Okay. You got the direction right with \$R_1\$ and the current source to the left of it!!! This is great!
Let me suggest some help now:
I didn't take everything you did in your update. Just the \$V_1\$ and \$R_1\$ parts. Notice that the two current sources cancel out. And since they have \$\infty\$ impedance you can just cut them out from the circuit, as shown above.
Now you can see that \$R_1\$ and \$R_3\$ (as you show) are in parallel. And can be combined into a single resistor of \$25\:\Omega\$:
But now \$R_2\$ and the newly created \$25\:\Omega\$ resistor can be combined, as they are in series with each other. So that's the same as:
And at this point, this slight modification may help a little further:
You should now see a voltage divider presented to \$V_2\$. And you should be able to work out the Thevenin equivalent of those three parts, making them into a new voltage source and just one resistor (instead of two.) Then you can combine that with what remains.
Does this help you get closer?
I really enjoyed seeing your edited addition to your question. That's the kind of thing that tells me you will do well. I'm excited for you and proud to see you moving forward like this. +1 for the added work!