Question 1179222
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At a local Brownsville play production, 420 tickets were sold. 
The ticket prices varied on the seating arrangements and cost $8, $10, or $12. 
The total income from ticket sales reached $3920. 
If the combined number of $8 and $10 priced tickets sold was 5 times the number of $12 tickets sold, 
how many tickets of each type were sold?
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<pre>
Let x be the number of the $8 tickets, and

let y be the number of the $10 tickets.


Then, from the condition, the number  (x+y)  is  5/6  of 420, i.e. x+y = 350,

      and the number of those who bought the $12 tickets was the rest  1/6 of 420, i.e. 70 persons.


Now we have the system of 2 (two) equations in two unknowns


     x +   y         =  350     (1)    

    8x + 10y + 12*70 = 3920     (2)    (total revenue)


We simplify this system to this EQUIVALENT strandard form


      x +   y        =  350     (3)  

     8x + 10y        = 3080     (4)  


Multiply equation (3) by 10 (both sides) and then subtract from it equation (4).  You will get then


    10x - 8x         = 3500 - 3080

       2x            =  420

        x            = 420/2 = 210.


Thus 210 persons bought $8 tickets;  hence,  350-210 = 140 bought $10 tickets;  and the rest  70 persons bought $12 tickets. 
</pre>

Solved.



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By the way, &nbsp;the problem' setup, &nbsp;presented in the post by @josgarithmetic,


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;IS TOTALLY WRONG,



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;so for your safety, &nbsp;you better ignore it . . .