Josh and Racquel sold cookies and brownies at the math club bake sale. The cookies cost $0.47 each, and the brownies cost $0.60 each. They sold a total of 603 items (cookies and brownies) and raised $323.84 Write a system of equations and determine the number of cookies sold and the number of brownies sold.
Could you please show the work on how to do this?
just put the facts into symbols, one item at a time:
cost of c cookies = .47c
cost of b brownies = .60b
c+b = 603
.47c + .60b = 323.84
now, you know that c = 603-b, so
.47(603-b) + .60b = 323.84
283.41 - .47b + .60b = 323.84
.13b = 40.43
b = 311
c = 603-311 = 292
To solve this problem, we can set up a system of equations to represent the given information.
Let's denote the number of cookies sold as "c" and the number of brownies sold as "b".
From the given information, we can write two equations:
Equation 1: The total number of items sold is 603 --> c + b = 603
Equation 2: The total amount raised is $323.84 --> (0.47 * c) + (0.60 * b) = 323.84
Now, we have a system of equations:
c + b = 603
0.47c + 0.60b = 323.84
To solve this system of equations, we can use either the substitution method or the elimination method. Let's use the elimination method in this case:
First, let's multiply Equation 1 by 0.47 to make the coefficients of "c" in both equations the same:
0.47(c + b) = 0.47(603)
0.47c + 0.47b = 283.41 --> Equation 3
Now, we can subtract Equation 3 from Equation 2 to eliminate "c":
(0.47c + 0.60b) - (0.47c + 0.47b) = 323.84 - 283.41
0.13b = 40.43
Divide both sides of the equation by 0.13 to solve for "b":
b = 40.43 / 0.13
b ≈ 311.00
Now, substitute the value of "b" back into Equation 1 to find the value of "c":
c + 311 ≈ 603
c ≈ 292.00
Therefore, approximately 292 cookies were sold and approximately 311 brownies were sold.