Question 1170573
It would be very to type the real solution.

You can actually remove some of the options using process of elimination. y must be a positive, as there cannot be a negative seat number. x is equal to the number of seats at the table. Let's start off with the first table, which has 21 seats. 21*3=63, and 21*4=84. We can eliminate options c and e, as they would give us a negative number as our seat. 

Now for the solution. Table A has 21 seats. Every 3rd person is eliminated. 3,6,9,12,15,18, and 21 will all be eliminated. Any equation that gives us a multiple of 3 as our answer cannot be true. Therefore, we can also eliminate option D as 3*21-57 would give us 6, which is a multiple of 3.

This leaves options A and B. Since they both satisfy table A, let's move to table B. Table B has 22 people, and we know y cannot equal a multiple of 3 or a negative number. 22*3-61=5, and 22*4-83=5 as well. They both satisfy table B, so move on to table C, which has 23 seats. 23*3-61=8, but 23*4-83=9.

Option B would give us a multiple of 3, so B cannot be the correct answer.
This leaves option A as the only possible answer. You will find that the equation in option A will never give a multiple of 3 for numbers 21-30.

So, y=3x-61 is the correct answer. You can solve this without using process of elimination as well, but that would be hard to type out.

Edit: If you simply cross out every third person from 1 to 21 and repeating the process until you get one number left, seat number 2 would win for table A, number 5 for table B, number 8 for table C, etc.