Question

The walker and Baldwin families ordered lunch at the refreshment stand at the basketball game. The walker family ordered 4 hot dogs and 3 cheeseburgers and paid 23.75. The Baldwin family ordered 6 hots dogs and 2 cheeseburgers and payed 25. Write a system of equations using h for hot dogs and c for cheeseburgers.

What are the prices of a hotdog and a cheeseburger?

Answer 1

Let's denote the price of a hot dog as \( h \) and the price of a cheeseburger as \( c \).

From the information given, we can set up a system of equations based on the orders of the two families.

1. The Walker family ordered 4 hot dogs and 3 cheeseburgers for a total of $23.75:
\[
4h + 3c = 23.75 \quad \text{(Equation 1)}
\]

2. The Baldwin family ordered 6 hot dogs and 2 cheeseburgers for a total of $25:
\[
6h + 2c = 25 \quad \text{(Equation 2)}
\]

Now we have the following system of equations:
\[
\begin{align*}
4h + 3c &= 23.75 \quad \text{(1)} \\
6h + 2c &= 25 \quad \text{(2)}
\end{align*}
\]

To solve this system, we can use the method of substitution or elimination. Here, we'll use the elimination method.

First, let's multiply Equation 1 by 2 and Equation 2 by 3 to align the coefficients of \( c \):
\[
\begin{align*}
8h + 6c &= 47.50 \quad \text{(Equation 3)} \\
18h + 6c &= 75 \quad \text{(Equation 4)}
\end{align*}
\]

Next, we can subtract Equation 3 from Equation 4:
\[
(18h + 6c) - (8h + 6c) = 75 - 47.50
\]
This simplifies to:
\[
10h = 27.50
\]
Now, divide both sides by 10:
\[
h = 2.75
\]

Now that we have the price of a hot dog \( h = 2.75 \), we can substitute this value back into either Equation 1 or Equation 2 to find the price of a cheeseburger \( c \). We'll use Equation 1:
\[
4(2.75) + 3c = 23.75
\]
This simplifies to:
\[
11 + 3c = 23.75
\]
Subtract 11 from both sides:
\[
3c = 12.75
\]
Now, divide by 3:
\[
c = 4.25
\]

Thus, the prices are:
- Price of a hot dog \( h = 2.75 \)
- Price of a cheeseburger \( c = 4.25 \)