Use superposition to find the voltage v.

First "kill" the 70 volt source.

Next, note that the 20 ohm and 2 ohm resistors are in parallel, so we can combine them. I will use mesh analysis to solve the resulting 2 mesh circuit.

Define the dependent source parameter in terms of mesh currents

i_b = I_A

Write KVL around each mesh.

A:

4 I_A + 2 i_b + 1.818 (I_A - I_B) = 0
4 I_A + 2 I_A + 1.818 (I_A - I_B) = 0
7.818 I_A - 1.818 I_B = 0 (equation 1)

B:

1.818 (I_B - I_A) - 2 i_b - 50 + 10 I_B = 0
1.818 (I_B - I_A) - 2 I_A + 10 I_B = 50
- 3.818 I_A + 11.818 I_B = 50 (equation 2)

Solving the two equations yields

I_A = 1.064
I_B = 4.575

Now we can determine v

v = - 50 + 10 I_B = - 50 + 10 (4.575)
v = - 4.255 V (due to 50 V source)

Define dependent source parameter in terms of node voltages.

i_b = (V_A - V_C) / 4

Both voltage sources directly connect two essential nodes. Express their values in terms of the node voltages and form supernodes.

V_A = 70

2 i_b = V_C - V_B
2 (V_A - V_C) / 4 = V_C - V_B

Insert V_A = 70 from above

2 (70 - V_C) / 4 = V_C - V_B

Multiply through by 2

70 - V_C = 2 V_C - 2 V_B
- 2 V_B + 3 V_C = 70 (equation 3)

Node A is connected to the reference node by a voltage source, so no KCL needed.

Nodes B and C are connected by a voltage source, thus form a supernode.

KCL at supernode B/C

(V_B - V_A) / 20 + (V_C - V_A) / 4 + V_B / 2 + V_C / 10 = 0

Multiply through by 20

(V_B - V_A) + 5 (V_C - V_A) + 10 V_B + 2 V_C = 0
- 6 V_A + 11 V_B + 7 V_C = 0

Insert V_A = 70 from above

- 6 (70) + 11 V_B + 7 V_C = 0
11 V_B + 7 V_C = 420 (equation 4)

Solving equations 3 and 4 gives

V_B = 16.38
V_C = 34.26

But v = V_C so

v = 34.26 (due to 70 V source)

v = -4.255 + 34.26
v = 30 V