SOLUTION: rob's school is selling tickets to a play.on the first day of ticket sales the school sold 9 adult tickets and 8 child tickets for a total of $145. the school took in $82 on the se

Question 1174417: rob's school is selling tickets to a play.on the first day of ticket sales the school sold 9 adult tickets and 8 child tickets for a total of $145. the school took in $82 on the second day by selling 2 adult tickets and 8 child tickets.what is the price each of one adult ticket and one child ticket?
Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!

Formally....

x = cost of each child ticket
y = cost of each adult ticket

9x+8y = 145 (1st day total sales)
2x+8y = 82 (2nd day total sales)

Solve by subtracting the second equation from the first, giving
7x = 63

Use that to find the cost of each adult ticket, x; then use that value with the total sales for either day to find the cost of each child ticket.

That's a simple example of an algebraic solution. You can get the answer informally by doing exactly the same calculations without the formal algebra.

9 adult tickets and 8 child tickets for $145, and 2 adult tickets and 8 child tickets for $82, means 9-2=7 adult tickets cost $145-$82 = $63.
7 adult tickets for $63 means $63/7 = $9 for each one.
2 adult tickets and $9 each, for a total of $18, plus 8 child tickets for a total of $82, means 8 child tickets for $82-$18 = $64
8 child tickets for $64 means $64/8 = $8 for each one.

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