mesh analysis with current sources + dependent current source
It's not complicated. Try it without bothering to do the Norton to Thevenin conversion:







simulate this circuit &ndash; Schematic created using CircuitLab

$$\begin{align*}
0\:\text{V} + V_\text{IS1} - R_1\cdot\left(I_1-I_2\right) &amp;=0\:\text{V}\\
0\:\text{V} - R_1\cdot\left(I_2-I_1\right)-R_2\,I_2-V_{1.5\,I_x}&amp;=0\:\text{V}\\
0\:\text{V}+V_{1.5\,I_x} - R_3\,I_3-R_4\cdot\left(I_3-I_4\right)&amp;=0\:\text{V}\\
0\:\text{V}- R_4\cdot\left(I_4-I_3\right)-V_\text{IS2}&amp;=0\:\text{V}\\\\
I_X&amp;=I_1-I_2\\
1.5\,I_X &amp;=I_3-I_2\\
I_1&amp;=2\:\text{A}\\
I_4&amp;=-5\:\text{A}
\end{align*}$$

I think you should be able to follow how I developed each of those, above. (I assumed that the voltage across each current source has its positive terminal on the top side and its negative terminal on the bottom side -- which doesn't have to be true for the above equations to work correctly. It just has to be consistently applied, is all.)

You certainly can, if you want, do the conversion first and reduce the number of loops by one. But there's no real need to do that.

If you use Sage (freely available to anyone), then you might set up the above like this:

var('i1 i2 i3 i4 r1 r2 r3 r4 ix vis1 vis2 vix')
eq1=Eq(vis1-r1*(i1-i2),0)
eq2=Eq(-r1*(i2-i1)-r2*i2-vix,0)
eq3=Eq(vix-r3*i3-r4*(i3-i4),0)
eq4=Eq(-r4*(i4-i3)-vis2,0)
eq5=Eq(ix,i1-i2)
eq6=Eq(1.5*ix,i3-i2)
ans=solve([eq1,eq2,eq3,eq4,eq5,eq6],[i2,i3,ix,vis1,vis2,vix])


It solves just fine.