Choose the two equations that can be solved as a system of equations to determine how many children and how many adults attended the fair.

Question

The admission fee at a county fair is $2 for children and $4 for adults. Suppose that on the last day, 1600 people enter the fair and $5000 is collected.

Choose the two equations that can be solved as a system of equations to determine how many children and how many adults attended the fair.

a − c = 1600

4a + 2c = 5000

4a − 2c = 5000

a + c = 1600

Answer 1

Hi! So the best way to solve questions like these is to break them down into smaller pieces. So the first thing we can do is recognize that the variable a=adults and c=children.

Next, as we read through the question, it is best to take a few notes. So we see that there were 1600 people at the fair which means that there was some mix of adults and children that added up to 1600. Also we see that the total amount that was collected at the fair was 5000$. From this we can conclude that some combination of adults (priced at 4$ each) and children (priced at 2$ each) added up to 5000$. The problem already tells us that we need to choose two equations which lets us know we will be dealing with a system of equations.

(***note: for each unknown variable, we need an equation to solve for it. So in this example we need to find the number of adults and children in attendance which means we need 2 equations)

Starting with the simpler of the two equations, let's first write out our equation for the number of people in attendance:

a+c=1600 --> So this equation is specifically for the amount of people in attendance. We this says that some number of adults (a) plus some number of children (c) sums up to 1600

4a+2c=5000 --> This equation is now for the amount of money that was collected. This is saying that some number of adults (a) at 4$ each plus some number of children (c) at 2$ adds up to 5000$

Now we create a system of equations to solve for the number of children and adults. It will look like this:

4a+2c=5000

a+c=1600

(note that either equation can be on top, I just chose the one what was simpler to work with)

So now to simplify, we realize we need to eliminate one of the variables in order to solve for the other. Lets start with eliminating (c). In order to do that, we need to multiply the whole bottom equation, a+c=1600, by a value that will allow c on the bottom to be opposite the value on top. That number is -2:

4a+2c=5000 4a+2c=5000

2(a+c=1600) = -2a-2c=-3200

Simplifying both these equations with each other (same as adding them together):

2a=1800

Now solve for a by isolating it. Divide both sides by 2 and we get a=900 which means there were 900 adults in attendance. Now to solve for the number of children, we refer back to our original equation that was meant to account for the number of total people: a+c=1600 and now we know that a=900 which means our equation now is: 900+c=1600. We can rearrange this equation by subtracting 900 from both sides. Now:

c=1600-900 so c=700

This is our final answer: There were 900 adults and 700 children. I hope this helps!

Answer 2

The total number of adults and children must equal 1600, so

A + C = 1600

The total number of dollars collected must equal 5000. With A adults each paying 4, and C children each paying 2,

4A + 2C = 5000