SOLUTION: A basketball team sells tickets that cost​ $10, $20,​ or, for VIP​ seats,​ $30. The team has sold 3203 tickets overall. It has sold 216 more​ $20 tickets than​

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: A basketball team sells tickets that cost​ $10, $20,​ or, for VIP​ seats,​ $30. The team has sold 3203 tickets overall. It has sold 216 more​ $20 tickets than​       Log On

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Question 1140206: A basketball team sells tickets that cost​ $10, $20,​ or, for VIP​ seats,​ $30. The team has sold 3203

tickets overall. It has sold 216

more​ $20 tickets than​ $10 tickets. The total sales are ​$61, 650
.
How many tickets of each kind have been​ sold?
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

let tickets be xthat cost​ $10 ,ytickets that cost​ $20 , and z tickets that cost​ $30
The team has sold+3203+tickets overall.
x%2By%2Bz=3203....eq.1...solve for z
z=3203-x-y......eq.1a
It has sold 216 more​ $20 tickets than​ $10 tickets.
=>y=x%2B216
The total sales are ​$61+650:

10x%2B20y%2B30z=61650....substitute y and z
10x%2B20%28x%2B216%29%2B30%283203-x-%28x%2B216%29%29=61650.......solve or x
10x%2B20x%2B4320%2B30%283203-x-x-216%29=61650
30x%2B4320%2B30%283203-2x-216%29=61650
30x%2B4320%2B96090-60x-6480=61650
93930-30x=61650
93930-61650=30x
32280=30x
x=32280%2F30
x=1076
go to y=x%2B216, plug in value for x
y=1076%2B216
y=1292
go to z=3203-x-y......eq.1a ,plug in value for x and y
z=3203-1076-1292
z=3203-2368
z=835

they sold:
+1076 tickets that cost​ $10
1292 tickets that cost​ $20
835 tickets that cost​ $30