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# help

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At a college production of a​ play, 420 tickets were sold. The ticket prices were​ \$8, \$10, and​ \$12, and the total income from ticket sales was ​\$3900. How many tickets of each type were sold if the combined number of​ \$8 and​ \$10 tickets sold was 5 times the number of​ \$12 tickets​ sold?

Apr 6, 2020

#1
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Apr 6, 2020
#3
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That link has 400 tickets, not 420.

Guest Apr 6, 2020
#2
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Let  x  represent the number of \$8.00 tickets sold

y  represent the number of \$10.00 tickets sold

z  represent the number of \$12.00 tickets sold

Since the combined number of \$8.00 tickets and \$10.00 tickets sold was 5 times the number of \$12.00 tickets sold:

x + y  =  5z   --->     z  =  (x + y)/5

Value of the tickets sold:  8x + 10y + 12(x + y)/5  =  3900

(multiply by 5)                  40x + 50y + 12x + 12y  =  19 500

52x + 62y  =  19 500

Number of tickets sold:  x + y + (x + y)/5  =  420

(multiply by 5)                 5x + 5y + x + y  =  2100

6x + 6y  =  2100

(divide by 6)                                     x + y  =  350

Combining the two equations:  52x + 62y  =  19500

x +     y  =  350

(multiplying the bottom equation by  -52:

52x + 62y  =   19500

-52x -  52y  =   -18200

y  =  130

x + y   = 350   --->   x + 130  =  350   --->   x  =  220

z  =  (x + y)/5   --->   z  =  (220 + 130) / 5   --->   z  =  70

Apr 6, 2020