Solving circuit with source transformation
Let's take the circuit and remember the two rules to make source transformation:

1- A current source sourcing a current \$I\$ in parallel with a resistor \$R\$ transforms into a voltage source with voltage \$R\,I\$ and the same resistor \$R\$ in series with this voltage source. That is the principle of finding the Thévenin Equivalent.

2- A voltage source with voltage \$V\$ in series with a resistor \$R\$ transforms into a current source with current \$\dfrac{V}{R}\$ and the same resistor \$R\$ in parallel with this voltage source. That is the principle of finding the Norton Equivalent.

Illustrating what has been aforementioned:







simulate this circuit &ndash; Schematic created using CircuitLab

Given this explanation the circuit then boils down to some steps.

First step: Voltage source to the left into a current source and current source to the right into a voltage source.







simulate this circuit

Now transform the voltage source to the right into a current source:







simulate this circuit

Finally, the current is easily calculated from this last figure:

\$ i=3.5\,\dfrac{6.667}{20+6.667} \approx 875\, mA \$