Voltage drop across a series with a source and a resistor
The system is complete within itself. You don't need to reference ground. KCL means that only \$i_x\$ flows around the loop. Voltage drop in resistors is \$i_xR\$ so each resistor drops \$2i_x\$ volts. So using KVL; $$12V- 2i_x -2i_x -6i_x -2i_x = 0V$$ $$ Ix = 12/12 = 1\ Amp.$$