To determine a viable solution for how many robotics kits (y) and chemistry sets (x) Cindy can purchase with her $50 winnings, we can create an equation based on the total cost:
\[ 10y + 8x = 50 \]
In this equation:
- \(y\) 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.
We are looking for non-negative integer solutions (0 or greater) for \(x\) and \(y\).
To find solutions, we can express \(y\) in terms of \(x\):
\[ 10y = 50 - 8x \] \[ y = \frac{50 - 8x}{10} \]
For \(y\) to be a non-negative integer, the right side of the equation must also be a non-negative integer. This means that \(50 - 8x\) must be non-negative and divisible by 10.
Let's explore some integer values for \(x\):
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If \(x = 0\): \[ y = \frac{50 - 8(0)}{10} = \frac{50}{10} = 5 \] Solution: (5 robotics kits, 0 chemistry sets)
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If \(x = 1\): \[ y = \frac{50 - 8(1)}{10} = \frac{42}{10} = 4.2 \quad \text{(not an integer)} \]
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If \(x = 2\): \[ y = \frac{50 - 8(2)}{10} = \frac{34}{10} = 3.4 \quad \text{(not an integer)} \]
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If \(x = 3\): \[ y = \frac{50 - 8(3)}{10} = \frac{26}{10} = 2.6 \quad \text{(not an integer)} \]
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If \(x = 4\): \[ y = \frac{50 - 8(4)}{10} = \frac{18}{10} = 1.8 \quad \text{(not an integer)} \]
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If \(x = 5\): \[ y = \frac{50 - 8(5)}{10} = \frac{10}{10} = 1 \] Solution: (1 robotics kit, 5 chemistry sets)
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If \(x = 6\): \[ y = \frac{50 - 8(6)}{10} = \frac{2}{10} = 0.2 \quad \text{(not an integer)} \]
From these possibilities, we find two viable integer solutions:
- 5 robotics kits and 0 chemistry sets: \((y, x) = (5, 0)\)
- 1 robotics kit and 5 chemistry sets: \((y, x) = (1, 5)\)
These combinations are valid as they satisfy the original monetary constraint.
Thus, two possible viable combinations are:
- 5 robotics kits and 0 chemistry sets.
- 1 robotics kit and 5 chemistry sets.
You can choose either of these as a viable solution.