2,200 + (-12-V)/5,600 + .005 = 0

(22-V)/2,200 + (-12-V)/5,600 + = -.005 | Bring the known current to the other side

(5,500)(22-V) + (2,200)(-12-V) = -.005(5,500)(2,200) | Multiply each side by (5,600)(2,200)

121,000 - 5,500V - 26,400 - 2,200V = -61,600 | Do the math

96,800 - 7,800V = -61,600 | Add like variables

-7,800V = -158,400 | Subtract 96,800 from both sides

V = ~20.308 | Divide each side by -7800

This is the voltage at the junction, now to get the voltage at the tip of the current source add the voltage drop across the 3.3k resistor (.005 * 3,300 = 16.5 V) so, 20.308 V + 16.5 = 36.808 V

Thevenin Voltage = 36.808 V