Example 2-25: Solve for I.. I, 2 kg 2 ΚΩ Supernode Vb 2 k2 Va 12 V 1+ 6V 1 ΚΩ +) -4 V Ž 2 k2 Ε 2 1. 12 Soln.: The curren

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The given solution uses KCL and ohm's law to obtain the answer.
However, I don't know how to get the yellow parts.
It seems that the yellow parts are calculaing the voltage difference between two points.
What are the rules to add or minus a voltage and which point minus which point?
For example, why not Va-6 but 6-Va, and Va+12 but not 12.
Can anyone give some explanation?
Thanks.
Example 2-25: Solve for I.. I, 2 kg 2 ΚΩ Supernode Vb 2 k2 Va 12 V 1+ 6V 1 ΚΩ +) -4 V Ž 2 k2 Ε 2 1. 12 Soln.: The current Ix through the 12 V voltage source cannot be expressed in terms of Va and Vb, but note that I1 = 12 + Ix = I2 + Io + 13 Hence, V, and Vb can be "considered" as a supernode, and V V +12 V, +12-(-4) 2k 1k 2k 2k 6 - V. = 2V + V +12+ V +16 6-V- a a + a + a a a 1 1 -22 v and V = I - V +12 - = 3.8mA 5 O 2k 2.