4h + 3c = 23.75 and 6h + 2c = 25
This is the correct system of equations using h for hot dogs and c for cheeseburgers.
The Walker and Baldwin families ordered lunch at the refreshment stand at the basketball game. The Walker family ordered 4 hot dogs and 3 cheeseburgers and paid $23.75. The Baldwin family ordered 6 hots dogs and 2 cheeseburgers and paid $25. Write a system of equations using h for hot dogs and c for cheeseburgers.(1 point)
Responses
h + c = 23.75 and h + c = 25
h + c = 23.75 and h + c = 25
7c = 23.75 and 8h = 25
7c = 23.75 and 8h = 25
4h + 3c = 23.75 and 6h + 2c = 25
4h + 3c = 23.75 and 6h + 2c = 25
4h − 3c = 23.75 and 6h − 2c = 25
3 answers
What are the prices of a hot dog and a cheeseburger?
This question requires you to show your work.
(2 points)
$4.25 $2.00 $5.00 $2.75
cheeseburger
hot dog
This question requires you to show your work.
(2 points)
$4.25 $2.00 $5.00 $2.75
cheeseburger
hot dog
To find the prices of a hot dog and a cheeseburger, we can solve the system of equations:
4h + 3c = 23.75
6h + 2c = 25
Let's first eliminate one variable by multiplying the first equation by 2 and the second equation by 3:
8h + 6c = 47.50
18h + 6c = 75
Subtract the first equation from the second equation to eliminate c:
18h + 6c - 8h - 6c = 75 - 47.50
10h = 27.50
h = 2.75
Now plug the value of h back into the first equation:
4(2.75) + 3c = 23.75
11 + 3c = 23.75
3c = 12.75
c = 4.25
Therefore, the price of a hot dog is $2.75 and the price of a cheeseburger is $4.25.
4h + 3c = 23.75
6h + 2c = 25
Let's first eliminate one variable by multiplying the first equation by 2 and the second equation by 3:
8h + 6c = 47.50
18h + 6c = 75
Subtract the first equation from the second equation to eliminate c:
18h + 6c - 8h - 6c = 75 - 47.50
10h = 27.50
h = 2.75
Now plug the value of h back into the first equation:
4(2.75) + 3c = 23.75
11 + 3c = 23.75
3c = 12.75
c = 4.25
Therefore, the price of a hot dog is $2.75 and the price of a cheeseburger is $4.25.