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?
Accepted Solution
A:
Answer:1,084 tickets were sold that cost $101,291 tickets were sold that cost $20767 tickets were sold that cost $30Step-by-step explanation:Letx ----> the number of tickets that cost $10 soldy ----> the number of tickets that cost $20 soldz ----> the number of tickets that cost $30 soldwe know thatx+y+z=3,142 -----> equation A10x+20y+30z=59,670 ----> equation By=x+207 ----> equation Csubstitute equation C in equation A and equation Bx+(x+207)+z=3,142 ----> 2x+z=2,935 ----> equation D10x+20(x+207)+30z=59,670 ---> 30x+30z=55,530 ----> equation ESolve the system of equations D and E by graphingThe solution is the intersection point both graphsThe solution is the point (1,084,767)sox=1,084, z=767see the attached figureFind the value of yy=x+207 ----> y=1,084+207=1,291therefore1,084 tickets were sold that cost $101,291 tickets were sold that cost $20767 tickets were sold that cost $30