Electronic – What use is the reference node in nodal analysis

First off, mesh analysis would be simpler because you would only have one unknown (the current in the right mesh).

As you know, in nodal analysis you are summing currents at the nodes. Current into the nodes positive and current out of the nodes negative. The way you have defined the problem you should have 3 equations and three unknowns. Your first equations seems to do this correctly for node B. It kind of falls apart after that. For example, node A equation should just be .005 = (Va-Vb)/3000. You already know that Vc = 16 volts.

So simplifying the node A equation gives you what Andy aka said, then just substitute it back into your node B equation.

In general, you don't create an equation for node 1. You take nodes 1, 2 and 3 together as a single node, called a generalized node, aka supernode, for current summing purposes, and create an equation for that. Like this: Let the voltages at nodes N1, N2 and N3 be called \$v_1\$, \$v_2\$ and \$v_3\$, respectively. You can sum the currents out of the supernode: $$ \frac{v_2}{ 2\mathrm{k\Omega}} + \frac{v_3}{ 5\mathrm{k\Omega} }+\frac{v_1}{ 4\mathrm{k\Omega} }= 0 $$ You now have three unknowns and one equation, but you know that \$ v_2 = v_1 - 30 \$ and \$v3 = v1 - 20\$, so you can back-substitute into the previous equation and solve it for \$v_1\$

Dependant sources are handled in the same manner.