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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              [1]
5146C + 2807A = 122570             [2]

To make the number in front of C the same on both we multiple [1] by 5146 and [2] by 4296 to give:

22107216C + 11027878A = 508002828  [3]
22107216C + 12058872A = 526560720  [4]

Now we can do [4] - [3] to give:

1030994A = 18557892

Divide throughout by 1030994 so that:

A = 18

Substituting A = 18 in [1] will give:

4296C + 2143 x 18 = 98718
4296C + 38574     = 98718
4296C             = 60144

C = 14

As required. QED.

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