Answer
$R_{120\Omega}$: $I_1=0.029A$ down
$R_{74\Omega}$: $I_1=0.029A$ right
$R_{110\Omega}$: $I_2=0.025A$ up
$R_{25\Omega}$: $I_2=0.025A$ left
$R_{56\Omega}$: $I_3=0.004A$ up
Work Step by Step
Name the currents:
$I_1$ from resistor $R_{120\Omega}$ to $R_{74\Omega}$
$I_2$ from resistor $R_{110\Omega}$ to $R_{25\Omega}$
$I_3$ from resistor $R_{56\Omega}$
$I_2=I_1+I_3$
$5.8V=194\Omega I_1-56\Omega I_3$
$3.0V=135\Omega I_2+56\Omega I_3$
$3.0V=135\Omega I_1+214\Omega I_3$
$I_3=0.014A-0.631I_1$
$5.8V=194\Omega I_1-56\Omega (0.014A-0.631I_1)$
$5.8V=229\Omega I_1-0.784V$
$I_1=0.029A$
$I_3=0.014A-0.631I_1=-0.004A$
$I_2=I_1+I_3=0.029A-0.004A=0.025A$