I can't solve this problem. Please help me.

 

There are four steps to obtaining the Thevenin equivalent, what have you done so far?

 

actually , i don't know how to start . I can solve the independent sources's questions. But there is a dependent current source . I've tried to solve this five or six times and different ways but i couldn't get close to the answer.

 

Try to use a superposition and find Vth voltage
http://forum.allaboutcircuits.com/t...rrent-and-voltage-sources.101488/#post-764893

Ix = 6V/(5 + 3 + 4) + 1.5Ix* 5/12 (current divider rule)
12Ix = 6 + 7.5Ix
4.5Ix = 6
Ix = 6/4.5 = 1.3333A

So Vth = Ix * 4 ohm = 5.33V

Next you need to find a Rth, so you need to short A with B and find Isc current.

Ixsc = 6V/(5 + 3 ) + 1.5Ixsc* 5/8

Ixsc = 12A

Rth = Vht/Ixsc = 5.333V/12A = 0.444Ohm