Using the Kirchhoff's voltage law

for loop $ABCDA , 2 i _1+5\left( i _1+ i _2\right)=12$ or $7 i _1+5 i _2=12 \ldots$ (1)

and for loop $EBCFE$, $2 i_2+5\left(i_1+i_2\right)=12$ or $5 i_1+7 i_2=12 \ldots(2)$

Now, (1) $\times 7,(2) \times 5 \Rightarrow 49 i_1+35 i_2=84 \ldots(3)$ and

$25 i_1+35 i_2=60 \ldots(4)$

(3) $-(4) \Rightarrow 2 i_1=24$ or $i_1=1 A$

Substituting the value of $i_1$ in (1) we get, $i_2=\frac{12-7}{5}=1 A$

Thus the current through $5 \Omega=i_1+i_2=1+1=2 A$