Question 1204986
x = price of a child ticket.
y = price of an adult ticket.


you have two equations that need to be solved simultaneously.
they are:
6x + 5y = 130
12x + 13y = 284


multiply both sides of the first equation by 2 and leave the second equation as is to get:
12x + 10y = 260
12x + 13y = 284


subtract the first equation from the second to get:
3y = 24
solve for y to get y = 8


replace y with 8 in the second equation to get:
12x + 13*8 = 284
simplify to get:
12x + 104 = 284
subtract 104 from both sides of the equation to get:
12x = 180
solve for x to get:
x = 180/12 = 15


you have x = 15 and y = 8


go back to your original equations to see if they are true when you replace x with 15 and y with 8.


the two original equations are:
6x + 5y = 130 becomes 6*15 + 5*8 = 130 which becomes 90 + 40 = 130 which is true.
12x + 13y = 284 becomes 12*15 + 13*8 = 284 which becomes 180 + 104 = 284 which is also true.


the problem is satisfied when x = 15 and y = 8.
that means the price of a child ticket is 15 and the price of an adult ticket is 8.


that's your solution.
this appears to be counter-intuitive as you would expect the child ticket to be less than the adult ticket, but that's the way it worked out.


here's what it looks like on a graph.


<img src = "http://theo.x10hosting.com/2023/111701.jpg">