Thevenin-Norton Conversion Using Thévenin-Norton equivalence, prove that the circuit below

is equivalent to the circuit below


The main point behind my confusion is that the current source has no parallel resistance. As a result, I am unable to apply the Thévenin-Norton equivalence to it. Can anyone give me a hint about what to do with the current source?
 A: There are two different things with similar names.
1) The Thevenin-Norton equivalence. Or the Thevenin-Norton transformation.
You can convert a Thevenin input into a Norton input, and vice-versa.
That means, [a voltage source in series with a resistance] is, as a block, equivalent to a [current source in parallel with anojther resistance].
You can convert from one to another very easily.
But there is also a different thing:
2) The Thevenin equivalent of a given circuit.
For a linear circuit, you can always reduce it to a Thevenin circuit.
That means, you can have a ver complicated circuit, with many resistors, capacitors, sources, and so on. 
Provided that it is linear, it can always be reduced to a [Voltage source + resistance in series].
In other words, if you replace the whole circuit with this Thevening one, nobody will note. The result is the same "as seen from" the terminals.
So, how to do it?
You'll find many other better explanations, but, in short, it is:
A) For the voltage source:  measure the voltage in open circuit
B) For the resistance: void all independent generators (not depednent ones!, only independent) And, after that, calculate teh equivalent resistance seen from the terminals.
When I say "measure", it can also be "calculate it".
Hope this helps
