voltage across 17ohms =25v

current through 17ohms(I17)= 25/17 =1.4705A

apply kcl at node3

I17+4=I25v

1.4705+4=I25v

I25v=5.407A

apply source transformation to 3A current

then we get 6v inseries with 2ohm

apply kcl at node1

(v-6)/(5+2)+v/10-I25v+I17ohm=0

(v-6)/7+v/10-5.47+1.47=0

0.142v-0.857+0.1v-4=0

0.242v=4.857

v=20.07v

i1=(v-6)/7=(20.07-6)/7=2.01A