I'm having a hard time finding V2 in terms of I2. I've tried various forms of KCL and KVL but I can't seem to find the right answer. I hope someone can enlighten me.

 

Can you show us your work?

Then we can further diagnose what the problem is and point you in the right direction.

 

\(I_2 = \frac{V_2-2V_x}{2} + \frac{V_2}{5} = \frac{V_2-2\frac{V_2}{4}}{2} + \frac{V_2}{5} = 0.45V_2\)

\(Z_{22} = \frac{V_2}{I_2} = \frac{V_2}{0.45V_2} = 2.222\)

This is clearly wrong since the answer is supposed to be 1.111 ohm.

 

\(I_2 = \frac{V_2-2V_x}{2} + \frac{V_2}{5} = \frac{V_2-2\frac{V_2}{4}}{2} + \frac{V_2}{5} = 0.45V_2\)

\(Z_{22} = \frac{V_2}{I_2} = \frac{V_2}{0.45V_2} = 2.222\)

This is clearly wrong since the answer is supposed to be 1.111 ohm.

My node equation is:
\(I_2 = \frac{V_2-(-2V_x)}{2} + \frac{V_2}{5} = \frac{V_2+2V_2(\frac{1}{1+4})}{2} + \frac{V_2}{5} = 0.9V_2\)

Current direction was chosen as leaving the node for both branches.
Note the polarity on the dependent voltage source.
Also, Vx can be replaced by the expression for a 1Ω and 4Ω voltage divider (making it 1/5 V2)

This gives 1.111ohms when used in the second equation.