Applying Kirchhoffs law in mash $(I)$

$2 i_1+6 i_1-6 i_2+8 i_1+0.5 i_1-15=0$

$16.5 i_1-6 i_2=15 \ldots . \text { (I) }$

Applying Kirchhoff's law in mash $(II)$

$7 i _2+ i _2+10 i _2+6 i _2-6 i _1=0$

$24 i _2-6 i _1=0$

$4 i _2- i _1=0$

$i _1=4 i _2 \ldots \ldots \text { (II) }$

After solving of equation $(I)$ and $(II)$

Then,

$16.5 \times 4 i _2-6 i _2=15$

$66 i _2-6 i _2=15$

$60 i _2=15$

$i _2=\frac{1}{4} A$

Now put the value of $i_2$ in equation $(I)$

$i_1=4 i_2$

$i_1=4 \times \frac{1}{4}$

$i_1=1\,\,A$

Now, the current from the battery in the circuit is $1\,\,A$.