How do you solve the following system?: x +3y = 3 , 2x +4y = -2

$x = - 9 , \setminus \setminus \setminus y = 4$

Explanation:

Given equations

$x + 3 y = 3 \setminus \ldots \ldots \left(1\right)$ &

$2 x = 4 y = - 2$

$x + 2 y = - 1 \setminus \ldots \ldots \ldots \left(2\right)$

Subtracting (2) from (1), we get

$x + 3 y - \left(x + 2 y\right) = 3 - \left(- 1\right)$

$y = 4$

setting the value of $y$ in (1), we get

$x + 3 \left(4\right) = 3$

$x = 3 - 12$

$x = - 9$

Hence the solution is $x = - 9 , \setminus \setminus \setminus y = 4$