The circuit is as follow:
va = R1*ia + 6
va = R3*ib + 4*va
Thus we get the following system:
- 2.103 *(ia +ib) = 103 *ia + 6
6. 103 *(ia +ib) = 3. 103 *ib
We have 2 equations for 2 unknowns so we can solve the system. We get
ia = 0.006 A and ib = - 0.012 = - 12 mA