How do you solve #1/2x+1/3y=-1/6# and #1/2x+2y=13/2# using substitution?
To simplify this process multiply each equation by appropriate constants to clear the fractions:
[2b} appears to be the easiest one to convert into a form for substitution:
Substituting this back into [1b]
We can now substitute
[as always, you should substitute these values back into the original equations to verify the results... assume that I've done this.]