Question

Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.

Two fifth grade classes are attending an amusement park as a field trip. Their teachers purchased the tickets for the class and the parent chaperones. Mrs. Chang purchased 21 child tickets and 5 adult tickets, which cost a total of $502. Mr. Bryant purchased 16 child tickets and 16 adult tickets, paying a total of $736. What are the ticket prices?

The price of a child ticket is $
and the price of an adult ticket is $
.

Answer 1

Let \( x \) be the price of a child ticket and \( y \) be the price of an adult ticket. We can set up the following system of equations based on the information provided:

1. From Mrs. Chang's purchases:
\[
21x + 5y = 502
\]

2. From Mr. Bryant's purchases:
\[
16x + 16y = 736
\]

Now we have the system of equations:
\[
\begin{cases}
21x + 5y = 502 \quad (1) \\
16x + 16y = 736 \quad (2)
\end{cases}
\]

Next, we can simplify equation (2):
\[
16x + 16y = 736 \implies x + y = 46 \quad (3) \quad \text{(by dividing the entire equation by 16)}
\]

Now we will express \( y \) in terms of \( x \) from equation (3):
\[
y = 46 - x
\]

Next, we substitute \( y \) into equation (1):
\[
21x + 5(46 - x) = 502
\]
Expanding this:
\[
21x + 230 - 5x = 502
\]
Combining like terms:
\[
16x + 230 = 502
\]
Subtracting 230 from both sides:
\[
16x = 272
\]
Dividing by 16:
\[
x = 17
\]

Now we substitute \( x \) back into equation (3) to find \( y \):
\[
y = 46 - 17 = 29
\]

Thus, the prices of the tickets are:
- The price of a child ticket is \( \boxed{17} \)
- The price of an adult ticket is \( \boxed{29} \)

To verify, we can substitute \( x \) and \( y \) back into the original equations.

1. For Mrs. Chang:
\[
21(17) + 5(29) = 357 + 145 = 502 \quad \text{(True)}
\]

2. For Mr. Bryant:
\[
16(17) + 16(29) = 272 + 464 = 736 \quad \text{(True)}
\]

Both equations are satisfied, confirming that our solution is correct.