The sum of twice the number of green marbles and the number of yellow marbles in a bag is more than 20. The total number of green and yellow marbles is less than 15. Which of the following is a viable solution to the system of inequalities, where x is the number of green marbles and y is the number of yellow marbles?(1 point)

Question

Responses

(6,7)
left parenthesis 6 comma 7 right parenthesis

(10,5)
left parenthesis 10 comma 5 right parenthesis

(8,6)
left parenthesis 8 comma 6 right parenthesis

(3,12)

Answer 1

To solve this problem, we will express the conditions as inequalities and then check each of the proposed solutions.

Let \( x \) represent the number of green marbles and \( y \) represent the number of yellow marbles. The two conditions given are:

1. The sum of twice the number of green marbles and the number of yellow marbles is more than 20: \[ 2x + y > 20 \]

2. The total number of green and yellow marbles is less than 15: \[ x + y < 15 \]

Now let's check each of the proposed solutions:

1. (6, 7):

   • Check \( 2x + y > 20 \): \[ 2(6) + 7 = 12 + 7 = 19 \quad (\text{not } > 20) \]
   • Check \( x + y < 15 \): \[ 6 + 7 = 13 < 15 \quad (\text{true}) \]
   • Conclusion: Not a viable solution.

2. (10, 5):

   • Check \( 2x + y > 20 \): \[ 2(10) + 5 = 20 + 5 = 25 \quad (> 20) \]
   • Check \( x + y < 15 \): \[ 10 + 5 = 15 \quad (\text{not } < 15) \]
   • Conclusion: Not a viable solution.

3. (8, 6):

   • Check \( 2x + y > 20 \): \[ 2(8) + 6 = 16 + 6 = 22 \quad (> 20) \]
   • Check \( x + y < 15 \): \[ 8 + 6 = 14 < 15 \quad (\text{true}) \]
   • Conclusion: Viable solution.

4. (3, 12):

   • Check \( 2x + y > 20 \): \[ 2(3) + 12 = 6 + 12 = 18 \quad (\text{not } > 20) \]
   • Check \( x + y < 15 \): \[ 3 + 12 = 15 \quad (\text{not } < 15) \]
   • Conclusion: Not a viable solution.

The only viable solution to the system of inequalities is (8, 6).