Neal takes $40 to the festival to spend on games and snacks. The games cost $3 each, and the snacks cost $4 each. Neal would like to be able to play and purchase at least 12 things.

Question

Let x=
number of games and y=
number of snacks. Which of the following sets of inequalities represents this situation?

(1 point)
Responses

x+y≥12

3x+4y≥40

x plus y is greater than or equal to 12 3 x plus 4 y is greater than or equal to 40

3x+4y≥12

x+y≤40

3 x plus 4 y is greater than or equal to 12 x plus y is less than or equal to 40

x+y≤12

3x+4y≤40

x plus y is less than or equal to 12 3 x plus 4 y is less than or equal to 40

x+y≥12

3x+4y≤40

x plus y is greater than or equal to 12 3 x plus 4 y is less than or equal to 40

Answer 1

To represent the situation with the correct set of inequalities, let's analyze the information provided:

1. Neal takes $40 to the festival.
2. Games cost $3 each (denoted as \(x\)).
3. Snacks cost $4 each (denoted as \(y\)).
4. Neal wants to be able to play and purchase at least 12 items total.

From the information given:

• The total number of items (games + snacks) must be at least 12: \[ x + y \geq 12 \]

• The total cost of games and snacks must not exceed $40: \[ 3x + 4y \leq 40 \]

Thus, the appropriate inequalities to represent the situation are:

1. \(x + y \geq 12\) (Neal wants to buy at least 12 items)
2. \(3x + 4y \leq 40\) (Neal cannot spend more than $40)

From the options given, the correct set of inequalities that matches this situation is:

\(x + y \geq 12\) and \(3x + 4y \leq 40\).

This corresponds to the fourth option:

• \(x + y \geq 12\)
• \(3x + 4y \leq 40\)