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\$
If you use KVL in the other loop (36v i5 and i2) then use that, the other kvl equations and two of the kcl equations it should work.
Adding all the kcl equations should give 0=0. They don't include the voltage so they can't actually solve for anything here.
Using star-delta transforms and mesh analysis can make it a bit easier to solve (transform the delta that doesn't have I1 in it to a star/wye).
Best Answer
Well, we are trying to analyze the following circuit:
simulate this circuit – Schematic created using CircuitLab
Using KCL, we can write:
$$\text{I}_1=\text{I}_2+\text{I}_3\tag1$$
Using KVL, we can write:
$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{a}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_2}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_1}{\text{R}_3}\\ \\ \text{V}_\text{x}=\text{V}_1-\text{V}_2=\alpha\text{I}_1 \end{cases}\tag2 $$
Using some Mathematica code:
Using Your values we get:
Which gives:
Which approximate to: