Something goes wrong in your last calculation The Electrician. See post no.17 for i3 and ib values.
I missed the minus sign on i3 in post #17. You had i3=-2.9117647 A.
I think it helps to leave spaces in appropriate places like this: i3 = -2.9117647 A
So I'll redo the calculation:
Check -25+30*i3+5*ua+20*ib+ua, has to be 0.
OK, let's evaluate all the terms using your values for i1, i2 and i3, using the fact that ua = 10*(i1-i2), ib = i3-i2
Listing the results, I get:
-25 = -25
30*i3 = -87.352941
5*ua = 108.82353
20*ib = -18.235294
ua = 21.764706
These add up to .000001, essentially zero.
For any value of i1, a value of i3 = -27/10 - (6*i1)/5 gives a value of zero. However, this is only the solution to one equation, and not a solution to the complete network. It is also necessary to satisfy the equation 3ib = i1-i3, but you assumed i1=3*ib-i3, which is incorrect.