How do you solve #(1/2)x - (1/3)y = -3# and #(1/8)x + (1/6)y = 0#?
The minimum common multiple between 2,3,6 and 8 is 24. So
let's convert all these fractions in something/24:
Then multiply all the equalities by 24:
add it to the other equation without change the result:
and applying the value