The figure below represents the given circuit.

Apply Kirchhoff voltage law in loop I,

$-9+6{\mathrm{I}}_1+12\left(I_1+I_2\right)=0$

$18{\mathrm{I}}_1+12{\mathrm{I}}_2=9$

$6{\mathrm{I}}_1+4{\mathrm{I}}_2=3 ....\left(1\right)$

Apply Kirchhoff law in loop II.

$-6+15{\mathrm{I}}_2+12\left(I_1+I_2\right)=0$

$12{\mathrm{I}}_1+27{\mathrm{I}}_2=6$

$4{\mathrm{I}}_1+9{\mathrm{I}}_2=2 ....\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$,

$I_1=0.5 \mathrm{A}$

$I_2=0$