Question

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

Answer 1

`x/2 - y/3 = -3` (1)

`x/8 + y/6 = 0` (2) -> y/3 = -x/4

(1) ->`x/2 + x/4 = -3 -> 3x = -12 -> x = -4`

(2) ->`-1/2 + y/6 = 0 -> y = 6/2 = 3`