Expert: Scott A Wilson - 4/25/2011

Question
A basketball field house seats 15,000. Courtside seats sell for $8, endzone for $6 and balcony for $5. Total for a sell out is %$86,000. If half the courtside and balcony and all endzone seats are sold, ticket sales total $49,000. How many of each type of seat are there?

Answer
The problem is a matrix of three variables and three unknowns.
Let the seats be x=courtside, y=endzone, and z=balcony.

It is given there are 15,000 sets, so x+y+z=15,000.

It is given that x is at 8, y is at 6, and z is at 5.

It is given if all seats are sold, it is 86,000,
and this is 8x + 6y + 5z = 86,000.

If half of the courtside, half of the balcony, and all of the endzone seats are sold,
it is 49,000.  Since x=courtside and z=balcony, that gives 4x + 6y + 2.5z = 49,000.

This generates the matrix
1   1   1   15000
8   6   5   86000
4   6   2.5   49000,
and these elements are in 4 columns at this end, but they don't come through.
If necessary, cut the data and paste it into a spreadsheet.  If this doesn't line them up
and if you desire to see it mailed, write back and request with the date it was sent on.


Since there is a one in the top row, 1st column, subtract 8 times this row from the 2nd
and 4 times this row from the 3rd.  This gives
1   1   1   15000
0   -2   -3   -34000
0   2   -1.5   -11000

Since there is a -2 in the 2nd row, 2nd position, divide that row by -2, add half of it to the 1st row, and add all to the 3rd row.  This gives
1   0   -0.5   -2000
0   1   1.5   17000
0   0   -4.5   -45000

Divide the last row by -4.5, add 1/3 of it to the 2nd row,
and subtract 1/9 of it from the 1st row.  This gives
1   0   0   3000
0   1   0   2000
0   0   1   10000

When the three variavles are looked back at, it is then known how many of each type there are since x=3000, y=2000, and z=10000.  In fact, I belive x was courtside seats, y was endzone seats,
and z was balcony seats.