This is a good, old-fashioned matrice problem.
Let x = # of unknown child tickets sold
Let y = # of unknown student tickets sold
Let z = # of unknown adult tickets sold
We know the following information:
Total tickets sold: (x + y + z) is (=) 273
Total of ticket sales: (5x + 7y + 12z) is (=) $1,972
Twice as many child tickets were sold as adult tickets; putting this in whole numbers: x - 2z = 0
Now, we can place this into a matrix; then solve with the respective eliminations:
1 1 1 = 273
5 7 12 = 1972
1 0 -2 = 0
I prefer to use a system that greatly reduces the chance of putting a non-Zero number where a ZERO exists; plus, my strategy is to get rid of items in a uniform, predictable manner. I assign circled numbers to where the items are to be eliminated, in the order that I wish to eliminate them:
1 at R3, C1
2 at R2, C1
3 at R3, C2
4 at R1, C3
5 at R2, C3
6 at R1, C2
Using this above system, I perform the following row operations:
R1 - R3 = new R3 [1]
-5R1 + R2 = new R2 [2]
R2 - 2R3 = new R3 [3]
R2 - 7R1 = new R1 [4]
-7R3 + R2 = new R2 [5]
5R2 + 2R1 = new R1 [6]
Then, I do the following row reductions:
-1/14 R1 = new R1
1/2 R2 = new R2
This provides me with the following matrix:
1 0 0 = 122
0 1 0 = 90
0 0 1 = 61
We do the following 3 checks to determine the validity:
(1) 122 + 90 + 61 = 273 (Check)
(2) 5 (122) + 7 (90) + 12 (61) = 610 + 630 + 732 = 1,972 (Check)
(3) 122 = 2(62) (Check)
This gives me the following:
