Write a system of equations to describe the situation below, solve using any method, and fill
in the blanks.
the members of a cooking club are making cakes, which they will sell at a street fair for $14
apiece. it cost $22 for a booth at the fair, and the ingredients for each cake cost $3. at some
point, the club members will sell enough cakes so that their sales cover their expenditures.
how much will the sales and expenditures be? how many cakes will they have sold?
the club's sales and expenditures will be $ x once they have sold y cakes.
submit

When club members will sell 2 cakes, sales will cover expenditure . sales for 2 cakes is $28 and expenditure for 2 cakes is $28. Required equations are  as follows  

x = 14y (equation for sale)  

x = 3y + 22 (equation for expenditure).

Solution:

Need to write a system of equations to describe given situation.

Given that  

Members of cooking club are making cakes to sell at street fair.

Selling price of a piece of cake = $14

Cost of booth at the fair = $22

Cost of ingredient for 1 cake = $3

we need to find a point where expenditure is equal to sale.

let say number of cakes be represented by variable y

So selling price of y cakes = selling price of 1 cake multiplied by y = 14 x y = 14y

COST OF y CAKES

Cost of ingredient of y cakes = cost of ingredient of 1 cake x y = 3 x y = 3y

Cost of the booth = $22

total cost of y cakes = Cost of ingredient + cost of booth = 3y + 22

Expenditure will cover when total cost is equal to selling price  

=> 14y = 3y + 22

On solving the above equation for y

=> 14y – 3y = 22

=> 11y = 22

=> y = 2  

So when club members will sell 2 cakes

Sale = 2 x 14 = $28

Expenditure (cost) = (3)2 + 22 = $28

So after selling 2 cakes expenditure will get covered.

Also given that club sale and expenditure will be $x once they have sold y cakes

=> x = 14y (equation for sale)  

And x = 3y + 22 (equation for expenditure).

hope this is what you were asking for!

step-by-step explanation:

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.