A snack bar sells two sizes of snack packs. A large snack 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 snacks 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 snacks
Let y = large snack packs
From the given information, we can write the following equations:
1) The total number of snack packs sold: x + y = 60
2) The total cost of all the snack packs sold: 3x + 5y = 220
Part B:
To solve the system of equations using the elimination method, we can multiply equation 1 by 3 to match the coefficient of x with equation 2:
3(x + y) = 3(60)
3x + 3y = 180
Now, we can subtract equation 1 from the new equation to eliminate x:
(3x + 3y) - (x + y) = 180 - 60
3x - x + 3y - y = 120
2x + 2y = 120
Next, we can divide equation 2 by 5 to make the coefficient of y match with the new equation:
(3x + 5y)/5 = 220/5
3x/5 + 5y/5 = 44
(3/5)x + y = 44
Now, we have two equations with equal coefficients of y:
2x + 2y = 120
(3/5)x + y = 44
Multiplying equation 2 by 2 to make the coefficients of y equal:
2((3/5)x + y) = 2(44)
(6/5)x + 2y = 88
Now, we can subtract equation 2 from the new equation to eliminate y:
(6/5)x + 2y - ((3/5)x + y) = 88 - 44
(6/5)x - (3/5)x + 2y - y = 44
(3/5)x + y = 44
We now have two equations:
2x + 2y = 120
(3/5)x + y = 44
We can multiply equation 2 by 2 to make the coefficients of y match:
2((3/5)x + y) = 2(44)
(6/5)x + 2y = 88
Now, we can subtract equation 2 from the new equation to eliminate y:
2x + 2y - ((6/5)x + 2y) = 120 - 88
2x - (6/5)x + 2y - 2y = 32
(4/5)x = 32
Now, we can solve for x:
x = (32 * 5) / 4
x = 40
Therefore, the snack bar sold 40 small snack packs.