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?

n = 12$    x = 8$   y = 10$     where n, x, and y are number of tickets

12 n + 8 x + 10 y = 3920     and n + x + y = 420

12n + 8 (x + y) + 2 y = 3920

12 n + 8 (5 n) + 2 y = 3920        since 5 (x + y) = n

52 n + 2 y = 3920   or  y = 1960 - 26 n

Also, n + x + y = 420   or n + 5 n = 420   since x + y = 5 n

n = 70    so 70 of the $12 were sold

And since y = 1960 - 26 n     we have y = 140 tickets

Now 12 * 70 + 8 x + 140 * 10 = 3920

This gives x = 210 tickets

Check:   210 + 140 + 70 = 420 tickets

Also, 12 * 70 + 210 * 8 + 140 * 10 = 3920

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step-by-step explanation:

it has to have an angle of 90 degrees