A basketball team sells tickets that cost $10, $20, or, for VIP seats, $30. The team has sold 3142 tickets overall. It has sold 207 more $20 tickets than $10 tickets. The total sales are $59,670. How many tickets of each kind have been sold?

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calculista calculista · Answer 1 · 2019-08-04T17:16:29+02:00

Answer:

1,084 tickets were sold that cost $10

1,291 tickets were sold that cost $20

767 tickets were sold that cost $30

Step-by-step explanation:

Let

x ----> the number of tickets that cost $10 sold

y ----> the number of tickets that cost $20 sold

z ----> the number of tickets that cost $30 sold

we know that

x+y+z=3,142 -----> equation A

10x+20y+30z=59,670 ----> equation B

y=x+207 ----> equation C

substitute equation C in equation A and equation B

x+(x+207)+z=3,142 ----> 2x+z=2,935 ----> equation D

10x+20(x+207)+30z=59,670 ---> 30x+30z=55,530 ----> equation E

Solve the system of equations D and E by graphing

The solution is the intersection point both graphs

The solution is the point (1,084,767)

so

x=1,084, z=767

see the attached figure

Find the value of y

y=x+207 ----> y=1,084+207=1,291

therefore

1,084 tickets were sold that cost $10

1,291 tickets were sold that cost $20

767 tickets were sold that cost $30