Answer
a) $I_1=0.426$
$I_2=0.314A$
$I_3=0.111A$
b) $V_{ter}=5.89V$
Work Step by Step
a) $I_1=I_2+I_3$
$24.0V=I_135\Omega+I_229\Omega$
$I_2=0.828A-1.207I_1$
$18.0V=I_135\Omega+I_328\Omega$
$I_3=0.643A-1.25I_1$
$I_1=0.828A-1.207I_1+0.643A-1.25I_1$
$I_1=0.426$
$I_2=0.828A-1.207I_1=0.314A$
$I_3=0.643A-1.25I_1=0.111A$
b) $V_{ter}=\mathcal{E}-I_3r=6.0V-(0.111A)(1.0\Omega)=5.89V$
