Question

the student council is having a bake sale. they spend $40 for the ingredients to make brownies and $1.50 to make a dozen cookies. the brownies sell for $6 per dozen and the cookies sell for $7.50 per dozen. the student council sells a total of $195 in bake goods

Answer 1

Where's the question?

Answer 2

e write a linear equation with two equations and two variables represent the situation

Answer 3

Oh.

Answer 4

Let's first define some variables to help us solve the problem:
- Let B be the number of dozens of brownies sold.
- Let C be the number of dozens of cookies sold.

From the problem statement, we know that:
- The cost of the ingredients for the brownies is $40.
- The cost of making a dozen cookies is $1.50.
- The selling price of a dozen brownies is $6.
- The selling price of a dozen cookies is $7.50.
- The total revenue from the bake sale is $195.

Using these variables and equations, we can set up a system of two equations to solve for the unknowns B and C:

Equation 1: B*6 + C*7.5 = 195
This equation represents the total revenue from the sale of brownies and cookies.

Equation 2: B*40/12 + C*1.5 = total cost
This equation represents the total cost of making the brownies and cookies.

To solve for B and C, we need to first find the total cost of making the brownies and cookies.

Total cost = cost of brownies + cost of cookies
Total cost = $40 + C*$1.50
Total cost = $40 + $1.5C

Substituting this expression for the total cost into Equation 2, we get:

B*40/12 + C*1.5 = $40 + $1.5C

Simplifying this equation, we get:

10B + 3C = 80

Now we have two equations with two unknowns:

10B + 3C = 80
6B + 7.5C = 195

We can solve this system of equations using any method of our choice (such as substitution or elimination). One possible method is to multiply the first equation by 2 and subtract it from the second equation:

6B + 7.5C = 195
- (20B + 6C = 160)