Voltage drop across a series with a source and a resistor

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.

KCL at the center node gives:

$$ \frac{-V_x-50}{10} + 5 + 0.1V_x + \frac{-V_x}{20} = 0 $$

Which is only an equation of \$V_x\$ and can be immediately solved to get \$V_x = 0\$.

KCL at the top node (with voltage \$V_T\$):

$$ \frac{V_T - 50}{20} - 5 + I_Y = 0 $$

Ohm's law on the 25 ohm resistor gives \$ I_Y = \frac{V_T - 5I_Y}{25} \implies V_T = 30I_Y\$. Using this in the preceeding equation gives:

$$ \frac{30I_Y - 50}{20} - 5 + I_Y = 0 \implies I_Y = 3 \text{A}$$

So, along with your own answer and Alfred Centauri's answer, we can conclude that your book's answer of \$V_X = 70 V\$ is incorrect.