How do you solve the system (1/2)x - (1/3)y = -3 and (1/8)x + (1/6)y = 0?

$\frac{x}{2} - \frac{y}{3} = - 3$ (1)
$\frac{x}{8} + \frac{y}{6} = 0$ (2) -> y/3 = -x/4
(1) ->$\frac{x}{2} + \frac{x}{4} = - 3 \to 3 x = - 12 \to x = - 4$
(2) ->$- \frac{1}{2} + \frac{y}{6} = 0 \to y = \frac{6}{2} = 3$