A snack bar sells two sizes of snack packs. A large snack pack is $5, and a small snack pack is $3. In one day, the snack bar sold 60 snack packs for a total of $220.
Part A
Write a system of equations that represents the scenario above. Let x = small snack packs and y = large snack packs.
Part B
How many small snack packs did the snack bar sell? Use the Elimination Method to solve.
Part A:
Let x = small snack packs
Let y = large snack packs
The price of a small snack pack is $3, so the revenue from selling small snack packs can be represented as:
3x
The price of a large snack pack is $5, so the revenue from selling large snack packs can be represented as:
5y
The snack bar sold a total of 60 snack packs, so the total number of snack packs sold can be represented as:
x + y = 60
The snack bar made a total of $220, so the total revenue can be represented as:
3x + 5y = 220
Therefore, the system of equations that represents the scenario above is:
x + y = 60
3x + 5y = 220
Part B:
To solve the system of equations using the Elimination Method, we can eliminate one variable by multiplying one equation by a certain number and subtracting it from the other equation. In this case, let's eliminate x by multiplying the first equation by 3:
3(x + y) = 3(60)
3x + 3y = 180
Now we subtract the second equation (3x + 5y = 220) from the multiplied first equation:
(3x + 3y) - (3x + 5y) = 180 - 220
3x + 3y - 3x - 5y = -40
-2y = -40
Dividing both sides by -2, we get:
y = (-40) / (-2)
y = 20
Therefore, the snack bar sold 20 large snack packs.