The admission fee at a fair is $3.25 for children and $6.75 for adults. On a certain day, 870 people enter the fair and $3,772.50 is collected. Write a system of linear equations that you can use to determine how many children and adults attended. Be sure to define your variables. Solve the system. SHOW YOUR WORK PLEASE!

Question

09-10-2018
Mathematics

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The admission fee at a fair is $3.25 for children and $6.75 for adults. On a certain day, 870 people enter the fair and $3,772.50 is collected. Write a system of linear equations that you can use to determine how many children and adults attended. Be sure to define your variables. Solve the system. SHOW YOUR WORK PLEASE!

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lublana lublana · Answer 1 · 2018-10-18T17:54:53+02:00

Given that admission fee for 1 child = $3.25

If there are x children then admission fee for x children = $3.25x

Given that admission fee for 1 adult = $6.25

If there are y adults then admission fee for y adults = $6.75y

Then total fee collected = 3.25x+6.75y

Given that On a certain day, 870 people enter the fair then equation will be

x+y=870

or y=870-x...(i)

And $3,772.50 is collected means we get equation:

3.25x+6.75y = 3772.50

or 325x+675y = 377250...(ii)

Hence required system of equation is {x+y=870, 3.25x+6.75y = 3772.50}

Now we solve both to find values of x and y

Plug (i) into (ii)

325x+675(870-x) = 377250

325x+587250-675x = 377250

587250-350x = 377250

-350x = 377250-587250

-350x = -210000

x=600

now plug value of x into (i)

y=870-x=870-600=270

Hence final answer is:

Number of children = 600

Number of adults = 270

carlosego carlosego · Answer 2 · 2018-10-18T17:58:10+02:00

For this case we have the following variables:

x: Represents the number of children at the fair

y: Represents the number of adults at the fair

If 870 people enter the fair we have:

[tex]x + y = 870[/tex]

If that day is collected 3772.50 dollars, we have:

[tex]3.25x + 6.75y = 3772.50[/tex]

So, we have two equations with two unknowns:

[tex]x + y = 870[/tex] -----> (1)

[tex]3.25x + 6.75y = 3772.50[/tex] -----> (2)

Clearance of (1):[tex]y = 870-x[/tex]

Substituting in 2:

[tex]3.25x + 6.75 (870-x) = 3772.50\\3.25x + 6.75 * 870-6.75x = 3772.50\\3.25x-6.75x = 3772.50- (6.75 * 870)\\3.25x-6.75x = 3772.50-5872.5\\-3.5x = -2100[/tex]

[tex]x = \frac{-2100}{-3.5}\\x = 600[/tex]

Thus, there were 600 children at the fair.

To know the number of adults, we cleared and from the equation (1):

[tex]y = 870-x\\y = 870-600\\y = 270[/tex]

Thus, there were 270 adults at the fair.

Answer:

600 children and 270 adults