Solve for current values in figure.

Explanation

Applying kirchhoff's first law at the junction B
we have `i_(1)+i_(2)=i_(3)` ..(1)
Applying Kirchhoff's second law to loop ABEFA
`-12+i_(2)xx1.5-i_(1)xx1+8=0`
`i_(1)-1.5i_(2)=-4` .(2)
From loop `BCDEB`
`-(i_(2)xx1.5)-(i_(3)xx9)+12=0`
`1.5i_(2)+9i_(3)=12` ..(3)
on solving `i_(1)=-1A`
and `i_(3)=1A`