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 100 hot dogs and 200 pretzels 100 hot dogs and 200 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 250 hot dogs and 50 pretzels

Answer 1

To solve this problem, let's assign variables to each quantity:
Let's call the number of hot dogs sold 'h' and the number of pretzels sold 'p'.

The total number of hot dogs and pretzels sold is given as 300, so we have the equation:
h + p = 300.

The income earned is given as $800, and we know that hot dogs are sold for $4 each and pretzels for $2 each. Therefore, we have the equation:
4h + 2p = 800.

Now we have a system of equations:
h + p = 300,
4h + 2p = 800.

To solve this system, we can use substitution or elimination.
Let's use elimination:
Multiply the first equation by (-2) to have -2h - 2p = -600.
Now, add this equation to the second equation:
(4h + 2p) + (-2h - 2p) = 800 + (-600), which simplifies to:
2h = 200.
Divide both sides by 2:
h = 100.

Now, substitute this value back into the first equation to solve for p:
100 + p = 300,
p = 200.

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

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