Answer: Child ticket: $8, Adult ticket: $6
Explanation:
Let's call the price of 1 adult ticket "a" and the price of 1 child ticket "c".
On the first day, the school sold 3 adult tickets for a total of 3a dollars, and 2 child tickets for a total of 2c dollars, for a total of 34 dollars. So, we have the equations:
3a + 2c = 34
On the second day, the school sold 3 adult tickets for a total of 3a dollars and 12 child tickets for a total of 12c dollars, for a total of 114 dollars. So, we have another equation:
3a + 12c = 114
We can now use these two equations to solve for the value of "a" and "c".
We can use substitution to solve for one variable, then substitute it back into either of the equations to solve for the other variable.
Let's solve for "a" by substituting the first equation into the second:
3a + 12c = 114
3a = 114 - 12c
a = (114 - 12c) / 3
Now that we have an equation for "a", we can substitute it back into the first equation to solve for "c":
3a + 2c = 34
3[(114 - 12c) / 3] + 2c = 34
114 - 12c + 2c = 34
-10c = -80
c = $8
So, the price of 1 adult ticket is
a = (114 - 12c) / 3
a = (114 - 12 * 8) / 3
a = (114 - 96) / 3
a = 18 / 3 = $6.