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3241004551 [841]
3 years ago
14

Tickets for the community fair cost $12 for adults and $5 dollars for children. On the first day of the fair, 312 tickets were s

old for a total of $2,204 . How many adult tickets and how many child tickets were sold?
Mathematics
1 answer:
givi [52]3 years ago
6 0

With this information we can set up 2 equations:

x + y = 312      (# of tickets sold for adults + # of tickets sold to adults = 312)

12x + 5y = 2204 ( # of tickets sold for adults times $12 + # of tickets sold to adults times $5  = $2204)

Where x is how many  tickets were sold to adults and y how many tickets were sold to children

Now we can solve this system of equations by substitution:

isolate y in the first equation to find its value and plug it in the second equation

x + y = 312

isolate y by subtracting x from both sides:

x - x + y = 312

y = 312 - x

Apply y = 312 - x to the second equation

12x + 5y = 2204

12x + 5( 312 - x) = 2204

12x + 1560 - 5x = 2204

7x + 1560 = 2204

Subtract 1560 from both sides to isolate x

7x + 1560 - 1560 = 2204 - 1560

7x = 644

Divide both sides by 7

7/7x = 644/7

x = 92

Now plugin 92 for x in the first equation to find the value of y

x + y = 312

92 + y = 312

subtract 92 from both sides

92 - 92 + y = 312 - 92

y = 220

x = 92, y = 220

92 tickets were sold to adults and 220 tickets were sold to children

Hope it helps :)

Branliest would be appreciated

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