A few days ago
bridget

# math question?

At the Golden Oldies Theater, tickets for adults cost \$5.50 an tickets for children cost \$3.50. How many of each kind of ticket was purchased if 21 tickets were bought for \$83.50?

Top 6 Answers
A few days ago
Agent Feyd

Favorite Answer

Solve for a.

5.5a + 3.5b = 83.5

a = (83.5 – 3.5b)/5.5

a + b = 21

(83.5 – 3.5b)/5.5 + b = 21 — substitute for a

83.5 – 3.5b + 5.5b = 115.5 — multiply by 5.5 to remove denominator

2b = 32 — combine the b’s and move 83.5 to the right side

b = 16

21 – b = 5 — find a using b

checking:

5.5 * 5 + 3.5 * 16 = 83.5 — taking the original replacing a and b with their found values

27.5 + 56 = 83.5 — perform the multiplications

83.5 = 83.5 — add.

Done.

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A few days ago
Lilmaster
let the number of tickets bought by adult = x

therefore, number of tickets bought by children = 21-x

now , 5.50(x)+3.50(21-x)=83.50

=> 5.50x + 73.5-3.50x=83.50

=>2x=83.5-73.5

=> x = 10/2 = 5

therefore, no. of tickets bought by children = 21-5 =16

and no. of tickets bought by adults = 5 -ans.

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A few days ago
Marley K
Let a = number of adult tickets

21 – a = number of child tickets

5.5 a + 3.5(21 – a) =83.5

5.5 a + 73.5 – 3.5 a = 83.5

2 a = 10

a = 5

5 adult tickets were purchased, and

16 children’s tickets were purchased

That’s it! 🙂

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A few days ago
Apex01
Well i’d start out by this: 5.50x + 3.50y= 83.50 maybe?? Where to go from there i do not know.
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A few days ago
ashley
5 adult tickets and 16 tickets for children

form two equations

adult tickets=x

children tickets=y

x+y=21……………………(1)

5.50x+3.50y=83.50…………………..(2)

5.50(x+y=21)

1(5.50x+3.50y=83.50)

5.50x+5.50y=115.5

5.50x+3.50y=83.50

subtract the equations

0+5.50y-3.50y=115.5-83.50

2y=32

y=16

and then find x

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A few days ago
Anonymous
there will be 10 adult tickets and 11 kid tickets
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