Answer
$V_R=2.4V$
Work Step by Step
$I_R=I_1+I_2$
$2.00V=0.35\Omega I_1+4.0\Omega I_R$
$I_R=0.5A-0.088I_1$
$1.00V=0.35\Omega(I_2-I_1)$
$1.00V=0.35\Omega(I_R-2I_1)$
$1.00V=0.35\Omega(0.5A-2.088I_1)$
$2.86A=0.5A-2.088I_1$
$I_1=-1.13A$
$I_R=0.5A-0.088(-1.13A)$
$I_R=0.60A$
$V_R=I_RR=(0.60A)(4.00\Omega)=2.4V$
