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 $80. . 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) 200 hot dogs and 100 pretzels 50 hot dogs and 250 pretzels 100 hot dogs and 200 pretzels 250 hot dogs and 50 pretzels
Let's say the number of hot dogs the boosters need to sell is x and the number of pretzels they need to sell is y.
The income from selling hot dogs is 4x dollars.
The income from selling pretzels is 2y dollars.
Since the total income they need to reach is 80 dollars, we have the equation 4x + 2y = 80.
We also know that the total number of hot dogs and pretzels they have in stock is 300, so we have the equation x + y = 300.
Now, we have a system of equations:
4x + 2y = 80
x + y = 300
To solve for x and y, we can multiply the second equation by -4 and add it to the first equation to eliminate x:
-4(x + y) + 4x + 2y = -4(300) + 80
-4y + 2y = -1200 + 80
-2y = -1120
y = -1120 / -2
y = 560
Substituting this value back into the second equation, we get:
x + 560 = 300
x = 300 - 560
x = -260 (This is not possible, as the number of hot dogs and pretzels cannot be negative.)
Therefore, there is no solution to this problem using positive integers.