I don't know how to proceed with this question or where to start.
If there is (net) negative feedback, then you proceed by setting the voltage across the op-amp input terminals equal to zero:
$$v_+ = v_-$$
Note that with zero volts across the input terminals, the 2k resistor in parallel with the current source is irrelevant; there is zero volts across it so there is zero current through it. You may remove it from the circuit without changing the solution.
This should get you started.
@AlfredCentauri I still don't see the bottom loop, do you mean the
loop v+ connected to VB then connected to the voltage source and then
the resistor and finally VA. Is that considered a loop even with the
op-amp? And when I do I still don't get your equation.
simulate this circuit – Schematic created using CircuitLab
This is the bottom-most loop and KVL clock-wise 'round the loop starting with the voltage across the 1k resistor is:
$$i_1 \cdot 1k\Omega -2V + V_B - V_A = 0 $$
rearranging yields
$$V_B = V_A - i_1 \cdot 1k\Omega + 2V$$
If the presence of the voltage source above is puzzling, recall that the output of the ideal op-amp is an ideal (controlled) voltage voltage source referenced to ground which I've shown explicitly here.
Best Answer
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:
$$\begin{align*} \frac{V_a}{4\Omega}+\frac{V_a}{2\Omega}+3A&=\frac{0V}{4\Omega}+\frac{V_b}{2\Omega} \\ V_a\cdot\left(\frac{1}{4\Omega}+\frac{1}{2\Omega}\right)+V_b\cdot\left(\frac{-1}{2\Omega}\right)&=-3A \end{align*}$$
Node 'b' now: $$\begin{align*} \frac{V_b}{2\Omega}+\frac{V_b}{3\Omega}&=4A+\frac{V_a}{2\Omega}+\frac{0V}{3\Omega} \\ V_a\cdot\left(\frac{-1}{2\Omega}\right)+V_b\cdot\left(\frac{1}{2\Omega}+\frac{1}{3\Omega}\right)&=4A \end{align*}$$
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:
$$\begin{align*} \frac{V_a}{4\Omega}+\frac{V_a-V_b}{2\Omega}+3A&=0 \\ \frac{V_b-V_a}{2\Omega}+\frac{V_b}{3\Omega}-4A&=0 \end{align*}$$
Or, in negated form:
$$\begin{align*} -\frac{V_a}{4\Omega}-\frac{V_a-V_b}{2\Omega}-3A&=0 \\ \frac{V_a-V_b}{2\Omega}-\frac{V_b}{3\Omega}+4A&=0 \end{align*}$$
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.