current-sourcevoltage-source
I've seen and solved many examples but I'm stuck here . Maybe I'm confused as to what is parallel and what is in series here. 
I can do the transformations up to the 1A current source but I can't isolate a parallel resistance for it so I can transform. It's most likely simple . Can I add it with the current source on the left as if it has zero resistance ?
PS:This is from the unsolved problems section of a textbook I'm studying and it is no homework.
A current source is the dual of a voltage source. An ideal voltage source has zero output impedance, so that the voltage doesn't drop under load. It shouldn't be shorted, because in theory there would flow an infinite current.
An ideal current source has infinite output impedance. This means that the load's impedance is negligible and won't influence the current flowing. Like voltage sources shouldn't be shorted, current sources shouldn't be left open. An open current source will still try to source the set current, and the theoretical current source will go to infinite voltage.
edit (following your comment)
Here you can read impedance as resistance. If the current source would have a limited resistance changes in load would change the current, because the total resistance would change. You don't want that. So if the current source's resistance is infinite the load can be ignored and the resistance always remains the same (infinite). Therefore the current will as well.
A practical current source may be constructed as follows:
One diode has the same voltage drop as the base-emitter junction, so the other diode sets the transistor's emitter to about 0.7V. A fixed voltage across a fixed resistor gives a fixed emitter current, which is about the same as the collector current if the transistor's \$H_{FE}\$ is high enough. (Strictly speaking this is a current sink rather than a current source, but the principle remains the same.)
Another current sink uses an opamp as control element:
The main thing you need to know about opamps in this configuration is that they will try to keep the voltage on both inputs equal. So suppose you set \$V_{SET}\$ to 1V, then the opamp will try to make the - input also 1V. It does so by inserting current into the transistor's base. This will cause a current through the load \$I_{LOAD}\$ which is (almost) equal to \$I_{SET}\$. And \$I_{SET}\$ is constant to get the 1V across \$R_{SET}\$, according to Ohm's Law:
\$ I_{SET} = \dfrac{V_{SET}}{R_{SET}} \$
Since \$V_{SET}\$ and \$R_{SET}\$ are constant, so will \$I_{SET}\$ be. QED.
Looks like a trick question. Circuit is shorted, so all the current will flow through the wire. Current through \$L_1\$ is 0. Ran the SPICE sim just to be sure. Results and netlist below :)
I1 0 0 10m
R1 0 0 15
R2 0 N001 47
L1 N001 0 10m
.tran 0 10m 0 1u
.backanno
.end
Best Answer
This looks like homework so I'm not going to give you a full solution, but I will tell you that the next step is to convert the 4A current source with parallel resistor into a voltage source with series resistor.
EDIT: After getting a 2A current source in parallel with 20 ohms, the next step is to combine the 2A and 1A sources. Current sources in parallel add, just like voltage sources in series.
By the way, the voltage is 40V, not 4V.