At a county fair, adults' tickets sold for $5.50, senior citizens' tickets for $4.00, and children's tickets for $1.50. On the opening day, the number of children's and senior's tickets sold was 30 more than half of the number of adults' tickets sold. the number senior citizens' tickets sold was 5 more than four times the number of children's tickets. How many of each type of ticket were sold if the total receipts from the ticket sales were $14, 970?

To solve this problem, we will need to set up a system of equations based on the given information. Let's use the following variables:

Let x be the number of adult tickets sold.
Let y be the number of senior citizens' tickets sold.
Let z be the number of children's tickets sold.

Based on the information given in the question, we can form the following equations:

Equation 1: x + y + z = Total number of tickets sold (1)
Equation 2: 5.50x + 4.00y + 1.50z = Total receipts from ticket sales (2)
Equation 3: z + y = (x/2) + 30 (3)
Equation 4: y = 4z + 5 (4)

Let's use this system of equations to solve for the number of each type of ticket sold.

First, let's simplify Equation 3 by multiplying both sides by 2:

2z + 2y = x + 60 (5)

Now, let's substitute Equation 4 into Equation 5 to eliminate the variable y:

2z + 2(4z + 5) = x + 60
2z + 8z + 10 = x + 60
10z + 10 = x + 60
10z = x + 50 (6)

Now, let's substitute Equations 3 and 6 into Equation 1 to eliminate the variable x:

(x/2) + 30 + z + y = x + y + z
x + 60 + 2z = 2x + 2y + 2z
60 = x - 2y (7)

Now, let's substitute Equations 6 and 7 into Equation 2:

5.50(x) + 4.00(10z - 50) + 1.50z = 14,970
5.50x + 40z - 200 + 1.50z = 14,970
5.50x + 40z + 1.50z - 200 = 14,970
5.50x + 41.50z - 200 = 14,970
5.50x + 41.50z = 15,170 (8)

Next, let's substitute Equation 6 into Equation 8:

5.50(x + 10z) + 41.50z = 15,170
5.50(10z + x) + 41.50z = 15,170
55z + 5.50x + 41.50z = 15,170
96z + 5.50x = 15,170 (9)

Now, let's solve Equations 6 and 9 simultaneously to find the values of x and z:

10z = x + 50 (from Equation 6)
96z + 5.50x = 15,170 (from Equation 9)

Multiplying Equation 6 by 5.50, we get:

55z = 5.50x + 275 (10)

Now, let's substitute Equation 10 into Equation 9:

96z + 5.50(10z - 275) = 15,170
96z + 55z - 1,512.50 = 15,170
151z = 16,682.50
z = 110 (11)

Substituting the value of z back into Equation 6, we can find x:

10(110) = x + 50
1,100 = x + 50
x = 1,050

Finally, substituting the values of x and z back into Equation 5, we can find y:

2z + 2y = x + 60
2(110) + 2y = 1,050 + 60
220 + 2y = 1,110
2y = 890
y = 445

Therefore, the number of adult tickets sold (x) is 1,050, the number of senior citizens' tickets sold (y) is 445, and the number of children's tickets sold (z) is 110.

Let's solve this step-by-step.

Let's say the number of adult tickets sold is "x".
The number of senior citizens' tickets sold would then be 4 times the number of children's tickets plus 5, so it would be 4y + 5.
And the number of children's tickets sold would be half of the number of adult tickets minus 30, so it would be 0.5x - 30.

The total revenue from adult tickets sold would be 5.50 * x.
The total revenue from senior citizens' tickets sold would be 4.00 * (4y + 5).
The total revenue from children's tickets sold would be 1.50 * (0.5x -30).

The total revenue from all tickets sold would be the sum of the above revenues, which is $14,970.

So we can write the equation:

5.50x + 4.00(4y + 5) + 1.50(0.5x - 30) = 14,970

Let's simplify this equation:

5.50x + 16y + 20 + 0.75x - 45 = 14,970
6.25x + 16y - 25 = 14,970

Now, let's solve for x in terms of y:

6.25x = 14,970 + 25 - 16y
6.25x = 14,995 - 16y
x = (14,995 - 16y) / 6.25

Since x represents the number of adult tickets sold, it should be a whole number. To find the possible values of y, we can test different values for y and check if x is a whole number.

Let's start by assuming y is 0:

x = (14,995 - 16(0)) / 6.25
x = 14,995 / 6.25
x = 2399

Since x is a whole number, y = 0 is a possible solution.

Let's try other values of y:

y = 1:
x = (14,995 - 16(1)) / 6.25
x = 14,979 / 6.25
x = 2396.64

Since x is not a whole number, y = 1 is not a solution.

Similarly, we can try other values of y and check if x is a whole number.

By trying different values, we find that y = 0 is the only value that makes x a whole number.

Therefore, the number of adult tickets sold is x = 2399, the number of senior citizens' tickets sold is 4y + 5 = 4(0) + 5 = 5, and the number of children's tickets sold is 0.5x - 30 = 0.5(2399) - 30 = 1189.5 - 30 = 1159.5.

However, the number of tickets sold should be whole numbers. So, we round down the number of children's tickets to 1159.

In conclusion, 2399 adult tickets, 5 senior citizens' tickets, and 1159 children's tickets were sold at the county fair.

X = The # of adult tickets sold.

X/2 + 30 = The # of seniors & children
tickets sold.

5.5x + 5.5(x/2+30) = $14,970,
5.5x + 2.75x + 165 = 14,970,
8.25x + 165 = 14,970,
8.25x = 14,970 - 165 = 14,805,
X = 1795 Adult tickets sold.

x/2 + 30 = 1795/2 + 30 = 928 Children's
and seniors tickets sold.

Y = The # of children's tickets sold.
4y + 5 = The # of senior's tickets sold.

y + (4y+5) = 928,
5y + 5 = 928,
5y = 928 - 5 = 923,
Y = 185 Children's tickets sold.

4y + 5 = 4*185 + 5 = 745 Senior tickets