Question

A group of friends will buy at most 8 snacks at a movie theater and spend no more than $42. They will pay $4.00 for each box of candy and $7.00 for each bag of popcorn. The system of inequalities graphed below represents this information.

Answer 1

Let x = candy , y = popcorn

so,

the cost of one box of candy = $4

The cost of one bag of popcorn = $7

so, the solution of part A

The system of inequalities represents the situation is as following:

`\begin{gathered} x+y\leq8 \\ 4x+7y\leq42 \end{gathered}`

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Part B:

We need to find which combination of candy and popcorn could the group buy:

a. 2 candy and 6 popcorn

check for the first inequality : 2 + 6 = 8

check for the second inequality : 2 * 4 + 7 * 6 = 8 + 42 = 50 > 42

So, this option is wrong

b. 3 candy and 4 popcorn

check for the first inequality : 3 + 4 = 7 < 8

check for the second inequality : 4 * 3 + 7 * 4 = 12 + 28 = 40 < 42

So, this option is true

c. 5 candy and 4 popcorn

check for the first inequality : 5 + 4 = 9 > 8

So, this option is wrong

d. 8 candy and 1 popcorn

check for the first inequality : 8 + 1 = 9 > 8

So, this option is wrong

so, the answer of part B is:

option b

the group could by 3 boxes of candy and 4 bags of popcorn