Determine the node voltages va and vb for the circuit
Just set it up and knock them down. The left side of the initial equation, for each node I set up, will be the currents leaving the node. The right side will be the currents entering the node.

Node 'a' first:

$$\begin{align*}
\frac{V_a}{4\Omega}+\frac{V_a}{2\Omega}+3A&amp;=\frac{0V}{4\Omega}+\frac{V_b}{2\Omega} \\
V_a\cdot\left(\frac{1}{4\Omega}+\frac{1}{2\Omega}\right)+V_b\cdot\left(\frac{-1}{2\Omega}\right)&amp;=-3A
\end{align*}$$

Node 'b' now:
$$\begin{align*}
\frac{V_b}{2\Omega}+\frac{V_b}{3\Omega}&amp;=4A+\frac{V_a}{2\Omega}+\frac{0V}{3\Omega} \\
V_a\cdot\left(\frac{-1}{2\Omega}\right)+V_b\cdot\left(\frac{1}{2\Omega}+\frac{1}{3\Omega}\right)&amp;=4A
\end{align*}$$

Those are your two equations in two unknowns. In this case, \$V_a=-1\tfrac{1}{3}V\$ and \$V_b=4V\$, if I did things right.

I can reconstitute my equation form into yours, more or less, as:

$$\begin{align*}
\frac{V_a}{4\Omega}+\frac{V_a-V_b}{2\Omega}+3A&amp;=0 \\
\frac{V_b-V_a}{2\Omega}+\frac{V_b}{3\Omega}-4A&amp;=0
\end{align*}$$

Or, in negated form:

$$\begin{align*}
-\frac{V_a}{4\Omega}-\frac{V_a-V_b}{2\Omega}-3A&amp;=0 \\
\frac{V_a-V_b}{2\Omega}-\frac{V_b}{3\Omega}+4A&amp;=0
\end{align*}$$

Perhaps you can see the differences here? This last equation pair looks like yours except for the sign of \$\tfrac{V_b}{3\Omega}\$. Probably, because you failed to realize it was \$\tfrac{0V-V_b}{3\Omega}\$, as a guess.

Let me know if you need me to walk slowly through how I set up the initial equations.