use Kirchhoff's Voltage Law to sum the voltages around our super mesh.

So looking at the super mesh, In the lower left hand corner and

going around it in a clockwise fashion back to the beginning,

we first encounter a three kilo-ohm resistor.

And the voltage drop for that is going to be 3k times I sub one,

since that's the only current flowing through the three kilo-ohm resistor.

Similarly at the top, we have a one kilo-ohm resistor.

The only current flowing through it is I sub 1, so

the voltage drop is 1K times I sub 1.

Continuing around our loop, we encounter our dependent source,

the negative polarity of that first.

So it's going to be a -2,000 I sub x prime for