Answer:
4, 16 and 10
Explanation:
Here we would like to work with a system of 3 equations, we need ti find them. Lets call s the 10oz coffees, m the 14oz and l the 20oz.
As Kyle sold $37.2 selling 3 different types of coffee at their respective prices we can say:
revenue = s*0.95 + m*1.15 + l*1.5 = 37.2 [equation 1]
And as he sold 30 cups:
s + m + l = 30 [equation 2]
And as the total volume of coffee is 464 oz:
s*10 + m*14 + l*20 = 464 [equation 3]
So, we have a system of 3 equations with three unknowns s, m and l.
Take the eq 2 and subtract m and l in both sides:
s = 30 - m - l
Now, take it and replace in the other 2:
(30 - m - l)*10 + 14m + 20l = 464 [eq 3]
0.95(30 - m - l) + 1.15m + 1.5l = 37.2 [eq 1]
Working with equation 3:
300 -10m -10l + 14m + 20l = 464
300 + 4m + 10l = 464
4m + 10l = 164 [equation 3']
Now with eq 1:
0.95(30 - m - l) + 1.15m + 1.5l = 37.2
28.5 - 0.95m - 0.95l + 1.15m + 1.5 l = 37.2
28.5 + 0.2m + 0.55l = 37.2
0.2m + 0.55l = 8.7 [equaiton 1']
Now we passed to a 2-equation system with eq 1' and eq 3'.
Take eq 3' and subtract 4m in both sides:
10l = 164 - 4m
Now divide by 10 in both sides
l = [164 - 4m]/10
Now, with this value for l replace in eq 1':
0.2m + 0.55 ( [164 - 4m]/10) = 8.7
0.2m + 0.055 (164 - 4m) = 8.7
0.2m + 9.02 - 0.22m = 8.7
- 0.02m = -0.32
m = -0.32/-0.02
m = 16 ----> 16 14oz coffees
Now replace m=16 in eq 1' or 3' to find l:
4m +10l = 164
64 + 10l = 164
10l = 100
l = 10 ----> 10 20oz coffees
So, finally we replace l=10 and m=16 in equaiton 2:
s + m + l = 30
s + 16 +10 = 30
s = 30 - 26
s = 4
So, Kyle filled 4 10oz coffes, 16 14oz coffees and 10 20oz coffees.