As the auditor for my local theme park, I noticed that on Saturday there were 4296 children and 2143 adults and the takings were £98,718.
However, on Sunday, there were 5146 children and 2807 adults and the takings were £122,570.
How much were the adults tickets and children's tickets?
This is easily solved using a common method for solving simultaneous equations.
First construct two algebraic equations, where C is the number of children, and A is the number of adults:
4296C + 2143A = 98718 
5146C + 2807A = 122570 
To make the number in front of C the same on both we multiple  by 5146 and  by 4296 to give:
22107216C + 11027878A = 508002828 
22107216C + 12058872A = 526560720 
Now we can do  -  to give:
1030994A = 18557892
Divide throughout by 1030994 so that:
A = 18
Substituting A = 18 in  will give:
4296C + 2143 x 18 = 98718
4296C + 38574 = 98718
4296C = 60144
C = 14
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