Question 5757
Let


A = number of advanced tickes sold
D = number of tickets sold at the door.


We are told that there were 800 tickets sold. So, we have our first equation:


{{{ A+D=800 }}}<----- NUMBER OF TICKETS EQUATION


Now we have to come up with the equation that computes the total cost. That would be price of advanced tickets times how many advanced tickets were sold PLUS number of door tickets times how many door tickets were solved = the amount of money collected.


{{{ 14.50A + 22.00D = 16640 }}} <---- COST EQUATION. We have two variables here and we can't solve this equation for a single value. However, we have the "number of tickets" equation that we can manage so that this cost equation we have will only be expressed with 1 variable.


The question is, what do we do with the first expression? The question does ask for how many tickets were sold at the door. So ultimately, we want to keep the D variable in our cost equation. This means that we can solve the "number of tickets" equation for A. That would bring us A = 800 - D. Now that we have this, we can substitute 800 - D for the A that shows up in the cost equation. We now have:


{{{ 14.50(800-D) + 22.00D = 16640 }}} <--- the cost equation rewritten with the substitution so that there 's only one variable we'll be solving for.


{{{ 11600 - 14.50D + 22.00D = 16640 }}} <--- Expanded


{{{ 11600 + 7.5D = 16640 }}} <---- Simplified


{{{ 7.5D = 5040 }}}


{{{ D = 672 }}} <---- So there were 672 door tickets sold.