Supernode with 2 voltage sources and 1 current source: node voltage analysis
As Photon points out and you already seemed to know, the arrangement means that the current source has no impact. But I wanted to add a redrawn schematic to help emphasize the fact. Ideal voltage sources have zero impedance. Looking at the redrawn schematic, it is pretty easy to see which path the current source's current magnitude will take: through \$V_1\$ and \$V_2\$. And that means it doesn't affect the voltages at any of the nodes.

This schematic also emphasizes that there really is only one unknown node, \$V_x\$. The others are determined from that. The only additional problems are working out the currents in the two voltage sources.







simulate this circuit &ndash; Schematic created using CircuitLab

Using nodal analysis this works out to:

$$\begin{align*}
\frac{V_x}{R_3} + I_{V_2} &amp;= 3\:\textrm{A} + \frac{0\:\textrm{V}}{R_3} ~~~\therefore~~~~ V_x=R_3\cdot\left( 3\:\textrm{A}-I_{V_2}\right)\\
 \\
\frac{V_x+9\:\textrm{V}}{R_1} + I_{V_1} &amp;= I_{V_2} + \frac{0\:\textrm{V}}{R_1} ~~~\therefore~~~~ V_x=R_1\cdot\left(I_{V_2}-I_{V_1}\right)-9\:\textrm{V}\\
 \\
\frac{V_x+30\:\textrm{V}}{R_2} + 3\:\textrm{A} &amp;= I_{V_1} + \frac{0\:\textrm{V}}{R_2} ~~~\therefore~~~~ V_x=R_2\cdot\left(I_{V_1}- 3\:\textrm{A}\right)-30\:\textrm{V}
\end{align*}$$

The above is three equations in three unknowns which works out to \$I_{V_1}=4.3\:\textrm{A}\$ and \$I_{V_2}=4.05\:\textrm{A}\$ and \$V_x=-10.5\:\textrm{V}\$.

If I had to guess what would happen when you remove \$I_1\$ from the circuit, I'd guess that \$I_{V_1}=1.3\:\textrm{A}\$ and \$I_{V_2}=1.05\:\textrm{A}\$ and \$V_x=-10.5\:\textrm{V}\$. Which would show that all of the current from \$I_1\$ goes through \$V_1\$ and \$V_2\$. (And a solution would show that this is, in fact, the correct result.)