Kirchhoff's current law can be either,

used either by combining the currents that are flowing into

a node or complimentary you can combine the currents out of that node.

So in this problem we're going to combine the currents or

add the currents that are flowing into the top node.

First of all we know that we have a two amp source which is flowing into the node.

We have I3, which is flowing out of the node.

We have point nine I3 flowing into the node and

we also have the current which is going down through

the 6 ohm resistor on the right hand side of the circuit.

And that current, we know, is going to be equal to, and

it's flowing out of the top node so we're going to put a negative sign.

It's going to be equal to V sub 1,

which is the voltage associated with the top node.

Divided by 6 ohms using Ohm's law and that's equal to 0.

So we've used Kirchhoff's current law which is the sum of the currents in to any

node equal 0 at any instant in time to come up with our first equation.

So that's equation number one.