How do you solve the following linear system:  2x+ 3y=-1 , 5x+y=8 ?

$x = \frac{25}{13} , \setminus y = - \frac{21}{13}$

Explanation:

Given that

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

$5 x + y = 8 \setminus \ldots \ldots . \left(2\right)$

Multiplying (2) by $3$ on both sides & subtracting from (1), we get

$2 x + 3 y - 3 \left(5 x + y\right) = - 1 - 3 \setminus \times 8$

$- 13 x = - 25$

$x = \frac{25}{13}$

setting $x = \frac{25}{13}$ in (2), we get

$5 \left(\frac{25}{13}\right) + y = 8$

$y = 8 - \frac{125}{13} = - \frac{21}{13}$