The admission fee at the small fair is $1.50 for children and $4.00 for adults. on a certain day, 2200 people enter the fair and $5050 is collected. how many children and how many adults attended?

Answers

You'll need to write an equation for this. Let's name our adults y, and the children x. So x+y=2200, as 2200 people went to the fair and people are either x or y. We also know that kid admission is 1.50, and adult admission is 4, so use y and x to make an equation for this. 1.5x+4y=5050 as 5050 was collected. So, you have the system of equations

x+y=2200
1.5x+4y=5050

I'll use substitution to solve. Isolate x in the top equation by subtracting y. so x+y-y=2200-y, and x=2200-y. Substitute this value for x into the second equation, so 1.5(2200-y)+4y=5050. Use the distributive property for the parentheses, so 3300-1.5y+4y=5050. Add together the y values of -1.5 and 4. You get 2.5, so 3300+2.5y=5050. Subtract 3300 from each side of the equation to begin isolating 2.5y, which gives you 1750. Divide 1750 by 2.5 to isolate y, which gives you 700. So you know y=700, so 700 adults attended. You know y=700 so put that into the first equation.

x+700=2200.
Isolate x by subtracting 700 from both sides, which gives you x=1500. So, 700 adults and 1500 children attended.
To solve this problem, you'll need to create a system of equations and then solve for the variables. For my two equations, x = # of children and y = # of adults

We know from the problem that the total amount of people who attended the fair is 2200, so that tells us that the number of children (x) plus the number of adults (y) will give us 2200. So, that's our first equation.

x + y = 2200 

We also know that the total amount of money collected was $5050. This tells us that the number of children's tickets sold (1.50 * x) plus the number of adult's tickets sold (4 * y) will give us $5050. So, that's our second equation.

1.50x + 4y = 5050

Now, you take both equations and solve for the variables.
x + y = 2200
1.50x + 4y = 5050

x + y = 2200
x = -y + 2200

1.50(-y + 2200) + 4y = 5050
-1.50y + 3300 +4y = 5050
3300 + 2.5y = 5050
2.5y = 1750
y = 700

Now that you know that y =700, you plug that information into either of the two equations and solve for x. I'm going to use the first equation because it's easier.
x + y = 2200
x + 700 = 2200
x = 1500

So, x = 1500 and y = 700
I would write systems of equations for this problem.

1.5x + 4y = 5050

x + y = 2200

From there solve the system of equations where x represents the number of children and y represent the number of adults.

x + y = 2200

x = 2200 - y

Plug "x" into the other system of equation.

1.5(2200-y) + 4y = 5050

3300 - 1.5y + 4y = 5050

2.5y = 1750

y = 700

Since you have the value for y, solve for x.

x + y = 2200

x + 700 = 2200

x = 1500

So, the solution is: 1500 children and 700 adults attended.

Just to check our answers:

1.5x + 4y = 5050

1.5(1500) + 4(700) = 5050

2250 + 2800 = 5050

5050 = 5050

1,100 kids 1,100 adults

Step-by-step explanation:

2,200 / 2 = 1,100

700 adults and 1500 children

Step-by-step explanation:

Let the number of adults be x and the number of children be y , then

x + y = 2200 equation 1

4x + 1.5y = 5050 equation 2

solving the system of linear equation by substitution method , from equation 1 make x the subject of the formula , that is

x = 2200 - y equation 3

substitute x = 2200 - y into equation 2 , that is

4 ( 2200 - y ) + 1.5y = 5050

8800 - 4y + 1.5y = 5050

8800 - 2.5y = 5050

2.5y = 8800 - 5050

2.5y = 3750

y = 3750/2.5

y = 1500

substitute y = 1500 into equation 3 , we have

x = 2200 - y

x = 2200 - 1500

x = 700

Therefore , 700 adults and 1500 children entered

$1.50C + $4A= $5050
 C + A = 2200

-4(C + A = 2200)
-4C - 4A = -8800
1.50C+4A = 5050

-2.5C =- 3750
-2.5C/-2.5 =- 3750/-2.5
C = 1500

C + A = 2200
1500 + A =2200
1500 - 1500 + A = 2200 - 1500
A = 700

CHECK
$1.50C + $4A= $5050
$1.50(1500) + $4(700)= $5050
$2250 +$2800 =$5050
$5050 = $5050

 C + A = 2200
1500 + 700 =2200
2200 = 2200

answer:

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step-by-step explanation:



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