Solution:
The answer is:
a = $60
e = $75
i = $40
o = $50
u = $25
To solve it, I set up 5 simultaneous linear equations:
(I used brackets to indicate each equation)
2a + i + o + 2u = 260 [1]
a + e + 2i + 2o + u = 340 [2]
a + e + 3i + = 255 [3]
2a + e + i + 4o = 435 [4]
a + i + o + 3u = 225 [5]
3*[2] - [5]: 2a + 3e + 5i + 5o = 795 [6]
-2*[2] + [1]: - 2e - 3i - 3o = -420 [7]
[4] - 2*[3]: - e - 5i + 4o = -75 [8]
[6] - 2*[3]: e - i + 5o = 285 [9]
[8] + [9]: - 6i + 9o = 210 [10]
[7] + 2*[9]: - 5i + 7o = 150 [11]
5*[10]: -30i + 45o = 1050 [12]
-6*[11]: 30i - 42o = -900 [13]
[12] + [13]: 3o = 150
o = 50
then substitute back to find i = 40, e = 75,
a = 60, and u = 25.
You could also solve this with Cramer's Rule or with a computer program:
Click here to see Keith Mealy's computer program