We have a Thevenin's resistance of 3 kilohms,

so it's a 3 kilohm resistor here and

we put our load back in and our load is 6 kilohms.

And we're looking for the voltage across that 6 kilohm resistor.

And using voltage division, this problem becomes very easy.

We know that Vout is equal to

6V times (6k divided by 6k plus 3k).

So our Vout in the end comes out to be 6 volts

times, I'm sorry.

Voc is not 6 volts, Voc is 9 volts from below.

I looked at the wrong number.

So Voc is 9 volts, we put the 9 volts in.

This should be 9 volts instead of 6 volts.

So we're going to take that and

remove that and put in a 9 and

then we end up with a Vout equal to 6 volts.

So, the circuit that we've drawn is the Thevenin's Equivalent Circuit for

our problem.

And we found that by first analyzing our circuit for Voc,

that's what this step was about.

And we analyzed the circuit for R Thevenin's, and

that's what this bottom part of our analysis was about.

We took those values, put them back into our circuit, recognizing that Voc

was 9 volts and then we solved for Vout using voltage division.

It was the 9 volts times the division of that voltage across the 6k and

the 3k resistors in series with another, and

this gives us the final Vout equal to 6 volts.