Use a series of source transformation to find the current in

Use a series of source transformation to find the current in the 8 ohm resistor.

In the given diagram a practical current source is there by using source transformation I am converting to practical voltage source

V=I*R=3*10=30V

R=10ohms

Here 8,10 ohms are in series result is 18 ohms.

By using node analysis

Let voltage across 6 ohms is V volts.

(V-30)/18 +(V/6)+(V-15)/3=0

V-30+3V+6V-90=0

V=12V.

Current through 8 ohm resistor is (30-12)/18=1Amp.