The school that Stephan attends is selling tickets to a choral performance.

On the first day of ticket sales, the school sold 45 senior tickets and 15 child tickets for a total of $405. On the second day, the school sold 52 senior tickets and 15 child tickets, taking in $461.

Find the price of one senior ticket and one child ticket.

By setting up and solving a system of equations representing the ticket sales over two days, we find that the price of a senior ticket is $8 and the price of a child ticket is $3.

Explanation:

To determine the price of one senior ticket and one child ticket from the choral performance, we can set up a system of equations based on the information given about the ticket sales over two days. Let's define x as the price of a senior ticket and y as the price of a child ticket.

From the first day, the equation would be:

45x + 15y = $405

From the second day, the equation would be:

52x + 15y = $461

To solve this system of equations, we can use the method of substitution or elimination. Let's use elimination:

First, multiply each term of the first equation by -1 so we can eliminate y when we add the equations:

-45x - 15y = -$405

Now we add both equations:

(-45x - 15y) + (52x + 15y) = -$405 + $461

This simplifies to:

7x = $56

Divide both sides by 7 to find the price of a senior ticket:

x = $8

Now that we have the price of a senior ticket, we can substitute this back into the first equation to find the price of a child ticket:

45($8) + 15y = $405

$360 + 15y = $405

Subtract $360 from both sides:

15y = $45

Divide both sides by 15:

y = $3

So, the price of a senior ticket is $8 and the price of a child ticket is $3.

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