Find the Thevenin equivalent.

NOTE: Blue items were added to solve the problem and are not part of the original diagram.

Express dependent source parameters in terms of mesh currents.

i_b = i₂

v₂ = 50 K i₃

Express current sources in terms of mesh currents.

Independent source:

i₁ = 500 mA (Equation 1)

Dependent source:

i₃ = - 80 i_b

i₃ = - 80 i₂ (Equation 2)

Now, erasing the current sources leaves only one mesh (2) around which to write KVL.

1310 i₂ + 4 X 10^-5 v₂ + 100 (i₂ - i₁) = 0

1310 i₂ + 4 X 10^-5 (50 K i₃) + 100 (i₂ - 500 mA) = 0

1410 i₂ + 2 i₃ = 50 mA (Equation 2)

Solving Equations 2 and 3 yields

i₂ = 40 mA

i₃ = - 3.2 mA

Finally

v_oc =v₂ = 50 K i₃ = 50 K (- 3.2 mA) = - 160 V

so

v_Th = - 160 V

Now, find i_sc.

We could do a full-blown mesh analysis as indicated by the mesh currents in the circuit below, but a little bit of reflection will lead to a quicker solution.

(If you are uncomfortable with the short-cuts at this point, don't worry - you can always fall back on the algorithmic techniques: Nodal Analysis and Mesh Analysis.)

First, note that the 50 KW resistor is shorted out, thus there is no current through it nor voltage across it (thus v₂ = 0).
We can thus remove the 50 KW resistor from the circuit without altering the function.
(Another way of seeing this is that we effectively have 50 KW in parallel with a wire (0 W). This gives an equivalent resistance of 0 W, thus we can replace the 50 KW and the wire with just a wire.)

Since v₂ = 0, the dependent voltage source has a value of zero also. We can thus replace the dependent voltage source with a wire without affecting the circuit.

Making the modifications to the circuit yields

We can easily get i_b using current division:

i_b = 500 m (100 / 1410) = 35.461 mA

Next,

i_sc = - 80 i_b = - 80 (35.461 m) = - 2.837 mA

Finally we can calculate R_Th

R_Th = v_oc / i_sc = - 160 / - 2.837 m
R_Th = 56.4 KW

so the Thevenin equivalent circuit is: