OK , Here is the basic method

1. Move the ground connection to node 1 , and just call the old ground node 4
2. Set the 40v source to zero volts, as a voltage source, it has zero ohms impedance, so this step basically shorts together nodes 1 and 3
3. Call nodes 1+3 node 0 (i.e. ground ), you then have a current source with the positive end connected to ground via 3//2ohms and the negative end connected to ground via 4//6ohms, you can then work out currents sources contributions to each resistors voltage, write this down somewhere.
4. Set the voltage source back to 40v , and set current source to zero , (i.e. disconnect it)
5. Now work out the voltages across the 2 and 3 ohm resistors, should be 2/5 and 3/5 of 40v , 16v and 24v ; and the 4 and 6 ohm resistors 4/10ths and 6/10ths of 40 or 24v and 16v , write these down somewhere.
6. Now simply add algebraically the voltages for each resistor from steps 2 and 5 , (the voltage on the 2ohm resistor will be slightly less than 16v and 3 ohm slightly more than 24v)
7. Now subtract the voltage across the 4ohm resistor from all nodes, this will result in node 4 being at 0v , i.e. ground potential, and all the other node voltages will be the wanted voltages.

This is not the only method, but shows one way to break the problem down to manageable steps. Another approach would be to work out the thevenin equivalent of the voltage source and resistors )this will give 2 thevenin voltages and resistance, then apply the current across the thevenin equivalent and work from there. You should really try both ways. You can also convert the current source to its equivalent thevenin voltage source , then paralle both thevenin sources. You will find if you try all three methods, you will find yourself seeing the same numbers in each step. There is even a boring yet very complex way of working it out by writing equations for each resistance .

The circuit shown is actually a Wheatstone bridge, more obvious if you rotate it 90 degrees CCW.

OK , Here is the basic method

1. Move the ground connection to node 1 , and just call the old ground node 4
2. Set the 40v source to zero volts, as a voltage source, it has zero ohms impedance, so this step basically shorts together nodes 1 and 3
3. Call nodes 1+3 node 0 (i.e. ground ), you then have a current source with the positive end connected to ground via 3//2ohms and the negative end connected to ground via 4//6ohms, you can then work out currents sources contributions to each resistors voltage, write this down somewhere.
4. Set the voltage source back to 40v , and set current source to zero , (i.e. disconnect it)
5. Now work out the voltages across the 2 and 3 ohm resistors, should be 2/5 and 3/5 of 40v , 16v and 24v ; and the 4 and 6 ohm resistors 4/10ths and 6/10ths of 40 or 24v and 16v , write these down somewhere.
6. Now simply add algebraically the voltages for each resistor from steps 2 and 5 , (the voltage on the 2ohm resistor will be slightly less than 16v and 3 ohm slightly more than 24v)
7. Now subtract the voltage across the 4ohm resistor from all nodes, this will result in node 4 being at 0v , i.e. ground potential, and all the other node voltages will be the wanted voltages.

This is not the only method, but shows one way to break the problem down to manageable steps. Another approach would be to work out the thevenin equivalent of the voltage source and resistors )this will give 2 thevenin voltages and resistance, then apply the current across the thevenin equivalent and work from there. You should really try both ways. You can also convert the current source to its equivalent thevenin voltage source , then paralle both thevenin sources. You will find if you try all three methods, you will find yourself seeing the same numbers in each step. There is even a boring yet very complex way of working it out by writing equations for each resistance .

The circuit shown is actually a Wheatstone bridge, more obvious if you rotate it 90 degrees CCW.

Getting back to your working , it is common to use subscripts to denote voltages so V1 is then $V_1$ and the voltage of node 2 wrt node one is $V_1_2$

OK , Here is the basic method

1. Move the ground connection to node 1 , and just call the old ground node 4
2. Set the 40v source to zero volts, as a voltage source, it has zero ohms impedance, so this step basically shorts together nodes 1 and 3
3. Call nodes 1+3 node 0 (i.e. ground ), you then have a current source with the positive end connected to ground via 3//2ohms and the negative end connected to ground via 4//6ohms, you can then work out currents sources contributions to each resistors voltage, write this down somewhere.
4. Set the voltage source back to 40v , and set current source to zero , (i.e. disconnect it)
5. Now work out the voltages across the 2 and 3 ohm resistors, should be 2/5 and 3/5 of 40v , 16v and 24v ; and the 4 and 6 ohm resistors 4/10ths and 6/10ths of 40 or 24v and 16v , write these down somewhere.
6. Now simply add algebraically the voltages for each resistor from steps 2 and 5 , (the voltage on the 2ohm resistor will be slightly less than 16v and 3 ohm slightly more than 24v)
7. Now subtract the voltage across the 4ohm resistor from all nodes, this will result in done 4 being at 0v , i.e. ground potential, and all the other node voltages will be the wanted voltages.

This is not the only method, but shows one way to break the problem down to manageable steps. Another approach would be to work out the thevenin equivalent of the voltage source and resistors )this will give 2 thevenin voltages and resistance, then apply the current across the thevenin equivalent and work from there. You should really try both ways. You can also convert the current source to its equivalent thevenin voltage source , then paralle both thevenin sources. You will find if you try all three methods, you will find yourself seeing the same numbers in each step. There is even a boring yet very complex way of working it out by writing equations for each resistance .

OK , Here is the basic method

1. Move the ground connection to node 1 , and just call the old ground node 4
2. Set the 40v source to zero volts, as a voltage source, it has zero ohms impedance, so this step basically shorts together nodes 1 and 3
3. Call nodes 1+3 node 0 (i.e. ground ), you then have a current source with the positive end connected to ground via 3//2ohms and the negative end connected to ground via 4//6ohms, you can then work out currents sources contributions to each resistors voltage, write this down somewhere.
4. Set the voltage source back to 40v , and set current source to zero , (i.e. disconnect it)
5. Now work out the voltages across the 2 and 3 ohm resistors, should be 2/5 and 3/5 of 40v , 16v and 24v ; and the 4 and 6 ohm resistors 4/10ths and 6/10ths of 40 or 24v and 16v , write these down somewhere.
6. Now simply add algebraically the voltages for each resistor from steps 2 and 5 , (the voltage on the 2ohm resistor will be slightly less than 16v and 3 ohm slightly more than 24v)
7. Now subtract the voltage across the 4ohm resistor from all nodes, this will result in node 4 being at 0v , i.e. ground potential, and all the other node voltages will be the wanted voltages.

This is not the only method, but shows one way to break the problem down to manageable steps. Another approach would be to work out the thevenin equivalent of the voltage source and resistors )this will give 2 thevenin voltages and resistance, then apply the current across the thevenin equivalent and work from there. You should really try both ways. You can also convert the current source to its equivalent thevenin voltage source , then paralle both thevenin sources. You will find if you try all three methods, you will find yourself seeing the same numbers in each step. There is even a boring yet very complex way of working it out by writing equations for each resistance .

The circuit shown is actually a Wheatstone bridge, more obvious if you rotate it 90 degrees CCW.