Answer
$I_1=1.77A$
$I_2=1.36A$
$I_3=0.41A$
$I_4=0.68A$
$I_5=-0.27A$
Work Step by Step
$I_1=I_3+I_2$
$I_3=I_4+I_5$
$I_1-I_2=I_4+I_5$
$I_1=I_2+I_4+I_5$
$0=-(6\Omega)I_2+5.0V+12V-(5\Omega)I_1$
$17V=(5\Omega)I_1+(6\Omega)I_2$
$17V=(5\Omega)I_1+(12\Omega)I_4$
$7V=(6.8\Omega)I_5+(5\Omega)I_1$
$I_2=\frac{17V-(5\Omega)I_1}{6V}$
$I_4=\frac{17V-(5\Omega)I_1}{12V}$
$I_5=\frac{7V-(5\Omega)I_1}{6.8V}$
$I_4=\frac{17-5I_1}{6}+\frac{17-5I_1}{12}+\frac{7-5I_1}{6.8}$
$I_1=1.77A$
$I_2=\frac{17-5(1.77)}{6}=1.36A$
$I_4=\frac{17-5(1.77)}{12}=0.68A$
$I_5=\frac{7-5(1.77)}{6.8}=-0.27A$
