Can you explain how to do 13 please

I must confess that I was about to pass this question by, but I was captured by the respectful and dignified way in which you asked for help. 

A careful reading of the problem gives you two equations in two unknowns, which you can then solve as simultaneous equations.  Here's how it looks:

Call 'C' the price of the senior Citizen ticket.
Call 'S' the price of the Student ticket.

On the first night . . .    10 C  +  12 S  =  208

On the second night . . .  8C  +  3 S  =  74

Those are your two simultaneous equations. Now the idea is to multiply or divide each side of one equation in such a way that when you add or subtract it from the other equation, one of the variables will become a zero quantity ...  you'll be left with an equation in one variable, which you can easily solve. THEN, knowing the value of one variable, you can put it back into one of the original equations,and find the value of the other variable.

This all sounds more complicated than it is.   Here's how it goes:
We have . . .

10 C  +  12 S  =  208  and
8C  +  3 S  =  74

I'm going to multiply each side of the second equation by  4, and then write it under the first one:

10 C  +  12 S  =  208
32 C  +  12 S  =  296

Now, subtract the lower equation from the upper one, and you get . . .

- 22 C  +  0  =   - 88

Divide each side of this one by  -22  and you have    C = $4.00 .

THAT's what you need, to blow the whole problem wide open.  Knowing
the value of 'C', let's substitute it into the equation for the first night:

10 C  +  12 S  =  208

10(4)  +  12 S  =  208

40  +  12 S  =  208

Subtract  40  from each side :    12 S  =  168

Divide each side by  12 :    S  =  $ 14.00 .

Finally, as we look over our results, and see that Students have to pay  $14  to see the show but Seniors can get in for only  $4 , we reflect on this ... or at least I do ... and realize that getting old is not necessarily all bad.