Answer
$I_1=0.60A$, left
$I_2=0.29A$, left
Work Step by Step
$V_3-I_1R_1+V_1=0$
$9.0V=I_1(22\Omega+1.4\Omega)+I_218\Omega$
$9.0V=I_1(23.4\Omega)+I_218\Omega$
$I_2=0.5A-1.3I_1$
$6.0V=I_11.4\Omega-I_218\Omega$
$6.0V=I_11.4\Omega-(0.5A-1.3I_1)18\Omega$
$6.0V=24.8I_1\Omega-9.0V$
$15.0V=24.8I_1\Omega$
$I_1=0.60A$
$I_2=0.5A-(1.3)(0.60A)=-0.29A$
