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Real-World Systems of Equations Quick Check

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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

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

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

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

50 hot dogs and 250 pretzels

1 answer

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

Let:

  • \( h \) = number of hot dogs sold
  • \( p \) = number of pretzels sold

From the information provided, we have the following two equations:

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

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

Let's solve the system of equations step by step.

Step 1: Solve the first equation for one of the variables

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

Step 2: Substitute into the second equation

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

Step 3: Simplify and solve for \( h \)

Distributing the 2: \[ 4h + 600 - 2h = 800 \] Combine like terms: \[ 2h + 600 = 800 \] Subtract 600 from both sides: \[ 2h = 200 \] Divide by 2: \[ h = 100 \]

Step 4: Substitute back to find \( p \)

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

Conclusion

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

So the correct response is: 100 hot dogs and 200 pretzels.