Help on superposition
Despite what most textbooks claim, superposition of dependent sources is valid if done correctly.

There are three sources in this circuit so there will be three terms in the superposition.

For the first term, the two current sources are zeroed (opened) so \$V_x\$ is given by voltage division:

\$V_x = 10V \cdot \dfrac{4}{4 + 20} = \dfrac{5}{3}V\$

For the second term, the voltage source is zeroed (shorted), so the two resistors are now in parallel, and the 2A source is activated.  Thus:

\$V_x = 2A \cdot 4\Omega || 20 \Omega = \dfrac{20}{3}V\$

Since the third, dependent source is in parallel with the 2A source, the last term has the same form:

\$V_x = 0.1 V_x \cdot 4\Omega || 20 \Omega = \dfrac{1}{3}V_x\$

Now, it's crucial at this point to not try and solve the previous equation (you'll only get \$V_x = 0\$ if you do.)

Rather, proceed with the superposition sum and then solve.

\$V_x = \dfrac{5}{3}V + \dfrac{20}{3}V + \dfrac{1}{3}V_x\$

Grouping terms:

\$V_x (1 - \frac{1}{3}) = \dfrac{25}{3}V\$

Solving:

\$V_x = 12.5V\$