Now how do those relate to I0?

because we're really looking for I0.

And so, I0 is the current, again, through the 6k resistor in the center.

Going downward through to that branch of the circuit.

We know that I0 is going to be equal to I1,

which is flowing the same direction as I0,- I2 which is flowing

at the opposite direction up through the center leg of the circuit.

So if that's the case, we could subtract I2 from I1,

and we get 0.75 mA for I0.

So we, again, summed up the voltages around the loops.

So a point of interest that makes this a little easier, in some cases,

to write your equations is looking at the 6k resistor in the center part

of the circuit.

We've identified that voltage drop as

plus to minus 6k(I1- I2), okay?

So it's 6k (I1- I2) for

that voltage drop, as we're going around the circuit in the left-hand loop.

As we go around the right-hand loop,

we had to add up the negative of that as part of our loop.

It's the very first term in our second equation.

And it was a minus 6K(I1- I2).

So I'm going to put a negative sign out here in front to reflect that.