1. If we will take a look at node N₁ from the circuit diagram, we can see the currents entering and leaving the node. If we will put or indicate signs for this, let's say the current entering a node is positive (+) and the current leaving the node as negative (-). Node rule says that the algebraic sum of the currents entering and leaving a node is equal to zero. So applying the signs to the currents, we should have an answer which is equal to zero. Remember that the simulated current values are on the table given. (I₁ = 1.30 A; I₂ = 1.41 A; I₃ = 2.71 A)

• I₁ and I₂ are currents leaving the node, so negative and I₃ is entering the node, that would be positive.
• - I₁ - I₂ + I₃ = 0
• - 1.30 A - 1.41 A + 2.71 A = 0
• 0 = 0 (Node rule verified)

Since we got the same answer as the prediction of the node rule which is 0 A therefore, we have verified the rule. 

2. Let's do the same thing as number 1 but we will use node N₂. 

• I₁ + I₂ - I₃ = 0 (I₁ and I₂ are positive because they are entering the node while I₃ is leaving the node, so it is negative.)
• 1.30 A + 1.41 A - 2.71 A = 0
• 0 = 0 (Node rule verified)

3. From what we got from 1 and 2, we used different nodes but at both nodes, same currents are entering and leaving, so it is not necessary to solve equations for both nodes. Both equations are independent in mathematical sense, because when coming up with the set of equations considering the circuit given, you can just use either equation 1 or equation 2 and still get the same answer. 

please see the attached file for complete solution.