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i made a simulation using Lt spice but i did get this 6,43 V that you mentionnedTotal Voltage will be 6V-voltage at the junction 5k&2k. for the analysis of 1.5V, the current will flow through the 2k & 5k only. so the current is -1.5/(2k+5k)=-0.214mA, so @ junction the voltage is -0.214mA*2k=-0.428V, hence the total voltage 6-(-0.428)= 6.428 or 6.43V.
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This only in the case of ideal circuit, assuming, there is no resistance on the 6V path, so that no current flow.