Question

A basketball team sells tickets that cost​ $10, $20,​ or, for VIP​ seats,​ $30. The team has sold 469 tickets overall. It has sold 112 more​ $20 tickets than​ $10 tickets. The total sales are ​$8300 How many tickets of each kind have been​ sold?

Answer 1

x + y + z = 469 (1)

x + 112 = y (2)

10x + 20y + 30z = 8300 (3)

Replace (x + 112) for y to (1) and (3) equations:

x + x + 112 + z = 469
⇒2x + z = 357

10x + 20( x + 112 ) + 30z = 8300
⇒ 30x + 30z = 6060
Simplify, divide equation by 30
⇒x + z = 202

Put these 2 equations together:
2x + z = 357
x + z = 202
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x = 155 $10 tickets sold

(2) 155 + 112= 267 $20 tickets sold

z - ( 115 + 267) = 469

z = 87 $30 tickets sold