I think I figured it out.....

For I1 I have : 2KI1 + 3K(I1-I2) = 12v

for I2 I have : 4KI1 + 2KI2 + 3K(I1-I2) = 0

Then from there I believe I will have to simplify and solve.

I1 : 2KI1 + 3KI1 - 3KI2 = 12 reduces to 5K1 - 3K2 = 12

I2 : 3KI1 - 3KI2 = 0

then

I2: -3k2 = -3K1 reduced to I2 = 1/1000 or 1/k amps


then using Ohms law take I2 multiply the 4K resistor on the right to get the voltage across it and it will be the same as Vo, which is 4V.