How do you solve this circuit with the node voltage method?
Redraw the schematic so that the whole thing isn't quite so confusing. (Positive at the top, negative at the bottom, signal input at left, signal output at right. Don't bus power rails around -- it's confusing to do that.)







simulate this circuit &ndash; Schematic created using CircuitLab

Note that I've used the schematic editor that is available to you when you create a question here? You should use it, too.

You should be able to see where I got the \$-5\:\text{V}\$ node point. I've labeled all of the remaining possible nodes (ground and \$-5\:\text{V}\$ are already labeled, obviously.)

At this point, I can see that you really only need to do nodal analysis on \$V_X\$. The other nodes are almost certainly unimportant.

Also, you should be able to see that \$R_4\$ is completely irrelevant since the current source has \$\infty\$ impedance. You would be perfectly in your rights to simply remove it and everything "behind" the current source. The only thing that matters is what it injects into the \$V_X\$ node.



Some notes here, assuming you were NOT doing nodal analysis:


You should be able to see that \$R_2\$ and \$R_3\$ can be summed up into an equivalent resistor value.
You should be able to see that once (1) is complete, that you are left with a simple voltage divider made up of two resistors. You should be able to Thevenize this almost immediately into an equivalent voltage source and an equivalent series resistance from it.
Given the current source, this then means you can work out the voltage at \$V_X\$, almost trivially.


Of course, if you do the above all you've done is avoid the nodal analysis.



At this point, you really should be able to perform your nodal analysis. You can include a statement for both \$V_X\$ and \$V_Y\$ and solve them simultaneously. Or you can lump \$R_2\$ and \$R_3\$ and only write one equation for \$V_X\$. Which method you apply depends entirely on what you feel is more appropriate to your circumstances.