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 !

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

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:

If that day is collected 3772.50 dollars, we have:

So, we have two equations with two unknowns:

-----> (1)

-----> (2)

Clearance of (1):

Substituting in 2:

Thus, there were 600 children at the fair.

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

Thus, there were 270 adults at the fair.

600 children and 270 adults

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18