How do you solve the system −3x + y = 2 and 4x + 2y = 4?

$- 3 x + y = 3 \implies y = 3 + 3 x$ ;
$4 x + 2 y = 4 \implies 4 x + 2 \cdot \left(3 + 3 x\right) = 4 \implies 4 x + 6 + 6 x = 4$
$\implies 10 x + 6 = 4 \implies 10 x = - 2 \implies x = - \frac{2}{10} = - \frac{1}{5}$
$y = 3 + 3 x \implies y = 3 + 3 \cdot \left(- \frac{1}{5}\right) \implies y = 3 - \frac{3}{5} \implies y = \frac{15}{5} - \frac{3}{5}$
$y = \frac{12}{5}$