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azamat
3 years ago
7

Tickets for a certain show cost ​$17​, ​$21​, ​or, for VIP​ seats, ​$40. If ten times as many ​$17 tickets were sold as VIP​ tic

kets, and the number of ​$17 tickets sold was 57 more than the sum of the number of ​$21 tickets and VIP​ tickets, sales of all three kinds of tickets would total ​$51 comma 471. How many of each kind of ticket would have been​ sold?
Mathematics
1 answer:
lawyer [7]3 years ago
8 0

Answer:

Tickets sold:

VIP =126

$17 tickets =1,260

$21 tickets =9\cdot 126+57=1,191

Step-by-step explanation:

Let x be the number of VIP tickets.  

If ten times as many ​$17 tickets were sold as VIP​ tickets, then the number of $17 tickets is 10x.

If the number of ​$17 tickets sold was 57 more than the sum of the number of ​$21 tickets and VIP​ tickets, then 10x+57=x+y and the number y of $21 tickets is 9x+57.

Amounts earned:

VIP tickets =\$40x

$17 tickets =\$17\cdot 10x=\$170x

$21 tickets =\$21\cdot (9x+57)=\$(189x+1,197)

Total =\$(40x+170x+189x+1,197)=\$(399x+1,197)

The sales of all three kinds of tickets would total ​$51,471, so

399x+1,197=51,471\\ \\399x=51,471-1,197\\ \\399x=50,274\\ \\x=126

Tickets sold:

VIP =126

$17 tickets =1,260

$21 tickets =9\cdot 126+57=1,191

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