Nodal Analysis - Finding voltage given the network
I wish you'd used the schematic editor. Then I'd have labels for everything. I'm not going to waste time adding such a schematic. (But I get that you also included some text and an answer in your picture.) So just examine the following using nodal analysis, assuming the bottom node is ground or \$0\:\text{V}\$:

$$\begin{align*}
\frac{V_1}{3\:\Omega}+\frac{V_1}{2\:\Omega}&amp;=10\:\text{A}+\frac{V_2}{3\:\Omega}+\frac{V_3}{2\:\Omega}\tag{Node 1}\\\\
\frac{V_2}{3\:\Omega}+\frac{V_2}{4\:\Omega}&amp;=4\cdot I_X+\frac{V_1}{3\:\Omega}+\frac{0\:\text{V}}{4\:\Omega}\tag{Node 2}\\\\
\frac{V_3}{2\:\Omega}+\frac{V_3}{6\:\Omega}+4\cdot I_X&amp;=\frac{V_1}{2\:\Omega}+\frac{0\:\text{V}}{6\:\Omega}\tag{Node 3}\\\\
I_X&amp;=\frac{V_2}{4\:\Omega}\tag{Compute $I_X$}
\end{align*}$$

Sympy tells me:

solve( [
....: Eq(v1/3+v1/2,10+v2/3+v3/2),
....: Eq(v2/3+v2/4,4*ix+v1/3+0/4),
....: Eq(v3/2+v3/6+4*ix,v1/2+0/6),
....: Eq(ix,v2/4) ],
....: [v1,v2,v3,ix] )
{v2: -64, ix: -16, v1: 80, v3: 156}


Does that help?

Here is how I developed the above equations:


  
  
  Equation rules:
  
  
  Keep all outgoing currents on the left side
  Keep all incoming currents on the right side
  If not already chosen by the problem, pick a convenient node and call
  it "ground." This allows you to assign \$0\:\text{V}\$ to that node and
  this will simplify some of the equations. Which node to select is up to
  you. But I will pick one that either helps make the schematic easier to
  understand, or else will simplify the equations more than other choices.
  
  
  I like to imagine a 3D chessboard -- those funny chessboards with tiny
  sub-chessboard areas that sit at different heights. Each node in the
  circuit is represented as one of these sub-areas (or "platforms"), sitting at some unknown
  elevation and held there on a long pole beneath it. This unknown elevation
  is the voltage you must find. One of these can be placed at "0 height." But
  that's just arbitrary, since then there still may be platforms above and/or
  below. So it's just a way of conveniently locating the "ground floor," so
  to speak.
  
  I treat the currents as water. Water flows onto and off of each platform.
  The left side of the equation, the outgoing currents, is represented by the
  water that is flowing off of the perimeter of the platform. The right side
  of the equation, the incoming currents, is represented by the water that is
  flowing onto the platform from other platforms above it. The amount of water
  falling onto such a platform must equal the water flowing off of it via the
  perimeter edge. (Of course, otherwise water would magically build up on the platform -- or the reverse, get created out of thin air. Both are impossible.)
  
  (No water is ever lost and all of it
  remains in play and the water cannot accumulate on the floors -- the net
  flow onto a floor must equal the net flow off of it -- KCL! -- and hence
  the justification for setting the left side equal to the right side.)
  
  


Look at \$N_1\$.

You only have two paths out of the node, via the two resistors. So the left side of the equation will have the voltage at the \$N_1\$ platform pouring away through these two resistors. That's the two fractions on the left side. You cannot go against a current source. So that is not a path out of the node.

You have three paths into the node, via the two resistors (of course, because they work both directions) and then also via the incoming current source. So the right side of the equation will have the current source value plus the two fractions posed by taking the voltage at those nearby platforms (\$N_2\$ and \$N_3\$) divided by their resistor paths into this node.

That's it. That's the first equation.

The rest are done just the same way.

I was first introduced to this method because that's how the formulas are generated by Spice in source code I examined. It took me only a moment to "get the idea" and let it replace all prior instructions I had regarding nodal analysis.

But a very, very good reason is also that the bookkeeping is easier. I'm a lot less likely now to make dumb mistakes. I can't count how many times I got the sign wrong when trying to work out "the voltage on this side minus the voltage on that side" mental gymnastics. This method completely saves me from such troubles. And I can write accurate equations as fast as I can write with a pen or pencil, that I know are right from simple inspection. And that is the final good reason. It's trivial for me to double-check the work, later, without really having to engage my brain much. With the "difference from this side to that side" gymnastics, I still have to do that same gymnastics if I review the equations. And while I may catch errors this next time, there is a far too high possibility that I might again just "agree" with myself over old errors.

All such problems are removed using this method.

And Spice likes it, too.



Let me redo the above equations more into your own form:

$$\begin{align*}
10\:\text{A}&amp;=\frac{V_1-V_3}{2\:\Omega}+\frac{V_1-V_2}{3\:\Omega}\tag{Node 1}\\\\
\frac{V_2-0\:\text{V}}{4\:\Omega}&amp;=\frac{V_1-V_2}{3\:\Omega}+4\cdot \frac{V_2-0\:\text{V}}{4\:\Omega}\tag{Node 2}\\\\
\frac{V_1-V_3}{2\:\Omega}&amp;=4\cdot \frac{V_2-0\:\text{V}}{4\:\Omega}+\frac{V_3-0\:\text{V}}{6\:\Omega}\tag{Node 3}
\end{align*}$$

In this case, I have used your approach. But notice that I came up with different equations.

Please examine the new equations above and compare them with yours.