To determine which of the given options is a viable solution for the number of robotics kits (y) and chemistry sets (x) that Cindy can buy with her $50 winnings, we can set up an equation based on her spending.
The cost of the robotics kits and chemistry sets can be represented by the equation:
\[ 10y + 8x = 50 \]
Now, we can test each of the provided options to see if they satisfy this equation.
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(5,1): Here, y = 5 (robotics kits) and x = 1 (chemistry set) \[ 10(5) + 8(1) = 50 + 8 = 58 \quad (\text{not valid}) \]
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(0,5): Here, y = 0 (robotics kits) and x = 5 (chemistry sets) \[ 10(0) + 8(5) = 0 + 40 = 40 \quad (\text{not valid}) \]
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(3,2.6): Here, y = 3 (robotics kits) and x = 2.6 (chemistry sets) \[ 10(3) + 8(2.6) = 30 + 20.8 = 50.8 \quad (\text{not valid}) \]
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(−5,9): Here, y = -5 (robotics kits) and x = 9 (chemistry sets) \[ 10(-5) + 8(9) = -50 + 72 = 22 \quad (\text{not valid}) \]
None of the provided options satisfies the equation \(10y + 8x = 50\).
To find viable integer solutions, we note that both y (the number of robotics kits) and x (the number of chemistry sets) should be non-negative integers. So, the only suitable combinations can be calculated to find potential integer solutions within the budget.
For instance:
- If \(y = 0\), then \(8x = 50\) leads to non-integer x.
- If \(y = 1\), thus \(10 + 8x = 50\) gives \(8x = 40\) or \(x = 5\).
- If \(y = 2\), then \(20 + 8x = 50\) gives \(8x = 30\) or non-integer.
- Continuing this process will lead to potential integer solutions.
The valid combination that satisfies this condition with whole numbers can be \( (1, 5) \).
Conclusively, all tested options are invalid; to summarize none of the options (5, 1), (0, 5), (3, 2.6), and (−5, 9) satisfy the budget criteria of spending the entire $50.