We also have the current flowing down through R2 which is V2- 0 for

the ground node, divided by R2.

We have the current flowing out

through R3 which is (V2- V3) / R3.

And we have a current flowing out through R4,

which is V2 again, minus V3 over R4 equal to zero.

So that's our second independent equation.

So we have these unknown nodal voltages V1 and V2 and V3.

So we have two independent equations, we need our third independent equation that

we get from node 3 in order to solve for our unknown nodal voltages V1, V2 and V3.

So we're going to sum the currents again, out of node 3.

So it's -I sub 2, because that current source is flowing into node 3.

And we're going to add to that the current which is

flowing out of that node through the resistance R4,

which is (V3- V2) / R4.

And we have the current which is flowing

out of node 3 through the resistance

R3 which is (V3- V2) / R3 = 0.