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Question

Real-World Systems of Equations Quick Check
5 of 55 of 5 Items
Question
The Baynard High School Boosters are selling hot dogs and pretzels at the concession stand. They have 300 hot dogs and pretzels in stock and want to earn an income of $800. They plan to sell the hot dogs for $4 each and the pretzels for $2 each. How many hot dogs and how many pretzels do the boosters need to sell to reach their goal of $800?(1 point)
Responses

50 hot dogs and 250 pretzels
50 hot dogs and 250 pretzels

250 hot dogs and 50 pretzels
250 hot dogs and 50 pretzels

200 hot dogs and 100 pretzels
200 hot dogs and 100 pretzels

100 hot dogs and 200 pretzels
100 hot dogs and 200 pretzels

Answer 1

To solve this problem, we can set up a system of equations based on the given information.

Let \( h \) represent the number of hot dogs sold and \( p \) represent the number of pretzels sold. We have the following conditions:

1. The total number of items (hot dogs and pretzels) is 300: \[ h + p = 300 \]

2. The total income from selling hot dogs and pretzels needs to be $800: \[ 4h + 2p = 800 \]

Now we can solve this system of equations.

First, let's solve the first equation for \( p \): \[ p = 300 - h \]

Next, we can substitute this expression for \( p \) into the second equation: \[ 4h + 2(300 - h) = 800 \]

Now simplify the equation: \[ 4h + 600 - 2h = 800 \] \[ 2h + 600 = 800 \] \[ 2h = 800 - 600 \] \[ 2h = 200 \] \[ h = 100 \]

Now that we have \( h \), we can find \( p \): \[ p = 300 - 100 = 200 \]

So the Baynard High School Boosters need to sell 100 hot dogs and 200 pretzels to reach their goal of $800.

The correct response is: 100 hot dogs and 200 pretzels.