Nodal Analysis verification
You already know the value of \$V_1=-5\:\text{V}\$. So there is no real point to developing an equation for it. That will complicate the work and increase the chances for introducing human mistakes. I recommend keeping it as simple as possible (but no simpler.)

I get the following:

$$\begin{align*}
\frac{V_2}{4\:\Omega}+\frac{V_2}{1\:\Omega}+2\cdot \left(-5\:\text{V}-V_2\right)&amp;=\frac{-5\:\text{V}}{4\:\Omega}+\frac{V_3}{1\:\Omega}\\\\
\frac{V_3}{1\:\Omega}+\frac{V_3}{2\:\Omega}&amp;=1\:\text{A}+\frac{V_2}{1\:\Omega}+\frac{3\:\text{V}}{2\:\Omega}
\end{align*}$$

I'll let you compare these to your own equations. But the above do solve out, correctly, for the two unknown voltages appearing in those equations. The rest from there is easy to work out, if you care about other currents or voltages.