Formulate a system of equations for the situation below and solve. For the openi
ID: 3169066 • Letter: F
Question
Formulate a system of equations for the situation below and solve. For the opening night at the Opera House, a total of 1000 tickets were sold. Front orchestra seats cost 590 apiece, rear orchestra seats cost $70 apiece, and front balcony seats cost $50 apiece. The combined number of tickets sold for the front archestra and rear orchestra exceeded twice the number of front balcony tickets sold by 400. The total receipts for the performance were $70,000. Determine how many tickets of each type were sold front orchestra rear orchestra front balconyExplanation / Answer
Let the number of tickets sold for the front orchestra seats, rear orchestra seats and the front balcony seats be x, y and z respectively. Since the total number of tickets sold is 1000, we have x+y+z= 1000…(1). Also, since the total receipts were $ 70000, hence 90x+70y+50z = 70000 or, 9x+7y+5z = 7000…(2). Further, the combined number tickets sold for the front orchestra seats and the rear orchestra seats exceeded twice the number of front balcony seats by 400, we have x+y = 2z+400 or, x+y-2z = 400…(3).
The augmented matrix for this system of linear equations is A =
1
1
1
1000
9
7
5
7000
1
1
-2
400
To solve the system of linear equations, we will reduce A to its RREF as under:
Add -9 times the 1st row to the 2nd row
Add -1 times the 1st row to the 3rd row
Multiply the 2nd row by -1/2
Multiply the 3rd row by -1/3
Add -2 times the 3rd row to the 2nd row
Add -1 times the 3rd row to the 1st row
Add -1 times the 2nd row to the 1st row
Then the RREF of A is
1
0
0
200
0
1
0
600
0
0
1
200
Thus, x = 200, y = 600 and z = 200. Therefore, The number of various types of tickets sold are vas under:
front orchestra seats-200
rear orchestra seats-600
front balcony seats-200
1
1
1
1000
9
7
5
7000
1
1
-2
400