A bookstore is placing an order for new books and magazines. Each book costs $15, and each magazine costs $5. The store can spend at most $1,000 on these books and magazines. Additionally, each book takes up 2 square feet of space, and each magazine takes up 1 square foot of space. The store has only 50 square feet of display space available for the new items. The following system of inequalities can be used to determine possible combinations of books, x, and magazines, y, the store can order: 15x+5y≤1000 2x+y≤50 Which combinations of books, x, and magazines, y, can the bookstore order? (1 point) Responses 10 books and 30 magazines 10 books and 30 magazines 40 books and 5 magazines 40 books and 5 magazines 50 books and 0 magazines 50 books and 0 magazines 20 books and 15 magazines

Answer 1

To find out which combinations of books \( x \) and magazines \( y \) the bookstore can order, we need to evaluate each response option against the given inequalities:

1. \( 15x + 5y \leq 1000 \) (cost constraint)
2. \( 2x + y \leq 50 \) (space constraint)

We will check each combination:

1. 10 books and 30 magazines: \[ 15(10) + 5(30) = 150 + 150 = 300 \leq 1000 \quad \text{(satisfied)} \] \[ 2(10) + 30 = 20 + 30 = 50 \leq 50 \quad \text{(satisfied)} \] Valid combination.

2. 40 books and 5 magazines: \[ 15(40) + 5(5) = 600 + 25 = 625 \leq 1000 \quad \text{(satisfied)} \] \[ 2(40) + 5 = 80 + 5 = 85 \not\leq 50 \quad \text{(not satisfied)} \] Invalid combination.

3. 50 books and 0 magazines: \[ 15(50) + 5(0) = 750 + 0 = 750 \leq 1000 \quad \text{(satisfied)} \] \[ 2(50) + 0 = 100 + 0 = 100 \not\leq 50 \quad \text{(not satisfied)} \] Invalid combination.

4. 20 books and 15 magazines: \[ 15(20) + 5(15) = 300 + 75 = 375 \leq 1000 \quad \text{(satisfied)} \] \[ 2(20) + 15 = 40 + 15 = 55 \not\leq 50 \quad \text{(not satisfied)} \] Invalid combination.

The only valid combination is 10 books and 30 magazines.