So, we immediately see that this problem lends itself to current division.

We have a two-milliamp source.

It's flowing upwards, through the center lag, and then it splits to the left, and

to the right.

And so, if could find the current, which is going to the right,

through the six kilo ohm resistor, then we can find V out prime.

because we know, V out prime in this case is going to be 6K times this current,

which is flowing through this 6K resistor, which is called I6K.

And so this current is I6K.

And we know I6K is in current division.

I6K is going to be two-milliamps.

The part of two milliamps which flows to the right.

And so that's going to be the sum of the voltages on the other side,

which is three kilo ohm.

1 kilo ohm plus 2 kilo ohm,

divided by that same resistance, 3K plus 6K.

That is the current flowing through the 6K resistor.

So we can plug that into our V out prime equation because

it's just a number now and we can solve for V out prime.

And if we do that,

we get a V out prime equal to four volts.

Now we will look at the contribution of the three-volt source, and

we're going to eliminate the two-milliamp source.

So we're going to redraw our circuit one more time and this time, for

the three-volt source and the contribution of that source to our output voltage.

So again, we draw our resistors as we had before.

Our 1K and 2K resistors, we still have

our 6K load resistance like this.

And we're measuring the output voltage across V out, double prime in this case.

We have a 1K resistor here, a 2K resistor at the top.

And our current source is open circuited.

So we have our volted source that we're looking for the contribution from.

We've taken away our current source, treating it as an open circuit and

we're looking for V out.