The problem describes the ticket sales for Darryl's school. On the first day, 1 senior citizen ticket and 5 child tickets were sold for a total of $77. On the second day, 8 senior citizen tickets and 10 child tickets were sold for a total of $196. We need to find the price of a senior citizen ticket and the price of a child ticket.

1. Problem Description

The problem describes the ticket sales for Darryl's school. On the first day, 1 senior citizen ticket and 5 child tickets were sold for a total of $

7

7. On the second day, 8 senior citizen tickets and 10 child tickets were sold for a total of $

1

9

6. We need to find the price of a senior citizen ticket and the price of a child ticket.

2. Solution Steps

Let $s$ be the price of a senior citizen ticket and $c$ be the price of a child ticket.

We can set up a system of two equations based on the given information:

$1s + 5c = 77$

$8s + 10c = 196$

We can solve this system of equations using substitution or elimination. Let's use elimination. Multiply the first equation by -2:

$-2(s + 5c) = -2(77)$

$-2s - 10c = -154$

Now we have:

$-2s - 10c = -154$

$8s + 10c = 196$

Add the two equations together:

$(-2s - 10c) + (8s + 10c) = -154 + 196$

$6s = 42$

Divide both sides by 6:

$s = \frac{42}{6}$

$s = 7$

Now that we have the price of a senior citizen ticket, we can substitute it back into one of the original equations to solve for $c$. Let's use the first equation:

$s + 5c = 77$

$7 + 5c = 77$

Subtract 7 from both sides:

$5c = 77 - 7$

$5c = 70$

Divide both sides by 5:

$c = \frac{70}{5}$

$c = 14$

3. Final Answer

The price of a senior citizen ticket is $7, and the price of a child ticket is$

1

4.