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?

Respuesta :

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 
------------------------------------------
x                                  = 155  $10 tickets sold 

(2)  155 + 112= 267 $20 tickets sold 

z - ( 115 + 267) = 469 

z                       = 87 $30 tickets sold