Real-World Systems of Equations Quick Check

Question

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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

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

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

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

100 hot dogs and 200 pretzels

Answer 1

To solve this problem, we can define two variables:

• Let \( h \) be the number of hot dogs sold.
• Let \( p \) be the number of pretzels sold.

We are given the following information:

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

2. The total income needed is $800, where hot dogs are sold at $4 each and pretzels are $2 each: \[ 4h + 2p = 800 \]

We have a system of equations:

1. \( h + p = 300 \)
2. \( 4h + 2p = 800 \)

Now, we can solve this system.

From the first equation, we can express \( p \) in terms of \( h \): \[ p = 300 - h \]

Now we can substitute \( p \) into the second equation: \[ 4h + 2(300 - h) = 800 \]

Simplifying this: \[ 4h + 600 - 2h = 800 \] \[ 2h + 600 = 800 \] \[ 2h = 200 \] \[ h = 100 \]

Now substitute \( h \) back into the equation for \( p \): \[ p = 300 - h = 300 - 100 = 200 \]

So, the boosters need to sell 100 hot dogs and 200 pretzels to reach their goal of $800.

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