**Answer:** Adult's tickets were £18 and children's were £14.

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.