Rangsk

Not a EE major but as a CS major I've taken some ECE classes on circuits...
From what I can tell, that dependent current source doesn't do anything, look at this:

The voltage across the R2 and R3 must be the same and V = IR, thus:
2*ia*R2 = ia*R3
2*R2 = R3
8 = 8

Thus, if that current source weren't there, it would behave the same way

With that in mind, you can simply do this:

(R2*R3/(R2 + R3) +R1) * ib = 42V
plug it in, do the math, and get ib = 9A

Not positive if this is right.

Now, if that current supply had been different, the problem would have been a whole lot harder.

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