Cindy won $50.00 for getting first place in the Science Fair. She spent her winnings on robotics kits and chemistry sets. Each robotics kit (y) costs $10.00, while each chemistry set costs $8.00 (x). Which of the following is a viable solution to the number of robotics kits and chemistry sets Cindy can purchase, assuming she spends her entire winnings?(1 point)

To determine which of the given responses represents a viable solution to the number of robotics kits (y) and chemistry sets (x) that Cindy can purchase, we need to set up a linear equation based on the amount she spent.

The equation representing Cindy's spending is:

\[
10y + 8x = 50
\]

where:
- \( y \) is the number of robotics kits,
- \( x \) is the number of chemistry sets,
- \( 10 \) is the cost of each robotics kit,
- \( 8 \) is the cost of each chemistry set, and
- \( 50 \) is the total amount she won.

Let's evaluate each option:

1. **(−5, 9)**:
\[
10(-5) + 8(9) = -50 + 72 = 22 \quad \text{(not viable)}
\]

2. **(0, 5)**:
\[
10(0) + 8(5) = 0 + 40 = 40 \quad \text{(not viable)}
\]

3. **(5, 1)**:
\[
10(5) + 8(1) = 50 + 8 = 58 \quad \text{(not viable)}
\]

4. **(3, 2.6)**:
\[
10(3) + 8(2.6) = 30 + 20.8 = 50.8 \quad \text{(not viable)}
\]

Since none of the provided pairs satisfy the equation \( 10y + 8x = 50 \) perfectly, it appears that none of the options are viable solutions under the interpretation that \( y \) and \( x \) must be non-negative integers.

However, if we relax the integer condition, we note that (3, 2.6) is close to meeting the equation closely. But please bear in mind that, typically, we might expect whole kit and set counts.

In conclusion, if strictly considering whole purchases, no solutions satisfy the original usage of \( 10y + 8x = 50 \). If the only requirement is that it should not exceed $50, then \( (3, 2.6) \) could be seen as the closest, although it technically exceeds.