Question 122035
This is the same as one of those coin problems where you have to figure out how many dimes and nickels you have.  You will have two equations in two variables, one of the equations shows the relationship between the numbers of the two kinds of tickets, and the other equation shows the relationship between their values.


Let {{{s}}} be the number of student tickets and {{{n}}} be the number of non-student tickets.


The first thing we know is that {{{s + n=812}}} because the problem says 812 tickets were sold.


Since each student ticket costs $2, the total value of the student tickets must be {{{2s}}}.  Likewise, the total value of non-student tickets must be {{{3n}}}.  We are also given that the total value of all the tickets is $1912, so we can write:


{{{2s+3n=1912}}}


Take the first equation and solve it for one of the variables, let's solve for {{{n}}}:


{{{s + n=812}}}
{{{s+n-s=812-s}}}
{{{n=812-s}}}


Now we have an expression for n that can be substituted into the second equation:


{{{cartoon(2s+3*red(n)=1912,2s+3*red((812-s))=1912)}}}


Distribute the 3 and collect like terms:
{{{2s+2436-3s=1912}}}
{{{2436-s=1912}}}
{{{-s=1912-2436}}}
{{{-s=-524}}}
{{{s=524}}}


Now we know that 524 student tickets were sold, which is what the question asked.


Check:

If 524 student tickets were sold, then {{{812 - 524 = 288}}} non-student tickets were sold.

{{{524*2+288*3=1048+864=1912}}} answer checks.