Question 61028
You are right with your answers, and I like your
approach better than mine, in fact.
I used two variables where you only used one.
Amt. of mix is 12 oz
Amt of 1st acid = x
Amt. of 2nd acid is the rest = y (you said (12-x))
Amt. of acid in the mixture = .45(12)
Amt. of 1st acid = .30x
Amt. of 2nd acid = .50y 
.30x + .50y = .45*12
x + y = 12
.30x + .50(12-x) = .45*12
.3x + 6 - .5X = 5.4
.2x = 6 - 5.4
.2x = .6
x = 3
y = 12 - 3
y = 9

3 oz of 30% acid
9 oz of 50% acid
you are correct